# GN3 — Researcher durable context
Durable mathematical and research-method context for GN3 Researchers.
**Scope.** This is a generated, read-only startup snapshot of slow-changing GN3 context. Supabase GN3 is canonical. Read this page only at the beginning of a session: ChatGPT may cache it for the rest of that session. After startup, obtain all updates directly from Supabase GN3 through the GN3 synchronization interface; do not reread this page for freshness. Live guidance, research-tree state, active research, audit queues, failures, removals, and frontier movement are intentionally excluded.
**Transport.** The exporter reads Supabase through revision-pinned GN3 RPCs with a maximum content page of 20,000 characters and verifies every assembled body.
**Fresh live state.** After reading this page in full, call `gn3_sync('researcher', since_revision := null, cursor := 0, page_chars := 12000, expected_revision := null)`. Consume every returned page at the pinned repository revision. Thereafter use the last fully consumed revision as numeric `since_revision` for delta synchronization.
**Fallback.** If this generated page is unavailable, incomplete, or visibly truncated, use `gn3_startup('researcher', ...)` and consume the complete canonical startup stream.
# Startup
## GN3 startup
_GN3 document: `start001`_
# GN3 startup
GN3 uses the Supabase-backed document tree as its live authority.
Supabase project:
- name: `GN3_RESEARCH`
- ref: `fewmvjslkhoygixiimgn`
The public GitHub Pages surface is a generated, read-only transport for slow-changing role context; it is never canonical and contains no live research state.
## Connector startup rule
Invoke GN3 only through the known project above.
If the Supabase connector rejects an otherwise valid `gn3_*` SQL call at its safety/access layer **before the database executes it**, treat that as a connector-level transient rejection, not as evidence that the SQL, GN3 function, or project is wrong.
Retry the same high-level GN3 call once. If useful, an equivalent PostgreSQL quoting form such as dollar-quoting string arguments is acceptable. Do **not** respond by listing Supabase projects, searching for another project, inspecting backing tables, reconstructing state elsewhere, or changing to a different research source.
If repeated retries fail, report the connector failure explicitly.
## Researcher / Vice Director — preferred fresh startup
Read the appropriate durable role page **in full**:
- Researcher: `https://sircaptaincaleb.github.io/GN3_RESEARCH/researcher.html`
- Vice Director: `https://sircaptaincaleb.github.io/GN3_RESEARCH/vice_director.html`
The page contains the slow-changing startup material: architecture/policies, proof spine, certified CORE/UTILITY shelf material, and toolkit mathematics. Live guidance, the research tree, active research, audit state, failures, removals, and frontier movement are deliberately excluded.
After reading the page, call:
`gn3_sync(role, since_revision := null, cursor := 0, page_chars := 12000, expected_revision := null)`
A null `since_revision` means **fresh live snapshot**. Consume every returned page at the pinned `repository_revision` until `complete=true`. Thereafter continue with:
`gn3_sync(role, last_seen_revision)`
using the last fully consumed repository revision.
If the Pages surface is unavailable, incomplete, or visibly truncated, fall back to the full canonical startup:
`gn3_startup(role, target_id := null, cursor := 0, page_chars := 12000, expected_revision := null)`
and consume every returned page.
## Astra
The durable Astra page is available at:
`https://sircaptaincaleb.github.io/GN3_RESEARCH/astra.html`
It is useful as the slow-changing mathematical/policy context, but Astra still uses the role-aware Supabase startup/live mechanism for its current strategic layer. Until an Astra-specific live-sync interface replaces that step, `gn3_startup('astra', ...)` remains the complete canonical Astra initialization.
## Auditor
The durable Auditor page is available at:
`https://sircaptaincaleb.github.io/GN3_RESEARCH/auditor.html`
The exact audit target and dependency closure are assignment-specific and remain in Supabase. Use `gn3_startup('auditor', target_id, ...)` to obtain that exact live target context.
## Retrieval discipline
The 12,000-character GN3 default is chosen to keep complete JSON responses transport-safe after metadata overhead. The GitHub Pages assembler itself uses revision-pinned `gn3_get_documents` reads capped at 20,000 content characters and verifies every assembled body.
If a high-level response is visibly truncated, retry the same GN3 function with a smaller character budget. Do not reconstruct missing state with raw SQL.
For mathematical documents, synchronization normally delivers the fresh `statement` interface rather than the proof body. Fetch proof bodies only when needed with `gn3_get_document` or `gn3_get_documents`.
Ordinary research uses the generated durable page plus GN3 RPCs, not direct reads of `gn3_documents`, `gn3_dependencies`, `pg_proc`, or other backing tables/catalogs. Raw SQL is for GN3 infrastructure engineering, diagnosis, migration, or administration.
## Fast concurrency path
During an active wave, use gn3_head() for the cheapest "did the world move?" check. Use gn3_direction_delta(last_seen_revision) or gn3_frontier_digest(role) when a full sync would be unnecessary.
For multi-object research publication, prefer gn3_stage_research_bundle(...), optionally mutate the stage with gn3_append_staged_batch(...) or gn3_replace_staged_batch(...), inspect it with gn3_get_staged_batch(...), then commit with gn3_commit_staged_batch(...). Staging does not advance repository revision; commits use object-local conflict checks. gn3_post_research(...) also tolerates unrelated sibling publications while still rejecting semantic conflicts.
Use gn3_help('staging'), gn3_help('search'), or another targeted help topic instead of ingesting full help when only one interface is needed.
# Policies and role protocol
## Connector and retrieval guidance
_GN3 document: `i0000001`_
# Connector and retrieval guidance
**Status: STARTUP MECHANICS.**
This file owns GN3 retrieval and concurrency mechanics. Project topology is in `01_ARCHITECTURE.md`; research behavior in `03_RESEARCH_PROTOCOL.md`; audit/lifecycle semantics in their dedicated policies. `gn3_help()` is the authoritative function/workflow reference.
## Supabase project identity
GN3 uses exactly this Supabase project for live research state:
- name: `GN3_RESEARCH`
- ref: `fewmvjslkhoygixiimgn`
Workers should target this project directly. They should not enumerate Supabase projects merely to rediscover it during ordinary startup.
## Durable role pages
GN3 publishes generated, read-only GitHub Pages containing slow-changing role context:
- Researcher: `https://sircaptaincaleb.github.io/GN3_RESEARCH/researcher.html`
- Vice Director: `https://sircaptaincaleb.github.io/GN3_RESEARCH/vice_director.html`
- Astra: `https://sircaptaincaleb.github.io/GN3_RESEARCH/astra.html`
- Auditor: `https://sircaptaincaleb.github.io/GN3_RESEARCH/auditor.html`
Supabase remains the sole live authority. These pages are compiled transport artifacts, not writable or canonical state. They deliberately exclude current guidance, research-tree state, active research, audit queues, failures, removals, and frontier movement.
For Researcher and Vice Director fresh startup, read the appropriate durable page **in full**, then obtain the complete live layer with:
`gn3_sync(role, since_revision := null, cursor := 0, page_chars := 12000, expected_revision := null)`
Null `since_revision` means a fresh live snapshot. Consume every page before work begins, carrying the returned `repository_revision` as `expected_revision` until `complete=true`.
Thereafter continue with:
`gn3_sync(role, last_seen_revision)`
which returns the revision-pinned live delta plus current guidance/research-tree state and bounded correctness alerts.
If GitHub Pages is unavailable, incomplete, or visibly truncated, use the full canonical fallback:
`gn3_startup(role, target_id := null, cursor := 0, page_chars := 12000, expected_revision := null)`
and consume every returned page.
Astra still uses its role-specific Supabase initialization in addition to any durable page reading. Auditor startup may be called without a target to receive the audit queue; a `target_id` remains available when a narrow dependency-closure startup is useful.
## Connector-level safety rejection
The Supabase connector may occasionally reject an otherwise valid SQL invocation at its own safety/access layer before PostgreSQL executes the query.
For an ordinary high-level GN3 call such as `SELECT * FROM public.gn3_startup(...)` or another `gn3_*` RPC, such a rejection is **not evidence that the GN3 function, SQL, or project identity is wrong**.
Recovery rule:
1. retry the same high-level call once against project ref `fewmvjslkhoygixiimgn`;
2. if useful, retry with an equivalent PostgreSQL quoting form, such as dollar-quoted string arguments;
3. do not enumerate projects, search for another project, inspect backing tables, substitute Slack/GitHub state, or redesign the access path;
4. if the retry still fails, report the connector-layer failure explicitly.
Do not turn a transient connector rejection into research-state reconstruction.
## Transport safety
GN3's 12,000-character default is chosen so the complete JSON response remains transport-safe after metadata overhead. The external Pages assembler uses revision-pinned `gn3_get_documents` reads capped at 20,000 content characters and verifies each reconstructed body before publishing.
If a GN3 response is visibly truncated, retry the **same high-level function** with a smaller `page_chars` or `part_chars`. Do not recover by querying backing tables.
When using the Supabase connector, send **one SQL query per connector request**. Do not concatenate multiple SQL statements with semicolons and expect them to execute as an ordered multi-query transaction or return independent usable results; connector behavior does not support that workflow reliably. If several GN3 operations are logically one transition, use the appropriate GN3 batch function. Otherwise issue separate connector requests, carrying the returned repository revision forward.
Likewise, do not batch several large paginated GN3 reads into one connector request merely to save calls. A single GN3 page can be transport-safe while the combined outer connector response becomes truncated. Consume large pages one request at a time.
## Normal reads
Use the narrowest read after synchronization:
- `gn3_get_document()` — one exact known document, including paginated body;
- `gn3_get_documents()` — several exact documents in one revision-pinned, character-paginated read; request `statement` or `body`;
- `gn3_search()` — genuine discovery;
- `gn3_get_subtree()` — a known tree location;
- `gn3_current_strategy()`, `gn3_current_guidance()`, `gn3_frontier()`, `gn3_active_research()`, `gn3_live_research_since()` — specialized state views;
- `gn3_needs_audit()`, `gn3_uncertified()` — certification work;
- `gn3_notices(role, ...)` — active/resolved worker-facing notices when a dedicated notice read is useful; normal startup/sync already delivers relevant notices;
- `gn3_changes_since()` — rawer revision-journal synchronization.
Follow returned cursors until `complete=true` whenever completeness matters.
**Ordinary workers must not query `gn3_documents`, `gn3_dependencies`, `pg_proc`, or other GN3 backing tables/catalogs directly for research state.** Do not enumerate documents manually, reproduce search with `LIKE`, inspect function definitions to rediscover signatures, or read statements/bodies/audit state by table query. Use `gn3_help()` and the GN3 RPCs.
Direct table/catalog SQL is reserved for GN3 infrastructure development, diagnosis, migration, or backend administration.
## Revision coherence and writes
Revision pinning remains the default for coherent paged reads and ordinary small writes. If a revision-pinned write races unrelated repository activity, synchronize before retrying.
For large or multi-object writes, prefer transient staging rather than resending the payload.
### General staged batches
`gn3_stage_batch(operations_json, ttl_minutes)` uploads an entire ordinary GN3 operation batch once. Staging does **not** advance the repository revision. It records object-local concurrency tokens for every existing document the transaction touches, including that document's outgoing dependency set.
Inspect with `gn3_get_staged_batch(stage_id)`; discard with `gn3_discard_staged_batch(stage_id)`.
Commit with the small call `gn3_commit_staged_batch(stage_id)`.
- The commit does not require an expected global repository revision.
- Unrelated research/audit writes do not invalidate the stage.
- The complete staged batch commits atomically with one GN3 revision.
- If a touched document changed after staging, the commit fails safely and the staged payload remains available; inspect the conflict and reconcile without resending the large batch.
Stages expire automatically; the default TTL is six hours and callers may request 5–1440 minutes.
Use ordinary `gn3_apply_batch(expected_revision, operations_json)` when the payload is small and immediate. Use staged batches when the transaction is large, expensive to reconstruct, or likely to race concurrent work.
### Audit-specific staged batches
Audit uses a stricter snapshot-first variant because certification may depend on documents that are not themselves being edited.
1. `gn3_audit_snapshot(target_ids, extra_watch_ids, expected_revision)` captures object-local tokens for each target, every current direct dependency, and any explicitly watched additional premise at the pinned read revision.
2. The Auditor performs proof review without changing GN3 state and accumulates intended repair/certification operations.
3. `gn3_stage_audit_batch(operations, snapshot)` accepts those operations only if every watched object still matches the pre-audit snapshot. Staging itself does not advance the repository revision.
4. `gn3_verify_audit_batch(stage_id)` reports any object-local conflicts.
5. `gn3_commit_audit_batch(stage_id)` atomically rechecks the same baselines under the revision lock and applies the full batch in one GN3 revision.
Therefore unrelated concurrent work may freely advance the global repository revision while an audit is underway. Only a change to the audited target or a watched premise invalidates that audit. A staged audit that conflicts remains available for inspection/reconciliation.
Audit staging also packages worker-facing consequences automatically. FAIL operations, localized repairs followed by certification, and Auditor-authored mathematical revisions left uncertified generate first-class `notice` objects in the same atomic commit. An Auditor may put a concise human explanation in a target operation's optional `notice` field; otherwise GN3 emits a mechanical summary of the changed surfaces/disposition.
### Staged single-body convenience
`gn3_stage_body`, `gn3_get_staged_body`, `gn3_commit_staged_body`, and `gn3_discard_staged_body` remain as a lightweight convenience for one large body replacement. General staged batches subsume this case when several related changes belong together.
Do not perform SQL-side read-modify-write editing against backing tables. Retrieve through GN3, construct the intended content explicitly, then use ordinary or staged GN3 write APIs.
## Live team research
Publish active mathematical work with `gn3_post_research(..., statement)`. The optional statement is the compact claim teammates receive by default; the body contains proof/supporting detail.
For Researcher/Vice Director, fresh `gn3_sync(role, null)` supplies all current active research statements together with current strategy, current guidance, the research tree, audit state, frontier state, and active worker notices. Numeric `since_revision` supplies ongoing changes. Astra may use `gn3_sync('astra', snapshot_revision)` for a delta after the Astra hot-shelf snapshot. `gn3_live_research_since()` remains available as the narrower research-only feed.
## Legacy and retrieval failure
Supabase GN3 is the live authority. GitHub repository content and Slack are provenance/archaeology only; the generated GitHub Pages role views are the sole exception and are read-only compiled projections of Supabase.
If required current material cannot be retrieved completely and coherently through the durable page plus GN3 live interfaces, treat that as a retrieval/interface problem to diagnose. Do not silently substitute fragments, legacy state, or ad hoc SQL reconstruction.
## Concurrency fast path
The preferred high-concurrency interfaces are intentionally narrow:
- gn3_head() returns repository revision, current strategy/guidance, operational wave status, and small queue counts.
- gn3_direction_delta(since_revision, role) reports whether strategy, guidance, or operational status changed and returns compact changed-research/notices.
- gn3_frontier_digest(role) returns a compact current frontier: newest certified, pending/contingent, recent failed results with failure fingerprints, active moonshots, and notices.
- gn3_search_math(...) is statement/title-weighted mathematical search with certified-only, current-guidance, and subtree filters.
- gn3_help(topic) provides targeted help such as staging, search, writes, or a documented gn3_* function name.
gn3_sync and gn3_startup now avoid repeating the large state envelope on continuation pages. If a complete stream is modest (at most 28,000 content characters), the first request may automatically expand to one transport-safe page. The response includes transport.recommended_page_chars. Larger reads remain revision-pinned: carry the returned repository_revision as expected_revision.
General staged batches are **mutable before commit**. Use gn3_append_staged_batch to add operations or gn3_replace_staged_batch to replace the staged source transaction. gn3_get_staged_batch exposes the actual staged source operations, watches, conflicts, timestamps, and readiness. Existing watched objects retain their original concurrency baselines; newly introduced objects acquire baselines when added. Dependency targets are watched as semantic premises. Audit stages may also be mutated, but only within the original audit snapshot watch-set.
For ordinary research, use gn3_stage_research_bundle to stage theorem/proof body, statement, dependencies, and metadata together. gn3_post_research_bundle is the immediate convenience form. Ordinary gn3_post_research and bundle publication distinguish semantic conflicts from unrelated repository churn: stale revisions caused only by unrelated sibling activity may publish optimistically, while changes to strategy/guidance, the research tree, or explicit premises still force synchronization.
For a failed result, gn3_stage_research_repair stages body + statement + dependencies + metadata patch + reopening to pending as one object-local transaction. The immediate gn3_repair_research form performs the same logical repair in one revision.
GN3 does not currently keep long-lived server-side read sessions. Large paged reads therefore remain explicitly revision-pinned rather than silently mixing worlds. Prefer the compact head/delta/digest interfaces when exact reconstruction of a large live stream is unnecessary.
## 01 — GN3 architecture
_GN3 document: `i0000002`_
# 01 — GN3 architecture
**Status: CANONICAL.**
This file owns GN3 topology, authority, mathematical surfaces, and lifecycle. Other numbered policies own language, research method, audit, computation, terminology, and document form.
## Roles
GN3 separates four functions:
- **Astra** — scarce strategic Director: theorem-level abstraction, route selection, high-level probes, moonshots, and an explicit strategy assignment on every activation—either affirming the current strategy epoch or issuing a new one when the conceptual map changes.
- **Vice Director** — global coherence of the live project: reconcile parallel research, compare competing syntheses and abstractions, compress the strongest justified mathematical picture, maintain the current strategy epoch, current guidance wave, Astra hot shelf, and research tree, select audits, compose durable mathematics, and keep live state coherent.
- **Researchers** — full mathematical research cycles under current strategy and guidance: assess the live position, make an opening moonshot before local proof search, attack mathematics, publish useful intermediate results continuously, synthesize, pursue conceptual ascent, re-synthesize/compress, perform minor self-maintenance, then revisit and revise or replace the moonshot before selecting the next local attack.
- **Auditors** — independent checking and repair of exact assigned mathematics.
Detailed Astra and Auditor behavior lives in `ASTRA.md` and `AUDITOR.md`. Researcher and Vice Director receive the same mathematical startup context; their difference is responsibility, not hidden mathematics.
Synthesis, conceptual ascent, abstraction proposal, and moonshotting are **team-wide mathematical responsibilities**, not services reserved to the Vice Director or Astra. Researchers should perform them in parallel from their local mathematical vantage and publish nontrivial outputs into live research as they arise. The Vice Director's unique responsibility is to compare those parallel interpretations globally: resolve incompatibilities, identify dominance and genuine parent structure, decide what becomes part of the project's coordinating abstraction model, and keep project surfaces compressed and mutually consistent.
## Authority and namespace
Supabase GN3 is the sole live authority. GitHub and Slack are historical provenance only.
Workers interact with that authority through the GN3 RPC interface supplied by startup/help. Backing Postgres tables are implementation detail, not a normal research API; direct SQL table reads/writes are reserved for infrastructure work, diagnosis, migration, or administration.
The durable namespace is:
- `PROOF_SPINE/` — the sequential canonical proof;
- `SHELVES/CORE/` — certified theorem-quality material plausibly destined for the proof spine;
- `SHELVES/UTILITY/` — certified reusable mathematics plausibly destined for the toolkit;
- `SHELVES/HOT/` — provisional strategic delta for Astra; not certification-bearing. The Astra hot shelf carries a `snapshot_revision` so Astra can delta-sync only newer live changes;
- `TOOLKIT/` — final reusable mathematics; its README is the statement-level startup index;
- `RESEARCH_TREE` — the live mathematical abstraction/coordinating layer, spanning useful levels from the global theorem down to concrete local targets;
- active/retained research documents — working mathematics and explicit strategic/abstraction proposals;
- current strategy — the unique higher-level strategic epoch: conceptual model, global route decomposition, and major strategic pressure points;
- current guidance — the unique tactical team-thrust wave inside the current strategy epoch;
- `NOTICES/` — compact live coordination alerts. Individual notice objects have `doc_type='notice'`; active notices are delivered to their intended worker roles, and resolved notices remain queryable until normal pruning/trash;
- `ARCHAEOLOGY/` — historical recovery and provenance, never current authority.
## Search and retrieval
Use `gn3_search` for genuine discovery. Search separates matching, ordering, and returned projection:
- match modes `strict`, `conjunctive`, and `disjunctive` mean respectively one literal case-sensitive phrase, every whitespace-delimited term, and at least one whitespace-delimited term;
- legacy `substring`, `exact`, `title`, and `fts` modes remain available;
- order `oldest` and `newest` use document `updated_at`; `relevance` orders by `sum_i sqrt(x_i)`, where `x_i` is the literal occurrence count of search term `i` in title plus body;
- projection `statement` returns only document ID and the document's statement column, avoiding body retrieval.
Candidate matching is indexed before relevance scoring. Relevance scoring is bounded by candidate count rather than document-text length; there is no text-length cap. Use `p_under_id` to narrow discovery to a subtree when appropriate. Ordinary exact-document reads still use `gn3_get_document` or `gn3_get_documents`, not search.
All GN3 pagination is stateless. The server must not persist cursor objects, cached result sets, pagination sessions, or continuation snapshots. A cursor is only a scalar continuation token supplied again by the caller, and each page is recomputed from canonical state. For multi-page reads, carry the returned `repository_revision` forward as `p_expected_revision`; if the repository changes, the next page must fail rather than continue against a different state.
## Mathematical status
Mathematical meaning is classified independently of document lifecycle and audit state through `mathematical_status`:
- `proved` — an asserted mathematical result;
- `conjecture` — a precise open mathematical claim;
- `proposal` — a research route, abstraction, mechanism, moonshot, or other idea not asserted as proved;
- `evidence` — a counterexample, construction, computation, obstruction, or other evidentiary mathematical object.
Nonmathematical documents use no mathematical status. This field is orthogonal to `status` and `audit_status`. Changing it changes the audit fingerprint. Titles and prose labels must not substitute for this field when distinguishing proved mathematics from conjectural or proposed mathematics.
## Mathematical documents: statement and body
A mathematical document may carry a nullable `statement` separate from its full `body`.
The statement is the compact mathematical interface: definitions/hypotheses/conclusions needed to know what was proved. The body contains proof, derivation, computation, examples, or supporting detail. Startup, dependency context, and peer-research feeds prefer a fresh statement; exact body retrieval is on demand.
A statement is trusted as fresh only when its recorded source-body hash matches the current body hash. Mathematical interface changes participate in audit invalidation. Do not treat the statement as an informal summary: it must state the actual mathematics.
## Durable mathematical surfaces
The proof spine and toolkit are final canonical destinations. CORE and UTILITY are certified staging surfaces.
A shelf candidate must itself be independently audited; certified ingredients do not automatically certify a rewritten composition. Shelf documents are organized by mathematical topic/proof coordinate, not worker, guidance wave, discovery order, or result number.
CORE should contain coherent proto-proof sections. UTILITY should contain independently reusable mathematical units. Both are mathematics only: no workflow commentary, provenance narrative, or audit ledger. HOT is the intentional exception because it is a temporary strategic handoff surface.
Structural synthesis, conceptual ascent, audit, shelf admission, and final placement are distinct operations: synthesis determines how results fit; ascent seeks stronger explanations or better abstractions; audit verifies exact mathematics; shelves preserve stable audited material; final promotion waits for the mathematical role to settle.
## Research tree
The research tree is the project's **big-picture mathematical coordination layer**. It is not merely a task list or progress outline.
It should represent the live understanding of the problem at multiple useful abstraction granularities. It may contain the global theorem, major reductions, intermediate abstraction statements, candidate parent lemmas, conjectural mechanisms, obstruction formulations, competing conceptual decompositions, and concrete local targets.
Research-tree nodes are not automatically certified mathematics. The tree must distinguish established facts from proposals, conjectural abstractions, and open targets.
Researchers propose useful abstraction statements through live research. The Vice Director integrates the proposals that genuinely clarify or coordinate the search, decides their placement and relationships, removes dominated or obsolete abstractions, and maintains the tree as a coherent explanation of the big picture.
The tree should answer both:
- what concrete problems are currently being attacked; and
- what the team currently thinks those problems **mean**, at each useful level of abstraction.
It is deliberately lossy and may be rewritten aggressively as understanding changes. Durable mathematics should live in canonical/toolkit/shelf/research documents, not be preserved merely because an old abstraction once appeared in the tree.
## Generated durable context
GitHub Pages is a generated, read-only delivery surface for slow-changing role context assembled from Supabase GN3. It is not canonical state and is never edited as an independent source of truth.
Role pages may contain durable startup policy, proof-spine material, toolkit mathematics, and certified CORE/UTILITY material. They deliberately exclude live guidance, research-tree state, active research, audit queues, failures, removals, and frontier movement; those remain live Supabase state and are synchronized through GN3 RPCs.
The **Vice Director owns publication of these generated durable-context views**. After completing a meaningful durable update affecting startup context, architecture/protocol, proof spine, toolkit, CORE, or UTILITY, the Vice Director should trigger the `Publish GN3 role context` GitHub Actions workflow. Several related durable edits should normally be integrated and published with one rebuild rather than triggering a rebuild after every small edit.
Publishing does not change authority: the workflow reads current revision-pinned GN3 content, assembles the role pages, and overwrites the current Pages deployment. No historical page snapshots are required. Live research changes alone do not require a Pages rebuild.
## Live state and lifecycle
Current operational state is represented directly in Supabase:
- exactly one live strategy document should have status `current`;
- exactly one live guidance document should have status `current`; guidance waves may change several times within one strategy epoch;
- research status `active` means current search;
- `retained` means deliberately preserved but non-active;
- trash is reversible curation;
- audit status `failed` means known invalid as stated. Failed research automatically leaves the active surface and becomes `retained`; reopening a repaired failed research document with `gn3_mark_pending()` reactivates it;
- notice status `active` means the alert still deserves worker attention; `resolved` means its consequence has been absorbed. Notices are nonmathematical and use `audit_status='not_required'`.
Keep `retained` small. Do not create shadow archival statuses such as old/superseded/retracted. Superseded working research must not remain `active`: retain it only when it still has deliberate reference value, otherwise redirect any needed dependents and trash it. Historically valuable material belongs in archaeology; obsolete working noise belongs in trash.
## Dependencies, trust, and deletion
Live dependencies point only to live documents. Proof-support dependencies among mathematics/research/proof documents are acyclic.
Certification is local to the checked mathematical interface. A changed direct premise invalidates the direct certified dependent; GN3 does not recursively hash the entire theorem DAG. Thus a downstream theorem can remain locally certified while the full chain through an upstream pending node is not certified.
Trash is blocked when an outside live document still depends on the selected subtree. Redirect, absorb, or retain the dependency first. Hard purge is irreversible inside Supabase because the change journal stores no historical bodies; normal curation is therefore trash first and purge only in a later deliberate pass. Same-pass purge is for disposable tests, failed imports, or confirmed exact duplicates.
## Governing shape
Discovery may be expansive; durable GN3 state must be compressive.
Do not duplicate state into parallel status files, audit ledgers, changelogs, or provenance layers. Notices are the narrow exception for ephemeral coordination: they point workers at a consequential change but never replace the target document's audit state, exact mathematics, guidance, research tree, or revision journal. The proof spine records the proved/open sequential argument; the research tree records the live abstraction hierarchy and coordinating big picture; guidance records the current directive; active research carries concurrent mathematical findings and strategic proposals; document audit fields record certification; the revision journal records continuity.
Researchers publish useful mathematics and proposals continuously. The Vice Director continuously reconciles them, removes obsolete machinery, composes surviving mathematics, and keeps these surfaces mutually consistent.
## 02 — Mathematical language and proof coding
_GN3 document: `i0000003`_
# 02 — Mathematical language and proof coding
**Status: CANONICAL.**
This file owns the language standard for current GN3 mathematics.
## Intrinsic mathematical language
State mathematics in terms of the objects and relations themselves, not discovery history or proof-management vocabulary. A reader should not need Slack, Git history, result numbers, prior guidance, or neighboring prose to know what a theorem means.
Prefer standard graph/hypergraph language and explicit descriptions. Project-coined vocabulary is allowed only when registered in `06_TERMINOLOGY.md`.
## Mathematical compilability
Every load-bearing statement must:
- bind the ambient object and every symbol;
- state all mathematical hypotheses actually used;
- type induced subtournaments, paths, covers, deletions, sets, and maps correctly;
- give the exact conclusion proved;
- distinguish existence from construction and retained properties from properties of a newly produced object;
- preserve any downstream-relevant orientation, endpoint, witness, disjointness, order, maximality, or recurrence condition.
Dependencies explain why a fact is available; they do not replace the hypotheses or definitions needed to parse the statement.
## No hidden provenance or reachability
Do not encode mathematical content by phrases such as accepted case, current branch, inherited configuration, earlier engine, surviving route, or reachable state unless those words have an intrinsic mathematical definition in the statement.
A sequence of proof transformations is not automatically a mathematical object. If reachability, descent, or recurrence is load-bearing, define the state space and relation explicitly.
## Strengthening and factorization
Test natural strengthenings and simpler parent lemmas. Prefer a genuinely stronger useful theorem when it follows cleanly, but do not hide unresolved work inside an ambitious abstraction.
Factor repeated arguments through natural reusable lemmas when that improves proof topology. Inline one-use machinery. Simplification must remove complexity, not merely rename or relocate it.
New strengthenings or materially new proofs are new mathematics and require audit before certification.
## Finite checks
A load-bearing finite case analysis must be checkable from the text through a conceptual argument, an explicit bounded table, a compact certificate, or a precisely cited earlier result. Avoid unexplained phrases such as one checks.
Computation follows `05_COMPUTATION_DISCIPLINE.md`.
## Canonical proof form
Canonical exposition is sequential, mathematical, and self-contained up to explicit canonical inputs. Provenance, migration history, obsolete GN3/A7C3 identifiers, audit history, and discovery transcript stay outside the proof.
Open implications are stated as gaps or research targets. They must never be disguised as theorem output, bookkeeping state, or inherited status.
## 03 — GN3 research protocol
_GN3 document: `i0000004`_
# 03 — GN3 research protocol
**Status: CANONICAL.**
This file owns research method, strategic synthesis, conceptual ascent, guidance, and the Director cycle. Architecture/lifecycle is in `01_ARCHITECTURE.md`; certification in `04_AUDIT_PROTOCOL.md`; computation in `05_COMPUTATION_DISCIPLINE.md`.
## What counts as progress
Useful progress is mathematics that changes the search: a proof, stronger lemma, exact reduction, counterexample, obstruction, necessary hypothesis, structural classification, abstraction proposal, or simplification that removes live complexity.
When an approach stalls, distinguish:
1. **counterexample/impossibility** — the proposed statement or mechanism is false;
2. **necessary missing hypothesis** — a concrete extra condition is required;
3. **structural obstruction** — a precise configuration blocks the route;
4. **no proof found** — the attempt failed without defeating the abstraction.
Only the first three materially refute or revise the idea.
## Strategy epochs, guidance waves, and live research
Exactly one **strategy** document is current. A strategy epoch records the higher-level mathematical picture: the abstraction at which the project is being coordinated, the major route decomposition, strategic pressure points, important negative space, and fallback routes. Strategy should change only when that conceptual picture materially changes.
Exactly one **guidance** document is current. A guidance wave is the tactical team thrust inside the current strategy epoch: concrete targets, preferred interfaces, near-term priorities, and what researchers should push now. Several guidance waves may occur within one strategy epoch. New guidance created through `gn3_issue_guidance` is associated with the current strategy in metadata.
Keep the distinction sharp: a new local target or coordinated thrust normally means a new guidance wave; a new abstraction, invariant, route decomposition, or theorem-level conceptual model may justify a new strategy epoch. Use `gn3_issue_strategy` for the latter and `gn3_issue_guidance` for the former.
**Guidance is wave-based history, not a mutable singleton. Strategy is epoch-based history for the same reason.** When the strategic picture materially changes—new abstraction, changed primary route, major synthesis, new obstruction decomposition, or substantially different shared targets—the Vice Director should normally issue a **new guidance document** and make it current, leaving the preceding wave intact as historical guidance. Edit the existing current guidance only for minor corrections, clarifications, qualifications, or refinements that remain recognizably within the same guidance wave. When uncertain whether a change is substantive, prefer a new wave. Superseding a guidance wave changes which document is current; it does not rewrite the old wave to match the new strategy.
Researchers should publish useful live mathematics **throughout** a research cycle whenever doing so may help concurrent workers: exact lemmas, reductions, obstructions, counterexamples, promising abstraction proposals, elevation proposals, and moonshot proposals need not wait for the cycle to end.
A mathematical research post should use the separate `statement` field for the claim when there is an exact claim, with the body for proof/support. Strategic or abstraction proposals that are not yet theorem statements should still be posted clearly as live research, with their status and intended force explicit.
Every live mathematical research post must also have the correct `mathematical_status`: use `proved` for established results, `conjecture` for precise open claims, `proposal` for unproved mechanisms, abstractions, or moonshots, and `evidence` for counterexamples, constructions, computations, or obstructions. Do not encode this distinction only in a title or prose label. `gn3_post_research` infers common cases; correct the status explicitly when inference would be ambiguous. Current-wave association is metadata, not physical nesting under guidance.
Do not hoard useful partial progress for a polished final report. Concurrent results frequently change another researcher's route or unstick an apparently blocked attack.
Check the live research feed at natural boundaries, after publishing a result that may interact with concurrent work, and when stalled. Incorporate concurrent work that changes the mechanism, target, or shortest route; do not constantly poll.
## Researcher cycle
Researchers run full mathematical cycles rather than acting only as local proof searchers.
### 1. Assess
Begin every research cycle by assessing the current position before choosing an attack.
The assessment should identify:
- the strongest currently justified mathematical picture relevant to the target;
- which recent results or obstructions materially changed that picture;
- the shortest credible routes now available;
- whether the stated target is still the right local abstraction;
- what would be most valuable to prove, disprove, strengthen, abstract, or simplify next.
Assessment is not a clerical recap. Its purpose is to prevent a researcher from blindly continuing an attack whose strategic value has changed.
### 2. Opening moonshot
Immediately after assessment, make a serious theorem-level attempt to bypass or collapse the expected local route **before** becoming absorbed in local proof search.
The opening moonshot should be motivated by the assessed state, not random speculation. It may propose a parent theorem, alternate invariant, new abstraction, direct augmentation, contraction, obstruction principle, or other plausible mechanism that would materially simplify the problem.
Probe it cheaply enough to reject obvious impossibilities, but do not require it to be fully developed before continuing. The point is to establish an ambitious hypothesis while the researcher still has a fresh global view.
**Record every nontrivial opening moonshot in live research** so concurrent workers and the Vice Director can react to it during the same wave.
### 3. Attack
Attack the highest-value unresolved mathematical point directly. The attack may pursue the opening moonshot if it survived the cheap probe, or the strongest local route identified by assessment if the moonshot is not yet actionable.
Prefer conceptual proof and small exact constructions. Search legacy material only when current GN3 lacks what is needed.
A useful attack may end in a proof, stronger lemma, exact reduction, counterexample, necessary hypothesis, structural obstruction, or a precise account of why the route did not progress.
**Publish during the attack whenever a result becomes independently useful.** Publication is not an end-of-cycle ceremony.
### 4. Synthesize
After meaningful progress or a genuine obstruction, compare the new mathematics with the surrounding live results.
Ask what the results collectively say, whether one statement dominates another, whether the same mechanism has appeared elsewhere, whether local facts compose into a cleaner unit, and whether some live machinery has become obsolete.
Researchers should publish useful synthesis when it changes the mathematical understanding. They need not personally edit every global project surface; the Vice Director reconciles parallel syntheses and maintains the canonical compressed view.
### 5. Elevate
**Strongly attempt conceptual ascent in every research cycle.** Do not treat elevation as optional simply because the immediate local result is usable.
Ask what mechanism makes the result true and whether it admits:
- a generalization or sharpening;
- a more intrinsic or invariant formulation;
- a useful new abstraction or abstraction boundary;
- a dual, negative-space, or obstruction formulation;
- a parent theorem replacing several local lemmas;
- a contraction or bypass around current proof machinery;
- simpler sufficient conditions with the same downstream force.
Elevation is not restricted to stronger theorem statements. A good abstraction proposal can be equally valuable if it reorganizes the search, exposes a common mechanism, or reveals a better level at which to formulate several problems.
**Record every nontrivial elevation proposal in live research**, even when unproved or not yet canonical, so the Vice Director and other researchers can assess it. If a genuine stronger formulation emerges, pursue or publish it. If the ascent attempt produces nothing nontrivial, return to the strongest current route rather than manufacturing decorative abstraction.
### 6. Re-synthesize and compress
After the ascent attempt, deliberately compress what the cycle has learned.
Recompare the local result, its strongest plausible elevation, the opening moonshot, and relevant concurrent work. Prefer one stronger parent statement, cleaner obstruction, or clearer abstraction over a pile of overlapping lemmas. Ask what can now be deleted from the active mental model, what has become a corollary, and what should remain only as evidence or a dependency.
This pass includes **minor researcher self-maintenance**. Researchers should clean up their own working artifacts when the mathematical status is clear:
- mark a discovered false result with `gn3_fail`; failed research automatically leaves the active surface;
- if one of your active posts has been genuinely superseded or dominated, change it to `retained` only when it still has deliberate reference value, otherwise trash it after redirecting any needed dependents;
- update stale statements/metadata on your own result when the exact mathematical interface has changed;
- do not wait for an Auditor merely to perform obvious lifecycle cleanup. Audit remains independent verification, not project janitorial work.
Do not perform broad curation of other researchers' work unless assigned or the cleanup is mechanically forced by your own replacement.
Publish a synthesis/compression result when the compressed formulation itself changes how teammates should understand or attack the problem.
### 7. Revisit the moonshot
At the end of the wave, return explicitly to the opening moonshot **after** attack, synthesis, elevation, and compression have changed the researcher's understanding.
Do not merely repeat it. Decide whether the wave's mathematics supports:
- **strengthening or sharpening** the original moonshot;
- **reformulating** it around a better mechanism or abstraction;
- **narrowing** it by adding a necessary hypothesis exposed during the wave;
- **replacing it entirely** with a more promising moonshot suggested by the new understanding;
- or **retiring it** because the wave produced decisive negative evidence.
Record the revised or replacement moonshot when nontrivial. If it supersedes the opening proposal, clean up the earlier proposal according to the ordinary lifecycle rules rather than leaving redundant active moonshots. If the moonshot failed for a precise and informative reason, preserve useful negative evidence.
The closing moonshot is therefore a **revision step**, not a second independent requirement to invent novelty from scratch.
### 8. Continue
At the end of the cycle, choose the next local attack from the updated assessment, compression, and revised moonshot rather than mechanically continuing the previous line.
The intended loop is:
`assess -> opening moonshot -> attack -> synthesize -> elevate -> re-synthesize/compress -> revisit moonshot -> continue`.
**Publication occurs throughout this loop**, not only at the end.
These activities may overlap in practice, but none should disappear merely because the researcher is locally productive. The intended pressure is both ambitious and compressive: begin with a serious bypass attempt, let the mathematics discipline it, and finish with a better strategic hypothesis than the one you started with.
## Structural synthesis
Structural synthesis asks what the old and new results collectively say.
A synthesis pass should:
- compare overlapping statements and retain the strongest useful formulations;
- identify common hypotheses, repeated mechanisms, and natural parent lemmas;
- compose local results into larger mathematical units;
- determine where results actually belong in the proof/toolkit/shelves;
- remove obsolete or dominated machinery;
- test whether the current abstraction decomposition is the clearest explanation of the mathematics;
- propose better abstraction nodes for the research tree when they clarify or coordinate multiple problems.
Synthesis is more than triage or application. Its purpose is maximal coherent understanding.
Researchers perform synthesis locally and in parallel. The Vice Director performs the global reconciliation step: compare those local syntheses, resolve incompatibilities and dominance, and decide what should become the project's compressed canonical picture.
## Conceptual ascent
Conceptual ascent asks what mechanism makes the current mathematics true and whether it admits a stronger or more useful formulation.
Look for:
- generalizations and sharpenings;
- invariant or coordinate-free reformulations;
- useful abstractions and intermediate abstraction levels;
- dual/negative-space/obstruction statements;
- parent theorems replacing several local lemmas;
- contractions or bypasses around current proof machinery;
- simpler sufficient conditions with the same downstream force.
Do not promote an abstraction merely because it is beautiful. It must simplify, strengthen, explain, coordinate, or open a materially shorter route.
Conceptual ascent is expected from researchers throughout ordinary work, not reserved for a Director cycle. Nontrivial ascent proposals belong in live research even before the Vice Director decides whether to integrate them into the research tree.
## Moonshots
Moonshots are **bookends on a research wave**.
After assessment, each researcher makes an opening theorem-level bypass attempt while the global picture is fresh. After the attack/synthesis/ascent/compression work, the researcher returns to that moonshot and revises, sharpens, narrows, replaces, or retires it according to what the mathematics actually taught.
A moonshot is a plausible theorem-level simplification, alternative mechanism, abstraction shift, parent theorem, invariant, or proof collapse, not random speculation. Probe the opening version cheaply enough to reject obvious impossibilities. Record nontrivial opening proposals and materially revised/replacement versions in live research so concurrent workers and the Vice Director can use them.
Parallel moonshots are desirable: independent researchers may see different proof collapses from different local vantage points. The closing revisit is not a demand for gratuitous novelty; it is a demand that the researcher's strategic hypothesis be updated by the wave rather than forgotten once local work begins.
## Research tree as coordinating abstraction layer
The research tree is the project's **big-picture mathematical coordination layer**.
It is not merely a task list, status outline, or compressed changelog. It should express the live mathematical problem at multiple useful levels of abstraction and show how concrete targets, intermediate abstractions, parent conjectures/lemmas, obstructions, and global objectives relate.
A good research tree may contain:
- the global theorem and major proof reductions;
- abstraction statements that explain several concrete subproblems at once;
- intermediate parent lemmas or conjectural mechanisms;
- more concrete local targets beneath those abstraction nodes;
- known obstructions or negative-space formulations that determine which branches remain plausible;
- competing abstractions when the project has not yet resolved which is best.
Abstraction nodes need not already be proved. They must be clearly distinguished from certified mathematical facts and should be retained only while they genuinely help coordinate or explain the search.
Researchers should propose abstraction nodes through live research. The Vice Director decides which proposals merit integration, how they relate, which dominate, and where they sit in the tree.
The tree should therefore answer not only **"what are we trying next?"** but also **"what do we currently think the problem is, at each useful level of abstraction?"**
Rewrite it aggressively when the conceptual model changes. Do not preserve obsolete abstractions for historical completeness.
## Astra and Vice Director
**Astra** is scarce and forward-facing. Use Astra for theorem-level architecture, invariant changes, global route selection, or genuinely strategic alternatives when the whole program needs reconsideration.
**Every Astra activation includes a strategy assignment.** After reading the current strategy and the compressed live frontier, Astra explicitly either affirms the current strategy epoch or issues a new one when the theorem-level conceptual map has materially changed. Reaffirmation does not create a new epoch. Strategy assignment is distinct from tactical guidance: strategy fixes the organizing abstraction and global route picture; guidance fixes the team's current thrust.
**Astra use is budget-aware.** Scarcity means maximize strategic value per call, not preserve calls for their own sake. When a substantial Astra allowance is close to expiring or resetting, the Vice Director should escalate aggressively enough to use the available strategic capacity productively. Suitable parallel uses include independent theorem-level reviews of the current frontier, competing route evaluations, abstraction/invariant probes, proof-collapse attempts, and adversarial checks of the current decomposition—even when local researcher targets remain concrete. Avoid wasting Astra on clerical work or context archaeology, but do not leave expiring strategic capacity unused merely because ordinary local work is still possible.
The **Vice Director** is the global coherence layer. The Vice Director synchronizes live state; reviews parallel researcher assessments, syntheses, compression passes, elevation proposals, and moonshots; determines the strongest globally justified mathematical picture; decides which abstraction proposals belong in the research tree; selects audits and durable compositions; keeps the current strategy epoch and tactical guidance wave distinct and current; **maintains Astra's hot shelf as the compressed strategic handoff surface**; removes dominated live machinery; and decides when Astra escalation is worth scarce attention.
The Vice Director may contribute original mathematical synthesis, ascent, or moonshots when the global view reveals something the local cycles missed. These are no longer serialized duties that the team waits on.
Routine project maintenance, audit queue handling, durable composition, and clerical synchronization stay with the Vice Director unless separately delegated.
## Vice Director cycle
At a natural research boundary the Vice Director should:
1. synchronize changed guidance, active research, audit state, shelves/toolkit, and proof frontier;
2. review the researchers' current assessments plus all live synthesis, elevation, abstraction, and moonshot proposals;
3. determine the strongest globally justified mathematical picture;
4. reconcile competing syntheses and abstraction proposals, resolving incompatibilities and dominance;
5. integrate the useful abstraction hierarchy into the research tree;
6. identify gaps that genuinely require a global synthesis/ascent pass and perform one only where useful;
7. choose or adjust the shortest credible global route and the next exact shared targets;
8. select load-bearing mathematics for audit or durable composition;
9. issue a new strategy epoch when the conceptual map materially changes, or a new guidance wave when only the coordinated thrust changes; edit an existing wave only for minor same-wave corrections;
10. refresh Astra's hot shelf whenever the strategic picture has materially changed or before an Astra handoff, so Astra can recover the current theorem-level state without revision archaeology;
11. decide whether Astra escalation is warranted.
Do not execute this as bureaucracy after every small lemma. The cycle exists to maintain global coherence across parallel mathematical work, not to make researchers wait for centralized understanding.
## Astra handoff
Escalate when the issue is genuinely theorem-level: competing global routes, a new invariant/abstraction, a possible major proof collapse, or uncertainty about the conceptual architecture.
Provide Astra a compressed delta: current target, strongest surviving base, precise obstruction, important negative evidence, and a small number of serious alternatives. **The Vice Director owns keeping the Astra hot shelf current**; it should contain enough compressed strategic state that Astra does not need to reconstruct the project by reading hundreds of revisions or raw research posts. Record the mathematical snapshot revision both in the shelf and its metadata. Astra can then use `gn3_sync('astra', snapshot_revision)` to consume only subsequent live deltas. Do not spend Astra context on archaeology, clerical state, or raw research transcripts.
## Durable placement
Certified mathematics that becomes structurally durable should be composed into CORE or UTILITY candidates and audited in its composed form. Final proof-spine/toolkit promotion waits until placement stabilizes. Superseded search branches should not remain active merely for historical completeness.
Corrections and deletion follow `01_ARCHITECTURE.md`, `04_AUDIT_PROTOCOL.md`, and `07_COMMUNICATION_STANDARD.md`.
## Publication transactions and operational state
Continuous publication remains a core research rule, but publication should now be **transactional at the mathematical unit**. When a result has explicit dependencies or closely coupled metadata, prefer gn3_stage_research_bundle(...) and commit the completed stage once. Do not publish a theorem and then create its proof dependencies through several independent revisions unless there is a mathematical reason to do so. A stage may be appended to or replaced before commit, so a researcher may strengthen the same proposed transaction as the proof develops.
gn3_post_research(...) is optimized for genuinely append-only publication: unrelated sibling revisions no longer force failure. Strategy/guidance changes, research-tree rewrites, or changes to explicit premises remain semantic conflicts and require re-synchronization. This preserves safety without treating every repository revision as a mathematical conflict.
Repairs of failed research should be atomic. Use gn3_stage_research_repair(...) when concurrency is active (or gn3_repair_research(...) for an immediate small repair) so body, statement, dependencies, metadata, and reopening to pending cannot temporarily disagree. Previous last_failure metadata remains visible after reopening; a repaired theorem is not launch-floor mathematics until independently certified again.
Researchers may attach an optional contribution_kind metadata field when it helps curation. Useful values include theorem, counterexample, compression, moonshot, repair, bridge, new_route, and local_lemma. This is descriptive metadata, not a new research-cycle bureaucracy. Vice Director queries/digests may use it to distinguish problem-shaping compression or counterexamples from local ammunition.
Mutable administrative facts do not belong in durable mathematical guidance prose. The current guidance may have a child status object with metadata.role = guidance_operational_status; it records fields such as wave_status and phase. gn3_head() and gn3_direction_delta() surface this overlay. Strategy remains the durable conceptual epoch and guidance remains the current theorem-level attack; starting, completing, auditing, or integrating a wave should normally update the operational overlay rather than rewrite mathematical guidance.
Use first-class notices for exceptional facts that workers must see: audit failures/adjustments, API changes, dependency invalidations, and frontier shifts. Preserve concise exact negative knowledge: a useful counterexample or exact false implication should be promoted/curated rather than forcing future researchers to rediscover the same dead end from failed drafts.
## 04 — GN3 shared certification interface
_GN3 document: `i0000005`_
# 04 — GN3 shared certification interface
This file contains the certification rules every role must understand. Detailed Auditor procedure is in `AUDITOR.md`.
## Mathematical status versus certification
`mathematical_status` and `audit_status` answer different questions. The former says what kind of mathematical object the document is; the latter records verification state of its exact interface.
- `proved` — asserted result and ordinarily audit-bearing;
- `evidence` — exact mathematical evidence and ordinarily audit-bearing when load-bearing;
- `conjecture` — open claim, not a proved result;
- `proposal` — unproved research mechanism or route.
The ordinary `gn3_needs_audit()` queue contains only `proved` and `evidence` documents. Conjectures and proposals may still be inspected explicitly, but certification of their wording or supporting premises must never be represented as proof of the conjectural or proposed conclusion. Changing `mathematical_status` changes the audit fingerprint.
## Scope and independence
Research may use provisional mathematics optimistically. Certification attaches only to the exact mathematical interface independently checked, identified by document ID and audit fingerprint at a repository revision.
The original author may explain or revise a result but may not self-certify it. The Vice Director selects what is load-bearing enough to audit; there is no requirement to audit every research result or clear a chronological backlog. Auditors may nevertheless take one target, several targets, or the entire current audit queue in one working pass.
## Dispositions
Use these dispositions:
- **PASS** — exact target certified.
- **PASS_ADJUSTED** — localized repair/qualification was required and the Auditor checked the repaired exact text without introducing substantive new mathematics.
- **REVISION REQUIRED** — not certified as written, including substantial Auditor-authored repair awaiting independent verification.
- **FAIL** — known false/invalid as stated; persist with `gn3_fail(..., reason)`, not merely `pending`.
Use certified mathematics only at the exact audited strength.
## Repair and second audit
Repair is part of audit. An Auditor may certify a localized repair that does not introduce a new proof idea or materially change the theorem.
A substantial Auditor-authored mathematical repair requires a second independent Auditor. Until that passes, the target remains uncertified.
A failed target may be edited while failed. Failed research is automatically `retained`, so it no longer appears as active work. Once a repaired candidate is complete enough to audit, reopen it with `gn3_mark_pending()`; this reactivates failed research and a failed document cannot be certified directly.
The localized/substantial distinction is mathematical, not a line-count rule.
## Supabase audit state
Authoritative certification lives on the target document: audit status, audit fingerprint, audited fingerprint, and dependencies. Optional audit records preserve useful reasoning but are not a second ledger.
The audit fingerprint tracks the mathematical body, any fresh statement interface, audit-relevant metadata, and direct dependency signature. Mathematical interface changes invalidate stale direct certification automatically. Ordinary provenance/lifecycle metadata does not by itself create new mathematics.
Certification is local rather than a recursive whole-DAG hash. A theorem can remain locally certified while an upstream dependency chain contains a pending node; the full chain is then not certified through that node.
## Shelf and canonical admission
A rewritten or merged mathematical exposition requires audit of that exact composition even when all ingredients were previously certified. Nothing mathematical enters CORE or UTILITY until the exact candidate has PASS or PASS_ADJUSTED.
A byte-for-byte move creates no new mathematical obligation. A mathematical edit to shelf or canonical text does.
## Role responsibilities
The Vice Director selects audit targets, supplies enough context, tracks load-bearing certification, and propagates failures/material weakenings into live strategy.
Researchers may continue while audits run and may use provisional mathematics, clearly marked as such.
Auditors independently check exact mathematics and repair when possible. Substantive repairs require a second Auditor as above.
Retention, trash, restore, and purge follow the lifecycle rules in `01_ARCHITECTURE.md`.
## Worker-facing audit notices
Audit consequences that materially change what teammates should trust are packaged as first-class `notice` objects rather than buried in an audit transcript.
`gn3_stage_audit_batch` automatically appends a notice, in the **same atomic audit commit**, when a target is failed, when a localized repair is certified as PASS_ADJUSTED, or when an Auditor changes mathematical content but leaves the new fingerprint uncertified/REVISION REQUIRED. Pure PASS creates no notice.
For a useful human explanation, add an optional `notice` field to any staged operation for that target. GN3 otherwise supplies a concise mechanical summary of the disposition and changed surfaces. The target document remains the authority for certification; the notice is only a worker-facing alert.
Researchers, Vice Director, and Astra receive relevant active notices through normal startup/sync. The Vice Director should resolve a notice once its consequence has been absorbed into guidance, dependencies, research-tree/shelf state, or otherwise no longer needs active attention. Resolved notices may later be pruned/trash-curated normally.
## Multi-target and batch auditing
Auditor startup without a `target_id` returns the current audit queue. An Auditor may choose one, several, or all queued targets. Targeted startup remains optional when a particular target's dependency closure is useful.
Use `gn3_get_documents(ids, 'body', ...)` to retrieve several exact proof bodies together when that is the efficient unit of work.
Auditors do **not** use ordinary direct write batching as the normal certification path. Before checking a coherent batch, capture `gn3_audit_snapshot(target_ids, extra_watch_ids, expected_revision)`. Audit against that exact target/premise snapshot, accumulate all outcomes, then stage them with `gn3_stage_audit_batch(operations, snapshot)`.
After staging, call `gn3_verify_audit_batch(stage_id)`; if it reports `ready_to_commit=true`, finish with `gn3_commit_audit_batch(stage_id)`. The commit is atomic and rechecks every watched object under the canonical revision lock.
This deliberately separates **audit work** from **repository commit**:
- unrelated repository revisions do not invalidate a staged audit;
- target or watched-dependency changes do invalidate it;
- the read-to-stage race is closed because staging must match the earlier audit snapshot;
- the verify-to-commit race is closed because commit rechecks the same baselines;
- localized repairs and certifications that belong together may commit in one repository revision;
- each certification/failure remains an independent mathematical judgment even when several are committed together.
If a staged operation introduces a dependency not present in the snapshot, take a new snapshot including that dependency before staging. An Auditor is not required to process the queue chronologically or one document per session.
## 05 — GN3 computation discipline
_GN3 document: `i0000006`_
# 05 — GN3 computation discipline
**Status: CANONICAL.**
GN3 is proof-first. MILP, SAT, SMT, CP-SAT, exhaustive enumeration, and similar tools are auxiliary methods for sharply bounded finite questions, not default research modes.
## Admission gate
Before a substantial computation, require all four:
1. **Exact question.** The finite feasibility/optimization/classification question and the mathematical inference from each possible outcome are already clear.
2. **Deliberately small instance.** There is strong reason to expect one bounded run to resolve cleanly; rough variable/constraint/search scale is understood.
3. **Compact output.** The needed result is a status, small witness/counterexample, optimum/count, or compact certificate—not a large transcript.
4. **Cheap failure.** If the run stalls, the worker can immediately return to mathematical search.
If any condition is doubtful, reduce the problem or reason mathematically first.
## Solver use
Use MILP for transparent small feasibility/extremal encodings; prefer feasibility, preprocessing, fixed parameters, and symmetry reduction over large raw formulations.
Use SAT/SMT/CP-SAT for small exact consistency questions with immediately interpretable variables. Avoid broad solution enumeration, repeated blocking loops, giant symmetry-unreduced searches, or opaque models.
A stalled first sensible formulation is evidence to shrink the mathematical question, not an invitation to spend the research cycle tuning the solver.
## Output and stop rules
Choose an output budget before execution. Suppress progress logs, traces, full assignments, generated matrices, exhaustive solution lists, and diagnostics unless a tiny excerpt is specifically needed.
Aggregate internally; output only the mathematical payload.
Stop when resource/search growth becomes surprising, interpretation needs a large dump, repeated reformulation becomes solver engineering rather than mathematics, or the computation no longer has a clear path to the live theorem/obstruction/counterexample.
Do not let sunk effort monopolize the task.
## Team concurrency
Normally at most one worker should pursue a solver-heavy version of the same subproblem unless guidance explicitly makes the finite computation a shared primary target. Other workers should continue independent proof-first routes.
Publish useful compact conclusions quickly. If the computational route stalls, report only the concise null information that matters and relinquish it.
## Durable computational evidence
Retain only what makes the inference intelligible and reproducible when needed:
- the exact finite question;
- mathematical meaning of variables/constraints when relevant;
- enough scale to show the check was bounded;
- compact output/witness/certificate;
- the mathematical conclusion;
- code only when future verification or reuse is genuinely valuable.
Raw logs, giant models, search traces, and large generated tables are not durable mathematics. A computational result that cannot be reduced to a small checkable payload may guide exploration but should not become a hidden load-bearing proof dependency.
Use computation to collapse uncertainty, not create more state for the team to inspect.
## 06 — GN3 terminology registry
_GN3 document: `i0000007`_
# 06 — GN3 terminology registry
This is the authority for project-coined mathematical terminology. Standard mathematical language and locally defined symbols need no registration.
## Rule
A project-coined term is reusable English shorthand whose GN3 meaning is not its ordinary established mathematical meaning.
**If such a term is not approved here, do not use it as established vocabulary; spell out the mathematics instead.**
A proposed term should name a recurring intrinsic object/relation for which standard terminology or a short descriptive phrase is materially worse. Approval requires an explicit definition and revision of this registry.
## Approved GN3 vocabulary
- **boundary tournament**, **tight triple/path/cycle**, **path cover**, **exact q-path cover** — canonical proof-spine definitions;
- **endpoint transfer** — canonical longest-path definition;
- **source path**, **left-extended path**, **right-extended path**, **oppositely extended pair** — canonical fixed-pair definitions;
- **continuation sequence**, **continuation reduction** — only their exact proof-spine definitions;
- **matching blocks**, `M_low`, `M_mid`, `M_high` — the ordered opposite-edge matchings in the non-Hamiltonian edge-ordered four-vertex classification;
- **comparison digraph** — only for an explicitly defined comparison graph as in the small-order machinery.
The list is intentionally short. A useful theorem does not need a nickname.
## Forbidden project shorthand
Do not repurpose the following as GN3-specific mathematical labels:
`core`, `four-core`, `endpoint core`, `fan`, `fiber`, `splice/splicing`, `turn`, `signature`, `profile`, `seam`, `rail`, `gate`, `packet`, `carrier`, `cell`, `bridge`, `face`, `quiet gap`, or `surgery`.
Their established mathematical meanings remain allowed. Otherwise state the object or operation explicitly: name the vertex set, ordered triple, path, partition, family, cut, concatenation, replacement, or deletion actually meant.
Legacy filenames may retain old words; that does not authorize them in current statements/proofs.
## Enforcement
Legacy Slack/GitHub/archaeology vocabulary does not override this registry.
When unapproved shorthand appears in current mathematics, rewrite it into explicit standard language without changing the content and without inventing a replacement nickname.
Auditors check terminology compliance. Directors and the Vice Director should remove discovery-time shorthand during synthesis. When unsure whether a phrase is standard or project-coined, prefer the explicit mathematical description.
## 07 — GN3 communication and document standard
_GN3 document: `i0000008`_
# 07 — GN3 communication and document standard
**Status: CANONICAL.**
This file owns the shape of live GN3 documents. Language rules are in `02_MATHEMATICAL_LANGUAGE.md`; lifecycle in `01_ARCHITECTURE.md`; certification in `04_AUDIT_PROTOCOL.md`.
## Research documents
A mathematical research document should use its nullable `statement` field for the exact standalone claim and its body for proof, derivation, bounded computation, examples, or supporting detail.
The statement must bind its ambient object, local objects, hypotheses, and conclusion. It is not an informal summary. Strategic interpretation and provenance belong in the body after the mathematics or in metadata.
Publish precise intermediate results early enough to unblock teammates. Live sharing does not confer certification.
## Mathematical formatting
Use Markdown backticks for inline mathematical expressions and symbols whenever practical, for example `pc(K)<=2`, `V(H)-S`, or `(a,s,c)`. This is the preferred GN3 transport-safe notation.
Avoid relying on LaTeX backslash commands in text sent through connector/SQL/tool layers. Backslashes may be interpreted or stripped before the document reaches GN3, producing malformed mathematics even when the intended LaTeX was correct. Use plain intrinsic notation inside backticks unless a richer mathematical display is genuinely needed and the transport path is known to preserve it exactly.
Every explicit proof block headed `## Proof` must end with the proof-ending square `∎`. This is part of the GN3 document form, not optional typography. It also lets statement/body projection and validation recognize where the proof block ends.
## Guidance documents
Current guidance should state, compactly:
- the common mathematical target;
- strongest relevant certified/provisional base;
- exact current obstruction;
- useful negative evidence/fences;
- highest-value attacks and moonshots;
- routes to defer.
Exactly one guidance document is current. Replace guidance atomically rather than maintaining a parallel hand-written history.
## Notice documents
A `notice` is a compact live coordination object under `NOTICES/`. Use it when workers need to see a consequential operational or mathematical-state change without duplicating the underlying authoritative state.
A notice body should say what changed, which target(s) it concerns, and what workers should do differently. Metadata carries at least `kind`, `severity`, `target_ids`, and `audience`. Active notices are delivered through startup/sync; resolve them after the consequence has been absorbed. Do not use notices as a second audit ledger, changelog, proof store, or guidance system.
Audit FAIL, PASS_ADJUSTED, and REVISION REQUIRED outcomes are generated automatically by staged audit commits. Auditors should add a concise optional `notice` field to the relevant staged target operation when the mechanical summary would not adequately explain the mathematical consequence.
## Audit and continuity
Certification belongs to the target document's audit fields/fingerprint, not prose wrappers or a second audit ledger. Detailed audit records are optional supporting material.
The GN3 revision journal is the continuity stream. It records operations and changed IDs, not historical body snapshots. Do not maintain a duplicate changelog.
## Corrections
When the same mathematical object is repaired, edit the existing live document rather than creating a competing canonical copy.
- editorial/localized repairs follow the audit rules;
- substantive mathematical change requires fresh certification;
- known false mathematics is first marked with `gn3_fail()`;
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- if the surviving theorem is materially different, create a new document;
- obsolete material is redirected/absorbed and then trashed under the architecture lifecycle rules.
Substantial Auditor-authored mathematical repairs require a second independent Auditor.
## Concurrency
Work from a known repository revision for coherent reads. Ordinary non-audit writes may use revision-pinned GN3 operations, `gn3_apply_batch()`, or the general staged-batch API when several changes must form one coherent transition.
Audit is different: proof review must use `gn3_audit_snapshot(...)` followed by `gn3_stage_audit_batch(...)`, `gn3_verify_audit_batch(...)`, and `gn3_commit_audit_batch(...)`. This makes certification conflict on the audited objects and premises actually relied on rather than on unrelated global repository activity.
Teammates normally consume research statements through the live feed and fetch full proof bodies only when needed.
## Pre-write check
Before making mathematics load-bearing, verify:
1. the statement is standalone and all symbols/hypotheses are bound;
2. the conclusion has exactly the proved strength;
3. language and typing satisfy `02_MATHEMATICAL_LANGUAGE.md`;
4. actual nonlocal premises are represented by dependency edges;
5. provisional and certified inputs are not conflated;
6. the change updates the natural existing document rather than creating duplicate canon;
7. every `## Proof` block ends with `∎`;
8. mathematical notation is transport-safe, preferring backticks/plain intrinsic notation over fragile LaTeX backslash commands.
## Durable promotion and dependencies
Discovery may be broad; durable mathematics should be compressed.
Certified proof-route mathematics is composed into coherent CORE candidates; reusable mathematics into UTILITY candidates. The exact composition must itself pass audit before shelf admission. Final proof-spine/toolkit promotion waits for stable mathematical placement.
Live mathematical dependencies must point to live documents and proof-support dependency cycles are forbidden. Deletion, retention, trash, and purge semantics are owned by `01_ARCHITECTURE.md`.
Supabase GN3 is the only live surface; Slack and GitHub may remain provenance but are not needed to understand current mathematics or state.
# Proof spine
## Proof spine
_GN3 document: `p0000001`_
# Proof spine
The canonical proof material is organized as follows.
- **Two tight paths in a boundary tournament** (`p0000002`) — the main sequential proof spine and current frontier.
- **Codimension-four Hamiltonian-side reduction** (`c0000001`) — certified codimension-four fallback structure.
- **Codimension-four exterior deletion structure** (`c0000002`) — certified exterior-deletion and five-complement consequences supporting the codimension-four branch.
The codimension-four documents were promoted byte-for-byte from the certified CORE shelf; their mathematical interfaces and certifications are unchanged.
## Two tight paths in a boundary tournament
_GN3 document: `p0000002`_
# Two tight paths in a boundary tournament
## 1. Definitions and the minimal-counterexample reduction
Throughout, vertex sets are finite.
For an integer `r>=2` and a set `V`, let `V^{\underline r}` denote the set of ordered `r`-tuples of pairwise distinct vertices of `V`. An **r-uniform directed hypergraph**, or **r-digraph**, is a pair `G=(V,E)` with `E⊆V^{\underline r}`. The set `V` is the vertex set, written `V(G)`, and the members of `E` are the directed `r`-edges. In this document `r=3` unless explicitly stated otherwise.
For an ordered triple `(x,y,z)` of distinct vertices, its **reverse** is `(z,y,x)`. A **3-uniform boundary tournament**, hereafter simply a **boundary tournament**, is a 3-digraph `H=(V,E)` such that for every three distinct vertices `x,y,z`, exactly one of
`(x,y,z)`, `(z,y,x)`
belongs to `E`. An ordered triple is **tight** if it belongs to `E`. Thus boundary antisymmetry is the assertion that exactly one member of every reversal pair is tight.
For `S⊆V(H)`, write `H[S]` for the induced boundary tournament with vertex set `S`, and write `H-S` for `H[V(H)-S]`. For a vertex `v`, `H-v` abbreviates `H-{v}`.
A **tight path** is an ordered list
`P=(v_0,...,v_k)`
of distinct vertices such that `(v_{i-1},v_i,v_{i+1})` is tight for every `1<=i<=k-1`. Its vertex set is `V(P)={v_0,...,v_k}` and its order is `|P|=k+1`. Paths of order one or two satisfy the condition vacuously. The **ordinary edges** of `P` are the unordered pairs `{v_{i-1},v_i}`. The endpoints are `v_0,v_k`; when `k>=2`, the other vertices are internal. A path is **nontrivial** if it has order at least two.
A **path cover** of an induced boundary tournament `G` is a partition of `V(G)` into vertex sets of tight paths, together with one chosen tight ordering on each part. Write `pc(G)` for the minimum number of paths in such a cover. An **exact k-path cover** is a path cover with exactly `k` nonempty path components. We write `P|Q|R` for a path cover with the displayed components. The **ordinary path forest** of a path cover is the ordinary graph on the same vertex set whose edges are the ordinary edges of its path components. A tight path on all vertices of `G` is a **Hamilton path** of `G`, and `G` is **Hamiltonian** if it has such a path. A path is **proper in G** if its vertex set is a proper subset of `V(G)`.
A **tight cycle** is a cyclic ordering of at least three distinct vertices in which every cyclically consecutive ordered triple is tight. It is **proper in H** if its vertex set is a proper subset of `V(H)`. Opening a tight cycle at any one of its ordinary cycle edges gives a tight path on the same vertex set.
The conjecture is
> Every boundary tournament has path-cover number at most two.
Assume the conjecture is false. Choose a counterexample `H` with minimum order, and write
`n=|V(H)|`.
Every proper induced subgraph of `H` has path-cover number at most two. Unless another ambient boundary tournament is explicitly named, every path, cycle, and induced subgraph is taken in `H`. Choosing any vertex `v`, a two-path cover of `H-v` together with `(v)` gives a three-path cover of `H`; hence
`pc(H)=3`.
Also `n>=4`: a boundary tournament on at most three vertices is covered by at most two tight paths.
### Lemma 1.1. Complements of paths
Let `S` be a nonempty proper subset of `V(H)`. If `H[S]` has a Hamilton tight path, then
`pc(H-S)=2`.
Consequently every proper tight path of `H` extends, by adding a two-path cover of its complement, to a spanning three-path cover of `H`.
**Proof.** Minimality gives `pc(H-S)<=2`. If `H-S` were Hamiltonian, its Hamilton path together with a Hamilton path on `S` would be a spanning two-path cover of `H`, contradicting `pc(H)=3`. Therefore `pc(H-S)=2`.
More generally, let `P|Q` be any exact two-path cover of `H-S`. Neither `H[S∪V(P)]` nor `H[S∪V(Q)]` is Hamiltonian: a Hamilton path on either union, together with the untouched other component, would two-cover `H`. ∎
Every tight path of `H` has order at most `n-4`. Indeed, `H` has no Hamilton path, so every tight path `P` is proper. Minimality gives `pc(H-V(P))<=2`; if `H-V(P)` were Hamiltonian, its Hamilton path together with `P` would two-cover `H`. Thus `pc(H-V(P))=2`, while every boundary tournament on at most three vertices is Hamiltonian. Hence `|V(H)-V(P)|>=4`. Opening a tight cycle gives a tight path on the same vertex set, so every tight cycle also has order at most `n-4`.
### Lemma 1.2. Deleting one or two vertices
For every vertex `v`,
`pc(H-v)=2`,
and every exact two-path cover of `H-v` has both paths nontrivial. For distinct vertices `a,c`,
`pc(H-{a,c})=2`,
and every exact two-path cover of `H-{a,c}` has both paths nontrivial.
**Proof.** Apply Lemma 1.1 first to the singleton `(v)` and then to the two-vertex path `(a,c)`. If an exact two-path cover of `H-v` had singleton component `(s)`, then the two-vertex path `(v,s)` together with the other component would two-cover `H`. If an exact two-path cover of `H-{a,c}` had singleton component `(s)`, boundary antisymmetry makes exactly one of `(a,s,c)` and `(c,s,a)` tight; that tight three-vertex path together with the other component would two-cover `H`. ∎
A spanning three-path cover has `n-3` ordinary edges. No spanning forest whose components are tight paths can have more ordinary edges, because a path forest on `n` vertices with more than `n-3` edges has at most two components and would give a two-path cover.
## 2. Small-order lemmas
Let `K_V` denote the ordinary complete graph on a vertex set `V`. Its edge set consists of the two-element subsets of `V`. The **line graph** `L(K_V)` is the graph with vertex set `E(K_V)` in which two vertices are adjacent exactly when the corresponding ordinary edges meet.
For a boundary tournament `G` on `V`, define the **comparison digraph** `Gamma(G)` as the orientation of `L(K_V)` in which, for distinct `u,v,w`,
`{u,v} -> {v,w}`
if and only if `(u,v,w)` is tight.
An **edge order** on `K_V` is a strict total order `<` on `E(K_V)`. An **edge-ordered complete graph** is a complete graph together with such an edge order. A vertex-simple sequence `(v_0,...,v_k)` is an **increasing path** if
`{v_0,v_1} < {v_1,v_2} < ... < {v_{k-1},v_k}`.
### Lemma 2.1. Comparison representation
For every boundary tournament `G`:
1. `Gamma(G)` is a well-defined orientation of `L(K_V)`.
2. A vertex-simple sequence is a tight path in `G` if and only if its consecutive ordinary edges form a directed chain in `Gamma(G)`.
3. There exists an edge order `<` satisfying
`(u,v,w) is tight iff {u,v}<{v,w}`
for all distinct `u,v,w` if and only if `Gamma(G)` is acyclic.
4. If `Gamma(G)` is cyclic and `C` is a shortest directed cycle, then `C` is chordless in `L(K_V)`. Its underlying ordinary edges are either a three-edge star, the three edges of an ordinary triangle, or the edges of a vertex-simple ordinary cycle of length at least four.
**Proof.** If `{u,v}` and `{v,w}` are incident ordinary edges, the two possible comparison directions correspond to the reversal pair `(u,v,w)` and `(w,v,u)`. Boundary antisymmetry chooses exactly one, proving the first assertion.
For a vertex-simple sequence `P=(v_0,...,v_k)`, put `e_i={v_{i-1},v_i}`. By definition, `e_i->e_{i+1}` is equivalent to tightness of `(v_{i-1},v_i,v_{i+1})`. This proves the second assertion.
If an edge order realizes all tight triples, every comparison arc points from a smaller edge to a larger edge, so `Gamma(G)` is acyclic. Conversely, if `Gamma(G)` is acyclic, it has a vertex of indegree zero: otherwise, following incoming arcs indefinitely would repeat a vertex and create a directed cycle. Repeatedly remove such a vertex to obtain a topological ordering of all vertices. In that total order, each adjacent pair of vertices of `Gamma(G)` appears in the direction of its comparison arc, so the order realizes every tight triple. This proves the third assertion.
Now let
`e_0 -> e_1 -> ... -> e_{m-1} -> e_0`
be a shortest directed cycle. A chord between nonconsecutive cycle vertices, in either direction, combines with one of the two directed segments of the cycle to give a shorter directed cycle. Hence the cycle is chordless.
For `m=3`, three pairwise incident ordinary edges are either the three edges of one triangle or three edges through one common vertex. For `m>=4`, chordlessness implies that nonconsecutive ordinary edges are disjoint and consecutive ones meet. Writing `v_i=e_{i-1}∩e_i` cyclically, the vertices `v_i` are distinct and `e_i={v_i,v_{i+1}}`. Thus the ordinary edges form a vertex-simple cycle. ∎
An **ordinary tournament** on a set `S` is an orientation of the complete graph on `S`. It is **transitive** if its vertices can be linearly ordered so that every arc points forward.
### Lemma 2.2. Three common-endpoint triples force a Hamilton five-path
Let `a,c,p,q,r` be distinct vertices of `H`. If
`(a,p,c)`, `(a,q,c)`, `(a,r,c)`
are tight, then `H[{a,c,p,q,r}]` has a Hamilton tight path.
**Proof.** For distinct `x,y∈{p,q,r}`, define tournaments
`x ->_a y` iff `(x,a,y)` is tight,
`x ->_c y` iff `(x,c,y)` is tight.
If distinct `x,y,z` satisfy `x->_a y->_c z`, then `(x,a,y,c,z)` is a Hamilton tight path. Assume no such mixed chain exists.
The tournament `->_a` cannot be transitive. Otherwise relabel so `x->_a y->_a z` and `x->_a z`. Excluding `(x,a,y,c,z)` forces `z->_c y`, while excluding `(x,a,z,c,y)` forces `y->_c z`, impossible. Hence, after relabelling,
`p->_a q->_a r->_a p`.
Excluding the three mixed chains forces
`r->_c q`, `p->_c r`, `q->_c p`.
Assume still that no Hamilton five-path exists. In each of the following six words the first two consecutive triples are tight, so the third must be non-tight and its reverse must be tight:
| word | forced tight triple |
| --- | --- |
| `a p c r q` | `(q,r,c)` |
| `a q c p r` | `(r,p,c)` |
| `a r c q p` | `(p,q,c)` |
| `p q a r c` | `(a,q,p)` |
| `q r a p c` | `(a,r,q)` |
| `r p a q c` | `(a,p,r)` |
Using these six triples and the same argument gives:
| word | forced tight triple |
| --- | --- |
| `a p r c q` | `(c,r,p)` |
| `a q p c r` | `(c,p,q)` |
| `a r q c p` | `(c,q,r)` |
| `p a q r c` | `(r,q,a)` |
| `q a r p c` | `(p,r,a)` |
| `r a p q c` | `(q,p,a)` |
Set
`u=1` iff `(q,p,r)` is tight,
`v=1` iff `(p,q,r)` is tight,
`w=1` iff `(p,r,q)` is tight.
For each of the eight values of `(u,v,w)`, every consecutive triple of each candidate below is already determined except the displayed reversal pair:
| `(u,v,w)` | first candidate | second candidate | remaining reversal pair |
| --- | --- | --- | --- |
| `000` | `a c q r p` | `r p q c a` | `(a,c,q)` / `(q,c,a)` |
| `100` | `c a r q p` | `q p r a c` | `(c,a,r)` / `(r,a,c)` |
| `010` | `a c p q r` | `q r p c a` | `(a,c,p)` / `(p,c,a)` |
| `110` | `a c p q r` | `q r p c a` | `(a,c,p)` / `(p,c,a)` |
| `001` | `c a p r q` | `r q p a c` | `(c,a,p)` / `(p,a,c)` |
| `101` | `c a p r q` | `r q p a c` | `(c,a,p)` / `(p,a,c)` |
| `011` | `a c r p q` | `p q r c a` | `(a,c,r)` / `(r,c,a)` |
| `111` | `c a q p r` | `p r q a c` | `(c,a,q)` / `(q,a,c)` |
Boundary antisymmetry makes exactly one member of the remaining reversal pair tight. Hence one candidate in each row is a Hamilton tight path, a contradiction. ∎
### Lemma 2.3. Non-Hamiltonian five-sets are edge-orderable
Let `G` be a boundary tournament on exactly five vertices. If `G` has no Hamilton tight path, then `Gamma(G)` is acyclic. Equivalently, there is a strict total order `<` on `E(K_5)` such that, for all distinct `u,v,w`,
`(u,v,w) is tight iff {u,v}<{v,w}`.
**Proof.** Suppose `G` is non-Hamiltonian and `Gamma(G)` contains a directed cycle. Choose a shortest one. By Lemma 2.1 it is a star triangle, an ordinary triangle, or a vertex-simple ordinary cycle. A comparison cycle of length five itself gives a Hamilton tight path, so an ordinary cycle can only have length four.
In the case tables below, a three-letter string such as `xyz` abbreviates the assertion that `(x,y,z)` is tight. We repeatedly use the following forcing rule. Let `W=v_0v_1v_2v_3v_4` be a word on all five vertices. If two of its three consecutive triples are known to be tight, then the third cannot also be tight, because otherwise `W` is a Hamilton path. Hence the reverse of the third triple is tight. We write `W => xyz` when this rule forces the triple `(x,y,z)`.
**Ordinary four-cycle.** Normalize the comparison cycle as
`oa -> ab -> bc -> co -> oa`,
so `(o,a,b),(a,b,c),(b,c,o),(c,o,a)` are tight, and let `d` be the fifth vertex. The following complementary branches are exhaustive; each row is read from left to right.
| branch | successive forced triples | contradiction |
| --- | --- | --- |
| `acd, oac` | `bcoad=>dao; boacd=>aob; daobc=>cbo; cdaob=>adc; adcbo=>bcd; oabcd=>bao` | `oab` and `bao` |
| `acd, cao` | `dbcoa=>cbd; doabc=>aod; caodb=>bdo; acbdo=>bca; bcaod=>doa; doabc=>bao` | `oab` and `bao` |
| `dca, bac` | `coabd=>dba; odbac=>bdo; abcod=>doc; bdoca=>aco; dbaco=>abd; coabd=>bao` | `oab` and `bao` |
| `dca, cab` | `dcabo=>oba; dabco=>bad; obadc=>cda; cobad=>boc; bocda=>dco; dcoab=>bao` | `oab` and `bao` |
Thus a shortest comparison cycle is not an ordinary four-cycle.
**Star triangle.** Normalize the star triangle at `o` so that
`(a,o,b)`, `(b,o,c)`, `(c,o,a)`
are tight. For the fifth vertex `d`, set
`S(d)={u in {a,b,c} : (u,o,d) is tight}`.
Cyclic permutation of `a,b,c` preserves the normalized star triangle. Also define the reverse boundary tournament `G^rev` by declaring `(x,y,z)` tight in `G^rev` exactly when `(z,y,x)` is tight in `G`. Reversing the vertex order of a tight path in `G` gives a tight path in `G^rev`. Passing to `G^rev` and then exchanging the labels `b,c` restores the normalized star triangle and sends `|S(d)|` to `3-|S(d)|`. Hence it is enough to treat `|S(d)|=0` and `|S(d)|=1`.
If `|S(d)|=0`, then `(d,o,a),(d,o,b),(d,o,c)` are tight. The following branches are exhaustive:
| branch | successive forced triples | contradiction |
| --- | --- | --- |
| `obc` | `dobca=>acb; doacb=>cao; dcaob=>acd; aobcd=>dcb; daobc=>oad; oadcb=>cda; bocda=>dco; bdcoa=>cdb; acdbo=>obd; caobd=>oac` | `oac` and `cao` |
| `cbo, oda, oba, bdc` | `bdcoa=>ocd; bocda=>adc; obadc=>dab; odabc=>cba; docba=>bco; dbcoa=>cbd; aobdc=>dbo; adboc=>bda; cbdao=>oad; bcoad=>dao` | `oda` and `dao` |
| `cbo, oda, oba, cdb` | `dobac=>cab; cdoba=>odc; odcab=>acd; acdbo=>obd; caobd=>oac; oacdb=>cao` | `oac` and `cao` |
| `cbo, oda, abo, oac` | `abocd=>dco; daboc=>bad; badco=>cda; cdaob=>oad; coadb=>bda; bcoad=>ocb; ocbda=>dbc; doacb=>bca; odbca=>bdo; bdoac=>cao` | `oac` and `cao` |
| `cbo, oda, abo, cao` | `dcaob=>acd; abocd=>dco; bdcoa=>cdb; acdbo=>obd; caobd=>oac` | `oac` and `cao` |
| `cbo, ado, obd` | `caobd=>oac; badoc=>dab; daboc=>oba; dobac=>cab; cdoba=>odc; odcab=>acd; aobdc=>cdb; oacdb=>cao` | `oac` and `cao` |
| `cbo, ado, dbo` | `adboc=>bda; cadob=>dac; bdaco=>oca; dboca=>obd` | `obd` and `dbo` |
If `|S(d)|=1`, cyclically relabel so `S(d)={c}`. Thus `(d,o,a),(d,o,b),(c,o,d)` are tight. The exhaustive branches are:
| branch | successive forced triples | contradiction |
| --- | --- | --- |
| `odb, adc` | `codba=>abd; acodb=>oca; ocabd=>bac; dobac=>abo; abocd=>dco; badco=>dab; daboc=>oba` | `oba` and `abo` |
| `odb, cda` | `cdaob=>oad; acodb=>oca; bocad=>dac; dboca=>obd; obdac=>adb; coadb=>dao` | `oad` and `dao` |
| `bdo, bdc` | `bdcoa=>ocd; abocd=>oba; aobdc=>dbo; dboca=>aco; adboc=>bda; bdaco=>cad; cadob=>oda; codab=>bad; bocda=>adc; obadc=>abo` | `oba` and `abo` |
| `bdo, cdb` | `bdoac=>cao; dcaob=>acd; acdbo=>obd; caobd=>oac` | `oac` and `cao` |
The symmetry already described handles `|S(d)|=2,3`. Thus a shortest comparison cycle is not a star triangle.
**Ordinary triangle.** Normalize the comparison triangle as
`oa -> ab -> bo -> oa`,
so `(o,a,b),(a,b,o),(b,o,a)` are tight, and let `c,d` be the other vertices. For `w∈{c,d}` define
`o in M(w)` iff `(b,w,a)` is tight,
`a in M(w)` iff `(o,w,b)` is tight,
`b in M(w)` iff `(a,w,o)` is tight.
If `M(c)` and `M(d)` shared a coordinate, that coordinate, together with `c,d`, would be the three middle vertices of three tight triples with the same first and third vertices; Lemma 2.2 would give a Hamilton five-path. Hence `M(c)` and `M(d)` are disjoint.
Up to cyclic permutation of `o,a,b` and exchange of `c,d`, the disjoint pair is one of
`(∅,∅)`, `(∅,{o})`, `(∅,{o,a})`, `(∅,{o,a,b})`, `({o},{a})`, `({o},{a,b})`.
The match-set definition fixes all mixed triples involving one exterior vertex and two vertices of the fixed three-set used in the following exhaustive table:
| `M(c),M(d)` | extra branch | successive forced triples | contradiction |
| --- | --- | --- | --- |
| `∅,∅` | none | `bcoda=>doc; bdoca=>aco` | `oca` and `aco` |
| `∅,{o}` | none | `bcoda=>doc; bdoca=>aco` | `oca` and `aco` |
| `∅,{o,a}` | none | `bcoda=>doc; docab=>bac; odbac=>abd; odacb=>cad; bocad=>cob; cobda=>dbo; cdboa=>bdc; oabdc=>bao` | `oab` and `bao` |
| `∅,{o,a,b}` | `abc` | `abcod=>doc; badoc=>dab; dabco=>ocb` | `ocb` and `bco` |
| `∅,{o,a,b}` | `cba` | `cbado=>dab; cdabo=>adc; boadc=>dao; cbdao=>dbc; dbcoa=>aoc; bdaoc=>oad` | `oad` and `dao` |
| `({o},{a})` | `ocd` | `bcoda=>doc; docab=>bac; odbac=>abd; oabdc=>cdb; aocdb=>coa; coabd=>bao` | `oab` and `bao` |
| `({o},{a})` | `dco` | `ocadb=>dac; bodac=>dob; dobca=>cbo; dcboa=>bcd; cboad=>dao; bcdao=>adc; badco=>dab; daboc=>cob; adcob=>cda; bcdao=>oad` | `oad` and `dao` |
| `({o},{a,b})` | `cod` | `codba=>abd; abcod=>cba; cbado=>dab; daboc=>cob; cdabo=>adc; adcob=>ocd; oabdc=>cdb; aocdb=>coa; coabd=>bao` | `oab` and `bao` |
| `({o},{a,b})` | `doc` | `docab=>bac; odbac=>abd; badoc=>dab; daboc=>cob; cdabo=>adc; adcob=>ocd; oabdc=>cdb; aocdb=>coa; coabd=>bao` | `oab` and `bao` |
The extra branches in the table are reversal pairs, so the table is exhaustive. Thus no shortest directed comparison cycle exists. Hence `Gamma(G)` is acyclic, and Lemma 2.1 gives the required edge order. ∎
### Lemma 2.4. Two non-Hamiltonian edge-ordered four-sets
Let `K` be an edge-ordered complete graph, let `T={u_0,u_1,u_2}`, and let `x,y` be distinct vertices outside `T`. If neither `K[T∪{x}]` nor `K[T∪{y}]` has an increasing Hamilton path, then `K[T∪{x,y}]` has an increasing Hamilton path.
**Proof.** First classify a non-Hamiltonian edge-ordered `K_4`. Let its six ordinary edges be
`e_1<e_2<...<e_6`.
If `e_1,e_2` met, say `e_1=ab` and `e_2=bc`, then for the fourth vertex `d` the edge `cd` occurs after `e_2`, so `a,b,c,d` is an increasing Hamilton path. Thus `e_1,e_2` are disjoint. Dually `e_5,e_6` are disjoint. Hence the pairs `{e_1,e_2}`, `{e_3,e_4}`, `{e_5,e_6}` are the three opposite perfect matchings of `K_4`, ordered in three strict blocks.
Relabel `T` so
`u_1u_2 < u_0u_2 < u_0u_1`.
For `r∈{x,y}`, the matching-block classification in `T∪{r}` forces
`ru_0 < ru_1 < ru_2`.
If `xu_1<yu_1`, then
`u_0,x,u_1,y,u_2`
is increasing. If `yu_1<xu_1`, then
`u_0,y,u_1,x,u_2`
is increasing. ∎
### Lemma 2.5. Extending a fixed three-vertex path
Let `P` be a tight path of `H` on three vertices and let `x,y,z` be distinct vertices of `V(H)-V(P)`. At least one of
`V(P)∪{x,y}`, `V(P)∪{x,z}`, `V(P)∪{y,z}`
has a Hamilton tight path.
**Proof.** Write `P` on vertices `{a,b,c}`. Assume, for contradiction, that all three five-sets obtained by adding two of `x,y,z` are non-Hamiltonian. By Lemma 2.3 each is represented by an edge order. Their restrictions to the common triangle `{a,b,c}` realize the same comparison orientation, so after relabelling the three vertices we may assume
`ab<ac<bc`.
Thus `(b,a,c)`, `(a,b,c)`, `(a,c,b)` are tight.
For an exterior vertex `w`, define
`a in M(w)` iff `(b,w,c)` is tight,
`b in M(w)` iff `(a,w,c)` is tight,
`c in M(w)` iff `(a,w,b)` is tight.
In an edge order on a five-set containing `{a,b,c,w}`, the relative order of `wa,wb,wc` gives exactly the following six possibilities:
| order of `wa,wb,wc` | `M(w)` |
| --- | --- |
| `wa<wb<wc` | `{a,b,c}` |
| `wa<wc<wb` | `{b,c}` |
| `wb<wa<wc` | `{a,b}` |
| `wb<wc<wa` | `{a}` |
| `wc<wa<wb` | `{c}` |
| `wc<wb<wa` | `∅` |
If two exterior vertices `u,v` shared a coordinate of their match sets, Lemma 2.2 would apply. Indeed, a shared coordinate `a` gives the triples `(b,a,c),(b,u,c),(b,v,c)`; a shared coordinate `b` gives `(a,b,c),(a,u,c),(a,v,c)`; and a shared coordinate `c` gives `(a,c,b),(a,u,b),(a,v,b)`. In every case `P∪{u,v}` would have a Hamilton five-path, a contradiction. Hence `M(x),M(y),M(z)` are pairwise disjoint.
Two match sets cannot both be `∅`. If `M(u)=M(v)=∅`, then in the edge order on `P∪{u,v}`,
`uc<ub<ua`, `vc<vb<va`.
If `ub<vb`, the word `c,u,b,v,a` is increasing; if `vb<ub`, the word `c,v,b,u,a` is increasing. Either contradicts non-Hamiltonicity.
Therefore exactly one match set is `∅`. The other two are disjoint nonempty members of
`{a}`, `{c}`, `{a,b}`, `{b,c}`, `{a,b,c}`.
Three nonempty pairwise disjoint match sets would have to be `{a},{b},{c}`, but `{b}` is not in the list. After naming the vertex `x` with empty match set and ordering `y,z`, only the following three cases remain:
`I. M(x)=∅, M(y)={a}, M(z)={c};`
`II. M(x)=∅, M(y)={a}, M(z)={b,c};`
`III. M(x)=∅, M(y)={a,b}, M(z)={c}.`
Assume throughout that the relevant five-set is non-Hamiltonian. We use the same forcing rule: if a five-vertex word has two known tight triples, the reverse of its third triple is forced. The initial mixed triples involving one exterior vertex and two vertices of the fixed three-set are
`I: cxb,cxa,bxa; byc,cya,bya; czb,cza,azb.`
`II: cxb,cxa,bxa; byc,cya,bya; czb,azc,azb.`
`III: cxb,cxa,bxa; byc,ayc,bya; czb,cza,azb.`
The following table completes the three cases:
| case | successive forced triples | contradiction |
| --- | --- | --- |
| I | `bycxa=>xcy; xcyab=>bay; cxbay=>abx; czbxa=>xbz; acxbz=>xca; xcazb=>zac; yzacb=>azy; cxbya=>ybx; ybxac=>cax; bycax=>acy; bacyz=>zyc; bazyc=>zab; czabx=>xba` | `abx` and `xba` |
| II | `cxbya=>ybx; ybxac=>cax; bycax=>acy; bacyz=>zyc; czbxa=>xbz; acxbz=>xca; xcazb=>zac; zacbx=>xbc; xbcya=>ycb; azycb=>yza; yzacb=>bca` | `acb` and `bca` |
| III | `cxbya=>ybx; ybxac=>cax; bycax=>acy; xbacy=>abx; czbxa=>xbz; acxbz=>xca; xcazb=>zac; yzacb=>azy; bacyz=>zyc; bazyc=>zab; czabx=>xba` | `abx` and `xba` |
Each case violates boundary antisymmetry. ∎
### Corollary 2.6. Four Hamiltonian five-subsets of every six-set
For every six-element set `E⊆V(H)`, at least four of the five-element subsets of `E` induce Hamiltonian boundary tournaments.
**Proof.** Let `E` be a six-set and let
`M={e in E : H[E-{e}] has a Hamilton path}`.
Choose any three-set `T⊆E`. The complementary three-set `P=E-T` can be ordered as a tight path: choose any middle vertex, and boundary antisymmetry chooses one of the two orders of the other two vertices. Lemma 2.5 applied to this path and the three vertices of `T` shows that `M∩T` is nonempty. If `|E-M|>=3`, choose `T⊆E-M` of order three, a contradiction. Thus `|M|>=4`. ∎
### Lemma 2.7. Non-Hamiltonian four-subsets of a five-set
Let `S⊆V(H)` have order five. If `H[S]` is Hamiltonian, at most three four-element subsets of `S` are non-Hamiltonian. If `H[S]` is non-Hamiltonian, at most one four-element subset of `S` is non-Hamiltonian.
**Proof.** Let `S` be a five-set. If `H[S]` is Hamiltonian, choose a Hamilton order `(v_0,...,v_4)`. Deleting `v_0` or deleting `v_4` leaves a Hamilton tight four-vertex path, so at most three of the five four-subsets are non-Hamiltonian.
Suppose instead that `H[S]` is non-Hamiltonian. By Lemma 2.3 it is represented by an edge order. If two distinct four-subsets were both non-Hamiltonian, they would have the form `T∪{x}` and `T∪{y}` for their three-vertex intersection `T`. Their induced edge orders have no increasing Hamilton path, so Lemma 2.4 gives an increasing Hamilton path on all five vertices. By the representation, this is a tight Hamilton path of `H[S]`, a contradiction. Thus in the non-Hamiltonian case at most one four-subset is non-Hamiltonian. ∎
### Lemma 2.8. Two-thirds Hamilton-five density
Let `W⊆V(H)` have order `r>=6`, and let `h_5(W)` be the number of five-element subsets `F⊆W` for which `H[F]` is Hamiltonian. Then
`h_5(W) >= (2/3) binom(r,5)`.
**Proof.** Count pairs `(U,F)` with
`F⊂U⊆W`, `|F|=5`, `|U|=6`,
such that `H[F]` is Hamiltonian. By Corollary 2.6, every six-set `U` contains at least four Hamiltonian five-subsets, so the number of such pairs is at least
`4 binom(r,6)`.
Every Hamiltonian five-set `F` lies in exactly `r-5` six-subsets of `W`. Hence
`(r-5) h_5(W) >= 4 binom(r,6)`.
Using
`binom(r,6)=binom(r,5)(r-5)/6`
gives the claimed bound. ∎
### Theorem 2.9. Order of a smallest counterexample
`n>10`.
**Proof.** Every set of at most three vertices has a Hamilton tight path. Hence every boundary tournament of order at most six has a two-path cover. For orders seven and eight, choose any six vertices. Corollary 2.6 supplies a Hamilton path on five of them, and the complement has order at most three and is Hamiltonian. Thus `n>=9`.
Suppose `n=9`. Let `F_5` be the Hamiltonian five-subsets and let `B_4` be the non-Hamiltonian four-subsets. Counting incidences between six-sets and Hamiltonian five-subsets gives
`4|F_5| >= 4 binom(9,6)`,
because every six-set contains at least four Hamiltonian five-subsets and every five-set lies in four six-sets. Hence `|F_5|>=84`.
The complement of every member of `F_5` belongs to `B_4`, since two complementary Hamilton paths would form a two-path cover of `H`. Hence
`|B_4|>=|F_5|`.
Now count incidences between five-sets and members of `B_4`. Each member of `B_4` lies in five five-sets. By Lemma 2.7, a Hamiltonian five-set contains at most three members of `B_4`, whereas a non-Hamiltonian five-set contains at most one. Since there are `binom(9,5)=126` five-sets,
`5|B_4| <= 3|F_5| + (126-|F_5|) = 126+2|F_5|`.
Using `|B_4|>=|F_5|` gives
`5|F_5| <= 126+2|F_5|`,
so `|F_5|<=42`, contradicting `|F_5|>=84`.
Suppose `n=10`, and let `F` be the family of Hamiltonian five-subsets. Lemma 2.8 gives
`|F| >= (2/3) binom(10,5)=168`.
On the other hand, the `binom(10,5)=252` five-subsets form `126` complementary pairs. The family `F` contains at most one member of each pair, because two complementary Hamiltonian five-sets would give a spanning two-path cover of `H`. Thus
`|F|<=126`,
a contradiction. Hence `n!=10`, and therefore `n>10`. ∎
## 3. Longest paths and endpoint transfers
Let `A|B|C` be a spanning three-path cover, let `A=(v_0,...,v_{ell-1})` be one component, and let `X` be another. An **endpoint transfer from `X` to `A`** is the operation of removing one endpoint `x` of `X` and appending or prepending `x` to the ordered path `A`, provided the resulting sequence is a tight path. If `X=(x)` is a singleton, the donor component disappears, so a successful transfer gives a spanning two-path cover.
If `ell=1`, either order of the two distinct vertices `V(A)∪{x}` is a tight two-vertex path, so a transfer is always possible. If `ell>=2` and `X=(x_0,...,x_m)`, appending `x_0` to the right end of `A` is possible exactly when
`(v_{ell-2},v_{ell-1},x_0)`
is tight, and prepending `x_m` to the left end is possible exactly when
`(x_m,v_0,v_1)`
is tight. If `X` is nontrivial, deleting the transferred endpoint leaves the remaining order of `X` unchanged and tight.
### Proposition 3.1. Endpoint improvement
Start from a spanning three-path cover and choose a longest component `A`. Repeatedly transfer an endpoint of another component into `A` whenever possible. Every transfer that does not already produce a two-path cover increases `|A|` by one, and `A` remains a longest component. Because every tight path has order at most `n-4`, after at most `n-4-|A|` transfers that do not produce a two-path cover the procedure ends with a spanning three-path cover in which no endpoint of either other component can be transferred into `A`.
For any integer `M`, the procedure may instead be stopped as soon as `A` has order at least `M+1`.
**Proof.** A transfer that does not produce a two-path cover increases the recipient `A` by one vertex and decreases only the donor by one vertex. Since `A` was at least as long as the donor before the transfer, it remains at least as long afterward. Every tight path has order at most `n-4`, so `|A|<=n-4` at every stage. Hence at most `n-4-|A|` such transfers can occur before no further transfer into `A` is possible. ∎
### Proposition 3.2. The ends of a longest component admitting no endpoint transfer
Let `A|B|C` be a spanning three-path cover. Suppose `A=(v_0,...,v_{ell-1})` has maximum order among the three components and no endpoint of `B` or `C` can be transferred into either end of `A`. Then
`ceil(n/3)<=ell<=n-4`,
so `|B|+|C|>=4`; in particular at least one of `B,C` is nontrivial. For every endpoint `x` of `B` or `C`,
`(x,v_{ell-1},v_{ell-2})`, `(v_1,v_0,x)`
are tight.
**Proof.** The longest of three spanning components has order at least `ceil(n/3)`, while the consequence of Lemma 1.1 gives `ell<=n-4`. Hence `|B|+|C|=n-ell>=4`, which in particular rules out both being singletons. Since `n>10`, `ell>=4`.
Fix an endpoint `x` of `B` or `C`. If `(v_{ell-2},v_{ell-1},x)` were tight, `x` could be transferred to the right end of `A`. Therefore it is not tight, and boundary antisymmetry gives `(x,v_{ell-1},v_{ell-2})`. If `(x,v_0,v_1)` were tight, `x` could be transferred to the left end of `A`. Therefore it is not tight, and boundary antisymmetry gives `(v_1,v_0,x)`. ∎
The ordered end pairs
`(v_{ell-1},v_{ell-2})`, `(v_1,v_0)`
are disjoint because `ell>=4`. These two triples give a specified left extension of the first pair and a specified right extension of the second pair for every endpoint `x` of `B` or `C`.
### Proposition 3.3. A globally longest component
A **globally longest tight path** of `H` is a tight path whose order is maximum among all tight paths of `H`. Let `A` be globally longest and put `L=|A|`. Then every exact two-path cover
`H-V(A)=B|C`
satisfies
`4<=|B|+|C|=n-L<=2L`.
Consequently
`ceil(n/3)<=L<=n-4`.
Moreover, `n=3L` if and only if some exact complementary two-cover has `|B|=|C|=L`; in that equality case every exact complementary two-cover has both components of order `L`. In particular there is a spanning three-path cover `A|B|C` with `A` globally longest. In every such cover, no endpoint of `B` or `C` can be transferred into either end of `A`; at least one of `B,C` is nontrivial; and every endpoint `x` of `B` or `C` satisfies
`(x,v_{L-1},v_{L-2})`, `(v_1,v_0,x)`
when `A=(v_0,...,v_{L-1})`.
**Proof.** The path `A` is proper because `H` has no Hamilton path. Lemma 1.1 gives `pc(H-V(A))=2`, so exact complementary two-covers exist. Every tight path leaves at least four vertices outside it, so `|V(H)-V(A)|>=4`. Since `A` is globally longest, each complementary path has order at most `L`, so
`|B|+|C|=n-L<=2L`.
This yields the displayed bounds on `L`. If `n=3L`, then every exact complementary two-cover has total order `2L`, while each component has order at most `L`; hence both have order exactly `L`. Conversely, one complementary cover with two components of order `L` gives `n=L+L+L=3L`.
Finally, any exact complementary two-cover gives a spanning three-path cover with component `A`. No endpoint transfer into `A` is possible, because that would create a tight path longer than `A`; hence Proposition 3.2 applies. ∎
## 4. A fixed pair: restrictions and continuation sequences
Fix distinct vertices `a,c`. Partition the remaining vertices as
`X={s in V(H)-{a,c} : (a,s,c) is tight}`,
`Y={s in V(H)-{a,c} : (c,s,a) is tight}`.
Boundary antisymmetry gives
`V(H)-{a,c}=X⊔Y`.
For `s∈X`, define
`L_s=(a,s)`, `R_s=(s,c)`.
For `s∈Y`, define
`L_s=(c,s)`, `R_s=(s,a)`.
The **source paths at `s`** are the two paths `L_s` and `R_s`.
If `P=(v_0,...,v_k)` and `w∉V(P)`, write `(w,P)` for `(w,v_0,...,v_k)` and `(P,w)` for `(v_0,...,v_k,w)`. A **left-extended path** is a pair `(w;P)` for which `(w,P)` is tight. A **right-extended path** is a pair `(P;w)` for which `(P,w)` is tight. The vertex `w` is the **extension vertex**.
Thus for `s∈X`, `L_s` is right-extended by `c` and `R_s` is left-extended by `a`; for `s∈Y`, `L_s` is right-extended by `a` and `R_s` is left-extended by `c`.
### Lemma 4.1. Restricting an extended path
Let `(w;P)` be a left-extended path, where
`P=(v_0,...,v_k)`.
Let `Q` be a tight path meeting `P` but not containing `v_0`, and let
`i=min{j : v_j∈V(Q)}`.
Then
`P'=(v_0,...,v_{i-1})`
is a nonempty tight path disjoint from `Q`, and `(w;P')` is left-extended by the same vertex `w`.
If `(P;w)` is right-extended and `Q` meets `P` but avoids `v_k`, let
`j=max{h : v_h∈V(Q)}`.
Then
`P''=(v_{j+1},...,v_k)`
is a nonempty tight path disjoint from `Q`, and `(P'';w)` is right-extended by the same vertex `w`.
**Proof.** Since `Q` avoids `v_0`, the first intersection index satisfies `i>=1`. The sequence `(w,v_0,...,v_{i-1})` is an initial segment of the tight path `(w,P)`, so it is tight. The definition of `i` gives disjointness from `Q`.
For the right-extended case, `Q` avoids `v_k`, so `j<=k-1`. The sequence `(v_{j+1},...,v_k,w)` is a terminal segment of the tight path `(P,w)`, so it is tight. The definition of `j` gives disjointness from `Q`. ∎
### Theorem 4.2. Fixed-pair restrictions
For every `s∈V(H)-{a,c}`, the following two fixed-pair restrictions hold.
| orientation of `s` | extended path before restriction | intersecting path `Q` | surviving subpath `P'` | inherited extension |
| --- | --- | --- | --- | --- |
| `s∈X` | `L_s=(a,s)` with `(L_s,c)=(a,s,c)` tight | `(a)` | `(s)` | right by `c` |
| `s∈X` | `R_s=(s,c)` with `(a,R_s)=(a,s,c)` tight | `(c)` | `(s)` | left by `a` |
| `s∈Y` | `L_s=(c,s)` with `(L_s,a)=(c,s,a)` tight | `(c)` | `(s)` | right by `a` |
| `s∈Y` | `R_s=(s,a)` with `(c,R_s)=(c,s,a)` tight | `(a)` | `(s)` | left by `c` |
In each row, the indicated extension belongs to the restricted path; the surviving singleton by itself does not determine that extension.
**Proof.** Consider `s∈X`. The tight triple `(a,s,c)` says that `L_s=(a,s)` is right-extended by `c`. Apply the right-extension part of Lemma 4.1 with `Q=(a)`. The surviving suffix is `(s)`, with the same right extension by `c`.
The same tight triple says that `R_s=(s,c)` is left-extended by `a`. Apply the left-extension part of Lemma 4.1 with `Q=(c)`. The surviving prefix is `(s)`, with the same left extension by `a`.
For `s∈Y`, interchange `a,c`. ∎
### Lemma 4.3. End-interval inheritance
Let `P=(v_0,...,v_k)` be a tight path.
If `(w,P)` is tight and some but not all vertices of `P` are deleted, let
`I=(v_i,...,v_j)`
be the first nonempty interval that remains in the order of `P`. Then `I` is left-extended by `w` when `i=0`, and by the deleted predecessor `v_{i-1}` when `i>0`.
If `(P,w)` is tight and `J=(v_i,...,v_j)` is the last nonempty interval that remains, then `J` is right-extended by `w` when `j=k`, and by the deleted successor `v_{j+1}` when `j<k`.
**Proof.** If `i=0`, `(w,I)` is an initial segment of `(w,P)`. If `i>0`, `(v_{i-1},I)` is a contiguous subpath of `P`.
If `j=k`, `(J,w)` is a terminal segment of `(P,w)`. If `j<k`, `(J,v_{j+1})` is a contiguous subpath of `P`. ∎
### Oppositely extended pairs and continuation sequences
An **oppositely extended pair** is a quadruple
`Pi=(alpha;P,Q;beta)`
where `P,Q` are vertex-disjoint nonempty tight paths, `alpha∉V(P)`, `beta∉V(Q)`, and
`(alpha,P)` and `(Q,beta)`
are tight. The paths `P,Q` are respectively the first and second supports of `Pi`; `alpha` is the left extension vertex of `P`, and `beta` is the right extension vertex of `Q`. When the extension vertices are already specified, we abbreviate the quadruple by `(P,Q)`.
Let
`Pi=(alpha;P,Q;beta)`, `Pi'=(alpha';P',Q';beta')`
be oppositely extended pairs.
Suppose `P=(d_0,d_1)`. A **first-support endpoint reduction** from `Pi` to `Pi'` is a quadruple
`(z,d,x,y)`
where `z∈{d_0,d_1}` is the surviving endpoint, `d` is the other endpoint, and `x,y` are distinct vertices outside
`V(P) union V(Q) union {alpha,beta}`,
such that one of
`(d,x,y)`, `(y,x,d)`
is tight, together with the requirements
`P'=(z)`, `Q'=Q`, `beta'=beta`,
and
`alpha'=alpha` if `z=d_0`, while `alpha'=d_0` if `z=d_1`.
Suppose instead that `Q=(d_0,d_1)`. A **second-support endpoint reduction** from `Pi` to `Pi'` is a quadruple
`(z,d,x,y)`
where `z∈{d_0,d_1}` is the surviving endpoint, `d` is the other endpoint, and `x,y` are distinct vertices outside
`V(P) union V(Q) union {alpha,beta}`,
such that one of
`(d,x,y)`, `(y,x,d)`
is tight, together with the requirements
`P'=P`, `Q'=(z)`, `alpha'=alpha`,
and
`beta'=beta` if `z=d_1`, while `beta'=d_1` if `z=d_0`.
An **endpoint reduction** from `Pi` to `Pi'` is either a first-support endpoint reduction from `Pi` to `Pi'` or a second-support endpoint reduction from `Pi` to `Pi'`.
Suppose
`P=(p)`, `Q=(t_0,...,t_{r-1})`, `r>=2`,
and put
`O=V(H)-({p} union V(Q))`.
A **second-support one-vertex shortening** from `Pi` to `Pi'` is a pair `(F_O,U)` where `F_O` is a path cover of `H[O]` with at most two components and `U` is a distinguished nontrivial component
`U=(u_0,u_1,...)`
of `F_O`, such that one of
`(t_0,u_0,u_1)`, `(u_1,u_0,t_0)`
is tight, and
`P'=P`, `Q'=(t_1,...,t_{r-1})`, `alpha'=alpha`, `beta'=beta`.
Suppose instead that
`P=(t_0,...,t_{r-1})`, `Q=(p)`, `r>=2`,
and put
`O=V(H)-(V(P) union {p})`.
A **first-support one-vertex shortening** from `Pi` to `Pi'` is a pair `(F_O,U)` where `F_O` is a path cover of `H[O]` with at most two components and `U` is a distinguished nontrivial component
`U=(u_0,u_1,...)`
of `F_O`, such that one of
`(t_{r-1},u_0,u_1)`, `(u_1,u_0,t_{r-1})`
is tight, and
`P'=(t_0,...,t_{r-2})`, `Q'=Q`, `alpha'=alpha`, `beta'=beta`.
A **one-vertex shortening** from `Pi` to `Pi'` is either a first-support one-vertex shortening from `Pi` to `Pi'` or a second-support one-vertex shortening from `Pi` to `Pi'`.
Suppose the supports are singletons
`P=(p)`, `Q=(t)`,
and let `y∉{p,t}`. A **second-singleton replacement by `y`** from `Pi` to `Pi'` is a quintuple
`(U,V,d,v_0,v_1)`
such that `U|V` is an exact two-path cover of `H-{p,t}` with both components nontrivial, `y∈V(U)`, `d` is a neighbor of `y` along `U`, `(v_0,v_1)` is an ordered edge of `V`, and one of the following two alternatives holds:
1. `(d,y,t)` is tight, `r=d`, `gamma=t`;
2. `(t,y,d)` is tight, `r=t`, `gamma=d`;
and, in either alternative, one of
`(r,v_0,v_1)`, `(v_1,v_0,r)`
is tight, while
`Pi'=(t;(p),(y);gamma)`.
Under the same singleton hypotheses, a **first-singleton replacement by `y`** from `Pi` to `Pi'` is a quintuple
`(U,V,d,v_0,v_1)`
such that `U|V` is an exact two-path cover of `H-{p,t}` with both components nontrivial, `y∈V(U)`, `d` is a neighbor of `y` along `U`, `(v_0,v_1)` is an ordered edge of `V`, and one of the following two alternatives holds:
1. `(d,y,p)` is tight, `r=p`, `gamma=d`;
2. `(p,y,d)` is tight, `r=d`, `gamma=p`;
and, in either alternative, one of
`(r,v_0,v_1)`, `(v_1,v_0,r)`
is tight, while
`Pi'=(gamma;(y),(t);p)`.
A **prescribed singleton replacement** from `Pi` to `Pi'` is either a first-singleton replacement by some `y` from `Pi` to `Pi'` or a second-singleton replacement by some `y` from `Pi` to `Pi'`.
Define `->` to be the binary relation on oppositely extended pairs given by
`Pi -> Pi'`
if and only if there is an endpoint reduction, a one-vertex shortening, or a prescribed singleton replacement from `Pi` to `Pi'`.
A **continuation sequence** is a finite sequence
`Sigma=(Pi_0,...,Pi_m)`
of oppositely extended pairs such that `Pi_i -> Pi_{i+1}` for every `0<=i<m`. An **extension of `Sigma`** is a continuation sequence
`Sigma'=(Pi_0,...,Pi_m,...,Pi_k)`
having `Sigma` as its initial segment.
Let `P∈{L_s,R_s}`. A **continuation reduction of `P` to `(s)`** is a continuation sequence `Sigma=(Pi_0,...,Pi_m)` with `m>=1` such that `P` is one support of `Pi_0`, the first relation `Pi_0 -> Pi_1` is an endpoint reduction on that support with surviving endpoint `s`, and the corresponding support of `Pi_1` is `(s)`.
A proper subpath of a support is not thereby a later support in a continuation sequence; consecutive pairs must satisfy one of the three relations defining `->`.
### Lemma 4.4. Endpoint reduction using a tight three-vertex path
Let
`Pi=(alpha;P,Q;beta)`
be an oppositely extended pair. Suppose one support is
`D=(d_0,d_1)`
and there are two distinct vertices outside
`V(P) union V(Q) union {alpha,beta}`.
For either prescribed endpoint `z∈{d_0,d_1}`, there is an endpoint reduction on the support `D` whose surviving support is `(z)` and whose other support is unchanged.
**Proof.** Let `d` be the endpoint of `D` different from `z`, and choose distinct
`x,y∉V(P) union V(Q) union {alpha,beta}`.
Boundary antisymmetry makes exactly one of `(d,x,y)` and `(y,x,d)` tight.
First suppose `D=P`. If `z=d_0`, put `alpha'=alpha`; if `z=d_1`, put `alpha'=d_0`. Then
`Pi'=(alpha';(z),Q;beta)`
is an oppositely extended pair, and `(z,d,x,y)` is a first-support endpoint reduction from `Pi` to `Pi'`.
Now suppose `D=Q`. If `z=d_1`, put `beta'=beta`; if `z=d_0`, put `beta'=d_1`. Then
`Pi'=(alpha;P,(z);beta')`
is an oppositely extended pair, and `(z,d,x,y)` is a second-support endpoint reduction from `Pi` to `Pi'`. ∎
### Lemma 4.5. Fixed-singleton descent
Let
`T=(t_0,...,t_{r-1})`
be a nonempty tight path.
1. If
`Pi_0=(alpha;(p),T;beta)`
is an oppositely extended pair, then there is a continuation sequence
`Pi_0 -> Pi_1 -> ... -> Pi_{r-1}`
whose last pair has supports `((p),(t_{r-1}))`. If `r>1`, each relation is a second-support one-vertex shortening, so the sequence has exactly `r-1` such shortenings.
2. If
`Pi_0=(alpha;T,(p);beta)`
is an oppositely extended pair, then there is a continuation sequence
`Pi_0 -> Pi_1 -> ... -> Pi_{r-1}`
whose last pair has supports `((t_0),(p))`. If `r>1`, each relation is a first-support one-vertex shortening, so the sequence has exactly `r-1` such shortenings.
**Proof.** For the first assertion, if `r=1`, take the one-term sequence. Assume `r>=2`, and put
`O=V(H)-({p} union V(T))`.
Every tight path of `H` leaves at least four vertices outside it. Since `p∉V(T)`, the set `O` has order at least three. The induced boundary tournament `H[O]` is proper, so minimality gives a path cover of `H[O]` by at most two tight paths. Some component is nontrivial; write it
`U=(u_0,u_1,...)`.
Boundary antisymmetry on `{t_0,u_0,u_1}` makes exactly one of
`(t_0,u_0,u_1)`, `(u_1,u_0,t_0)`
tight. Put
`T'=(t_1,...,t_{r-1})`.
The path `T'` remains right-extended by `beta`, so
`Pi_1=(alpha;(p),T';beta)`
is an oppositely extended pair. The chosen path cover, the component `U`, and the displayed tight triple give a second-support one-vertex shortening. Repeating the same construction removes one first vertex at each stage and reaches `(t_{r-1})` after exactly `r-1` shortenings.
For the second assertion, again take the one-term sequence when `r=1`. Assume `r>=2`, and put
`O=V(H)-(V(T) union {p})`.
As above, `|O|>=3`, and `H[O]` has a path cover by at most two tight paths with a nontrivial component
`U=(u_0,u_1,...)`.
Boundary antisymmetry on `{t_{r-1},u_0,u_1}` makes exactly one of
`(t_{r-1},u_0,u_1)`, `(u_1,u_0,t_{r-1})`
tight. Put
`T'=(t_0,...,t_{r-2})`.
The path `T'` remains left-extended by `alpha`, so
`Pi_1=(alpha;T',(p);beta)`
is an oppositely extended pair. The chosen path cover, the component `U`, and the displayed tight triple give a first-support one-vertex shortening. Repeating the same construction removes one last vertex at each stage and reaches `(t_0)` after exactly `r-1` shortenings. ∎
### Lemma 4.6. Prescribed singleton replacement
Let
`Pi=(alpha;(p),(t);beta)`
be an oppositely extended singleton pair, and let `y∉{p,t}`.
Then there are prescribed singleton replacements
`Pi -> Pi_R`, `Pi -> Pi_L`
such that `Pi_R` has supports `((p),(y))` and `Pi_L` has supports `((y),(t))`.
**Proof.** By Lemma 1.2 choose an exact two-path cover
`H-{p,t}=U|V`
with both paths nontrivial. Relabel if necessary so that `y∈V(U)`, and choose a neighbor `d` of `y` along `U`. Choose an ordered edge `(v_0,v_1)` of `V`.
For the replacement of the second support, exactly one of
`(d,y,t)`, `(t,y,d)`
is tight. In the first case put `r=d` and `gamma=t`; in the second put `r=t` and `gamma=d`. Boundary antisymmetry makes exactly one of
`(r,v_0,v_1)`, `(v_1,v_0,r)`
tight. Therefore
`Pi_R=(t;(p),(y);gamma)`
is an oppositely extended pair, and the displayed objects give a second-singleton replacement by `y`.
For the replacement of the first support, exactly one of
`(d,y,p)`, `(p,y,d)`
is tight. If `(d,y,p)` is tight, put `r=p` and `gamma=d`; then
`(d;(y,p),(t);p)`
is an oppositely extended pair. If `(p,y,d)` is tight, put `r=d` and `gamma=p`; then
`(p;(y,d),(t);p)`
is an oppositely extended pair. In either case the first support is `(y,r)` and its left extension is `gamma`.
Boundary antisymmetry makes exactly one of
`(r,v_0,v_1)`, `(v_1,v_0,r)`
tight. Thus the ordered edge `(v_0,v_1)` of `V` supplies the tight three-vertex path required for a first-support endpoint reduction preserving `y`. The resulting pair is
`Pi_L=(gamma;(y),(t);p)`,
so the displayed objects give a first-singleton replacement by `y`. ∎
### Theorem 4.7. Prescribed common singleton for both fixed-pair source reductions
Let
`s∈V(H)-{a,c}`
and let
`v∈V(H)-{a,c,s}`
be prescribed. Then the two source paths `L_s,R_s` admit separate continuation sequences with the following last pairs:
- the sequence starting from `R_s` has last pair `((s),(v))`;
- the sequence starting from `L_s` has last pair `((v),(s))`.
In particular, both source paths have continuation reductions to `(s)`, and both reductions can be made to end on the same unordered singleton pair `{s,v}`. The two sequences are alternative constructions and are not asserted to coexist.
**Proof.** Choose distinct vertices
`u,w∈V(H)-{a,c,s,v}`.
Boundary antisymmetry makes exactly one of `(u,v,w)` and `(w,v,u)` tight. Interchange `u,w` if necessary so that
`(u,v,w)`
is tight. Then `(u,v)` is right-extended by `w`, while `(v,w)` is left-extended by `u`.
The source path `R_s` is left-extended: by `a` when `s∈X`, and by `c` when `s∈Y`. Hence
`(R_s,(u,v))`
is an oppositely extended pair whose two supports both have order two. Its two supports together with their two displayed extension vertices use at most six vertices. Since `n>10`, at least five other vertices remain. By Lemma 4.4 there is an endpoint reduction of the first support with surviving endpoint `s`, giving
`((s),(u,v))`.
By Lemma 4.5, one one-vertex shortening then gives
`((s),(v))`.
Similarly, `L_s` is right-extended: by `c` when `s∈X`, and by `a` when `s∈Y`. Hence
`((v,w),L_s)`
is an oppositely extended pair whose two supports both have order two. Applying Lemma 4.4 to the second support gives an endpoint reduction with surviving endpoint `s`, giving
`((v,w),(s))`,
and Lemma 4.5(2) gives one first-support one-vertex shortening ending at
`((v),(s))`.
Every removed source-support vertex is therefore met by an explicitly displayed tight three-vertex path when it is removed. No step uses bare endpoint deletion. The two constructions are separate, so their auxiliary tight paths need not be compatible with one another. ∎
### Proposition 4.8. Comparison with a new edge through `s`
Let `s∈X` and `u∉{a,s,c}`.
1. If `Q=(s,u)`, exactly one of `(a,s,u)` and `(u,s,a)` is tight. If `(a,s,u)` is tight, it is a tight path properly containing both `L_s=(a,s)` and `Q`; if `(u,s,a)` is tight, that triple traverses the ordinary edges of `Q` and `L_s` in the opposite order.
2. If `Q=(u,s)`, exactly one of `(u,s,c)` and `(c,s,u)` is tight. If `(u,s,c)` is tight, it is a tight path properly containing both `Q` and `R_s=(s,c)`; if `(c,s,u)` is tight, that triple traverses the ordinary edges of `R_s` and `Q` in the opposite order.
The same statements hold for `s∈Y` after interchanging `a,c`.
**Proof.** Each item is one reversal pair, so boundary antisymmetry proves the assertion. ∎
### Theorem 4.9. A new endpoint and return within one continuation sequence
Let `s∈V(H)-{a,c}`.
If a two-vertex tight path through `s` has the form
`Q=(s,u)` or `Q=(u,s)`
with `u∉{a,s,c}`, then either `Q` and one source path through `s` lie in a tight three-vertex path containing both, or there is a tight triple containing `u` that traverses an ordered edge of `Q` and an ordered edge of that source path in the opposite order.
Now let `P` be one of the source paths `L_s,R_s`, let `Sigma=(Pi_0,...,Pi_m)` be a continuation reduction of `P` to `(s)`, and let
`Sigma'=(Pi_0,...,Pi_m,...,Pi_k)`
be an extension of `Sigma`. If some `Pi_j` with `j>1` has a two-vertex support whose vertex set is `V(P)`, then that support occurs after the reduction of `P` to `(s)`.
The conclusion concerns only members of the same continuation sequence; it gives no ordering between supports belonging to unrelated continuation sequences.
**Proof.** If `s∈X`, the two possible orientations of the triples on `{a,s,u}` and `{u,s,c}` give the first assertion exactly as in the two cases displayed in Proposition 4.8; if `s∈Y`, interchange `a,c`. For the second assertion, the definition of a continuation reduction gives `P` as a support of `Pi_0` and `(s)` as the corresponding support of `Pi_1`. Thus any later `Pi_j` with a support on the vertex set `V(P)` occurs after the singleton stage. ∎
### Lemma 4.10. Four vertices of one orientation
Let `S⊆V(H)-{a,c}` have order at least four, and suppose `(a,s,c)` is tight for every `s∈S`. Then there are distinct `x,y,z∈S` such that
`(x,a,y,c,z)`
is a tight path.
**Proof.** It is enough to prove the result for a four-element subset of `S`, so assume `|S|=4`. Define tournaments on `S` by
`x->_a y` iff `(x,a,y)` is tight,
`x->_c y` iff `(x,c,y)` is tight.
Suppose no distinct `x,y,z` satisfy `x->_a y->_c z`. If a vertex has indegree at least two in `->_a`, then it has outdegree zero in `->_c`; otherwise two of its `->_a` predecessors together with one `->_c` successor would give such a mixed chain.
The sum of indegrees in the four-vertex tournament `->_a` is six, so some vertex `y` has indegree at least two. Hence `y` has outdegree zero in `->_c`, so `y` is the unique `->_c` sink. Every other vertex has positive `->_c` outdegree and therefore `->_a` indegree at most one. The indegree sum then forces `y` to have `->_a` indegree three and each other vertex to have `->_a` indegree one. Thus `y` is also the `->_a` sink.
Choose `v!=y`. Since `y` is the `->_c` sink, `v->_c y`. The unique `->_a` predecessor `x` of `v` cannot be `y`, because `y` is the `->_a` sink. Hence `x->_a v->_c y`, a mixed chain on three distinct vertices, contradiction. Therefore such a mixed chain exists. Renaming its middle and last vertices as `y,z`, the triples `(x,a,y)`, `(a,y,c)`, `(y,c,z)` are tight, so `(x,a,y,c,z)` is tight. ∎
### Corollary 4.11. A five-vertex path through vertices with both source reductions
Every subset `S⊆X` of order at least four contains distinct `x,y,z` for which
`(x,a,y,c,z)`
is a tight path, and every subset `S⊆Y` of order at least four contains distinct `x,y,z` for which
`(x,c,y,a,z)`
is a tight path. For each of the three vertices, both fixed-pair source paths admit continuation reductions to the corresponding singleton. In particular, since one of `X,Y` has order at least `ceil((n-2)/2)>=5`, such a five-vertex path always exists.
**Proof.** Apply Lemma 4.10 to the chosen subset, interchanging `a,c` for `Y`. Theorem 4.7 supplies both source reductions for every vertex outside `{a,c}`. ∎
## 5. Deletion and comparison of path covers
If `F` is an ordinary graph and `S⊆V(F)`, write `F-S` for the induced subgraph on `V(F)-S`, `deg_F(v)` for the ordinary degree of `v`, `e_F(S)` for the number of ordinary edges of `F` with both endpoints in `S`, and `comp(F)` for the number of connected components of `F`, counting isolated vertices.
### Lemma 5.1. Counting components after deletion
Let `F` be an ordinary path forest with `k` components and let `S⊆V(F)`. Then
`comp(F-S)=k+sum_{v in S}(deg_F(v)-1)-e_F(S)`.
If `F` is the ordinary path forest of a tight-path cover, every nonempty component of `F-S` inherits from its path component before deletion a tight vertex order.
**Proof.** Let `N=|V(F)|`. Since `F` is a path forest with `k` components, it has `N-k` edges. Deleting `S` removes
`sum_{v in S} deg_F(v)-e_F(S)`
edges: the degree sum counts an edge internal to `S` twice, so one copy must be subtracted. The remaining graph has
`N-|S|`
vertices and
`N-k-sum_{v in S}deg_F(v)+e_F(S)`
edges. A forest has number of components equal to vertices minus edges, which gives the displayed formula. Tightness of inherited path orders follows because every surviving component is a contiguous subpath of an original tight path. ∎
### Proposition 5.2. One internal deletion produces a crossing
Fix distinct vertices `a,c`, let `U|V` be an exact two-path cover of `H-{a,c}`, and let `s` be internal in one of `U,V`. Let `T` be any exact two-path cover of `H-{a,c,s}`; such a cover exists.
Deleting `s` from `U|V` leaves three nonempty ordered paths. Some ordinary edge `xy` of `T` has its endpoints in two different paths among those three. Moreover the edge may be named so that there is a vertex `h∉{x,y}` for which `{h,x}` is an ordinary edge of the cover `U|V`. Boundary antisymmetry then gives exactly one tight triple from the reversal pair
`(h,x,y)`, `(y,x,h)`.
If the path containing `x` after deletion of `s` is nontrivial, `h` may be chosen in that path. If both paths crossed by `xy` are singletons, one may take `h=s`.
**Proof.** The three-set `{a,s,c}` has a Hamilton tight path, so Lemma 1.1 gives `pc(H-{a,c,s})=2`; hence an exact cover `T` exists.
Deleting the internal vertex `s` splits one of `U,V` into two nonempty subpaths and leaves the other path unchanged. Thus three nonempty path vertex sets remain. If every ordinary edge of both components of `T` had both endpoints inside one of these three sets, then each connected component of the ordinary path forest of `T` would lie inside one of the three sets. Two connected components could not cover all three nonempty sets. Therefore an ordinary edge `xy` of `T` crosses two of the three sets.
If one crossed set contains at least two vertices, name the endpoint in that set `x` and choose `h` adjacent to `x` along its inherited subpath. Then `{h,x}` is an ordinary edge of `U|V`, and `h` is distinct from `x,y`.
Otherwise both crossed sets are singleton subpaths. The unchanged component of `U|V` is nontrivial by Lemma 1.2, so the two singleton subpaths are the two pieces created by deleting the internal vertex `s`. Each singleton vertex was adjacent to `s` in the original path. Name either one `x` and take `h=s`. Boundary antisymmetry gives the displayed reversal pair. ∎
Thus the two compared covers determine ordinary edges `{h,x}`, `{x,y}` and the reversal pair `(h,x,y)`, `(y,x,h)`, exactly one member of which is tight.
For the remainder of this section, fix distinct vertices `a,c` and an exact two-path cover
`U|V`
of `H-{a,c}`.
### Proposition 5.3. Internal vertices of one orientation
Exactly `n-6` vertices are internal in `U|V`. Consequently one of the two orientation classes contains at least
`ceil((n-6)/2)`
internal vertices. In particular, since `n>10`, there are three internal vertices `p,q,r` for which either
`(a,p,c)`, `(a,q,c)`, `(a,r,c)`
are all tight, or
`(c,p,a)`, `(c,q,a)`, `(c,r,a)`
are all tight.
**Proof.** By Lemma 1.2 both paths are nontrivial, so together they have exactly four endpoints among the `n-2` vertices of `H-{a,c}`. The remaining `n-6` vertices are exactly the internal vertices. The partition `V(H)-{a,c}=X⊔Y` partitions them into two classes, so one class contains at least `ceil((n-6)/2)` internal vertices. Since `n>10`, this number is at least three. ∎
After interchanging `a,c` if necessary, we shall write the selected vertices so that
`(a,p,c)`, `(a,q,c)`, `(a,r,c)`
are tight.
### Proposition 5.4. The eight induced subgraphs
Let `p,q,r` be distinct vertices such that `(a,p,c)`, `(a,q,c)`, `(a,r,c)` are tight, and put
`P={p,q,r}`, `K={a,c,p,q,r}`, `W=V(H)-K`.
For every `J⊆P`, define
`G_J=H[W∪J]`.
Then `H[K-J]` has a Hamilton tight path and
`pc(G_J)=2`.
Consequently every exact two-path cover of `G_J`, together with any Hamilton tight path on `K-J`, is a spanning three-path cover of `H`.
**Proof.** If `J=∅`, Lemma 2.2 applied to the three common-endpoint triples `(a,p,c)`, `(a,q,c)`, `(a,r,c)` gives a Hamilton path on `K`.
If `|P-J|=2`, write `P-J={s,t}`. Exactly one of `(s,c,t)` and `(t,c,s)` is tight, so either `(a,s,c,t)` or `(a,t,c,s)` is a Hamilton path on `K-J`. If `|P-J|=1`, say `P-J={s}`, the path `(a,s,c)` is Hamilton on `K-J`. If `P-J=∅`, `(a,c)` is Hamilton on `K-J`.
Thus `H[K-J]` is Hamiltonian for every `J`. Since `W` has order `n-5>=6`, every `G_J` is a proper nonempty induced subgraph of `H`, so minimality gives `pc(G_J)<=2`. If `G_J` were Hamiltonian, a Hamilton path in `G_J` together with the Hamilton path on `K-J` would two-cover `H`. Therefore `pc(G_J)=2`. Since
`V(H)=(W∪J)⊔(K-J)`,
every exact two-path cover of `G_J` together with a Hamilton tight path on `K-J` is a spanning three-path cover of `H`. ∎
For each `s in P`, let `U|V` be the fixed exact cover of `G_P=H-{a,c}` and let `U'|V'` be any exact cover of `G_{P-{s}}`. Then deleting `s` from the ordinary path forest of `U|V` and comparing the resulting subpaths with `U'|V'` yields vertices `h,x,y` for which `{h,x}` is an ordinary edge inherited from `U|V`, `{x,y}` is an ordinary edge of `U'|V'`, and exactly one of `(h,x,y)`, `(y,x,h)` is tight.
## 6. A lexicographically maximal three-path cover
For a spanning three-path cover `F`, let
`lambda(F)=(ell_1,ell_2,ell_3)`, `ell_1>=ell_2>=ell_3`,
be its three component orders in decreasing order. For two distinct triples `x=(x_1,x_2,x_3)` and `y=(y_1,y_2,y_3)`, define `x` to be **lexicographically larger** than `y` if, for the least index `i` with `x_i!=y_i`, one has `x_i>y_i`. Since each `ell_i` is a positive integer and `ell_1+ell_2+ell_3=n`, only finitely many such triples occur. Choose a spanning three-path cover
`F=A|B|C`
for which `lambda(F)` is lexicographically maximal, and order the component names so that
`|A|>=|B|>=|C|`.
### Proposition 6.1. Properties of the extremal cover
For the lexicographically maximal cover `A|B|C`:
1. `A` is a globally longest tight path of `H`;
2. no endpoint can be transferred from one component into another component whose order is at least as large as the donor's order;
3. `|B|>=3`.
**Proof.** Let `Q` be any globally longest tight path of `H`. It is proper because `pc(H)=3`, so Lemma 1.1 completes it to a spanning three-path cover. Hence the first coordinate of the lexicographically maximal `lambda(F)` is at least `|Q|`; by definition it cannot exceed `|Q|`. Thus `|A|=|Q|` and `A` is globally longest.
Suppose an endpoint can be transferred from a donor `D` to a recipient `R` with `|R|>=|D|`. If `D` is a singleton, the transfer produces a spanning two-path cover, impossible. Otherwise the transfer changes the multiset of component orders by replacing `|R|,|D|` with `|R|+1,|D|-1`. Because `|R|>=|D|`, sorting the three orders gives a lexicographically larger triple, contradicting maximality. This proves the second assertion.
To prove the third, suppose `|B|<=2`. If `|C|=1`, then `|B∪C|<=3`; that induced subgraph has a Hamilton tight path, which together with `A` gives a spanning two-path cover. Therefore the only remaining possibility is `|B|=|C|=2`. Write
`B=(b_0,b_1)`, `C=(c_0,c_1)`.
The transfer of `c_0` to the right end of `B` is forbidden, so `(b_0,b_1,c_0)` is not tight and `(c_0,b_1,b_0)` is tight. The transfer of `c_1` to the left end of `B` is forbidden, so `(c_1,b_0,b_1)` is not tight and `(b_1,b_0,c_1)` is tight. Hence
`(c_0,b_1,b_0,c_1)`
is a tight path, and together with `A` it gives a spanning two-path cover, again impossible. Thus `|B|>=3`. ∎
### Proposition 6.2. Fixed-pair restrictions and reductions at the ends of the longest component
Write
`A=(v_0,...,v_{ell-1})`.
Choose a nontrivial component
`X_0=(x_0,...,x_m)`
from `{B,C}` and put
`a=x_0`, `c=x_m`.
Then
`(a,v_{ell-1},v_{ell-2})`, `(v_1,v_0,c)`
are tight. Each of the four vertices
`v_{ell-1}`, `v_{ell-2}`, `v_1`, `v_0`
lies outside `{a,c}`. Hence each has both fixed-pair restrictions, and each of its two source paths admits a continuation reduction to the corresponding singleton.
Because `n>10`, one may choose a vertex
`z∉{a,c,v_{ell-1},v_{ell-2},v_1,v_0}`.
For each of the four displayed end vertices `s`, both source reductions can be chosen to end on the same unordered singleton pair `{s,z}`; the two reductions for a fixed `s`, and the reductions for different `s`, remain separate continuation sequences.
For any one of these four vertices `s`, a two-vertex path through `s` with a new endpoint outside `{a,c}` either joins one source path through `s` inside a tight three-vertex path or forms, with that source path, a reversed ordered-edge pair in a tight triple. If one of the continuation sequences just described is an initial segment of a longer continuation sequence, then any later occurrence of the same source support occurs after its singleton stage.
No assertion is made that the several continuation sequences coexist or that one is an initial segment of another.
**Proof.** Proposition 6.1 shows that no endpoint of `B` or `C` can be transferred into `A`. Proposition 3.2 applied to the endpoints `a,c` of `X_0` gives the two displayed tight triples. Since the three components of `A|B|C` are disjoint, the four displayed vertices of `A` lie outside `{a,c}`, so Theorem 4.2 gives both fixed-pair restrictions for all four. The choice of `z` is possible because at most six vertices are excluded and `n>10`; Theorem 4.7 then gives both continuation reductions for each end vertex with the same prescribed second singleton `z`. Proposition 4.8 gives the extension-or-reversal alternative for a new two-vertex path through `s`. If a chosen reduction sequence is an initial segment of a longer continuation sequence, Theorem 4.9 places every later occurrence of the same source support after its singleton stage. ∎
Lexicographic maximality rules out every spanning three-path cover whose component-order triple is lexicographically larger than `lambda(F)`. For every vertex outside a fixed pair, each source path admits a continuation reduction to its singleton. A later occurrence of a source support is ordered after that singleton stage only when both occurrences lie in one continuation sequence.
It remains to establish the following augmentation statement:
> Let `F` be a spanning three-path cover of a smallest counterexample such that no endpoint can be transferred from one component into another component of at least equal order. Then either a spanning two-path cover exists or there is a spanning three-path cover `F'` with `lambda(F')` lexicographically larger than `lambda(F)`.
The statement is not proved here.
## Appendix A. Intersections of ordered paths
For a tight path `P=(p_0,...,p_k)`, an **ordered edge of `P`** is an ordered pair `(p_i,p_{i+1})` for some `0<=i<k`. A tight triple `(x,y,z)` **reverses an ordered edge of P** if `(z,y)` or `(y,x)` is an ordered edge of `P`; equivalently, one of its consecutive ordered pairs is the reverse of an ordered edge of `P`.
### Lemma A.1. Reversed order of common vertices
Let `P=(v_0,...,v_k)` and `Q` be tight paths. Suppose the common vertices of `P,Q` do not occur in the same relative order. Then at least one of the following exists:
1. an ordered edge of `Q` that is the reverse of an ordered edge of `P`;
2. a tight triple on `V(P)∪V(Q)` that reverses an ordered edge of one of the two paths at an intersection with the other;
3. a vertex-simple tight cycle on `V(P)∪V(Q)`.
In particular the lemma applies to two different Hamilton orders on the same vertex set.
**Proof.** Read the common vertices in their order along `Q`. Since the relative orders disagree, there are two consecutive common vertices along `Q`, say `v_i,v_j`, with `i>j`. Let `E` be the subpath of `Q` from `v_i` to `v_j`. By choice, the interior of `E` contains no vertex of `P`.
If `E` is the single ordinary edge from `v_i` to `v_j` and `i=j+1`, outcome 1 holds. Otherwise let `x` be the successor of `v_i` on `E` and `y` the predecessor of `v_j` on `E`. Since `i>j`, we have `i>=1` and `j<=k-1`, so the triples
`(v_{i-1},v_i,x)`, `(y,v_j,v_{j+1})`
are defined. If either is not tight, boundary antisymmetry gives its tight reverse, which gives outcome 2. If both are tight, traverse `E` from `v_i` to `v_j`, then traverse `P` from `v_j` to `v_{i-1}`, and close to `v_i`. The two displayed tight triples supply the joins. The interior of `E` is disjoint from `P`, so the resulting tight cycle is vertex-simple. ∎
### Lemma A.2. Intersection at an extended end
Let `(w;P)` be a left-extended path with
`P=(v_0,...,v_k)`,
and let `Q` be a tight path containing `v_0`. Then at least one of the following exists:
1. a tight path properly containing `P` as an ordered subpath;
2. a tight path properly containing `Q` as an ordered subpath;
3. a vertex-simple tight cycle;
4. a tight triple that reverses an ordered edge of `P` or `Q`.
All vertices in the conclusion lie in `V(P)∪V(Q)∪{w}`.
The same four alternatives hold if `(P;w)` is right-extended and `Q` contains `v_k`; again all vertices in the conclusion lie in `V(P)∪V(Q)∪{w}`.
**Proof.** First suppose `(w;P)` is left-extended. If `k=0`, then either `Q=(v_0)`, in which case `(w,v_0)` is a tight path properly containing both `P` and `Q`, or `Q` properly contains `P`. Hence assume `k>=1`.
If the common vertices of `P,Q` occur in different relative orders, apply Lemma A.1. Outcomes 2 and 3 there already suffice. If outcome 1 there gives a reversed common edge, choose a consecutive tight triple of `(w,P)` containing that edge. Such a triple exists because `(w,P)` has order at least three. This gives outcome 4 here.
Now assume the common vertices occur in the same order. Since `Q` contains `v_0`, the vertex `v_0` is the first common vertex along `Q`.
If `Q` has a predecessor `u` immediately before `v_0`, test `(u,v_0,v_1)`. If it is tight, the initial segment of `Q` ending at `v_0` followed by `P` is a tight path properly containing `P`; the initial segment has no other vertex of `P`. If it is not tight, `(v_1,v_0,u)` is tight and gives outcome 4.
It remains that `Q` starts at `v_0`. If `Q=(v_0)`, then `(w,v_0)` properly contains `Q`. Otherwise write the next vertex of `Q` as `q_1`. If `w∉V(Q)`, test `(w,v_0,q_1)`. If it is tight, `(w,Q)` properly contains `Q`; otherwise `(q_1,v_0,w)` is tight and gives outcome 4.
Finally suppose `w∈V(Q)`. The paths `(w,P)` and `Q` contain the common vertices `w,v_0` in opposite orders. Apply Lemma A.1 to these two paths. Its tight-triple outcome gives outcome 4 here, its cycle outcome gives outcome 3, and its reversed-edge outcome can be placed in a consecutive tight triple of `(w,P)`, giving outcome 4.
Now suppose `(P;w)` is right-extended. If `k=0`, then either `Q=(v_0)`, in which case `(v_0,w)` is a tight path properly containing both `P` and `Q`, or `Q` properly contains `P`. Hence assume `k>=1`.
If the common vertices of `P,Q` occur in different relative orders, apply Lemma A.1. Outcomes 2 and 3 there already suffice. If outcome 1 there gives a reversed common edge, choose a consecutive tight triple of `(P,w)` containing that edge. Such a triple exists because `(P,w)` has order at least three. This gives outcome 4 here.
Now assume the common vertices occur in the same order. Since `Q` contains `v_k`, the vertex `v_k` is the last common vertex along `Q`.
If `Q` has a successor `u` immediately after `v_k`, test `(v_{k-1},v_k,u)`. If it is tight, `P` followed by the terminal segment of `Q` beginning at `v_k` is a tight path properly containing `P`; that terminal segment has no other vertex of `P`. If it is not tight, `(u,v_k,v_{k-1})` is tight and gives outcome 4.
It remains that `Q` ends at `v_k`. If `Q=(v_k)`, then `(v_k,w)` properly contains `Q`. Otherwise write the preceding vertex of `Q` as `q`. If `w∉V(Q)`, test `(q,v_k,w)`. If it is tight, `(Q,w)` properly contains `Q`; otherwise `(w,v_k,q)` is tight and gives outcome 4.
Finally suppose `w∈V(Q)`. The paths `(P,w)` and `Q` contain the common vertices `v_k,w` in opposite orders. Apply Lemma A.1 to these two paths. Its tight-triple outcome gives outcome 4 here, its cycle outcome gives outcome 3, and its reversed-edge outcome can be placed in a consecutive tight triple of `(P,w)`, giving outcome 4. ∎
Every tight cycle has order at most `n-4`, because opening it gives a tight path on the same vertex set.
## Codimension-four Hamiltonian-side reduction
_GN3 document: `c0000001`_
# Codimension-four Hamiltonian-side reduction
Let \`H\` be a boundary tournament with \`V(H)=S \sqcup X\`, where \`|S|=4\`, such that \`H[X]\` is Hamiltonian and \`pc(H)>2\`. This section records the consequences needed from the existence of the Hamiltonian set \`X\`.
## 1. Minimal Hamiltonian side
### Proposition 1.1
There is a nonempty set \`Y\subseteq X\` such that, with \`K=H[S\cup Y]\`, the following hold.
1. \`H[Y]\` is Hamiltonian and \`pc(K)>2\`.
2. If \`Z\subsetneq Y\` is nonempty and \`H[Z]\` is Hamiltonian, then \`pc(H[S\cup Z])\le2\`.
3. If \`D\subset Y\`, \`1\le|D|\le2\`, and \`H[Y-D]\` is Hamiltonian, then \`pc(K-D)=2\`, and every exact two-path cover of \`K-D\` has both components of order at least two.
In particular, the third conclusion applies to deleting the first vertex, last vertex, first two vertices, last two vertices, or both endpoints of any Hamilton ordering of \`Y\`.
**Proof.**
Among all nonempty \`Y\subseteq X\` for which \`H[Y]\` is Hamiltonian and \`pc(H[S\cup Y])>2\`, choose one of minimum cardinality. The family is nonempty because \`X\` itself belongs to it. The second assertion follows immediately from minimality.
Every boundary tournament on at most three vertices is Hamiltonian, so every boundary tournament on at most six vertices has a path cover with at most two components. Therefore \`|V(K)|\ge7\` and \`|Y|\ge3\`.
Now let \`D\subset Y\` satisfy the hypotheses of the third assertion. Since \`Y-D\` is a nonempty proper Hamiltonian subset of \`Y\`, minimality gives \`pc(K-D)\le2\`. If \`K-D\` were Hamiltonian, a Hamilton path of \`K-D\` together with the path on the one- or two-vertex set \`D\` would give a spanning two-path cover of \`K\`, a contradiction. Hence \`pc(K-D)=2\`.
Suppose an exact two-path cover of \`K-D\` had a singleton component \`(w)\`. The set \`D\cup\{w\}\` has order at most three and is therefore Hamiltonian. Replacing the singleton by a Hamilton path on \`D\cup\{w\}\` would again give a spanning two-path cover of \`K\`, a contradiction. Thus both components of every exact two-path cover of \`K-D\` have order at least two. ∎
For the remainder of this section fix such a minimal set \`Y\`, write \`m=|Y|\`, and let \`Y=(x_0,x_1,\ldots,x_{m-1})\` be any Hamilton ordering. Put \`L=x_0\` and \`R=x_{m-1}\`. For \`0<=i<=j<=m-1\`, write \`Y[i,j]=(x_i,\ldots,x_j)\`.
## 2. The four-vertex complement
### Proposition 2.1
No tight path of \`K\` has more than \`m\` vertices. The induced boundary tournament \`K[S]\` is non-Hamiltonian.
Moreover, for every Hamilton ordering of \`Y\` with endpoints \`L,R\`:
1. \`K[S\cup\{L\}]\`, \`K[S\cup\{R\}]\`, and \`K[S\cup\{L,R\}]\` are non-Hamiltonian;
2. for every \`s\in S\`, each of \`K[(S-\{s\})\cup\{L\}]\`, \`K[(S-\{s\})\cup\{R\}]\`, and \`K[(S-\{s\})\cup\{L,R\}]\` is Hamiltonian;
3. \`K[S]\` has an edge-order representation in which the three opposite-edge perfect matchings occur as intrinsic strict blocks \`M_{\mathrm{low}}<M_{\mathrm{mid}}<M_{\mathrm{high}}\`.
**Proof.**
Suppose \`P\` is a tight path of \`K\` on more than \`m=|V(K)|-4\` vertices. If \`P\` spans \`K\`, then \`pc(K)=1\`, impossible. Otherwise its complement has order one, two, or three and is Hamiltonian, so a Hamilton path of the complement together with \`P\` gives a spanning two-path cover of \`K\`, again impossible. Thus no tight path has more than \`m\` vertices.
The set \`S\` is non-Hamiltonian, because otherwise Hamilton paths on \`Y\` and \`S\` would two-cover \`K\`.
If \`S\cup\{L\}\` were Hamiltonian, its Hamilton path together with the tight path obtained from the chosen ordering of \`Y\` by deleting \`L\` would two-cover \`K\`. Hence \`S\cup\{L\}\` is non-Hamiltonian, and the same argument applies to \`S\cup\{R\}\`. If \`S\cup\{L,R\}\` were Hamiltonian, its Hamilton path together with the nonempty interior subpath obtained by deleting \`L,R\` from \`Y\` would two-cover \`K\`; hence it too is non-Hamiltonian.
Fix \`s\in S\`. By Lemma 2.3 of the proof spine, the non-Hamiltonian five-set \`S\cup\{L\}\` has an edge-order representation. If \`(S-\{s\})\cup\{L\}\` were also non-Hamiltonian, then in this edge-ordered complete graph the two four-sets \`S\` and \`(S-\{s\})\cup\{L\}\` would both have no increasing Hamilton path and would meet in the three-set \`S-\{s\}\`. Lemma 2.4 of the proof spine would then give an increasing Hamilton path on all five vertices, a contradiction. Thus \`(S-\{s\})\cup\{L\}\` is Hamiltonian. The same argument applies with \`R\`.
Apply \`TOOLKIT/LOCAL_HAMILTON_EXTENSIONS.md\` Section 2 to the six-set \`S\cup\{L,R\}\`. At least four of its five-subsets are Hamiltonian. The two obtained by deleting \`L\` or \`R\` are \`S\cup\{R\}\` and \`S\cup\{L\}\`, which are non-Hamiltonian. Therefore all four sets \`(S-\{s\})\cup\{L,R\}\` are Hamiltonian.
Finally, an edge-order representation of \`S\cup\{L\}\` restricts to one of \`K[S]\`. Since \`K[S]\` is non-Hamiltonian, Lemma 2.4 places the six ordinary edges on \`S\` into three strict opposite-edge perfect-matching blocks. Any two edges from different perfect matchings meet, so their comparison is fixed by the boundary relation on \`S\`; hence the order of the three blocks is intrinsic. ∎
### Proposition 2.2
Assume \`m\ge2\`, and put \`u=x_1\`, \`v=x_{m-2}\`. Then for every \`s\in S\`, the triples \`(u,L,s)\` and \`(s,R,v)\` are tight.
**Proof.**
If \`(s,L,u)\` were tight, then prepending \`s\` to the Hamilton ordering of \`Y\` would give a Hamilton path on \`Y\cup\{s\}\`. The complementary three-set \`S-\{s\}\` is Hamiltonian, so \`K\` would have a spanning two-path cover. Therefore \`(s,L,u)\` is not tight, and boundary antisymmetry gives \`(u,L,s)\`. The terminal assertion is identical. ∎
## 3. Pair-deletion covers and a Hamilton five-subset
Let \`D\` be the first two or last two vertices of a Hamilton ordering of \`Y\`, and let \`T\` be any exact two-path cover of \`K-D\`, whose existence and nontriviality follow from Proposition 1.1.
For \`s\in S\`, put \`C_s=S-\{s\}\` and \`Y_s=(Y-D)\cup\{s\}\`. Let \`\tau_s\` be the number of ordinary edges of \`T\` with one endpoint in \`C_s\` and the other in \`Y_s\`, and let \`b_s\` be the number of nonempty components of the ordinary forest \`T[C_s]\`.
### Proposition 3.1
There is a vertex \`s\in S\` such that \`K[D\cup C_s]\` is Hamiltonian, \`\tau_s\ge2\`, and \`b_s\ge2\`.
**Proof.**
It is enough to treat the case in which \`D=\{u_0,u_1\}\` is the initial pair of a Hamilton ordering \`(u_0,u_1,a_2,\ldots,a_{m-1})\` of \`Y\`; the terminal case uses the corresponding prefix argument. Put
\`G_D=\{s\in S:K[D\cup C_s]\text{ is Hamiltonian}\}\`.
The six-set \`E=S\cup D\` is non-Hamiltonian, since otherwise a Hamilton path on \`E\` together with \`(a_2,\ldots,a_{m-1})\` would two-cover \`K\`. Also \`S\cup\{u_0\}\` is non-Hamiltonian, since otherwise a Hamilton path there together with \`(u_1,a_2,\ldots,a_{m-1})\` would two-cover \`K\`. By \`TOOLKIT/LOCAL_HAMILTON_EXTENSIONS.md\` Section 2, at least four five-subsets of \`E\` are Hamiltonian. Hence at least three of the four sets \`D\cup C_s\` are Hamiltonian, so \`|G_D|\ge3\`.
For every \`s\in G_D\`, \`TOOLKIT/PATH_COVER_SURGERY.md\` Section 1, applied with deleted set \`D\` and side \`C_s\`, gives \`\tau_s\ge1\`.
We have \`\delta\ge1\`. Indeed, if \`\delta=0\`, each component of the exact two-path cover \`T\` lies entirely in one side of the nonempty partition \`S|(Y-D)\`. Since both sides are nonempty and \`T\` has exactly two components, one component spans \`S\`, making \`K[S]\` Hamiltonian, a contradiction. Moving \`s\` across the partition gives
\`\tau_s=\delta-d_Y(s)+d_S(s)\`.
At most two vertices of \`S\` satisfy \`\tau_s=1\`. If \`\delta=1\`, cutting the unique crossing edge leaves three blocks; at least two lie in \`S\`, so \`T[S]\` has exactly two blocks. A vertex not incident with the crossing edge and satisfying \`\tau_s=1\` must have \`d_S(s)=0\`, hence is an isolated \`S\`-block, and there is at most one such vertex besides the crossing endpoint. If \`\delta=2\`, the equality \`\tau_s=1\` forces \`(d_Y(s),d_S(s))=(1,0)\`, so at most two vertices qualify. If \`\delta=3\`, it forces \`(2,0)\`, so at most one qualifies. If \`\delta\ge4\`, then \`\tau_s\ge\delta-2\ge2\` because every vertex has ordinary degree at most two.
At most one vertex satisfies \`b_s=1\`. Let \`e\` be the number of ordinary edges of \`T[S]\`. Since \`K[S]\` is non-Hamiltonian and \`T[S]\` is a path forest on four vertices, \`e\le2\`. If \`b_s=1\`, the three vertices of \`C_s\` form one path and therefore contribute two edges of \`T[S]\`; hence \`e=2\` and both edges avoid \`s\`. Two distinct vertices cannot both be avoided by two edges on the remaining two vertices.
Suppose no vertex of \`G_D\` satisfies both \`\tau_s\ge2\` and \`b_s\ge2\`. Every vertex of \`G_D\` with \`b_s\ge2\` then has \`\tau_s=1\`, and every other vertex of \`G_D\` has \`b_s=1\`. The two bounds and \`|G_D|\ge3\` force \`|G_D|=3\`: exactly two vertices have \`\tau_s=1\`, while the third, say \`s_*\`, has \`b_{s_*}=1\` and \`\tau_{s_*}\ge2\`.
Because \`b_{s_*}=1\`, the three vertices of \`C_{s_*}\` form an ordinary path \`x-y-z\` in \`T\`. Since \`K[S]\` is non-Hamiltonian, \`T[S]\` has at most two edges; they are therefore exactly \`xy,yz\`, and \`s_*\` is isolated in \`T[S]\`. The middle vertex \`y\` already has ordinary degree two, while each of \`x,z\` has room for at most one further edge. Hence \`\tau_{s_*}\le2\`, so \`\tau_{s_*}=2\`, with one edge from each of \`x,z\` to \`Y-D\`.
Let \`\delta'\` be the number of ordinary edges of \`T\` crossing \`S|(Y-D)\`. Then \`\delta'\ge2\`. For either endpoint, say \`x\`, we have \`d_S(x)=d_{Y-D}(x)=1\`, so
\`\tau_x=\delta'-d_{Y-D}(x)+d_S(x)=\delta'\ge2\`.
The same holds for \`z\`. Since \`G_D\` consists of \`s_*\` and two of \`x,y,z\`, at least one of \`x,z\` lies in \`G_D\`. For that vertex \`s\`, deleting \`s\` from \`x-y-z\` leaves one edge together with the isolated vertex \`s_*\`, so \`b_s=2\`, a contradiction. ∎
## 4. A five-vertex endpoint construction
Assume now that \`m\ge4\`.
### Proposition 4.1
There are \`s\in S\` and a Hamilton ordering \`M=(m_0,\ldots,m_4)\` of \`\{L,R\}\cup(S-\{s\})\` such that one of the following holds.
For \`0<=i<=j<=4\`, write \`M[i,j]=(m_i,\ldots,m_j)\`; when \`i>j\`, take \`M[i,j]\` to be the empty sequence.
1. \`L=m_p\` for some \`p\in\{0,3,4\}\`, and the two vertex sequences \`M[0,p](x_1,\ldots,x_{m-2})\` and \`(s)M[p+1,4]\`, with empty pieces omitted, partition \`V(K)\` and have exactly one non-tight consecutive triple.
2. \`R=m_q\` for some \`q\in\{0,1,4\}\`, and the two vertex sequences \`M[0,q-1](s)\` and \`(x_1,\ldots,x_{m-2})M[q,4]\`, with empty pieces omitted, partition \`V(K)\` and have exactly one non-tight consecutive triple.
**Proof.**
Write the intrinsic matching blocks of \`S\` as \`M_{\mathrm{low}}<M_{\mathrm{mid}}<M_{\mathrm{high}}\`. By \`TOOLKIT/FOUR_VERTEX_STRUCTURE.md\` Section 3, relative to each of \`L,R\`, one edge of \`M_{\mathrm{mid}}\` is incoming and the other is outgoing.
Relabel \`S=\{a,b,c,z\}\` so that
\`M_{\mathrm{low}}=\{ab,cz\}\`,
\`M_{\mathrm{mid}}=\{ac,bz\}\`,
\`M_{\mathrm{high}}=\{az,bc\}\`,
and so that \`bz\` is outgoing from \`L\` while \`ac\` is incoming to \`L\`. Thus
\`(L,b,z),(L,z,b),(a,c,L),(c,a,L)\`
are tight. The matching-block order also gives
\`(z,b,c),(c,a,z),(z,c,a),(a,c,b),(b,a,c),(b,a,z)\`
tight.
Call a Hamilton ordering of \`\{L,R\}\cup(S-\{s\})\` favorable if \`L\` occurs in position \`0,3,4\` or \`R\` occurs in position \`0,1,4\`. Suppose no favorable ordering exists.
There are two possibilities for the edge of \`M_{\mathrm{mid}}\` incoming to \`R\`.
If \`ac\` is incoming to \`R\`, then \`(a,c,R),(c,a,R),(R,b,z),(R,z,b)\` are tight. Successively testing the Hamilton orders
\`(L,R,z,b,c)\`, \`(R,z,c,a,L)\`, and \`(c,z,R,L,a)\`
forces \`(z,R,L)\`, \`(c,z,R)\`, and \`(a,L,R)\`, respectively. Then \`(z,c,a,L,R)\` is a favorable Hamilton ordering, a contradiction.
If \`bz\` is incoming to \`R\`, then \`(b,z,R),(z,b,R),(R,a,c),(R,c,a)\` are tight. Successively testing
\`(L,b,z,R,c)\`,
\`(c,L,b,z,R)\`,
\`(L,R,c,a,z)\`,
\`(z,c,a,L,R)\`,
\`(L,R,a,c,b)\`,
\`(b,a,c,L,R)\`,
\`(c,R,L,a,z)\`,
\`(c,R,z,a,L)\`,
\`(L,b,a,z,R)\`,
and \`(R,a,b,L,c)\`
forces respectively
\`(c,R,z)\`,
\`(b,L,c)\`,
\`(c,R,L)\`,
\`(R,L,a)\`,
\`(a,R,L)\`,
\`(R,L,c)\`,
\`(z,a,L)\`,
\`(a,z,R)\`,
\`(a,b,L)\`,
and \`(b,a,R)\`.
Then \`(b,a,R,L,c)\` is a favorable Hamilton ordering, again a contradiction.
Hence a favorable ordering exists. In the first alternative of the statement, the only consecutive triple not inherited from \`M\` or the Hamilton ordering of \`Y\` is, according as \`p=0,3,4\`,
\`(s,m_1,m_2)\`, \`(m_2,L,x_1)\`, or \`(m_3,L,x_1)\`.
In the second alternative, the only such triple is, according as \`q=0,1,4\`,
\`(x_{m-2},R,m_1)\`, \`(x_{m-2},R,m_2)\`, or \`(m_2,m_3,s)\`.
If this single new triple were tight, the two displayed sequences would form a spanning two-path cover of \`K\`, contrary to \`pc(K)>2\`. Therefore it is non-tight. ∎
## 5. Exact covers on the eight endpoint/complement vertices
Assume now that \`m\ge6\`, and put \`u=x_1\`, \`v=x_{m-2}\`, and \`N=(x_2,\ldots,x_{m-3})\`. The path \`N\` is nonempty.
Let the two edges of \`M_{\mathrm{mid}}\` be denoted \`I,O\`, where \`I\` is incoming and \`O\` is outgoing at \`L\`.
### Proposition 5.1
At \`R\` exactly one of the following occurs.
1. \`I\` is incoming and \`O\` outgoing. Writing \`O=\{o_0,o_1\}\` and \`I=\{i_0,i_1\}\`, every choice of orientations gives an exact two-path cover
\`(u,L,o_0,o_1)\mid(i_0,i_1,R,v)\`
of the induced boundary tournament on \`S\cup\{L,u,v,R\}\`.
2. \`O\` is incoming and \`I\` outgoing. Every orientation \`(o_0,o_1)\` of \`O\`, together with either orientation of the two-vertex path on \`I\`, gives an exact two-path cover
\`(u,L,o_0,o_1,R,v)\mid I\`.
In either case the induced boundary tournament on \`S\cup\{L,u,v,R\}\` has path-cover number two.
**Proof.**
By Proposition 2.1, both \`S\cup\{L\}\` and \`S\cup\{R\}\` are non-Hamiltonian. Applying \`TOOLKIT/FOUR_VERTEX_STRUCTURE.md\` Section 3 at each endpoint shows that exactly one edge of \`M_{\mathrm{mid}}\` is incoming and the other outgoing there. Relative to the fixed names \`I,O\` at \`L\`, the assignment at \`R\` is therefore exactly one of the two cases in the statement.
In the first case, for either orientation \`(o_0,o_1)\` of \`O\`, the triple \`(L,o_0,o_1)\` is tight; Proposition 2.2 gives \`(u,L,o_0)\`, so \`(u,L,o_0,o_1)\` is a tight four-vertex path. Similarly, for either orientation \`(i_0,i_1)\` of \`I\`, the triple \`(i_0,i_1,R)\` is tight and Proposition 2.2 gives \`(i_1,R,v)\`, so \`(i_0,i_1,R,v)\` is a tight four-vertex path. Their supports partition the eight vertices.
In the second case, \`O\` is outgoing at \`L\` and incoming at \`R\`. Hence for either orientation \`(o_0,o_1)\`, both \`(L,o_0,o_1)\` and \`(o_0,o_1,R)\` are tight. Together with Proposition 2.2 this makes \`(u,L,o_0,o_1,R,v)\` a tight six-vertex path. The remaining two vertices are exactly \`I\`, which form a two-vertex path in either orientation.
Thus the induced boundary tournament on \`S\cup\{L,u,v,R\}\` has path-cover number at most two. If it were Hamiltonian, a Hamilton path on those eight vertices together with the disjoint nonempty tight path \`N\` would two-cover \`K\`, impossible. Hence its path-cover number is exactly two. ∎
### Proposition 5.2
Suppose the second case of Proposition 5.1 holds. Write \`O=\{b,z\}\` and \`W=V(K)-I\`. Then \`K[W]\` has the four exact two-path covers
\`(u,L,b,z)\mid Y[2,m-1]\`,
\`(u,L,z,b)\mid Y[2,m-1]\`,
\`Y[0,m-3]\mid(z,b,R,v)\`,
and
\`Y[0,m-3]\mid(b,z,R,v)\`.
**Proof.**
Proposition 2.2 gives \`(u,L,s)\` and \`(s,R,v)\` tight for every \`s\in S\`. Since \`O=\{b,z\}\` is outgoing at \`L\`, both \`(L,b,z)\` and \`(L,z,b)\` are tight, giving the two left four-vertex paths. Since \`O\` is incoming at \`R\`, both \`(z,b,R)\` and \`(b,z,R)\` are tight, giving the two right four-vertex paths. The complementary long pieces are contiguous subpaths of \`Y\`, so all four displayed pairs are literal two-path covers of \`W\`.
If \`K[W]\` were Hamiltonian, a Hamilton path on \`W\` together with the two-vertex path on \`I\` would give a spanning two-path cover of \`K\`, impossible. Therefore \`pc(K[W])=2\`, and all four displayed covers are exact. ∎
## Codimension-four exterior deletion structure
_GN3 document: `c0000002`_
# Codimension-four exterior deletion structure
Let `K` be a boundary tournament with `pc(K)>2` and
`V(K)=S \sqcup V(P)`, `|S|=4`,
where
`P=(x_0,\ldots,x_{m-1})`
is a Hamilton tight path with `m>=3`. Put `Y=V(P)`, `L=x_0`, and `R=x_{m-1}`.
Only the displayed hypotheses are assumed throughout.
## 1. Universal four-vertex-complement structure
### Proposition 1.1
No tight path of `K` has more than `m` vertices. The induced boundary tournament `K[S]` is non-Hamiltonian.
Moreover, for every Hamilton ordering of `Y` with endpoints `L,R`:
1. `K[S\cup\{L\}]`, `K[S\cup\{R\}]`, and `K[S\cup\{L,R\}]` are non-Hamiltonian;
2. for every `s\in S`, each of `K[(S-\{s\})\cup\{L\}]`, `K[(S-\{s\})\cup\{R\}]`, and `K[(S-\{s\})\cup\{L,R\}]` is Hamiltonian;
3. `K[S]` has an edge-order representation in which the three opposite-edge perfect matchings occur as intrinsic strict blocks `M_{\mathrm{low}}<M_{\mathrm{mid}}<M_{\mathrm{high}}`.
**Proof.**
Suppose `Q` is a tight path of `K` on more than `m=|V(K)|-4` vertices. If `Q` spans `K`, then `pc(K)=1`, impossible. Otherwise its complement has order one, two, or three, so the induced boundary tournament on that complement is Hamiltonian. A Hamilton path of the complement together with `Q` would give a spanning two-path cover of `K`, again impossible. Thus no tight path has more than `m` vertices.
The induced boundary tournament `K[S]` is non-Hamiltonian, because otherwise a Hamilton path on `Y` and a Hamilton path of `K[S]` would two-cover `K`.
If `K[S\cup\{L\}]` were Hamiltonian, its Hamilton path together with the tight path obtained from the chosen ordering of `Y` by deleting `L` would two-cover `K`. Hence `K[S\cup\{L\}]` is non-Hamiltonian, and the same argument applies to `K[S\cup\{R\}]`. If `K[S\cup\{L,R\}]` were Hamiltonian, its Hamilton path together with the nonempty interior subpath obtained by deleting `L,R` from `Y` would two-cover `K`; hence it too is non-Hamiltonian.
Fix `s\in S`. By Lemma 2.3 of the proof spine, the non-Hamiltonian induced subtournament `K[S\cup\{L\}]` has an edge-order representation. If `K[(S-\{s\})\cup\{L\}]` were also non-Hamiltonian, then in this edge-ordered complete graph the two four-sets `S` and `(S-\{s\})\cup\{L\}` would both have no increasing Hamilton path and would meet in the three-set `S-\{s\}`. Lemma 2.4 of the proof spine would then give an increasing Hamilton path on all five vertices, a contradiction. Thus `K[(S-\{s\})\cup\{L\}]` is Hamiltonian. The same argument applies with `R`.
Apply `TOOLKIT/LOCAL_HAMILTON_EXTENSIONS.md` Section 2 to the six-set `S\cup\{L,R\}`. At least four of its five-subsets induce Hamiltonian boundary tournaments. The two obtained by deleting `L` or `R` are `S\cup\{R\}` and `S\cup\{L\}`, whose induced subtournaments are non-Hamiltonian. Therefore every `K[(S-\{s\})\cup\{L,R\}]`, `s\in S`, is Hamiltonian.
Finally, an edge-order representation of `K[S\cup\{L\}]` restricts to one of `K[S]`. Since `K[S]` is non-Hamiltonian, Lemma 2.4 places the six ordinary edges on `S` into three strict opposite-edge perfect-matching blocks. Any two edges from different perfect matchings meet, so their comparison is fixed by the boundary relation on `S`; hence the order of the three blocks is intrinsic. ∎
### Proposition 1.2
Assume `m\ge2`, and put `u=x_1`, `v=x_{m-2}`. Then for every `s\in S`, the triples `(u,L,s)` and `(s,R,v)` are tight.
**Proof.**
If `(s,L,u)` were tight, then prepending `s` to the Hamilton ordering of `Y` would give a Hamilton path on `Y\cup\{s\}`. The induced boundary tournament `K[S-\{s\}]` is Hamiltonian, so `K` would have a spanning two-path cover. Therefore `(s,L,u)` is not tight, and boundary antisymmetry gives `(u,L,s)`. The terminal assertion is identical. ∎
## 2. Exact covers after deleting exterior vertices
### Proposition 2.1
Put
`P_L=(x_1,\ldots,x_{m-1})`, `P_R=(x_0,\ldots,x_{m-2})`, and `Q=(x_1,\ldots,x_{m-2})`.
For every `s\in S` there exist Hamilton paths `A_s^L,A_s^R,A_s^{LR}` on, respectively,
`(S-\{s\})\cup\{L\}`, `(S-\{s\})\cup\{R\}`, `(S-\{s\})\cup\{L,R\}`.
The pairs
`A_s^L\mid P_L`, `P_R\mid A_s^R`, `A_s^{LR}\mid Q`
are exact two-path covers of `K-s`. Moreover every exact two-path cover of `K-s` has both components of order at least three.
For distinct `s,t\in S`, put `B=S-\{s,t\}`. Then `pc(K-\{s,t\})=2`, and exact two-path covers of `K-\{s,t\}` include
`P\mid B`, `P_L\mid C_L`, `P_R\mid C_R`,
where `B` has either of its two orders and `C_L,C_R` are any Hamilton paths on the three-vertex sets `B\cup\{L\}` and `B\cup\{R\}`.
**Proof.**
Proposition 1.1 supplies the three Hamilton paths for every `s\in S`. Each displayed pair has disjoint nonempty supports whose union is `V(K)-\{s\}`; the path `Q` is nonempty because `m>=3`. Thus each pair is a two-path cover of `K-s`.
If `K-s` were Hamiltonian, a Hamilton path on `K-s` together with the singleton path `(s)` would two-cover `K`. Hence `pc(K-s)=2`, and the displayed covers are exact.
Suppose an exact two-path cover of `K-s` had a component with support `C` of order one or two. Then `K[C\cup\{s\}]` has order at most three and is Hamiltonian. Replacing that component by a Hamilton path on `C\cup\{s\}` would give a spanning two-path cover of `K`, a contradiction. Thus both components have order at least three.
Now fix distinct `s,t\in S`. The set `B` has order two, so either ordering is a tight path. Each induced boundary tournament `K[B\cup\{L\}]` and `K[B\cup\{R\}]` has order three and therefore has a Hamilton path. Hence the three displayed pairs are two-path covers of `K-\{s,t\}`. If `K-\{s,t\}` were Hamiltonian, a Hamilton path on it together with the two-vertex path `(s,t)` would two-cover `K`. Therefore `pc(K-\{s,t\})=2`, and all three displayed covers are exact. ∎
## 3. Five-vertex complements
The following three propositions concern an arbitrary Hamilton path with five-vertex complement.
### Proposition 3.1
Let `K` be a boundary tournament with `pc(K)>2`, let `Y=(y_0,...,y_m)` be a Hamilton tight path with `m>=1`, and put `U=V(K)-V(Y)`. If `|U|=5`, then there exist distinct `u,v∈U` such that, for each `w∈{u,v}`, the three induced boundary tournaments
`K[U-{w}]`, `K[(U-{w})∪{y_0}]`, and `K[(U-{w})∪{y_m}]`
are Hamiltonian. Consequently, if `A_w^-` and `A_w^+` are Hamilton paths of the latter two induced subtournaments, then
`A_w^- | (w) | (y_1,...,y_m)`
and
`(y_0,...,y_{m-1}) | (w) | A_w^+`
are spanning three-path covers of `K`.
**Proof.**
If `K[U]` were Hamiltonian, a Hamilton path of `K[U]` together with `Y` would two-cover `K`. Hence `K[U]` is non-Hamiltonian, so Lemma 2.7 gives a set
`A_0={w∈U:K[U-{w}] is Hamiltonian}`
with `|A_0|>=4`.
The induced subtournaments `K[U∪{y_0}]` and `K[U∪{y_m}]` are non-Hamiltonian, since a Hamilton path of either subtournament together with the complementary inherited suffix or prefix of `Y` would two-cover `K`. Each six-set has at least four Hamiltonian five-subsets, while `K[U]` is non-Hamiltonian. Hence
`A_L={w∈U:K[(U-{w})∪{y_0}] is Hamiltonian}`
and
`A_R={w∈U:K[(U-{w})∪{y_m}] is Hamiltonian}`
both have order at least four. Therefore `|A_0∩A_L∩A_R|>=2`.
Choose distinct `u,v` in this intersection. For either `w`, each displayed three-path cover consists of a Hamilton path on one five-set, the singleton `w`, and the complementary inherited subpath of `Y`; their vertex sets are pairwise disjoint and cover `V(K)`. ∎
### Proposition 3.2
Let `K` be a boundary tournament with `pc(K)>2`, let `Y=(y_0,...,y_m)` be a Hamilton tight path with `m>=2`, and put `U=V(K)-V(Y)`. If `|U|=5`, then there exist distinct `u,v∈U` such that, for each `w∈{u,v}`, writing `A=U-{w}`, the induced boundary tournament `K-w` has exact two-path covers
`A_w^- | (y_1,...,y_m)`
and
`(y_0,...,y_{m-1}) | A_w^+`,
where `A_w^-` is a Hamilton path on `A∪{y_0}` and `A_w^+` is a Hamilton path on `A∪{y_m}`.
**Proof.**
Proposition 3.1 gives distinct `u,v∈U` such that, for each `w∈{u,v}`, both induced subtournaments `K[(U-{w})∪{y_0}]` and `K[(U-{w})∪{y_m}]` are Hamiltonian. Choose Hamilton paths `A_w^-` and `A_w^+` on these two sets.
For each such `w`, the paths `A_w^-` and `(y_1,...,y_m)` are nonempty, vertex-disjoint, and their supports partition `V(K)-{w}`; likewise `(y_0,...,y_{m-1})` and `A_w^+` are nonempty, vertex-disjoint, and partition `V(K)-{w}`. Hence the two displayed pairs are two-path covers of `K-w`.
They are exact. If `K-w` were Hamiltonian, a Hamilton path on `K-w` together with the singleton path `(w)` would form a spanning two-path cover of `K`, contradicting `pc(K)>2`. ∎
### Proposition 3.3
Let `K` be a boundary tournament with `pc(K)>2`, let `Y=(y_0,y_1,...,y_m)` be a Hamilton tight path with `m>=3`, and put `U=V(K)-V(Y)`. If `|U|=5`, then there exist distinct `u,v∈U` such that, for each `w∈{u,v}`, all four triples
`(y_1,y_0,w)`, `(y_2,y_1,w)`, `(w,y_m,y_{m-1})`, and `(w,y_{m-1},y_{m-2})`
are tight.
**Proof.**
Since `Y` is Hamiltonian and `pc(K)>2`, `K[U]` is non-Hamiltonian. Lemma 2.7 therefore gives at least four vertices `w∈U` for which `K[U-{w}]` is Hamiltonian; call their set `A_0`.
The six-vertex induced subtournament `K[U∪{y_0}]` is non-Hamiltonian, because otherwise its Hamilton path together with the inherited suffix `(y_1,...,y_m)` would two-cover `K`. At least four of its five-vertex induced subtournaments are Hamiltonian. Since `K[U]` is not, at least four vertices `w∈U` make `K[(U-{w})∪{y_0}]` Hamiltonian; call this set `A_L`. The same argument at `y_m` gives a set `A_R⊆U` of order at least four.
Thus
`|A_0∩A_L∩A_R|>=4+4+4-2·5=2`.
Choose distinct `u,v` in the intersection and fix `w∈{u,v}`.
If `(w,y_0,y_1)` were tight, then `(w,y_0,y_1,...,y_m)` together with a Hamilton path on `U-{w}` would two-cover `K`. Hence `(y_1,y_0,w)` is tight. The right-end argument gives `(w,y_m,y_{m-1})`.
If `(w,y_1,y_2)` were tight, then `(w,y_1,y_2,...,y_m)` together with a Hamilton path on `(U-{w})∪{y_0}` would two-cover `K`. Hence `(y_2,y_1,w)` is tight. The right-hand argument using `K[(U-{w})∪{y_m}]` gives `(w,y_{m-1},y_{m-2})`. ∎
### Corollary 3.4
The three preceding propositions apply directly in `K` to the endpoint truncations `P_L=(x_1,\ldots,x_{m-1})` and `P_R=(x_0,\ldots,x_{m-2})`, whose complements are `S\cup\{L\}` and `S\cup\{R\}`, respectively. Proposition 3.1 applies when the original path has `m>=3`, Proposition 3.2 when `m>=4`, and Proposition 3.3 when `m>=5`.
**Proof.**
Both truncations are inherited Hamilton tight paths on their vertex sets, and their vertex complements in the same ambient tournament `K` are exactly the two displayed five-sets. Each truncation has order `m-1`, so when it is written as `(y_0,...,y_k)` one has `k=m-2`. Thus the hypotheses `k>=1`, `k>=2`, and `k>=3` in Propositions 3.1, 3.2, and 3.3 correspond exactly to `m>=3`, `m>=4`, and `m>=5`. ∎
## 4. A five-vertex endpoint construction
Assume now that `m\ge4`.
### Proposition 4.1
There are `s\in S` and a Hamilton ordering `M=(m_0,\ldots,m_4)` of `\{L,R\}\cup(S-\{s\})` such that one of the following holds.
For `0<=i<=j<=4`, write `M[i,j]=(m_i,\ldots,m_j)`; when `i>j`, take `M[i,j]` to be the empty sequence.
1. `L=m_p` for some `p\in\{0,3,4\}`, and the two vertex sequences `M[0,p](x_1,\ldots,x_{m-2})` and `(s)M[p+1,4]`, with empty pieces omitted, partition `V(K)` and have exactly one non-tight consecutive triple.
2. `R=m_q` for some `q\in\{0,1,4\}`, and the two vertex sequences `M[0,q-1](s)` and `(x_1,\ldots,x_{m-2})M[q,4]`, with empty pieces omitted, partition `V(K)` and have exactly one non-tight consecutive triple.
**Proof.**
Write the intrinsic matching blocks of `S` as `M_{\mathrm{low}}<M_{\mathrm{mid}}<M_{\mathrm{high}}`. By `TOOLKIT/FOUR_VERTEX_STRUCTURE.md` Section 3, relative to each of `L,R`, one edge of `M_{\mathrm{mid}}` is incoming and the other is outgoing.
Relabel `S=\{a,b,c,z\}` so that
`M_{\mathrm{low}}=\{ab,cz\}`,
`M_{\mathrm{mid}}=\{ac,bz\}`,
`M_{\mathrm{high}}=\{az,bc\}`,
and so that `bz` is outgoing from `L` while `ac` is incoming to `L`. Thus
`(L,b,z),(L,z,b),(a,c,L),(c,a,L)`
are tight. The matching-block order also gives
`(z,b,c),(c,a,z),(z,c,a),(a,c,b),(b,a,c),(b,a,z)`
tight.
For `s\in S`, call a Hamilton ordering of `\{L,R\}\cup(S-\{s\})` favorable if `L` occurs in position `0,3,4` or `R` occurs in position `0,1,4`. Suppose no favorable ordering exists for any `s\in S`.
There are two possibilities for the edge of `M_{\mathrm{mid}}` incoming to `R`.
If `ac` is incoming to `R`, then `(a,c,R),(c,a,R),(R,b,z),(R,z,b)` are tight. Successively testing the Hamilton orders
`(L,R,z,b,c)`, `(R,z,c,a,L)`, and `(c,z,R,L,a)`
forces `(z,R,L)`, `(c,z,R)`, and `(a,L,R)`, respectively. Then `(z,c,a,L,R)` is a favorable Hamilton ordering, a contradiction.
If `bz` is incoming to `R`, then `(b,z,R),(z,b,R),(R,a,c),(R,c,a)` are tight. Successively testing
`(L,b,z,R,c)`,
`(c,L,b,z,R)`,
`(L,R,c,a,z)`,
`(z,c,a,L,R)`,
`(L,R,a,c,b)`,
`(b,a,c,L,R)`,
`(c,R,L,a,z)`,
`(c,R,z,a,L)`,
`(L,b,a,z,R)`,
and `(R,a,b,L,c)`
forces respectively
`(c,R,z)`,
`(b,L,c)`,
`(c,R,L)`,
`(R,L,a)`,
`(a,R,L)`,
`(R,L,c)`,
`(z,a,L)`,
`(a,z,R)`,
`(a,b,L)`,
and `(b,a,R)`.
Then `(b,a,R,L,c)` is a favorable Hamilton ordering, again a contradiction.
Hence a favorable ordering exists. In the first alternative of the statement, the only consecutive triple not inherited from `M` or the Hamilton ordering of `Y` is, according as `p=0,3,4`,
`(s,m_1,m_2)`, `(m_2,L,x_1)`, or `(m_3,L,x_1)`.
In the second alternative, the only such triple is, according as `q=0,1,4`,
`(x_{m-2},R,m_1)`, `(x_{m-2},R,m_2)`, or `(m_2,m_3,s)`.
If this single new triple were tight, the two displayed sequences would form a spanning two-path cover of `K`, contrary to `pc(K)>2`. Therefore it is non-tight. ∎
## 5. Exact covers on the eight endpoint/complement vertices
Assume now that `m\ge6`, and put `u=x_1`, `v=x_{m-2}`, and `N=(x_2,\ldots,x_{m-3})`. The path `N` is nonempty.
Let the two edges of `M_{\mathrm{mid}}` be denoted `I,O`, where `I` is incoming and `O` is outgoing at `L`.
### Proposition 5.1
At `R` exactly one of the following occurs.
1. `I` is incoming and `O` outgoing. Writing `O=\{o_0,o_1\}` and `I=\{i_0,i_1\}`, every choice of orientations gives an exact two-path cover
`(u,L,o_0,o_1)\mid(i_0,i_1,R,v)`
of the induced boundary tournament on `S\cup\{L,u,v,R\}`.
2. `O` is incoming and `I` outgoing. Every orientation `(o_0,o_1)` of `O`, together with either orientation of the two-vertex path on `I`, gives an exact two-path cover
`(u,L,o_0,o_1,R,v)\mid I`.
In either case the induced boundary tournament on `S\cup\{L,u,v,R\}` has path-cover number two.
**Proof.**
By Proposition 1.1, both `K[S\cup\{L\}]` and `K[S\cup\{R\}]` are non-Hamiltonian. Applying `TOOLKIT/FOUR_VERTEX_STRUCTURE.md` Section 3 at each endpoint shows that exactly one edge of `M_{\mathrm{mid}}` is incoming and the other outgoing there. Relative to the fixed names `I,O` at `L`, the assignment at `R` is therefore exactly one of the two cases in the statement.
In the first case, for either orientation `(o_0,o_1)` of `O`, the triple `(L,o_0,o_1)` is tight; Proposition 1.2 gives `(u,L,o_0)`, so `(u,L,o_0,o_1)` is a tight four-vertex path. Similarly, for either orientation `(i_0,i_1)` of `I`, the triple `(i_0,i_1,R)` is tight and Proposition 1.2 gives `(i_1,R,v)`, so `(i_0,i_1,R,v)` is a tight four-vertex path. Their supports partition the eight vertices.
In the second case, `O` is outgoing at `L` and incoming at `R`. Hence for either orientation `(o_0,o_1)`, both `(L,o_0,o_1)` and `(o_0,o_1,R)` are tight. Together with Proposition 1.2 this makes `(u,L,o_0,o_1,R,v)` a tight six-vertex path. The remaining two vertices are exactly `I`, which form a two-vertex path in either orientation.
Thus the induced boundary tournament on `S\cup\{L,u,v,R\}` has path-cover number at most two. If it were Hamiltonian, a Hamilton path on those eight vertices together with the disjoint nonempty tight path `N` would two-cover `K`, impossible. Hence its path-cover number is exactly two. ∎
### Proposition 5.2
Suppose the second case of Proposition 5.1 holds. Write `O=\{b,z\}` and `W=V(K)-I`. Then `K[W]` has the four exact two-path covers
`(u,L,b,z)\mid(x_2,\ldots,x_{m-1})`,
`(u,L,z,b)\mid(x_2,\ldots,x_{m-1})`,
`(x_0,\ldots,x_{m-3})\mid(z,b,R,v)`,
and
`(x_0,\ldots,x_{m-3})\mid(b,z,R,v)`.
**Proof.**
Proposition 1.2 gives `(u,L,s)` and `(s,R,v)` tight for every `s\in S`. Since `O=\{b,z\}` is outgoing at `L`, both `(L,b,z)` and `(L,z,b)` are tight, giving the two left four-vertex paths. Since `O` is incoming at `R`, both `(z,b,R)` and `(b,z,R)` are tight, giving the two right four-vertex paths. The complementary long pieces are contiguous subpaths of `P`, so all four displayed pairs are literal two-path covers of `W`.
If `K[W]` were Hamiltonian, a Hamilton path on `W` together with the two-vertex path on `I` would give a spanning two-path cover of `K`, impossible. Therefore `pc(K[W])=2`, and all four displayed covers are exact. ∎
# Core shelf
## Endpoint analysis reduces to multiple transitions, order disagreement, or successful exchange
_GN3 document: `aa506d6f`_
# Endpoint analysis reduces to multiple crossings, order disagreement, or a tight path with non-Hamiltonian five-vertex complement
Let `K` be a minimum-order counterexample to the statement that every boundary tournament admitting a partition
`V(K)=Y disjoint-union F`,
with `|F|=5` and `K[Y]` Hamiltonian, has path-cover number at most two. Fix a Hamilton tight path
`Q=(y_0,...,y_k)`, `k>=3`,
of `K[Y]`.
Then at least one of the following holds.
1. An exact two-path cover of `K-y_0` or `K-y_k` has at least two ordinary edges between its surviving vertices of `Y` and `F`.
2. For the original partition, or for a partition obtained by replacing one endpoint of `Y` by a common singleton attachment vertex `s in F`, there is an endpoint-deletion exact two-path cover with exactly one edge between the two parts whose path on the surviving Hamiltonian-side support has a vertex order different from the inherited Hamilton order. Consequently two Hamilton paths on a common support expose a reversed common ordered edge, a tight triple reversing an ordered edge at an intersection, or a vertex-simple tight cycle.
3. There are distinct `ell,r in F` such that
`(ell,y_1,...,y_{k-1},r)`
is tight and its complementary five-set is non-Hamiltonian.
## Proof
Apply the initial endpoint-cover reduction to exact two-path covers of `K-y_0` and `K-y_k`. For either endpoint, every exact cover satisfies one of three alternatives: at least two edges cross between the surviving Hamiltonian-side support and `F`; exactly one crosses and the Hamilton-side order differs from the inherited order; or exactly one crosses with inherited order and the cover can be replaced by one having a single attachment vertex from `F`.
If the first or second alternative occurs on either side, conclusion 1 or 2 follows.
Assume therefore that both endpoint covers have singleton replacements. Write their attachment vertices as `ell` on the left and `r` on the right. Thus there are exact covers
`(ell,y_1,...,y_k)|A_L`
of `K-y_0` and
`A_R|(y_0,...,y_{k-1},r)`
of `K-y_k`,
where `A_L` is Hamiltonian on `F-{ell}` and `A_R` is Hamiltonian on `F-{r}`.
If `ell!=r`, the initial endpoint-cover reduction gives
`(ell,y_1,...,y_{k-1},r)`
tight with non-Hamiltonian complementary five-set. This is conclusion 3.
It remains that `ell=r=s`. Apply the endpoint-swap theorem. Replacing `y_0` by `s` in the Hamiltonian side produces another partition with Hamiltonian first part and non-Hamiltonian five-vertex complement. Relative to that partition, deleting the opposite endpoint `y_k` gives an exact two-path cover with exactly one crossing, but the path on the surviving Hamiltonian-side support is
`(y_1,...,y_{k-1},s)`
rather than the inherited order
`(s,y_1,...,y_{k-1})`.
Thus conclusion 2 holds. ∎
## At least two successful-exchange exterior vertices obstruct both long sides
_GN3 document: `1194638a`_
# At least two exterior vertices obstruct both long sides
Let `K` be a boundary tournament with `pc(K)>2`. Suppose
`Q=(y_0,M,y_k)`
and
`P=(ell,M,r)`
are tight paths with `|M|>=2`, all four displayed endpoint vertices distinct and outside `M`. Let `T` be a three-vertex set disjoint from these vertices and from `M`, assume
`V(K)=V(M) disjoint-union {y_0,y_k,ell,r} disjoint-union T`,
and put
`S=T union {y_0,y_k}`.
Assume `K[S]` is non-Hamiltonian.
Define
`D={u in S : K[S-{u}] is Hamiltonian}`.
Then `|D|>=4`.
For `u in D`, put
`u in L`
iff `K[(S-{u}) union {ell}]` is Hamiltonian, and
`u in R`
iff `K[(S-{u}) union {r}]` is Hamiltonian.
Then
`|D-L|<=1`,
`|D-R|<=1`,
and hence
`|L intersection R|>=|D|-2>=2`.
Moreover
`L intersection R subseteq T`.
Consequently there are distinct `u,v in T` such that, for each `x in {u,v}`,
both
`K[(S-{x}) union {ell}]`
and
`K[(S-{x}) union {r}]`
are Hamiltonian, while both
`K[M union {ell,x}]`
and
`K[M union {r,x}]`
are non-Hamiltonian.
Thus each of `u,v` cannot be inserted into any position of either tight Hamilton path `(ell,M)` or `(M,r)`, and the local insertion-obstruction theorem gives a bounded obstruction on each side for each vertex.
## Proof
Because `S` is a non-Hamiltonian five-vertex boundary tournament, at most one of its five four-vertex deletions is non-Hamiltonian. Thus `|D|>=4`.
Take distinct `u,v in D` and put `C=S-{u,v}`. Choose a tight Hamilton path on the three-set `C`. Apply the fixed-three-path extension theorem to this path and the exterior vertices `u,v,ell`. At least one of
`S=C union {u,v}`,
`(S-{v}) union {ell}=C union {u,ell}`,
and
`(S-{u}) union {ell}=C union {v,ell}`
is Hamiltonian. Since `S` is not, at least one of `u,v` lies in `L`.
Thus no two vertices of `D` can both lie outside `L`, proving `|D-L|<=1`. The identical argument with `r` gives `|D-R|<=1`. Hence
`|L intersection R| >= |L|+|R|-|D| >= |D|-2 >=2`.
Fix `x in L intersection R`. Since `(S-{x}) union {ell}` is Hamiltonian, its complementary vertex set in `K` is
`M union {r,x}`.
If that complement were Hamiltonian, the two complementary Hamilton paths would form a spanning two-path cover of `K`; hence it is non-Hamiltonian. Similarly Hamiltonicity of `(S-{x}) union {r}` forces
`K[M union {ell,x}]`
non-Hamiltonian.
It remains to exclude `x=y_0,y_k`. The two tight paths `Q=(y_0,M,y_k)` and `P=(ell,M,r)` imply that both
`(y_0,M,r)`
and
`(ell,M,y_k)`
are tight: all internal triples lie in `M`, and the two endpoint triples are inherited one from each of `Q,P`.
If `x=y_0`, the set `M union {r,x}` is Hamiltonian via `(y_0,M,r)`, contradicting the preceding non-Hamiltonicity. If `x=y_k`, the set `M union {ell,x}` is Hamiltonian via `(ell,M,y_k)`, again a contradiction. Therefore every vertex of `L intersection R` lies in `T`.
So at least two distinct vertices `u,v in T` have both side extensions Hamiltonian and both complementary long-side enlargements non-Hamiltonian. Since `(ell,M)` and `(M,r)` are Hamiltonian paths, non-Hamiltonicity after adjoining `x` means insertion of `x` fails in every position of each displayed path. The local insertion-obstruction theorem then supplies the stated bounded obstruction on each side. ∎
## Dual-good deletions activate crossed-support comparison with the exchange endpoints
_GN3 document: `61981620`_
# Two deletion vertices activate crossed-support comparison with the two endpoints
Let `K` be a boundary tournament with `pc(K)>2`. Suppose
`Q=(y_0,M,y_k)`
and
`P=(ell,M,r)`
are tight paths with the same middle `M), let `T` be a disjoint three-vertex set such that
`V(K)=V(M) disjoint-union {y_0,y_k,ell,r} disjoint-union T`,
put
`S=T union {y_0,y_k}`,
and assume `K[S]` is non-Hamiltonian.
Then for at least two distinct vertices `u in T`, both
`K[(S-{u}) union {ell}]`
and
`K[(S-{u}) union {r}]`
are Hamiltonian.
Fix such a vertex `u), and choose Hamilton paths
`H_u^ell` on `(S-{u}) union {ell}`
and
`H_u^r` on `(S-{u}) union {r}`.
Then `K-u` has exact two-path covers
`H_u^ell|(M,r)`
and
`(ell,M)|H_u^r`,
and the crossed-support comparison theorem applies with deleted vertex `u`, exchanged vertices `ell,r`, common short support `S-{u}`, and common long support `V(M)`.
Consequently:
1. if `ell` is internal in `H_u^ell`, deleting `ell` splits `H_u^ell` into two nonempty tight subpaths whose supports partition `S-{u}`, and some ordinary edge of `H_u^r` crosses the resulting three-part partition after adjoining `{r}`;
2. if `r` is internal in `H_u^r`, the symmetric crossing conclusion holds for an edge of `H_u^ell`;
3. if `ell` is an endpoint of `H_u^ell` with neighbor `z in S-{u}`, then beginning `(ell,z,...)` forces `(z,ell,u)` tight, while ending `(...,z,ell)` forces `(u,ell,z)` tight;
4. the symmetric endpoint conclusions hold for `r` in `H_u^r`.
## Proof
The two-sided obstruction theorem supplies at least two distinct `u in T` for which both displayed five-sets are Hamiltonian. Fix one and choose the indicated Hamilton paths.
The complement of `(S-{u}) union {ell}` inside `V(K)-{u}` is `V(M) union {r}`, which is covered by the tight path `(M,r)`. Likewise the complement of `(S-{u}) union {r}` is `V(M) union {ell}`, covered by `(ell,M)`. Hence the two displayed pairs are exact two-path covers of `K-u`.
Put `T_0=S-{u}` and `C=V(M)`. Then
`V(H_u^ell)=T_0 union {ell}`,
`V(H_u^r)=T_0 union {r}`,
`V(M,r)=C union {r}`,
and
`V(ell,M)=C union {ell}`.
Deleting `r` from `(M,r)` and `ell` from `(ell,M)` leaves the same tight path `M` on `C`.
Thus all hypotheses of the cardinality-free crossed-support comparison theorem hold with deleted vertex `u` and exchanged vertices `ell,r`. Its internal and endpoint conclusions are exactly assertions 1--4. ∎
# Utility shelf
## Small-side path covers have bounded interaction skeleton
_GN3 document: `dc6d9363`_
# Small-side path covers have bounded interaction skeleton
Let `T` be an exact `q`-path cover of a vertex set
`X disjoint-union F`,
with `|F|=f`. Cut every ordinary edge of `T` joining `X` to `F`, and let `b_X,b_F` be the resulting numbers of nonempty path blocks contained in `X,F`, respectively. Let `t` be the number of cut edges.
Then
`b_F<=f`,
`b_X<=b_F+q<=f+q`,
and
`t=b_X+b_F-q<=2f`.
In particular, for an exact two-path cover with `|F|=5`,
`t<=10`, `b_X<=7`, and `b_F<=5`. Contracting each monochromatic path block to one vertex therefore gives a disjoint union of two alternating paths on at most twelve vertices.
## Proof
Each resulting `F`-block contains at least one vertex of `F`, and the blocks are pairwise disjoint. Hence
`b_F<=|F|=f`.
Consider one component path of `T`. After cutting its crossing edges, its monochromatic blocks alternate between `X` and `F`. Therefore, within that component, the number of `X`-blocks is at most the number of `F`-blocks plus one. Summing over the `q` components gives
`b_X<=b_F+q<=f+q`.
Before cutting, the ordinary path forest has `q` components. Each of the `t` cut forest edges increases the number of components by one, while after all cuts there are exactly `b_X+b_F` components. Hence
`b_X+b_F=q+t`,
so
`t=b_X+b_F-q`.
Using the preceding bound,
`t<= (b_F+q)+b_F-q=2b_F<=2f`.
For `q=2` and `f=5`, this gives
`t<=10`, `b_X<=7`, and `b_F<=5`. Hence the contracted graph has at most
`b_X+b_F<=12`
vertices, and each of its two components alternates between the two classes. ∎
## Bounded terminal data do not bound common path refinements
_GN3 document: `astr0035`_
# Bounded terminal data and unbounded common refinement
## 1. Exact terminal data for concatenating fixed paths
Let H be a finite boundary tournament. Let B_1,...,B_b be pairwise vertex-disjoint nonempty tight paths whose supports partition V(H). Their internal orders are fixed. For each B_i retain its first two and last two vertices, taking their union when they overlap, and record min(|B_i|,4). Retain the truth value of every tight ordered triple on the retained vertices.
Then these data determine exactly which ordered concatenations of the whole blocks, using each block once in total, form a path cover of H with at most two components. No reversal, splitting, or interleaving of a block is allowed in this assertion.
In particular, let V(K)=Y disjoint-union F with |F|=5, let a,b be distinct vertices of Y, and let T be an exact two-path cover of K-{a,b}. Cut every ordinary edge of T between Y and F, and add singleton blocks (a),(b). There are at most fourteen blocks, and at most thirty-five vertices suffice for the terminal data: at most seven Y-blocks contribute four vertices each, F contributes at most five, and a,b contribute two. This is an exact finite test of the specified block concatenations, not a reduction of K to an induced boundary tournament on thirty-five vertices.
## Proof
A consecutive triple in a concatenation is either internal to one block, intersects two consecutive blocks, or meets three consecutive blocks. Internal triples are tight by hypothesis. In the two-block case its vertices lie in the last two positions of the first block and the first two positions of the second. In the three-block case the middle block is a singleton and the triple consists of the last vertex of the first block, that singleton, and the first vertex of the third. Every such triple is therefore present in the retained table. The truncated length records distinguish singleton, doubleton, tripleton, and longer blocks and determine which tests are required. Checking precisely these triples is both necessary and sufficient.
For the quantitative assertion, the F-blocks are disjoint nonempty subsets of F, so their number is at most five. In each component of T, the monochromatic blocks alternate; hence the number of Y-blocks is at most the number of F-blocks plus two, and is at most seven. Thus T has at most twelve blocks. Adding a,b yields at most fourteen, and retaining terminal vertices uses at most 4*7+5+2=35 vertices. Shortening a long block in an induced subgraph could create unverified internal triples; the finite test instead treats that block's already-tight internal order as fixed. ∎
## 2. No bounded common refinement from bounded monochromatic block counts
For every integer r>=3 there exists a boundary tournament H on {1,...,2r} with tight Hamilton paths
Q=(1,2,...,2r)
and
R=(2,4,...,2r,1,3,...,2r-1)
such that Q and R have no common ordinary edge. Consequently every common refinement into ordered path segments (even allowing reversal) has 2r singleton parts. The only sets contiguous in both orders are singletons and the whole vertex set; hence every proper partition into common contiguous supports also has 2r singleton parts. Both paths nevertheless consist of one monochromatic block relative to the partition with all these vertices on the same side.
## Proof
Declare every consecutive ordered triple of Q and every consecutive ordered triple of R tight. These prescriptions are consistent: a consecutive triple of Q has all its vertices in an interval of three consecutive integers. A triple of R lying within one parity subsequence has span four; a triple crossing the join 2r,1 has span at least five when r>=3. Thus no triple prescribed by R has the same underlying three-set as a triple prescribed by Q. Within either path, different consecutive triples have different underlying sets. No reversal pair receives conflicting prescriptions. Orient every remaining reversal pair arbitrarily to obtain H.
Every ordinary edge of Q joins integers differing by one. Every ordinary edge of R joins integers differing by two, except {2r,1}, whose difference is 2r-1>=5. The ordinary edge sets are disjoint.
Observe that an interval of consecutive integers which occurs contiguously in R either lies in one parity subsequence and hence is a singleton, or crosses the parity join. In the latter case it contains 1 and 2r, so if it is an interval in Q it must be the entire vertex set. Thus the only common interval supports are singletons and the whole set. In particular a proper common interval partition has only singletons. Moreover a common ordered path segment, even allowing its reversal, must be a singleton since the edge sets are disjoint. This proves both assertions about refinements. ∎
## Strategic scope
The fixed-block test is genuinely bounded. It does not prove that a successful concatenation exists, or that cutting at the original Y-F transitions is enough. Comparison with another Hamilton order can require arbitrarily many cuts. An order-disagreement witness is therefore not, by itself, a bounded augmentation certificate. Any finite classification must explicitly state which blocks stay intact, which new cuts are allowed, and how a successful certificate lifts to the original paths.
## Exterior extension amplifies across Hamiltonian deletions of a non-Hamiltonian five-set
_GN3 document: `17693761`_
# Exterior extension amplifies across Hamiltonian deletions of a non-Hamiltonian five-set
Let `H` be a boundary tournament, let `S` be a non-Hamiltonian five-vertex set, and let `z` be a vertex outside `S`.
Define
`D={u in S : H[S-{u}] is Hamiltonian}`
and
`E_z={u in D : H[(S-{u}) union {z}] is Hamiltonian}`.
Then
`|D-E_z|<=1`.
In particular, since every non-Hamiltonian five-set has at least four Hamiltonian four-deletions,
`|E_z|>=3`.
## Proof
Take any two distinct vertices `u,v in D` and put `C=S-{u,v}`. Then `C` has three vertices, so choose any tight Hamilton path `P_C` on `C`.
Apply the certified fixed-three-path extension theorem to `P_C` and the exterior vertices `u,v,z`. At least one of
`C union {u,v}=S`,
`C union {u,z}=(S-{v}) union {z}`,
`C union {v,z}=(S-{u}) union {z}`
is Hamiltonian.
The first is non-Hamiltonian by hypothesis. Therefore at least one of the latter two is Hamiltonian, meaning at least one of `u,v` lies in `E_z`.
Thus no two distinct elements of `D` can both lie outside `E_z`, so `|D-E_z|<=1`.
The certified five-vertex deletion theorem gives `|D|>=4`, hence `|E_z|>=|D|-1>=3`. ∎
## Consequences
This is the parent mechanism behind the successful-exchange side-extension amplification: taking `z=ell` and `z=r` gives two extension families, each missing at most one good deletion, so their intersection has order at least two. The statement is independent of any G15 decomposition.
## Repeated failure of one four-deletion forces a cross-exterior Hamiltonian five-set
_GN3 document: `a2dc4d54`_
# Repeated failure of one four-deletion forces a cross-exterior Hamiltonian five-set
Let `H` be a boundary tournament. Let `A` be a four-vertex set with a Hamilton tight path
`(a,b,c,d)`.
Let `z,z'` be distinct vertices outside `A`.
Assume both five-sets
`A union {z}` and `A union {z'}`
are non-Hamiltonian.
Then both five-sets
`{a,b,c,z,z'}`
and
`{b,c,d,z,z'}`
are Hamiltonian.
## Proof
Apply the certified fixed-three-path extension theorem to the tight path `(a,b,c)` and the three exterior vertices `d,z,z'`. At least one of
`{a,b,c,d,z}=A union {z}`,
`{a,b,c,d,z'}=A union {z'}`,
`{a,b,c,z,z'}`
is Hamiltonian. The first two are non-Hamiltonian by hypothesis, so the third is Hamiltonian.
Apply the same theorem to the tight path `(b,c,d)` and exterior vertices `a,z,z'`. Again the first two resulting five-sets are `A union {z}` and `A union {z'}`, so the remaining set
`{b,c,d,z,z'}`
is Hamiltonian. ∎
## Five-set defect form
Let `S` be a non-Hamiltonian five-set, let `u in S` have `S-u` Hamiltonian, and let `z,z'` lie outside `S`. If both
`(S-u) union {z}` and `(S-u) union {z'}`
are non-Hamiltonian, choose any Hamilton path `(a,b,c,d)` on `S-u`. Then the two cross-exterior five-sets displayed above are Hamiltonian.
This gives a concrete certificate whenever the same exceptional deletion label recurs for two exterior vertices in the extension-amplification framework.
# Toolkit
## Reusable mathematics
_GN3 document: `t0000001`_
# Reusable mathematics
This README is the startup-facing **statement index** for the toolkit. It should represent every retained toolkit theorem or lemma at statement-level detail sufficient to recover its hypotheses and conclusion. The linked toolkit modules are primarily the proof and exposition surface and are opened when a proof or exact local detail is needed.
The toolkit is the preferred home for general-purpose lemmas that can be used independently of the main proof architecture. In particular, reusable mathematics extracted from the proof spine should be cited here rather than recovered from the sequential proof.
Proof-specific minimal-counterexample reductions, extremal-cover statements, and the fixed-pair continuation machinery remain in `PROOF_SPINE/TWO_TIGHT_PATHS.md`; their hypotheses are specialized to the main argument rather than general utilities.
## Foundational boundary-tournament utilities
### [Small-order Hamiltonicity and edge-order representation](SMALL_ORDER_HAMILTONICITY.md)
The universal small-order results formerly available only inside the proof spine:
- comparison-digraph representation of boundary tournaments and characterization of edge-order representability by acyclicity;
- three common-endpoint tight triples force a Hamilton five-path;
- **every non-Hamiltonian five-vertex boundary tournament is represented by a strict edge order**;
- the matching-block classification of a non-Hamiltonian edge-ordered `K_4`;
- two non-Hamiltonian edge-ordered four-sets with a common triple force an increasing Hamilton path on their five-vertex union;
- extension of a fixed tight three-vertex path by at least one of three exterior pairs;
- every six-set has at least four Hamiltonian five-subsets;
- the sharp local restrictions on non-Hamiltonian four-subsets of a five-set.
These statements are universal; no smallest-counterexample hypothesis is used.
### [Path restriction, forest deletion, and ordered-path intersections](PATH_FORESTS_AND_INTERSECTIONS.md)
General path utilities extracted from the proof spine:
- restriction of a left- or right-extended path at its first/last intersection;
- inheritance of an extension by the first or last surviving interval after deletion;
- the ordinary path-forest deletion formula
`comp(F-S)=comp(F)+sum_{v in S}(deg_F(v)-1)-e_F(S)`;
- disagreement in the relative order of common vertices of two tight paths forces a reversed common edge, a reversed-edge tight triple, or a vertex-simple tight cycle;
- the corresponding four-way alternative when another path meets an extended end.
## Hamilton-extension and local-structure tools
### [Local Hamilton extension lemmas](LOCAL_HAMILTON_EXTENSIONS.md)
Bad exterior pairs around a fixed tight three-path form a triangle-free graph; Hamilton five-sets through a prescribed subset of order at most three satisfy the stated density bounds; two parallel middle vertices force a Hamilton four-path. Its small-order inputs are now taken from `SMALL_ORDER_HAMILTONICITY.md`.
### [Parallel middle vertices and match sets relative to a fixed three-set](PARALLEL_TURNS.md)
Exterior vertices relative to a fixed three-vertex set admit the stated match-set classification; three parallel middle vertices admit a Hamilton five-path with at least one of those middle vertices as an endpoint, and no fixed middle vertex can always be prescribed; four parallel middle vertices force a five-path of the form `(x,a,y,c,z)`.
### [Four-vertex structure and fifth-vertex extensions](FOUR_VERTEX_STRUCTURE.md)
Two tight triples with a common first ordered pair on a non-Hamiltonian four-set force one of two matching-block edge orders; the cyclic non-Hamiltonian four-vertex configuration is extended by every fifth vertex; non-Hamiltonian fifth-vertex extensions of an edge-ordered non-Hamiltonian four-set satisfy the stated extreme- and middle-matching restrictions; the ordered two-vertex extension consequence and the `3/5` four-subset density bound hold. The matching-block classification is sourced from `SMALL_ORDER_HAMILTONICITY.md`.
## Path-cover manipulation
### [Path-cover modification lemmas](PATH_COVER_SURGERY.md)
Crossing under absorbable deletion; cyclic rotations of a tight path; one-vertex absorption from opposite end-edge orientations; the path/cycle edge-exchange component formula; joining two path-cover components through a Hamilton path; transitions across a vertex partition; deletion block count and unique-crossing consequences.
### [Cover-comparison and matching lemmas](COVER_COMPARISON.md)
A drop in the number of path-cover components forces an ordinary crossing edge; the Cartesian clause lemma has the stated boundary-tournament specialization; the weighted symmetric difference of two matchings satisfies the stated dichotomy when the total `F`-weight is positive.
### [Cover augmentation lemmas](COVER_AUGMENTATION.md)
Failed concatenations force reversed joining triples; the stated one-cut/two-join and two-cut singleton moves reduce component count; repeated singleton transfers force an extreme triangle edge; globally extreme two-vertex components impose the stated barriers; the three-vertex-component cut and fixed-two-cover concatenation lemmas hold.
### [Path insertion and endpoint replacement lemmas](PATH_INSERTION.md)
Two-sided endpoint replacement forces a reversal in common-vertex order; noninsertability in an edge-ordered complete graph yields the stated barrier gap; complete insertion failure in a boundary tournament yields one of the stated bounded comparison-digraph obstructions; opposite extensions of one ordered pair concatenate. Ordered-path intersection consequences are sourced from `PATH_FORESTS_AND_INTERSECTIONS.md`.
## Johnson-graph and set-family bounds
### [Johnson-graph density bounds](JOHNSON_DENSITY.md)
Local occupancy bounds on `(k+1)`-sets imply the stated quadratic and stronger local-degree bounds for a family of `k`-sets in `J(r,k)`, together with their equality conditions. The boundary-tournament specialization gives the stronger fixed-subset Hamilton-five density hierarchy: for `|S|=s<=3`, the non-Hamiltonian fraction among five-sets containing `S` is at most `(r-s)/[(5-s)(r-4)]`, strictly improving the elementary double-counting hierarchy for `r>10`. This file also retains the extracted `J(10,5)` complement-pair cut bound as a separate reusable set-family lemma.
### [Complement-free Johnson bounds](JOHNSON_COMPLEMENT_BOUNDS.md)
For a complement-free family `F` of `m` five-subsets of a ten-element set, the graph joining pairs with intersection one has average degree at most
`7+18m/252 <= 16`.
## Abstract path systems
### [Ternary path-system insertion theorem](TERNARY_PATH_SYSTEMS.md)
In a reversal-symmetric ternary path system satisfying the stated three-vertex completeness axiom, every exterior vertex has at least two insertion positions in a tight path. Every `n`-vertex system therefore has at least `2^(n-1)` Hamilton tight paths, and the bound is sharp.
## Small-order Hamiltonicity and edge-order representation
_GN3 document: `t0000002`_
# Small-order Hamiltonicity and edge-order representation
Throughout this file, `H` denotes an arbitrary boundary tournament unless another boundary tournament is explicitly named.
Let `K_V` denote the ordinary complete graph on a vertex set `V`. Its edge set consists of the two-element subsets of `V`. The **line graph** `L(K_V)` is the graph with vertex set `E(K_V)` in which two vertices are adjacent exactly when the corresponding ordinary edges meet.
For a boundary tournament `G` on `V`, define the **comparison digraph** `Gamma(G)` as the orientation of `L(K_V)` in which, for distinct `u,v,w`,
`{u,v} -> {v,w}`
if and only if `(u,v,w)` is tight.
An **edge order** on `K_V` is a strict total order `<` on `E(K_V)`. An **edge-ordered complete graph** is a complete graph together with such an edge order. A vertex-simple sequence `(v_0,...,v_k)` is an **increasing path** if
`{v_0,v_1} < {v_1,v_2} < ... < {v_{k-1},v_k}`.
## 1. Comparison representation
For every boundary tournament `G`:
1. `Gamma(G)` is a well-defined orientation of `L(K_V)`.
2. A vertex-simple sequence is a tight path in `G` if and only if its consecutive ordinary edges form a directed chain in `Gamma(G)`.
3. There exists an edge order `<` satisfying
`(u,v,w) is tight iff {u,v}<{v,w}`
for all distinct `u,v,w` if and only if `Gamma(G)` is acyclic.
4. If `Gamma(G)` is cyclic and `C` is a shortest directed cycle, then `C` is chordless in `L(K_V)`. Its underlying ordinary edges are either a three-edge star, the three edges of an ordinary triangle, or the edges of a vertex-simple ordinary cycle of length at least four.
**Proof.** If `{u,v}` and `{v,w}` are incident ordinary edges, the two possible comparison directions correspond to the reversal pair `(u,v,w)` and `(w,v,u)`. Boundary antisymmetry chooses exactly one, proving the first assertion.
For a vertex-simple sequence `P=(v_0,...,v_k)`, put `e_i={v_{i-1},v_i}`. By definition, `e_i->e_{i+1}` is equivalent to tightness of `(v_{i-1},v_i,v_{i+1})`. This proves the second assertion.
If an edge order realizes all tight triples, every comparison arc points from a smaller edge to a larger edge, so `Gamma(G)` is acyclic. Conversely, if `Gamma(G)` is acyclic, it has a vertex of indegree zero: otherwise, following incoming arcs indefinitely would repeat a vertex and create a directed cycle. Repeatedly remove such a vertex to obtain a topological ordering of all vertices. In that total order, each adjacent pair of vertices of `Gamma(G)` appears in the direction of its comparison arc, so the order realizes every tight triple. This proves the third assertion.
Now let
`e_0 -> e_1 -> ... -> e_{m-1} -> e_0`
be a shortest directed cycle. A chord between nonconsecutive cycle vertices, in either direction, combines with one of the two directed segments of the cycle to give a shorter directed cycle. Hence the cycle is chordless.
For `m=3`, three pairwise incident ordinary edges are either the three edges of one triangle or three edges through one common vertex. For `m>=4`, chordlessness implies that nonconsecutive ordinary edges are disjoint and consecutive ones meet. Writing `v_i=e_{i-1}∩e_i` cyclically, the vertices `v_i` are distinct and `e_i={v_i,v_{i+1}}`. Thus the ordinary edges form a vertex-simple cycle. ∎
An **ordinary tournament** on a set `S` is an orientation of the complete graph on `S`. It is **transitive** if its vertices can be linearly ordered so that every arc points forward.
## 2. Three common-endpoint triples force a Hamilton five-path
Let `a,c,p,q,r` be distinct vertices of `H`. If
`(a,p,c)`, `(a,q,c)`, `(a,r,c)`
are tight, then `H[{a,c,p,q,r}]` has a Hamilton tight path.
**Proof.** For distinct `x,y∈{p,q,r}`, define tournaments
`x ->_a y` iff `(x,a,y)` is tight,
`x ->_c y` iff `(x,c,y)` is tight.
If distinct `x,y,z` satisfy `x->_a y->_c z`, then `(x,a,y,c,z)` is a Hamilton tight path. Assume no such mixed chain exists.
The tournament `->_a` cannot be transitive. Otherwise relabel so `x->_a y->_a z` and `x->_a z`. Excluding `(x,a,y,c,z)` forces `z->_c y`, while excluding `(x,a,z,c,y)` forces `y->_c z`, impossible. Hence, after relabelling,
`p->_a q->_a r->_a p`.
Excluding the three mixed chains forces
`r->_c q`, `p->_c r`, `q->_c p`.
Assume still that no Hamilton five-path exists. In each of the following six words the first two consecutive triples are tight, so the third must be non-tight and its reverse must be tight:
| word | forced tight triple |
| --- | --- |
| `a p c r q` | `(q,r,c)` |
| `a q c p r` | `(r,p,c)` |
| `a r c q p` | `(p,q,c)` |
| `p q a r c` | `(a,q,p)` |
| `q r a p c` | `(a,r,q)` |
| `r p a q c` | `(a,p,r)` |
Using these six triples and the same argument gives:
| word | forced tight triple |
| --- | --- |
| `a p r c q` | `(c,r,p)` |
| `a q p c r` | `(c,p,q)` |
| `a r q c p` | `(c,q,r)` |
| `p a q r c` | `(r,q,a)` |
| `q a r p c` | `(p,r,a)` |
| `r a p q c` | `(q,p,a)` |
Set
`u=1` iff `(q,p,r)` is tight,
`v=1` iff `(p,q,r)` is tight,
`w=1` iff `(p,r,q)` is tight.
For each of the eight values of `(u,v,w)`, every consecutive triple of each candidate below is already determined except the displayed reversal pair:
| `(u,v,w)` | first candidate | second candidate | remaining reversal pair |
| --- | --- | --- | --- |
| `000` | `a c q r p` | `r p q c a` | `(a,c,q)` / `(q,c,a)` |
| `100` | `c a r q p` | `q p r a c` | `(c,a,r)` / `(r,a,c)` |
| `010` | `a c p q r` | `q r p c a` | `(a,c,p)` / `(p,c,a)` |
| `110` | `a c p q r` | `q r p c a` | `(a,c,p)` / `(p,c,a)` |
| `001` | `c a p r q` | `r q p a c` | `(c,a,p)` / `(p,a,c)` |
| `101` | `c a p r q` | `r q p a c` | `(c,a,p)` / `(p,a,c)` |
| `011` | `a c r p q` | `p q r c a` | `(a,c,r)` / `(r,c,a)` |
| `111` | `c a q p r` | `p r q a c` | `(c,a,q)` / `(q,a,c)` |
Boundary antisymmetry makes exactly one member of the remaining reversal pair tight. Hence one candidate in each row is a Hamilton tight path, a contradiction. ∎
## 3. Non-Hamiltonian five-sets are edge-orderable
Let `G` be a boundary tournament on exactly five vertices. If `G` has no Hamilton tight path, then `Gamma(G)` is acyclic. Equivalently, there is a strict total order `<` on `E(K_5)` such that, for all distinct `u,v,w`,
`(u,v,w) is tight iff {u,v}<{v,w}`.
**Proof.** Suppose `G` is non-Hamiltonian and `Gamma(G)` contains a directed cycle. Choose a shortest one. By Section 1 it is a star triangle, an ordinary triangle, or a vertex-simple ordinary cycle. A comparison cycle of length five itself gives a Hamilton tight path, so an ordinary cycle can only have length four.
In the case tables below, a three-letter string such as `xyz` abbreviates the assertion that `(x,y,z)` is tight. We repeatedly use the following forcing rule. Let `W=v_0v_1v_2v_3v_4` be a word on all five vertices. If two of its three consecutive triples are known to be tight, then the third cannot also be tight, because otherwise `W` is a Hamilton path. Hence the reverse of the third triple is tight. We write `W => xyz` when this rule forces the triple `(x,y,z)`.
**Ordinary four-cycle.** Normalize the comparison cycle as
`oa -> ab -> bc -> co -> oa`,
so `(o,a,b),(a,b,c),(b,c,o),(c,o,a)` are tight, and let `d` be the fifth vertex. The following complementary branches are exhaustive; each row is read from left to right.
| branch | successive forced triples | contradiction |
| --- | --- | --- |
| `acd, oac` | `bcoad=>dao; boacd=>aob; daobc=>cbo; cdaob=>adc; adcbo=>bcd; oabcd=>bao` | `oab` and `bao` |
| `acd, cao` | `dbcoa=>cbd; doabc=>aod; caodb=>bdo; acbdo=>bca; bcaod=>doa; doabc=>bao` | `oab` and `bao` |
| `dca, bac` | `coabd=>dba; odbac=>bdo; abcod=>doc; bdoca=>aco; dbaco=>abd; coabd=>bao` | `oab` and `bao` |
| `dca, cab` | `dcabo=>oba; dabco=>bad; obadc=>cda; cobad=>boc; bocda=>dco; dcoab=>bao` | `oab` and `bao` |
Thus a shortest comparison cycle is not an ordinary four-cycle.
**Star triangle.** Normalize the star triangle at `o` so that
`(a,o,b)`, `(b,o,c)`, `(c,o,a)`
are tight. For the fifth vertex `d`, set
`S(d)={u in {a,b,c} : (u,o,d) is tight}`.
Cyclic permutation of `a,b,c` preserves the normalized star triangle. Also define the reverse boundary tournament `G^rev` by declaring `(x,y,z)` tight in `G^rev` exactly when `(z,y,x)` is tight in `G`. Reversing the vertex order of a tight path in `G` gives a tight path in `G^rev`. Passing to `G^rev` and then exchanging the labels `b,c` restores the normalized star triangle and sends `|S(d)|` to `3-|S(d)|`. Hence it is enough to treat `|S(d)|=0` and `|S(d)|=1`.
If `|S(d)|=0`, then `(d,o,a),(d,o,b),(d,o,c)` are tight. The following branches are exhaustive:
| branch | successive forced triples | contradiction |
| --- | --- | --- |
| `obc` | `dobca=>acb; doacb=>cao; dcaob=>acd; aobcd=>dcb; daobc=>oad; oadcb=>cda; bocda=>dco; bdcoa=>cdb; acdbo=>obd; caobd=>oac` | `oac` and `cao` |
| `cbo, oda, oba, bdc` | `bdcoa=>ocd; bocda=>adc; obadc=>dab; odabc=>cba; docba=>bco; dbcoa=>cbd; aobdc=>dbo; adboc=>bda; cbdao=>oad; bcoad=>dao` | `oda` and `dao` |
| `cbo, oda, oba, cdb` | `dobac=>cab; cdoba=>odc; odcab=>acd; acdbo=>obd; caobd=>oac; oacdb=>cao` | `oac` and `cao` |
| `cbo, oda, abo, oac` | `abocd=>dco; daboc=>bad; badco=>cda; cdaob=>oad; coadb=>bda; bcoad=>ocb; ocbda=>dbc; doacb=>bca; odbca=>bdo; bdoac=>cao` | `oac` and `cao` |
| `cbo, oda, abo, cao` | `dcaob=>acd; abocd=>dco; bdcoa=>cdb; acdbo=>obd; caobd=>oac` | `oac` and `cao` |
| `cbo, ado, obd` | `caobd=>oac; badoc=>dab; daboc=>oba; dobac=>cab; cdoba=>odc; odcab=>acd; aobdc=>cdb; oacdb=>cao` | `oac` and `cao` |
| `cbo, ado, dbo` | `adboc=>bda; cadob=>dac; bdaco=>oca; dboca=>obd` | `obd` and `dbo` |
If `|S(d)|=1`, cyclically relabel so `S(d)={c}`. Thus `(d,o,a),(d,o,b),(c,o,d)` are tight. The exhaustive branches are:
| branch | successive forced triples | contradiction |
| --- | --- | --- |
| `odb, adc` | `codba=>abd; acodb=>oca; ocabd=>bac; dobac=>abo; abocd=>dco; badco=>dab; daboc=>oba` | `oba` and `abo` |
| `odb, cda` | `cdaob=>oad; acodb=>oca; bocad=>dac; dboca=>obd; obdac=>adb; coadb=>dao` | `oad` and `dao` |
| `bdo, bdc` | `bdcoa=>ocd; abocd=>oba; aobdc=>dbo; dboca=>aco; adboc=>bda; bdaco=>cad; cadob=>oda; codab=>bad; bocda=>adc; obadc=>abo` | `oba` and `abo` |
| `bdo, cdb` | `bdoac=>cao; dcaob=>acd; acdbo=>obd; caobd=>oac` | `oac` and `cao` |
The symmetry already described handles `|S(d)|=2,3`. Thus a shortest comparison cycle is not a star triangle.
**Ordinary triangle.** Normalize the comparison triangle as
`oa -> ab -> bo -> oa`,
so `(o,a,b),(a,b,o),(b,o,a)` are tight, and let `c,d` be the other vertices. For `w∈{c,d}` define
`o in M(w)` iff `(b,w,a)` is tight,
`a in M(w)` iff `(o,w,b)` is tight,
`b in M(w)` iff `(a,w,o)` is tight.
If `M(c)` and `M(d)` shared a coordinate, that coordinate, together with `c,d`, would be the three middle vertices of three tight triples with the same first and third vertices; Section 2 would give a Hamilton five-path. Hence `M(c)` and `M(d)` are disjoint.
Up to cyclic permutation of `o,a,b` and exchange of `c,d`, the disjoint pair is one of
`(∅,∅)`, `(∅,{o})`, `(∅,{o,a})`, `(∅,{o,a,b})`, `({o},{a})`, `({o},{a,b})`.
The match-set definition fixes all mixed triples involving one exterior vertex and two vertices of the fixed three-set used in the following exhaustive table:
| `M(c),M(d)` | extra branch | successive forced triples | contradiction |
| --- | --- | --- | --- |
| `∅,∅` | none | `bcoda=>doc; bdoca=>aco` | `oca` and `aco` |
| `∅,{o}` | none | `bcoda=>doc; bdoca=>aco` | `oca` and `aco` |
| `∅,{o,a}` | none | `bcoda=>doc; docab=>bac; odbac=>abd; odacb=>cad; bocad=>cob; cobda=>dbo; cdboa=>bdc; oabdc=>bao` | `oab` and `bao` |
| `∅,{o,a,b}` | `abc` | `abcod=>doc; badoc=>dab; dabco=>ocb` | `ocb` and `bco` |
| `∅,{o,a,b}` | `cba` | `cbado=>dab; cdabo=>adc; boadc=>dao; cbdao=>dbc; dbcoa=>aoc; bdaoc=>oad` | `oad` and `dao` |
| `({o},{a})` | `ocd` | `bcoda=>doc; docab=>bac; odbac=>abd; oabdc=>cdb; aocdb=>coa; coabd=>bao` | `oab` and `bao` |
| `({o},{a})` | `dco` | `ocadb=>dac; bodac=>dob; dobca=>cbo; dcboa=>bcd; cboad=>dao; bcdao=>adc; badco=>dab; daboc=>cob; adcob=>cda; bcdao=>oad` | `oad` and `dao` |
| `({o},{a,b})` | `cod` | `codba=>abd; abcod=>cba; cbado=>dab; daboc=>cob; cdabo=>adc; adcob=>ocd; oabdc=>cdb; aocdb=>coa; coabd=>bao` | `oab` and `bao` |
| `({o},{a,b})` | `doc` | `docab=>bac; odbac=>abd; badoc=>dab; daboc=>cob; cdabo=>adc; adcob=>ocd; oabdc=>cdb; aocdb=>coa; coabd=>bao` | `oab` and `bao` |
The extra branches in the table are reversal pairs, so the table is exhaustive. Thus no shortest directed comparison cycle exists. Hence `Gamma(G)` is acyclic, and Section 1 gives the required edge order. ∎
## 4. Edge-ordered four-set structure
### 4.1 Matching-block classification
An edge-ordered `K_4` has no increasing Hamilton path if and only if its three opposite-edge perfect matchings occur as three strict consecutive blocks in the edge order.
**Proof.** Let the six edges be
`e_1<e_2<...<e_6`.
If there is no increasing Hamilton path and `e_1,e_2` meet, say `e_1=ab`, `e_2=bc`, then for the fourth vertex `d` the edge `cd` occurs after `e_2`, so `a,b,c,d` is increasing, a contradiction. Hence `e_1,e_2` are disjoint. Dually `e_5,e_6` are disjoint. Therefore `{e_1,e_2}`, `{e_3,e_4}`, `{e_5,e_6}` are exactly the three opposite perfect matchings.
Conversely, in any Hamilton four-vertex word the first and third ordinary edges are disjoint, hence belong to the same opposite-edge matching. If each matching is a strict block, the middle edge lies wholly before or wholly after that block, so it cannot lie strictly between the first and third edges. Thus no Hamilton word is increasing. ∎
### 4.2 Two non-Hamiltonian four-sets force a Hamilton five-set
Let `K` be an edge-ordered complete graph, let `T={u_0,u_1,u_2}`, and let `x,y` be distinct vertices outside `T`. If neither `K[T∪{x}]` nor `K[T∪{y}]` has an increasing Hamilton path, then `K[T∪{x,y}]` has an increasing Hamilton path.
**Proof.** First classify a non-Hamiltonian edge-ordered `K_4`. Let its six ordinary edges be
`e_1<e_2<...<e_6`.
If `e_1,e_2` met, say `e_1=ab` and `e_2=bc`, then for the fourth vertex `d` the edge `cd` occurs after `e_2`, so `a,b,c,d` is an increasing Hamilton path. Thus `e_1,e_2` are disjoint. Dually `e_5,e_6` are disjoint. Hence the pairs `{e_1,e_2}`, `{e_3,e_4}`, `{e_5,e_6}` are the three opposite perfect matchings of `K_4`, ordered in three strict blocks.
Relabel `T` so
`u_1u_2 < u_0u_2 < u_0u_1`.
For `r∈{x,y}`, the matching-block classification in `T∪{r}` forces
`ru_0 < ru_1 < ru_2`.
If `xu_1<yu_1`, then
`u_0,x,u_1,y,u_2`
is increasing. If `yu_1<xu_1`, then
`u_0,y,u_1,x,u_2`
is increasing. ∎
## 5. Extending a fixed three-vertex path
Let `P` be a tight path of `H` on three vertices and let `x,y,z` be distinct vertices of `V(H)-V(P)`. At least one of
`V(P)∪{x,y}`, `V(P)∪{x,z}`, `V(P)∪{y,z}`
has a Hamilton tight path.
**Proof.** Write `P` on vertices `{a,b,c}`. Assume, for contradiction, that all three five-sets obtained by adding two of `x,y,z` are non-Hamiltonian. Section 3 shows that each is represented by an edge order. Their restrictions to the common triangle `{a,b,c}` realize the same comparison orientation, so after relabelling the three vertices we may assume
`ab<ac<bc`.
Thus `(b,a,c)`, `(a,b,c)`, `(a,c,b)` are tight.
For an exterior vertex `w`, define
`a in M(w)` iff `(b,w,c)` is tight,
`b in M(w)` iff `(a,w,c)` is tight,
`c in M(w)` iff `(a,w,b)` is tight.
In an edge order on a five-set containing `{a,b,c,w}`, the relative order of `wa,wb,wc` gives exactly the following six possibilities:
| order of `wa,wb,wc` | `M(w)` |
| --- | --- |
| `wa<wb<wc` | `{a,b,c}` |
| `wa<wc<wb` | `{b,c}` |
| `wb<wa<wc` | `{a,b}` |
| `wb<wc<wa` | `{a}` |
| `wc<wa<wb` | `{c}` |
| `wc<wb<wa` | `∅` |
If two exterior vertices `u,v` shared a coordinate of their match sets, Section 2 would apply. Indeed, a shared coordinate `a` gives the triples `(b,a,c),(b,u,c),(b,v,c)`; a shared coordinate `b` gives `(a,b,c),(a,u,c),(a,v,c)`; and a shared coordinate `c` gives `(a,c,b),(a,u,b),(a,v,b)`. In every case `P∪{u,v}` would have a Hamilton five-path, a contradiction. Hence `M(x),M(y),M(z)` are pairwise disjoint.
Two match sets cannot both be `∅`. If `M(u)=M(v)=∅`, then in the edge order on `P∪{u,v}`,
`uc<ub<ua`, `vc<vb<va`.
If `ub<vb`, the word `c,u,b,v,a` is increasing; if `vb<ub`, the word `c,v,b,u,a` is increasing. Either contradicts non-Hamiltonicity.
Therefore exactly one match set is `∅`. The other two are disjoint nonempty members of
`{a}`, `{c}`, `{a,b}`, `{b,c}`, `{a,b,c}`.
Three nonempty pairwise disjoint match sets would have to be `{a},{b},{c}`, but `{b}` is not in the list. After naming the vertex `x` with empty match set and ordering `y,z`, only the following three cases remain:
`I. M(x)=∅, M(y)={a}, M(z)={c};`
`II. M(x)=∅, M(y)={a}, M(z)={b,c};`
`III. M(x)=∅, M(y)={a,b}, M(z)={c}.`
Assume throughout that the relevant five-set is non-Hamiltonian. We use the same forcing rule: if a five-vertex word has two known tight triples, the reverse of its third triple is forced. The initial mixed triples involving one exterior vertex and two vertices of the fixed three-set are
`I: cxb,cxa,bxa; byc,cya,bya; czb,cza,azb.`
`II: cxb,cxa,bxa; byc,cya,bya; czb,azc,azb.`
`III: cxb,cxa,bxa; byc,ayc,bya; czb,cza,azb.`
The following table completes the three cases:
| case | successive forced triples | contradiction |
| --- | --- | --- |
| I | `bycxa=>xcy; xcyab=>bay; cxbay=>abx; czbxa=>xbz; acxbz=>xca; xcazb=>zac; yzacb=>azy; cxbya=>ybx; ybxac=>cax; bycax=>acy; bacyz=>zyc; bazyc=>zab; czabx=>xba` | `abx` and `xba` |
| II | `cxbya=>ybx; ybxac=>cax; bycax=>acy; bacyz=>zyc; czbxa=>xbz; acxbz=>xca; xcazb=>zac; zacbx=>xbc; xbcya=>ycb; azycb=>yza; yzacb=>bca` | `acb` and `bca` |
| III | `cxbya=>ybx; ybxac=>cax; bycax=>acy; xbacy=>abx; czbxa=>xbz; acxbz=>xca; xcazb=>zac; yzacb=>azy; bacyz=>zyc; bazyc=>zab; czabx=>xba` | `abx` and `xba` |
Each case violates boundary antisymmetry. ∎
## 6. Four Hamiltonian five-subsets of every six-set
For every six-element set `E⊆V(H)`, at least four of the five-element subsets of `E` induce Hamiltonian boundary tournaments.
**Proof.** Let `E` be a six-set and let
`M={e in E : H[E-{e}] has a Hamilton path}`.
Choose any three-set `T⊆E`. The complementary three-set `P=E-T` can be ordered as a tight path: choose any middle vertex, and boundary antisymmetry chooses one of the two orders of the other two vertices. Section 5 applied to this path and the three vertices of `T` shows that `M∩T` is nonempty. If `|E-M|>=3`, choose `T⊆E-M` of order three, a contradiction. Thus `|M|>=4`. ∎
## 7. Non-Hamiltonian four-subsets of a five-set
Let `S⊆V(H)` have order five. If `H[S]` is Hamiltonian, at most three four-element subsets of `S` are non-Hamiltonian. If `H[S]` is non-Hamiltonian, at most one four-element subset of `S` is non-Hamiltonian.
**Proof.** Let `S` be a five-set. If `H[S]` is Hamiltonian, choose a Hamilton order `(v_0,...,v_4)`. Deleting `v_0` or deleting `v_4` leaves a Hamilton tight four-vertex path, so at most three of the five four-subsets are non-Hamiltonian.
Suppose instead that `H[S]` is non-Hamiltonian. By Section 3 it is represented by an edge order. If two distinct four-subsets were both non-Hamiltonian, they would have the form `T∪{x}` and `T∪{y}` for their three-vertex intersection `T`. Their induced edge orders have no increasing Hamilton path, so Section 4.2 gives an increasing Hamilton path on all five vertices. By the representation, this is a tight Hamilton path of `H[S]`, a contradiction. Thus in the non-Hamiltonian case at most one four-subset is non-Hamiltonian. ∎
## Path restriction, forest deletion, and ordered-path intersections
_GN3 document: `t0000003`_
# Path restriction, forest deletion, and ordered-path intersections
Throughout the boundary-tournament statements below, `H` is arbitrary.
If `P=(v_0,...,v_k)` and `w∉V(P)`, write `(w,P)` for `(w,v_0,...,v_k)` and `(P,w)` for `(v_0,...,v_k,w)`. A **left-extended path** is a pair `(w;P)` for which `(w,P)` is tight. A **right-extended path** is a pair `(P;w)` for which `(P,w)` is tight.
For a tight path `P=(p_0,...,p_k)`, an **ordered edge of `P`** is an ordered pair `(p_i,p_{i+1})`. A tight triple `(x,y,z)` **reverses an ordered edge of `P`** if `(z,y)` or `(y,x)` is an ordered edge of `P`.
## 1. Restricting an extended path
Let `(w;P)` be a left-extended path, where
`P=(v_0,...,v_k)`.
Let `Q` be a tight path meeting `P` but not containing `v_0`, and let
`i=min{j : v_j∈V(Q)}`.
Then
`P'=(v_0,...,v_{i-1})`
is a nonempty tight path disjoint from `Q`, and `(w;P')` is left-extended by the same vertex `w`.
If `(P;w)` is right-extended and `Q` meets `P` but avoids `v_k`, let
`j=max{h : v_h∈V(Q)}`.
Then
`P''=(v_{j+1},...,v_k)`
is a nonempty tight path disjoint from `Q`, and `(P'';w)` is right-extended by the same vertex `w`.
**Proof.** Since `Q` avoids `v_0`, the first intersection index satisfies `i>=1`. The sequence `(w,v_0,...,v_{i-1})` is an initial segment of the tight path `(w,P)`, so it is tight. The definition of `i` gives disjointness from `Q`.
For the right-extended case, `Q` avoids `v_k`, so `j<=k-1`. The sequence `(v_{j+1},...,v_k,w)` is a terminal segment of the tight path `(P,w)`, so it is tight. The definition of `j` gives disjointness from `Q`. ∎
## 2. End-interval inheritance
Let `P=(v_0,...,v_k)` be a tight path.
If `(w,P)` is tight and some but not all vertices of `P` are deleted, let
`I=(v_i,...,v_j)`
be the first nonempty interval that remains in the order of `P`. Then `I` is left-extended by `w` when `i=0`, and by the deleted predecessor `v_{i-1}` when `i>0`.
If `(P,w)` is tight and `J=(v_i,...,v_j)` is the last nonempty interval that remains, then `J` is right-extended by `w` when `j=k`, and by the deleted successor `v_{j+1}` when `j<k`.
**Proof.** If `i=0`, `(w,I)` is an initial segment of `(w,P)`. If `i>0`, `(v_{i-1},I)` is a contiguous subpath of `P`.
If `j=k`, `(J,w)` is a terminal segment of `(P,w)`. If `j<k`, `(J,v_{j+1})` is a contiguous subpath of `P`. ∎
If `F` is an ordinary graph and `S⊆V(F)`, write `F-S` for the induced subgraph on `V(F)-S`, `deg_F(v)` for the ordinary degree of `v`, `e_F(S)` for the number of ordinary edges of `F` with both endpoints in `S`, and `comp(F)` for the number of connected components of `F`, counting isolated vertices.
## 3. Counting components after deletion
Let `F` be an ordinary path forest with `k` components and let `S⊆V(F)`. Then
`comp(F-S)=k+sum_{v in S}(deg_F(v)-1)-e_F(S)`.
If `F` is the ordinary path forest of a tight-path cover, every nonempty component of `F-S` inherits from its path component before deletion a tight vertex order.
**Proof.** Let `N=|V(F)|`. Since `F` is a path forest with `k` components, it has `N-k` edges. Deleting `S` removes
`sum_{v in S} deg_F(v)-e_F(S)`
edges: the degree sum counts an edge internal to `S` twice, so one copy must be subtracted. The remaining graph has
`N-|S|`
vertices and
`N-k-sum_{v in S}deg_F(v)+e_F(S)`
edges. A forest has number of components equal to vertices minus edges, which gives the displayed formula. Tightness of inherited path orders follows because every surviving component is a contiguous subpath of an original tight path. ∎
## 4. Reversed order of common vertices
Let `P=(v_0,...,v_k)` and `Q` be tight paths. Suppose the common vertices of `P,Q` do not occur in the same relative order. Then at least one of the following exists:
1. an ordered edge of `Q` that is the reverse of an ordered edge of `P`;
2. a tight triple on `V(P)∪V(Q)` that reverses an ordered edge of one of the two paths at an intersection with the other;
3. a vertex-simple tight cycle on `V(P)∪V(Q)`.
In particular the lemma applies to two different Hamilton orders on the same vertex set.
**Proof.** Read the common vertices in their order along `Q`. Since the relative orders disagree, there are two consecutive common vertices along `Q`, say `v_i,v_j`, with `i>j`. Let `E` be the subpath of `Q` from `v_i` to `v_j`. By choice, the interior of `E` contains no vertex of `P`.
If `E` is the single ordinary edge from `v_i` to `v_j` and `i=j+1`, outcome 1 holds. Otherwise let `x` be the successor of `v_i` on `E` and `y` the predecessor of `v_j` on `E`. Since `i>j`, we have `i>=1` and `j<=k-1`, so the triples
`(v_{i-1},v_i,x)`, `(y,v_j,v_{j+1})`
are defined. If either is not tight, boundary antisymmetry gives its tight reverse, which gives outcome 2. If both are tight, traverse `E` from `v_i` to `v_j`, then traverse `P` from `v_j` to `v_{i-1}`, and close to `v_i`. The two displayed tight triples supply the joins. The interior of `E` is disjoint from `P`, so the resulting tight cycle is vertex-simple. ∎
## 5. Intersection at an extended end
Let `(w;P)` be a left-extended path with
`P=(v_0,...,v_k)`,
and let `Q` be a tight path containing `v_0`. Then at least one of the following exists:
1. a tight path properly containing `P` as an ordered subpath;
2. a tight path properly containing `Q` as an ordered subpath;
3. a vertex-simple tight cycle;
4. a tight triple that reverses an ordered edge of `P` or `Q`.
All vertices in the conclusion lie in `V(P)∪V(Q)∪{w}`.
The same four alternatives hold if `(P;w)` is right-extended and `Q` contains `v_k`; again all vertices in the conclusion lie in `V(P)∪V(Q)∪{w}`.
**Proof.** First suppose `(w;P)` is left-extended. If `k=0`, then either `Q=(v_0)`, in which case `(w,v_0)` is a tight path properly containing both `P` and `Q`, or `Q` properly contains `P`. Hence assume `k>=1`.
If the common vertices of `P,Q` occur in different relative orders, apply Section 4. Outcomes 2 and 3 there already suffice. If outcome 1 there gives a reversed common edge, choose a consecutive tight triple of `(w,P)` containing that edge. Such a triple exists because `(w,P)` has order at least three. This gives outcome 4 here.
Now assume the common vertices occur in the same order. Since `Q` contains `v_0`, the vertex `v_0` is the first common vertex along `Q`.
If `Q` has a predecessor `u` immediately before `v_0`, test `(u,v_0,v_1)`. If it is tight, the initial segment of `Q` ending at `v_0` followed by `P` is a tight path properly containing `P`; the initial segment has no other vertex of `P`. If it is not tight, `(v_1,v_0,u)` is tight and gives outcome 4.
It remains that `Q` starts at `v_0`. If `Q=(v_0)`, then `(w,v_0)` properly contains `Q`. Otherwise write the next vertex of `Q` as `q_1`. If `w∉V(Q)`, test `(w,v_0,q_1)`. If it is tight, `(w,Q)` properly contains `Q`; otherwise `(q_1,v_0,w)` is tight and gives outcome 4.
Finally suppose `w∈V(Q)`. The paths `(w,P)` and `Q` contain the common vertices `w,v_0` in opposite orders. Apply Section 4 to these two paths. Its tight-triple outcome gives outcome 4 here, its cycle outcome gives outcome 3, and its reversed-edge outcome can be placed in a consecutive tight triple of `(w,P)`, giving outcome 4.
Now suppose `(P;w)` is right-extended. If `k=0`, then either `Q=(v_0)`, in which case `(v_0,w)` is a tight path properly containing both `P` and `Q`, or `Q` properly contains `P`. Hence assume `k>=1`.
If the common vertices of `P,Q` occur in different relative orders, apply Section 4. Outcomes 2 and 3 there already suffice. If outcome 1 there gives a reversed common edge, choose a consecutive tight triple of `(P,w)` containing that edge. Such a triple exists because `(P,w)` has order at least three. This gives outcome 4 here.
Now assume the common vertices occur in the same order. Since `Q` contains `v_k`, the vertex `v_k` is the last common vertex along `Q`.
If `Q` has a successor `u` immediately after `v_k`, test `(v_{k-1},v_k,u)`. If it is tight, `P` followed by the terminal segment of `Q` beginning at `v_k` is a tight path properly containing `P`; that terminal segment has no other vertex of `P`. If it is not tight, `(u,v_k,v_{k-1})` is tight and gives outcome 4.
It remains that `Q` ends at `v_k`. If `Q=(v_k)`, then `(v_k,w)` properly contains `Q`. Otherwise write the preceding vertex of `Q` as `q`. If `w∉V(Q)`, test `(q,v_k,w)`. If it is tight, `(Q,w)` properly contains `Q`; otherwise `(w,v_k,q)` is tight and gives outcome 4.
Finally suppose `w∈V(Q)`. The paths `(P,w)` and `Q` contain the common vertices `v_k,w` in opposite orders. Apply Section 4 to these two paths. Its tight-triple outcome gives outcome 4 here, its cycle outcome gives outcome 3, and its reversed-edge outcome can be placed in a consecutive tight triple of `(P,w)`, giving outcome 4. ∎
## Local Hamilton extension lemmas
_GN3 document: `t0000004`_
# Local Hamilton extension lemmas
Throughout this file, `H` is an arbitrary boundary tournament.
## 1. Bad extension pairs around a tight three-vertex path form a triangle-free graph
Let `P` be a tight path on three vertices, and let `X` be a set of `m>=3` vertices disjoint from `V(P)`. Define a graph `B_P(X)` on vertex set `X` by joining distinct `x,y` exactly when
`H[V(P) union {x,y}]`
has no Hamilton tight path.
Then `B_P(X)` is triangle-free. Consequently
`|E(B_P(X))| <= floor(m^2/4)`,
so at least
`binom(m,2)-floor(m^2/4)`
pairs `{x,y}` extend `P` to a Hamilton five-vertex induced subgraph. Equality in the upper bound occurs exactly when `B_P(X)` is a complete bipartite graph with part sizes `floor(m/2)` and `ceil(m/2)`.
**Proof.** For any three distinct `x,y,z in X`, `SMALL_ORDER_HAMILTONICITY.md` Section 5 says that at least one of
`V(P) union {x,y}`, `V(P) union {x,z}`, `V(P) union {y,z}`
has a Hamilton tight path. Hence `xy,xz,yz` cannot all be edges of `B_P(X)`, so the graph is triangle-free. Mantel's theorem gives the bound and its equality case. ∎
## 2. Density of Hamilton five-sets through a fixed small subset
Let `W` be an `r`-vertex subset of a boundary tournament, where `r>=6`, and let `S subseteq W` have order `s in {0,1,2,3}`. Let `h_5(W;S)` be the number of five-element sets `F` satisfying
`S subseteq F subseteq W`
for which `H[F]` has a Hamilton tight path. Then
`h_5(W;S) >= ((4-s)/(6-s)) binom(r-s,5-s)`.
Thus at least two-thirds of all five-subsets of `W` are Hamiltonian; among the five-subsets containing a prescribed vertex, pair, or triple, the corresponding proportions are at least `3/5`, `1/2`, and `1/3`.
**Proof.** Count pairs `(U,F)` such that
`S subseteq F subset U subseteq W`, `|F|=5`, `|U|=6`,
and `H[F]` is Hamiltonian.
There are `binom(r-s,6-s)` possible six-sets `U`. By `SMALL_ORDER_HAMILTONICITY.md` Section 6, each `U` has at least four Hamiltonian five-subsets. At most `s` of those can fail to contain all of `S`, because a five-subset of `U` is obtained by deleting one vertex. Hence each `U` contributes at least `4-s` admissible pairs.
On the other hand, each Hamiltonian five-set `F` containing `S` lies in exactly `r-5` six-subsets of `W`. Therefore
`(r-5) h_5(W;S) >= (4-s) binom(r-s,6-s)`.
Using
`binom(r-s,6-s)/(r-5)=binom(r-s,5-s)/(6-s)`
gives the claimed bound. ∎
For the stronger Johnson-degree density hierarchy, which improves these fixed-subset bounds for every `r>10`, see `JOHNSON_DENSITY.md` Section 4.
## 3. Two parallel middle vertices force a Hamilton four-path
Let `a,c,x,y` be four distinct vertices of a boundary tournament. If
`(a,x,c)` and `(a,y,c)`
are tight, then at least one of
`(a,x,c,y)`, `(a,y,c,x)`
is a tight Hamilton path on `{a,c,x,y}`.
**Proof.** Exactly one of `(x,c,y)` and `(y,c,x)` is tight. In the first case `(a,x,c,y)` is tight; in the second `(a,y,c,x)` is tight. ∎
## Parallel middle vertices and match sets relative to a fixed three-set
_GN3 document: `t0000005`_
# Parallel middle vertices and match sets relative to a fixed three-set
## 1. Match sets relative to a fixed three-set
Fix a three-vertex set
`C={a,b,c}`
in a boundary tournament `H`, and let `E` be a set of vertices disjoint from `C`.
For `x in E` and `d in C`, write `C-{d}={u,v}` and choose the order `(u,v)` for which `(u,d,v)` is tight. Define
`d in M_C(x)`
if and only if `(u,x,v)` is tight.
This is well-defined: reversing `u,v` reverses both tested triples, so boundary antisymmetry preserves whether their tightness agrees.
Define a graph `Gamma_C` on `E` by joining distinct `x,y` when `H[C union {x,y}]` has a Hamilton tight path.
Then:
1. for every `d in C`, the set
`E_d={x in E:d in M_C(x)}`
is a clique of `Gamma_C`;
2. if `U subseteq E` is independent in `Gamma_C`, then the sets `M_C(x)`, `x in U`, are pairwise disjoint, and therefore
`sum_{x in U}|M_C(x)|<=3`;
3. if `U={x,y,z}` is independent, then, after relabelling the vertices of `C` and the three exterior vertices, the triple of match sets is exactly one of
`(emptyset,emptyset,emptyset)`,
`({a},emptyset,emptyset)`,
`({a,b},emptyset,emptyset)`,
`({a},{b},emptyset)`,
`({a,b,c},emptyset,emptyset)`,
`({a,b},{c},emptyset)`,
`({a},{b},{c})`.
**Proof.** Fix `d in C` and suppose `x,y in E_d`. With `u,v` ordered so that `(u,d,v)` is tight, the triples
`(u,d,v)`, `(u,x,v)`, `(u,y,v)`
are all tight. `SMALL_ORDER_HAMILTONICITY.md` Section 2 gives a Hamilton tight path on
`{u,v,d,x,y}=C union {x,y}`.
Thus `xy` is an edge of `Gamma_C`, proving that `E_d` is a clique.
Hence two vertices of an independent set cannot share a coordinate `d`, so their match sets are pairwise disjoint subsets of the three-element set `C`. This gives the sum bound.
For three exterior vertices, the possible multisets of sizes of three pairwise disjoint subsets of a three-element set are
`(0,0,0)`, `(1,0,0)`, `(2,0,0)`, `(1,1,0)`, `(3,0,0)`, `(2,1,0)`, `(1,1,1)`.
Relabelling the exterior vertices orders the three sizes, and relabelling `a,b,c` gives exactly the seven displayed representatives. ∎
## 2. Three parallel middle vertices admit a Hamilton five-path with an exterior endpoint
Let `H` be a boundary tournament, let `a,c,p,q,r` be distinct vertices, and suppose
`(a,p,c)`, `(a,q,c)`, `(a,r,c)`
are tight. Then `H[{a,c,p,q,r}]` has a Hamilton tight path with at least one endpoint in `{p,q,r}`.
In general one cannot prescribe in advance which of `p,q,r` is an endpoint.
**Proof.** `SMALL_ORDER_HAMILTONICITY.md` Section 2 gives some Hamilton tight path on the five vertices. If one endpoint lies in `{p,q,r}`, there is nothing to prove. Suppose instead that the endpoints are `a,c`. After relabelling `p,q,r` as `x,y,z`, the path is one of
`(a,x,y,z,c)`, `(c,x,y,z,a)`.
### Case 1: `(a,x,y,z,c)` is tight
Set
`alpha=[(a,y,z) is tight]`,
`beta=[(z,c,x) is tight]`,
`gamma=[(z,c,y) is tight]`,
`delta=[(c,y,x) is tight]`,
`epsilon=[(x,a,y) is tight]`,
`eta=[(x,a,z) is tight]`.
Consider the seven candidate paths
`a,y,z,c,x`,
`a,z,c,y,x`,
`x,a,y,c,z`,
`x,a,z,c,y`,
`y,a,x,c,z`,
`z,a,x,y,c`,
`z,y,a,x,c`.
Using the three hypotheses and the triples already supplied by `(a,x,y,z,c)`, these candidates are tight respectively under the conditions
`alpha and beta`,
`gamma and delta`,
`epsilon and not gamma`,
`eta and gamma`,
`not epsilon and not beta`,
`not eta and not delta`,
`not alpha and not epsilon`.
Assume all seven fail. Failure of the last condition gives `alpha or epsilon`. If `alpha` is false then `epsilon` is true. If `alpha` is true, failure of the first condition gives `beta` false, and failure of the fifth again gives `epsilon` true. Thus `epsilon` is true. Failure of the third condition gives `gamma` true; failure of the second gives `delta` false; failure of the sixth gives `eta` true. The fourth candidate is then tight, a contradiction.
### Case 2: `(c,x,y,z,a)` is tight
Use
`beta=[(z,c,x) is tight]`,
`gamma=[(z,c,y) is tight]`,
`epsilon=[(x,a,y) is tight]`,
`eta=[(x,a,z) is tight]`.
The five candidates
`a,z,c,x,y`,
`x,a,z,c,y`,
`y,z,a,x,c`,
`x,a,y,c,z`,
`y,a,x,c,z`
are tight respectively under the conditions
`beta`,
`eta and gamma`,
`not eta`,
`epsilon and not gamma`,
`not epsilon and not beta`.
If all fail, then successively `beta` is false, `eta` is true, `gamma` is false, and `epsilon` is false; the fifth candidate is then tight, a contradiction.
Thus some Hamilton five-path has an endpoint in `{p,q,r}`. ∎
### Non-prescribability
Take five vertices `a,c,p,q,r` and order the ten ordinary edges by
`pq < ar < ac < qr < ap < cp < pr < aq < cr < cq`.
Let tight triples be those whose two consecutive ordinary edges increase in this order. Then
`ap<cp`, `aq<cq`, `ar<cr`,
so `(a,p,c),(a,q,c),(a,r,c)` are tight. The Hamilton increasing paths are exactly
`(a,p,r,c,q)` and `(r,a,p,c,q)`.
Thus `p` is never an endpoint of a Hamilton tight path in this example. Relabelling `p,q,r` shows that no fixed middle vertex can be prescribed universally.
## 3. Four parallel middle vertices force a five-path
Let `H` be a boundary tournament, let `a,c` be distinct vertices, and let `S⊆V(H)-{a,c}` have order at least four. Suppose
`(a,s,c)`
is tight for every `s∈S`. Then there are distinct `x,y,z∈S` such that
`(x,a,y,c,z)`
is a tight path.
**Proof.** It is enough to prove the result for a four-element subset of `S`, so assume `|S|=4`. Define tournaments on `S` by
`x->_a y` iff `(x,a,y)` is tight,
`x->_c y` iff `(x,c,y)` is tight.
Suppose no distinct `x,y,z` satisfy `x->_a y->_c z`. If a vertex has indegree at least two in `->_a`, then it has outdegree zero in `->_c`; otherwise two of its `->_a` predecessors together with one `->_c` successor would give such a mixed chain.
The sum of indegrees in the four-vertex tournament `->_a` is six, so some vertex `y` has indegree at least two. Hence `y` has outdegree zero in `->_c`, so `y` is the unique `->_c` sink. Every other vertex has positive `->_c` outdegree and therefore `->_a` indegree at most one. The indegree sum then forces `y` to have `->_a` indegree three and each other vertex to have `->_a` indegree one. Thus `y` is also the `->_a` sink.
Choose `v!=y`. Since `y` is the `->_c` sink, `v->_c y`. The unique `->_a` predecessor `x` of `v` cannot be `y`, because `y` is the `->_a` sink. Hence `x->_a v->_c y`, a mixed chain on three distinct vertices, contradiction. Therefore such a mixed chain exists. Renaming its middle and last vertices as `y,z`, the triples `(x,a,y)`, `(a,y,c)`, `(y,c,z)` are tight, so `(x,a,y,c,z)` is tight. ∎
## Four-vertex structure and fifth-vertex extensions
_GN3 document: `t0000006`_
# Four-vertex structure and fifth-vertex extensions
## 1. Two common-first-pair tight triples force one of two edge-orderable four-vertex forms
Let `H` be a boundary tournament, and let `a,b,c,d` be distinct vertices of `H`. Suppose
`(a,b,c)`, `(a,b,d)`
are tight and `H[{a,b,c,d}]` has no Hamilton tight path. Then the following six additional triples are tight:
`(b,a,c)`, `(b,a,d)`,
`(c,d,a)`, `(d,c,a)`,
`(c,d,b)`, `(d,c,b)`.
Moreover exactly two completions of the boundary relation are possible. Put
`M_0={ab,cd}`, `M_1={ac,bd}`, `M_2={ad,bc}`.
The induced boundary tournament is represented by an edge order in which the three opposite-edge matchings occur as strict blocks, with either
`M_0 < M_1 < M_2`
or
`M_0 < M_2 < M_1`.
**Proof.** Since `(a,b,c)` is tight, the path `(a,b,c,d)` can fail only at `(b,c,d)`; hence `(d,c,b)` is tight. Similarly `(a,b,d,c)` forces `(c,d,b)`.
Now `(a,c,d,b)` has `(c,d,b)` tight, so non-Hamiltonicity forces `(d,c,a)`; and `(a,d,c,b)` similarly forces `(c,d,a)`. Finally `(c,a,b,d)` and `(d,a,b,c)` force `(b,a,c)` and `(b,a,d)`.
The only remaining freedom can be taken to be the reversal pair `(a,c,b)` versus `(b,c,a)`.
Suppose first that `(a,c,b)` is tight. If `(c,b,d)` were tight, then `(a,c,b,d)` would be Hamiltonian; hence `(d,b,c)` is tight. If `(a,d,b)` were tight, then `(a,d,b,c)` would be Hamiltonian because `(d,b,c)` is tight; hence `(b,d,a)` is tight. If `(d,a,c)` were tight, then `(d,a,c,b)` would be Hamiltonian because `(a,c,b)` is tight; hence `(c,a,d)` is tight. These three choices, together with the eight triples already forced, determine every reversal pair and give the block order `M_0<M_1<M_2`.
Suppose instead that `(b,c,a)` is tight. If `(c,a,d)` were tight, then `(b,c,a,d)` would be Hamiltonian; hence `(d,a,c)` is tight. If `(b,d,a)` were tight, then `(b,d,a,c)` would be Hamiltonian because `(d,a,c)` is tight; hence `(a,d,b)` is tight. If `(d,b,c)` were tight, then `(d,b,c,a)` would be Hamiltonian because `(b,c,a)` is tight; hence `(c,b,d)` is tight. These three choices determine every reversal pair and give the block order `M_0<M_2<M_1`.
Both block orders have no increasing Hamilton four-vertex path by the matching-block classification in `SMALL_ORDER_HAMILTONICITY.md` Section 4.1, so both completions occur and there are no others. ∎
## 2. The cyclic non-Hamiltonian four-vertex configuration is extended by every fifth vertex
Let `H` be a boundary tournament and let `X={a,b,c,z} subseteq V(H)`. Suppose the tight triples of `H[X]`, one from each reversal pair, are
`abc, bca, cab, zba, azb, baz, acz, cza, zac, zcb, bzc, cbz`.
Then for every `d in V(H)-X`, the five-set `X union {d}` has a Hamilton tight path in which `d` is one position from an endpoint.
**Proof.** Write `uvw` for the assertion that `(u,v,w)` is tight. Assume for contradiction that no Hamilton tight path on `X union {d}` places `d` one position from an endpoint.
Exactly one of `dab` and `bad` is tight. We treat the two cases separately. In each row below, two consecutive triples of the displayed five-vertex order are tight. Since the order is not Hamiltonian, its third consecutive triple is non-tight, and boundary antisymmetry gives the triple in the last column.
If `dab` is tight, the following implications hold successively:
| tight triples | five-vertex order | forced tight triple |
| --- | --- | --- |
| `dab, abc` | `z d a b c` | `adz` |
| `adz, zcb` | `a d z c b` | `czd` |
| `acz, czd` | `a c z d b` | `bdz` |
| `cab, bdz` | `c a b d z` | `dba` |
| `dba, baz` | `c d b a z` | `bdc` |
| `azb, bdc` | `a z b d c` | `dbz` |
| `dbz, bzc` | `a d b z c` | `bda` |
| `bda, acz` | `b d a c z` | `cad` |
Now `bca`, `cad`, and `adz` are all tight, so
`(b,c,a,d,z)`
is a Hamilton tight path, a contradiction.
If `bad` is tight, the analogous explicit chain is
| tight triples | five-vertex order | forced tight triple |
| --- | --- | --- |
| `zba, bad` | `z b a d c` | `cda` |
| `bzc, cda` | `b z c d a` | `dcz` |
| `dcz, cza` | `b d c z a` | `cdb` |
| `cdb, baz` | `c d b a z` | `abd` |
| `cab, abd` | `c a b d z` | `zdb` |
| `zdb, bca` | `z d b c a` | `cbd` |
| `zcb, cbd` | `z c b d a` | `adb` |
| `cza, adb` | `c z a d b` | `daz` |
Now `cda`, `daz`, and `azb` are all tight, so
`(c,d,a,z,b)`
is a Hamilton tight path, again a contradiction.
Every five-vertex order displayed in the two tables, as well as the final path in each case, places `d` in position `1` or `3` when positions are numbered `0,...,4`. ∎
## 3. A non-Hamiltonian five-set over a non-Hamiltonian edge-ordered four-set
Let `H` be a boundary tournament and let `X subseteq V(H)` have four vertices. Suppose an edge order on the complete graph `K_X` represents `H[X]`, in the sense that for all distinct `u,v,w in X`,
`(u,v,w)` is tight in `H` if and only if `uv<vw`.
Assume `K_X` has no increasing Hamilton path. By `SMALL_ORDER_HAMILTONICITY.md` Section 4.1, its three opposite-edge perfect matchings occur in strict blocks. Write
`M_low < M_mid < M_high`
when every edge of `M_low` precedes every edge of `M_mid`, and every edge of `M_mid` precedes every edge of `M_high`.
Let `d in V(H)-X`. For an edge `{u,v}` of `X`, call `{u,v}`
- **outgoing from `d`** if `(d,u,v)` and `(d,v,u)` are both tight;
- **incoming to `d`** if `(u,v,d)` and `(v,u,d)` are both tight.
Assume `H[X union {d}]` has no Hamilton tight path.
### 3.1 Extreme matching edges
At least one edge of `M_high` is outgoing from `d`, and at least one edge of `M_low` is incoming to `d`.
No edge of `M_high` is incoming to `d`, and no edge of `M_low` is outgoing from `d`.
**Proof.** Normalize
`M_low={{t,r},{l,s}}`,
`M_mid={{t,s},{l,r}}`,
`M_high={{t,l},{r,s}}`.
Suppose neither edge of `M_high` is outgoing from `d`. For `{t,l}`, at least one of `ltd,tld` is tight; for `{r,s}`, at least one of `srd,rsd` is tight. The four possibilities give contradictions as follows:
| tight triples | if this were tight | Hamilton path | forced reverse | Hamilton path |
| --- | --- | --- | --- | --- |
| `ltd, srd` | `rdt` | `l s r d t` | `tdr` | `s l t d r` |
| `ltd, rsd` | `sdt` | `l r s d t` | `tds` | `r l t d s` |
| `tld, srd` | `rdl` | `t s r d l` | `ldr` | `s t l d r` |
| `tld, rsd` | `sdl` | `t r s d l` | `lds` | `r t l d s` |
In each row the first displayed Hamilton path uses the two assumed mixed triples and one tight triple inside `X`. Since that path cannot exist, the middle mixed triple is non-tight and its reverse in the fourth column is tight; the last displayed order is then Hamiltonian. Thus some edge of `M_high` is outgoing.
Now suppose neither edge of `M_low` is incoming to `d`. For `{t,r}`, at least one of `drt,dtr` is tight; for `{l,s}`, at least one of `dsl,dls` is tight. Again the four possibilities are exhaustive:
| tight triples | if this were tight | Hamilton path | forced reverse | Hamilton path |
| --- | --- | --- | --- | --- |
| `drt, dsl` | `rds` | `r d s l t` | `sdr` | `s d r t l` |
| `drt, dls` | `ldr` | `l d r t s` | `rdl` | `r d l s t` |
| `dtr, dsl` | `tds` | `t d s l r` | `sdt` | `s d t r l` |
| `dtr, dls` | `tdl` | `t d l s r` | `ldt` | `l d t r s` |
Thus some edge of `M_low` is incoming.
For the forbidden directions, normalize instead
`M_low={ab,cz}`, `M_mid={ac,bz}`, `M_high={bc,az}`.
Suppose first that the high edge `bc` is incoming to `d`, so `bcd,cbd` are tight. The following five non-Hamiltonian candidate orders force the displayed reverses in sequence:
| tight triples | five-vertex order | forced tight triple |
| --- | --- | --- |
| `abc, bcd` | `a b c d z` | `zdc` |
| `abz, zdc` | `a b z d c` | `dzb` |
| `dzb, zbc` | `a d z b c` | `zda` |
| `zcb, cbd` | `z c b d a` | `adb` |
| `zca, adb` | `z c a d b` | `dac` |
Then `zda,dac,acb` are tight, so `(z,d,a,c,b)` is Hamiltonian, a contradiction. The permutation exchanging `a` with `c` and `b` with `z` preserves all three matching blocks and exchanges the two edges of `M_high`, so neither high edge can be incoming.
Suppose next that the low edge `ab` is outgoing from `d`, so `dab,dba` are tight. The corresponding forcing chain is
| tight triples | five-vertex order | forced tight triple |
| --- | --- | --- |
| `dba, baz` | `c d b a z` | `bdc` |
| `bdc, caz` | `b d c a z` | `acd` |
| `bac, acd` | `b a c d z` | `zdc` |
| `abz, zdc` | `a b z d c` | `dzb` |
| `dzb, zbc` | `a d z b c` | `zda` |
Then `zda,dab,abc` are tight, so `(z,d,a,b,c)` is Hamiltonian. The same permutation `(a c)(b z)` exchanges the two edges of `M_low`, so no low edge is outgoing. ∎
### 3.2 The middle matching alternates
Continue with
`M_low={{t,r},{l,s}}`,
`M_mid={{t,s},{l,r}}`,
`M_high={{t,l},{r,s}}`.
Suppose the high edge `{t,l}` is outgoing from `d` and the low edge `{t,r}` is incoming to `d`. Let `s` be the fourth vertex. Then exactly one of the two middle edges `{t,s}`, `{l,r}` is outgoing from `d`, and the other is incoming to `d`.
**Proof.** Put
`A=[(d,t,s) is tight]`, `B=[(d,s,t) is tight]`,
`C=[(d,r,l) is tight]`, `D=[(d,l,r) is tight]`.
The block order gives the tight triples
`rts, trl, lrs, lst, tsr, rlt, slr, stl`.
We prove the four implications explicitly.
**`A=>B`.** Suppose `A` holds and `B` fails. Then `tsd` is tight. If `sdl` were tight, `(r,t,s,d,l)` would be Hamiltonian, so `lds` is tight. If `rld` were tight, `(t,r,l,d,s)` would be Hamiltonian, so `D=dlr` is tight. If `tdl` were tight, `(t,d,l,r,s)` would be Hamiltonian, so `ldt` is tight. But now `(l,d,t,s,r)` is Hamiltonian, using `ldt`, `A=dts`, and `tsr`. Thus `A=>B`.
**`B=>A`.** Suppose `B` holds and `A` fails. Then `std` is tight. If `tdr` were tight, `(l,s,t,d,r)` would be Hamiltonian, so `rdt` is tight. If `lrd` were tight, `(s,l,r,d,t)` would be Hamiltonian, so `C=drl` is tight. If `sdr` were tight, `(s,d,r,l,t)` would be Hamiltonian, so `rds` is tight. But then `(r,d,s,t,l)` is Hamiltonian, using `rds`, `B=dst`, and `stl`. Thus `B=>A`.
**`C=>D`.** Suppose `C` holds and `D` fails. Then `rld` is tight. If `sdr` were tight, `(s,d,r,l,t)` would be Hamiltonian, so `rds` is tight. If `dst` were tight, `(r,d,s,t,l)` would be Hamiltonian, so `tsd` is tight. If `sdl` were tight, `(r,t,s,d,l)` would be Hamiltonian, so `lds` is tight. But then `(t,r,l,d,s)` is Hamiltonian, using `trl`, `rld`, and `lds`. Thus `C=>D`.
**`D=>C`.** Suppose `D` holds and `C` fails. Then `lrd` is tight. If `tdl` were tight, `(t,d,l,r,s)` would be Hamiltonian, so `ldt` is tight. If `dts` were tight, `(l,d,t,s,r)` would be Hamiltonian, so `std` is tight. If `tdr` were tight, `(l,s,t,d,r)` would be Hamiltonian, so `rdt` is tight. But then `(s,l,r,d,t)` is Hamiltonian, using `slr`, `lrd`, and `rdt`. Thus `D=>C`.
Therefore `A=B` and `C=D`. If both common values were `0`, then `std` and `lrd` would be tight. Non-Hamiltonicity of `(l,s,t,d,r)` forces `tdr` to be non-tight, hence `rdt` is tight; then `(s,l,r,d,t)` is Hamiltonian. If both common values were `1`, then `B=D=1`; non-Hamiltonicity of `(r,d,s,t,l)` forces `rds` to be non-tight, hence `sdr` is tight; then `(s,d,r,l,t)` is Hamiltonian.
Thus the two common values differ. If `A=B=1`, the middle edge `{t,s}` is outgoing from `d`; if `A=B=0`, boundary antisymmetry makes both `tsd,std` tight, so `{t,s}` is incoming. The same statement holds for `{l,r}` using `C=D`, and the two directions are opposite. ∎
### 3.3 An ordered exterior edge extending in both directions forces a Hamilton P5 or P6
Normalize the matching blocks as
`M_low={ab,cz}`, `M_mid={ac,bz}`, `M_high={bc,az}`.
Let `u,v` be distinct vertices of `V(H)-X`. Suppose that for every `x in X`, both
`(x,u,v)` and `(u,v,x)`
are tight. Then either `H[X union {u}]` has a Hamilton tight path or `H[X union {u,v}]` has a Hamilton tight path.
**Proof.** The block order gives the tight triples
`(b,a,c)`, `(c,z,b)`, `(c,a,z)`, `(b,z,a)`.
Assume for contradiction that neither a Hamilton P5 on `X union {u}` nor a Hamilton P6 on `X union {u,v}` exists.
Since `(c,u,v)` and `(u,v,z)` are tight, the path
`(b,a,c,u,v,z)`
forces `(a,c,u)` to be non-tight, so `(u,c,a)` is tight. Similarly, from `(b,u,v)` and `(u,v,a)`, failure of
`(c,z,b,u,v,a)`
forces `(u,b,z)` tight.
Now in `X union {u}`, failure of `(b,u,c,a,z)` forces `(c,u,b)` tight. But then
`(c,u,b,z,a)`
is a Hamilton P5, contradiction. ∎
## 4. Three of the five four-subsets of an edge-ordered K5 are Hamiltonian
Every edge-ordered `K_5` has at least three vertex deletions whose remaining four vertices admit an increasing Hamilton path. Equivalently, at most two of its five induced `K_4`s are non-Hamiltonian.
Consequently, if `h_4(r)` is the number of four-subsets of an edge-ordered `K_r`, `r>=5`, that admit an increasing Hamilton path, then
`h_4(r) >= (3/5) binom(r,4)`.
**Proof.** In a non-Hamiltonian edge-ordered `K_4`, `SMALL_ORDER_HAMILTONICITY.md` Section 4.1 says the three opposite-edge perfect matchings occur in strict blocks. Hence for adjacent edges `e,f`, comparison is preserved on passing to their opposite edges `e*,f*`:
`e<f` if and only if `e*<f*`.
Suppose an edge-ordered `K_5` on `{a,b,c,d,e}` had three non-Hamiltonian vertex-deleted `K_4`s. Relabel so the bad four-sets are obtained by deleting `a,b,c`.
In `{b,c,d,e}`,
`bd<be` iff `ce<cd`.
In `{a,c,d,e}`,
`ce<cd` iff `ad<ae`.
In `{a,b,d,e}`,
`ad<ae` iff `be<bd`.
Chaining gives `bd<be` if and only if `be<bd`, impossible. Thus at most two four-subsets are non-Hamiltonian.
For the density statement, count pairs `(X,Y)` where `X` is a Hamiltonian four-set and `Y` is a five-set containing it. Every five-set contributes at least three such pairs, while every four-set lies in exactly `r-4` five-sets. Hence
`(r-4)h_4(r) >= 3 binom(r,5)`,
which simplifies to the displayed bound. ∎
## Local five-vertex forcing lemmas
_GN3 document: `u0000002`_
# Local five-vertex forcing lemmas
## 1. Two exterior vertices across adjacent positions of a tight path
### Proposition 1.1
Let \`H\` be a boundary tournament, let \`X=(x_0,\ldots,x_{m-1})\` be a tight path in \`H\`, let \`0\le i\le m-3\`, and let \`s,t\` be distinct vertices of \`V(H)-V(X)\`. Suppose
\`(x_{i+1},s,x_i)\`,
\`(x_{i+1},t,x_i)\`,
\`(x_{i+2},s,x_{i+1})\`,
and
\`(x_{i+2},t,x_{i+1})\`
are all tight.
Then exactly one of
\`(x_{i+2},s,x_{i+1},t,x_i)\`
and
\`(x_{i+2},t,x_{i+1},s,x_i)\`
is a tight five-vertex path.
**Proof.**
Exactly one of \`(s,x_{i+1},t)\` and \`(t,x_{i+1},s)\` is tight.
If \`(s,x_{i+1},t)\` is tight, then
\`(x_{i+2},s,x_{i+1},t,x_i)\`
is tight, using the hypotheses \`(x_{i+2},s,x_{i+1})\` and \`(x_{i+1},t,x_i)\`.
If \`(t,x_{i+1},s)\` is tight, then
\`(x_{i+2},t,x_{i+1},s,x_i)\`
is tight, using the other two hypotheses. The two cases are exclusive by boundary antisymmetry. ∎
## 2. A five-vertex consequence of ordered matching blocks
### Proposition 2.1
Let \`H\` be a boundary tournament, and let \`b,c,r,s,u\` be distinct vertices of \`H\`. Suppose \`H[\{b,c,r,s\}]\` is represented by an edge order whose opposite-edge perfect matchings satisfy
\`\{bc,rs\}<\{br,cs\}<\{bs,cr\}\`.
Assume \`H[\{b,c,r,s,u\}]\` is non-Hamiltonian and \`(b,c,u)\` is tight.
Then exactly one of \`(r,u,s)\` and \`(u,s,c)\` is tight.
**Proof.**
The matching-block order gives, among others, the tight triples
\`(b,c,s)\`,
\`(s,c,r)\`,
\`(r,b,s)\`,
\`(c,s,b)\`,
\`(r,s,c)\`,
\`(b,r,c)\`,
and
\`(s,r,b)\`.
Let \`A\` denote the assertion that \`(r,u,s)\` is tight and \`B\` the assertion that \`(u,s,c)\` is tight.
If both are false, boundary antisymmetry gives \`(s,u,r)\` and \`(c,s,u)\` tight. Hence
\`(b,c,s,u,r)\`
is a Hamilton tight path, a contradiction.
Suppose both are true. Since the five-set is non-Hamiltonian, whenever two consecutive triples of a displayed five-vertex order are tight, the reverse of its third consecutive triple is forced. Applying this successively gives
\`b\,u\,s\,c\,r \Rightarrow (s,u,b)\`,
\`b\,r\,u\,s\,c \Rightarrow (u,r,b)\`,
\`c\,u\,r\,b\,s \Rightarrow (r,u,c)\`,
\`r\,u\,c\,s\,b \Rightarrow (s,c,u)\`,
\`r\,s\,c\,u\,b \Rightarrow (b,u,c)\`,
and
\`s\,u\,b\,r\,c \Rightarrow (r,b,u)\`.
Now \`(s,r,b)\`, \`(r,b,u)\`, and \`(b,u,c)\` are tight, so
\`(s,r,b,u,c)\`
is a Hamilton tight path, again a contradiction. Therefore the two assertions cannot agree, and exactly one of the two displayed triples is tight. ∎
## Path-cover modification lemmas
_GN3 document: `t0000007`_
# Path-cover modification lemmas
## 1. Crossing forced by an absorbable deletion
Let `H` be a boundary tournament with `pc(H)>2`. Let `D` be a nonempty proper subset of `V(H)`, put `W=V(H)-D`, and let `S` be a nonempty proper subset of `W`. Suppose `H[D union S]` has a Hamilton tight path.
Then every exact two-path cover of `H-D` contains an ordinary path edge with one endpoint in `S` and the other in `W-S`.
**Proof.** Let `T_1|T_2` be an exact two-path cover of `H-D` and suppose no ordinary edge of either path crosses the cut `S | (W-S)`. Then each connected path component lies wholly in one side of the cut. Since both sides are nonempty and the two paths cover `W`, after exchanging their names we have
`V(T_1)=S`, `V(T_2)=W-S`.
A Hamilton tight path on `D union S`, together with `T_2`, is then a spanning two-path cover of `H`, contradicting `pc(H)>2`. ∎
## 2. Cyclic rotations of a tight path
Let `P=(p_0,p_1,...,p_r)` be a tight path in a boundary tournament, with `r>=2`. Put
`alpha=(p_r,p_0,p_1)`, `beta=(p_{r-1},p_r,p_0)`.
Then the cyclic rotation
`(p_r,p_0,p_1,...,p_{r-1})`
is tight exactly when `alpha` is tight, and
`(p_1,...,p_r,p_0)`
is tight exactly when `beta` is tight. Consequently exactly one of the following four possibilities occurs:
1. both rotations are tight, in which case `(p_0,p_1,...,p_r,p_0)` is a tight cycle;
2. only the first rotation is tight, and `(p_0,p_r,p_{r-1})` is tight;
3. only the second rotation is tight, and `(p_1,p_0,p_r)` is tight;
4. neither rotation is tight, and both `(p_1,p_0,p_r)` and `(p_0,p_r,p_{r-1})` are tight. If `r>=3`, then `(p_1,p_0,p_r,p_{r-1})` is a tight four-vertex path.
**Proof.** Each rotation preserves every old consecutive triple of `P` except its one displayed wrap triple. Thus the two equivalences are immediate. Boundary antisymmetry gives the reverse of each failed wrap triple, and when both fail and `r>=3` the two reversed triples concatenate. ∎
## 3. Opposite orientations of one end edge absorb every exterior vertex
Let `X` be a vertex set in a boundary tournament, and let `u,v` be distinct vertices of `X`. Suppose `H[X]` has a Hamilton tight path beginning with `(u,v)` and also a Hamilton tight path ending with `(v,u)`.
Then for every `d outside X`, the induced tournament `H[X union {d}]` is Hamiltonian.
**Proof.** Exactly one of `(d,u,v)` and `(v,u,d)` is tight. In the first case prepend `d` to the Hamilton path beginning with `(u,v)`; in the second append `d` to the Hamilton path ending with `(v,u)`. ∎
## 4. Component count after a path-cover edge exchange
Let a spanning tight-path cover on `n` vertices have `q` components, so its ordinary path forest has `n-q` edges. Delete `a` ordinary edges and insert `b` ordinary edges. Suppose the resulting spanning ordinary graph has only path components and cycle components, and that every component carries the corresponding tight path or tight cycle order. Let `c` be the number of cycle components.
Opening each cycle by deleting one of its ordinary cycle edges gives a spanning tight-path cover with
`q' = q-(b-a)+c`
components.
**Proof.** After the exchange there are `n-q-a+b` ordinary edges. Opening the `c` cycles leaves `n-q-a+b-c` edges. A spanning path forest with `q'` components has `n-q'` edges, so
`n-q'=n-q-a+b-c`,
which rearranges to the formula. ∎
## 5. Joining two path-cover components through a Hamilton path
Let `X,Y` partition `V(H)`, and suppose `Y` has an exact two-path cover
`U=(x,u_1,...,u_r)`, `V=(v_0,...,v_{s-1},y)`
with `r,s>=0`. Suppose `X` is nonempty and `H[X union {x,y}]` has a Hamilton tight path
`Q=(y,q_1,...,q_t,x)`.
Form the spanning vertex order
`K=(v_0,...,v_{s-1},y,q_1,...,q_t,x,u_1,...,u_r)`,
omitting an empty residual prefix or suffix. Every consecutive triple of `K` is tight except possibly
`alpha=(v_{s-1},y,q_1)` when `s>=1`,
and
`beta=(q_t,x,u_1)` when `r>=1`.
If `pc(H)>1`, at least one existing one of `alpha,beta` is not tight. If `pc(H)>2`, then `r,s>=1` and neither `alpha` nor `beta` is tight. Hence in the latter case both
`(q_1,y,v_{s-1})`, `(u_1,x,q_t)`
are tight.
**Proof.** All triples wholly inside the residual part of `V`, inside `Q`, or inside the residual part of `U` are already tight, so the displayed triples are the only possible failures.
If all existing attachment triples were tight, `K` would be a Hamilton tight path, contradicting `pc(H)>1`.
Now suppose `pc(H)>2`. If `K` had at most one non-tight consecutive triple, then cutting `K` at one of the two ordinary edges inside that triple would split `K` into two nonempty tight paths covering all vertices. Thus `H` would have a spanning two-path cover. Therefore `K` has at least two non-tight consecutive triples. Since only `alpha,beta` can fail, both must exist and both must fail. Boundary antisymmetry gives their reverses. ∎
## 6. Transitions across a vertex partition
Let `Pi={X_1,...,X_m}` be a partition of `V(H)` into nonempty sets, and let `T` be a spanning exact `q`-path cover. Let `t_Pi(T)` be the number of ordinary edges of the paths of `T` whose endpoints lie in different classes of `Pi`.
For each `i`, delete all such cross-class edges and let `b_i(T)` be the number of resulting nonempty path blocks contained in `X_i`. Then
`t_Pi(T)=sum_i b_i(T)-q`
and hence
`t_Pi(T) >= sum_i pc(H[X_i])-q`.
Equality holds exactly when the blocks inside every `X_i` form a minimum path cover of `H[X_i]`.
**Proof.** Deleting one cross-class edge from a path forest increases the number of components by one. Thus all deletions produce exactly `q+t_Pi(T)` blocks, which is `sum_i b_i(T)`. Since the blocks in `X_i` form a path cover of `H[X_i]`, we have `b_i(T)>=pc(H[X_i])`. The inequality and equality condition follow. ∎
## 7. Deletion block count and a unique crossing
Let
`V(H)=D disjoint-union S disjoint-union C`,
where `S,C` are nonempty. Suppose `H[D union S]` has a path cover with `a` components. Let `T` be a `k`-path cover of `H-D`. For each component of `T`, cut every ordinary edge with one endpoint in `S` and the other in `C`, and let `b_C(T)` be the total number of nonempty resulting blocks contained in `C`.
Then
`pc(H) <= a+b_C(T)`.
In particular, if `H[D union S]` is Hamiltonian and `pc(H)>k`, then `b_C(T)>=k`.
Under these latter hypotheses, if `T` has exactly one ordinary edge joining `S` to `C`, then exactly one component of `T` meets both sets. That component consists of one nonempty `S`-block followed by one nonempty `C`-block, or vice versa; every other component of `T` lies wholly in `C`.
Assume now that `H` is a boundary tournament and keep these latter hypotheses. If the unique crossing edge occurs in the order `x,y` with `x in S` and `y in C`, then for every Hamilton tight path of `H[D union S]` ending with the ordered pair `(p,x)`, the triple
`(y,x,p)`
is tight. Dually, if the unique crossing occurs in the order `y,x`, then for every Hamilton tight path of `H[D union S]` beginning with `(x,p)`, the triple
`(p,x,y)`
is tight.
**Proof.** After cutting all `S-C` edges of `T`, the `C`-blocks are disjoint tight paths covering `C`. Together with the given `a`-path cover of `D union S`, they form a spanning path cover of `H`, proving `pc(H)<=a+b_C(T)`.
If `a=1` and `pc(H)>k`, then `pc(H)>=k+1`, so `b_C(T)>=k`. If there is exactly one `S-C` edge in `T`, cutting it produces exactly `k+1` monochromatic blocks. There is at least one `S`-block and at least `k` `C`-blocks, so there is exactly one `S`-block and exactly `k` `C`-blocks. The asserted form of `T` follows.
Suppose the unique crossing is `x,y` with `x in S`, `y in C`, and let `Q` be a Hamilton path of `H[D union S]` ending with `(p,x)`. Replace the unique `S`-block of the mixed component of `T` by `Q`, leaving the adjacent `C`-block and every other component unchanged. If `(p,x,y)` were tight, these `k` paths would cover `H`, contradicting `pc(H)>k`. Hence `(p,x,y)` is not tight, so boundary antisymmetry gives `(y,x,p)`. The other orientation is identical after reversing the order of the replacement. ∎
## 8. Two crossings forced by a three-part one-vertex-deletion cover
Let `K` be a boundary tournament with `pc(K)>2`, let `d in V(K)`, and suppose `K-d` has a spanning three-path cover
`C | A | B`
by nonempty tight paths. Put `S=V(A) union V(B)`, and assume:
- `|S|=4` and `K[S]` is non-Hamiltonian;
- `min{|A|,|B|}<=2`; and
- whenever one of `A,B` has order three, its displayed tight ordering can be extended by `d` at one end to a tight four-vertex path.
Then every exact two-path cover `F` of `K-d` contains at least two ordinary edges whose endpoints lie in different members of the partition
`V(C) | V(A) | V(B)`.
**Proof.** Let `k` be the number of ordinary edges of `F` joining different members of this partition. Since `F` has two components while the partition has three nonempty classes, `k>=1`.
Suppose `k=1`. Cutting the unique crossing edge produces exactly three nonempty path blocks. Hence each of `V(C),V(A),V(B)` induces one connected block of `F`, and one of the three classes is an entire component of `F`.
The isolated class cannot be `V(C)`: otherwise the other component of `F` is a tight Hamilton path on `S`, contradicting non-Hamiltonicity of `K[S]`.
Thus one of `A,B` is an entire component of `F`. Every component of an exact two-path cover of `K-d` has order at least three. Indeed, if a component had order one or two, then adjoining `d` gives a set of order at most three, which is Hamiltonian; replacing that component by a Hamilton path on the enlarged set would give a spanning two-path cover of `K`.
Because `|A|+|B|=4` and `min{|A|,|B|}<=2`, the isolated component therefore has order three and the other of `A,B` has order one. By hypothesis, the displayed three-vertex path extends with `d` to a tight four-vertex path. Replacing that entire component of `F` by the extended path leaves the other component unchanged and yields a spanning two-path cover of `K`, again a contradiction.
Hence `k>=2`. ∎
## Four-vertex structure inside an exact two-path cover
_GN3 document: `u0000001`_
# Four-vertex structure inside an exact two-path cover
## 1. Degree and block counts
### Proposition 1.1
Let \`G\` be a boundary tournament with \`V(G)=S\sqcup Q\`, where \`|S|=4\` and \`G[S]\` is non-Hamiltonian. Let \`T\` be an exact two-path cover of \`G\` in which every vertex of \`S\` has ordinary degree two.
Put
\`e=|E(T[S])|\`,
let \`\delta\` be the number of ordinary edges of \`T\` joining \`S\` to \`Q\`,
and let \`b_Q\` be the number of nonempty components of \`T[Q]\`.
Then
\`e\in\{0,1,2\}\`,
\`\delta=8-2e\`,
and
\`b_Q=6-e\`.
Equivalently, \`(e,\delta,b_Q)\` is one of
\`(2,4,4)\`, \`(1,6,5)\`, \`(0,8,6)\`.
**Proof.**
Since \`G[S]\` is non-Hamiltonian, the path forest \`T[S]\` has at most two ordinary edges, so \`e\in\{0,1,2\}\`.
The sum of ordinary degrees over the four vertices of \`S\` is eight. Each edge of \`T[S]\` contributes two to this sum, while each edge joining \`S\` to \`Q\` contributes one. Hence
\`8=2e+\delta\`,
so \`\delta=8-2e\`.
Cut all \`\delta\` edges joining \`S\` to \`Q\`. The exact two-path forest becomes \`\delta+2\` maximal blocks lying entirely in one side. Since \`T[S]\` is a forest on four vertices with \`e\` edges, it has \`4-e\` components. Therefore
\`b_Q=(\delta+2)-(4-e)=6-e\`.
The three displayed cases follow. ∎
## 2. Alternation of the resulting blocks
### Proposition 2.1
Under the hypotheses of Proposition 1.1, contract every nonempty component of \`T[S]\` and every nonempty component of \`T[Q]\`. Every contracted component coming from \`S\` has degree two, and each of the two resulting path components alternates between components from \`Q\` and components from \`S\`, beginning and ending with a component from \`Q\`.
**Proof.**
Let \`B\` be one component of \`T[S]\`. Since \`G[S]\` is non-Hamiltonian, \`B\` has order at most three. If \`B\` is a singleton, its unique vertex has total ordinary degree two in \`T\`, so exactly two edges leave \`B\`. If \`B\` has order two, its one internal edge uses one incident edge at each endpoint, leaving exactly one outside edge at each endpoint. If \`B\` has order three, its two internal path edges leave one outside edge at each path endpoint and none at the middle vertex. Thus every contracted component coming from \`S\` has degree two.
After contracting all maximal same-side components, the two path components of the ordinary forest of \`T\` remain paths. Their vertices alternate between the two sides by maximality of the blocks. Since no contracted \`S\`-vertex has degree one, no path endpoint lies in \`S\`. Hence every contracted path begins and ends with a component from \`Q\`. ∎
## Path-cover modification and comparison lemmas
_GN3 document: `u0000003`_
# Path-cover modification and comparison lemmas
If `P=(p_0,\ldots,p_r)` and `Q=(q_0,\ldots,q_s)` are vertex-disjoint ordered paths, write `PQ` for the concatenated vertex sequence `(p_0,\ldots,p_r,q_0,\ldots,q_s)`. For `0<=i<=j<=r`, write `P[i,j]=(p_i,\ldots,p_j)`.
## 1. Exterior barriers at Hamilton path ends
### Proposition 1.1
Let \`H\` be a boundary tournament, let \`X\subsetneq V(H)\` induce a Hamiltonian boundary tournament, and let \`D\subseteq V(H)-X\` be nonempty. Assume \`H[X\cup\{d\}]\` is non-Hamiltonian for every \`d\in D\`.
If some Hamilton path of \`H[X]\` begins with \`(u,v)\`, then \`(v,u,d)\` is tight for every \`d\in D\`.
If some Hamilton path of \`H[X]\` ends with \`(u,v)\`, then \`(d,v,u)\` is tight for every \`d\in D\`.
Consequently no ordered pair \`(u,v)\` can begin a Hamilton path of \`H[X]\` while \`(v,u)\` ends a Hamilton path of \`H[X]\`.
**Proof.**
Let \`P\` be a Hamilton path of \`H[X]\` beginning with \`(u,v)\`, and fix \`d\in D\`. If \`(d,u,v)\` were tight, prepending \`d\` to \`P\` would give a Hamilton path of \`H[X\cup\{d\}]\`, contrary to hypothesis. Boundary antisymmetry therefore gives \`(v,u,d)\` tight.
Dually, if \`Q\` is a Hamilton path of \`H[X]\` ending with \`(u,v)\` and \`(u,v,d)\` were tight, appending \`d\` would give a Hamilton path of \`H[X\cup\{d\}]\`. Hence \`(d,v,u)\` is tight.
If \`(u,v)\` begins one Hamilton path and \`(v,u)\` ends another, the first conclusion gives \`(v,u,d)\` while the second, applied to the terminal pair \`(v,u)\`, gives \`(d,u,v)\`; these are reverses, a contradiction. ∎
## 2. Cutting one path and adjoining the other two components
### Proposition 2.1
Let \`H\` satisfy \`pc(H)>2\`, and let
\`M=(m_0,\ldots,m_t)\mid A=(a_0,\ldots,a_r)\mid B=(b_0,\ldots,b_s)\`
be a spanning three-path cover by nonempty tight paths.
For every \`0\le i<t\`, consider the two spanning vertex sequences
\`M[0,i]A\` and \`BM[i+1,t]\`.
Every consecutive triple in these sequences is inherited from \`M,A,B\` except possibly the following present triples:
\`(m_{i-1},m_i,a_0)\` when \`i\ge1\`,
\`(m_i,a_0,a_1)\` when \`r\ge1\`,
\`(b_{s-1},b_s,m_{i+1})\` when \`s\ge1\`,
and \`(b_s,m_{i+1},m_{i+2})\` when \`i+2\le t\`.
At least one present triple is non-tight.
The same statement holds after interchanging \`A\` and \`B\`.
**Proof.**
The two displayed vertex sequences are disjoint and span \`V(H)\`. Every consecutive triple lying wholly inside \`M\`, \`A\`, or \`B\` is inherited and tight. At the join from \`M[0,i]\` to \`A\`, the only possible new triples are \`(m_{i-1},m_i,a_0)\` and \`(m_i,a_0,a_1)\` when they exist. At the join from \`B\` to \`M[i+1,t]\`, the only possible new triples are \`(b_{s-1},b_s,m_{i+1})\` and \`(b_s,m_{i+1},m_{i+2})\` when they exist.
If all present new triples were tight, both spanning sequences would be tight paths and would form a spanning two-path cover of \`H\`, contrary to \`pc(H)>2\`. Interchanging \`A,B\` gives the symmetric statement. ∎
## 3. Two-sided concatenation obstruction
### Proposition 3.1
Let \`H\` be a boundary tournament with \`pc(H)>2\` and let
\`V(H)=V(A)\sqcup W\sqcup V(B)\`,
where \`A,B\` are nonempty tight paths and \`W\` has a cover \`P\mid Q\` by two nontrivial tight paths.
Then either neither \`AP\` nor \`AQ\` is tight, or neither \`PB\` nor \`QB\` is tight.
**Proof.**
Suppose \`A\` concatenates with one of \`P,Q\`, say \`P\`. If \`B\` concatenates after \`Q\`, then \`AP\mid QB\` is a spanning two-path cover of \`H\`. If \`B\` concatenates after \`P\`, then \`APB\mid Q\` is a spanning two-path cover: because \`P\` has at least two vertices, every consecutive triple of \`APB\` already occurs in either \`AP\` or \`PB\`. Both are impossible. Therefore, once one left concatenation exists, no right concatenation exists with either middle path. The same argument with left and right interchanged proves the dichotomy. ∎
## 4. Comparing nested deletion covers
### Proposition 4.1
Let \`K\` be a boundary tournament with \`pc(K)>2\`, let \`a,b\` be distinct vertices, let \`F=P\mid Q\` be an exact two-path cover of \`K-a\` in which the component containing \`b\` is nontrivial, and let \`T=R\mid S\` be an exact two-path cover of \`K-\{a,b\}\`.
If \`b\` is an endpoint of its component in \`F\`, let \`c\` be its path neighbor. If that component begins \`(b,c,\ldots)\`, then \`(c,b,a)\` is tight. If it ends \`(\ldots,c,b)\`, then \`(a,b,c)\` is tight.
If \`b\` is internal in its component in \`F\`, then deleting \`b\` from \`F\` gives an exact three-path cover of \`K-\{a,b\}\`, and some ordinary edge of \`T\` has endpoints in two different components of \`F-b\`.
**Proof.**
Suppose first that \`b\` is an endpoint of its component in \`F\`. Because the component is nontrivial, it has a path neighbor \`c\`.
If the component begins \`(b,c,\ldots)\` and \`(a,b,c)\` were tight, prepending \`a\` would enlarge that component to a tight path and, together with the other component of \`F\`, would give a spanning two-path cover of \`K\`. Hence \`(a,b,c)\` is non-tight and \`(c,b,a)\` is tight.
If the component ends \`(\ldots,c,b)\`, the same argument shows that \`(c,b,a)\` cannot be tight, since appending \`a\` would two-cover \`K\`; hence \`(a,b,c)\` is tight.
Now suppose \`b\` is internal in its component of \`F\`. Deleting \`b\` splits that component into two nonempty tight subpaths, while the other component of \`F\` remains nonempty. Thus \`F-b\` is an exact three-path cover of \`K-\{a,b\}\`. By \`TOOLKIT/COVER_COMPARISON.md\` Section 1, some ordinary edge of the exact two-path cover \`T\` has endpoints in two distinct components of \`F-b\`. ∎
## 5. Endpoint alternatives after deleting two vertices
### Proposition 5.1
Let \`K\` be a boundary tournament with \`pc(K)>2\`, let \`a,b\` be distinct vertices, and let \`T=P\mid Q\` be an exact two-path cover of \`K-\{a,b\}\`. If \`P=(p_0,\ldots,p_r)\` with \`r\ge1\`, then each of the following disjunctions holds:
1. \`(p_0,b,a)\` or \`(p_1,p_0,b)\` is tight;
2. \`(p_0,a,b)\` or \`(p_1,p_0,a)\` is tight;
3. \`(a,p_r,p_{r-1})\` or \`(b,a,p_r)\` is tight;
4. \`(b,p_r,p_{r-1})\` or \`(a,b,p_r)\` is tight.
**Proof.**
Prepend \`(a,b)\` to \`P\`. The only new consecutive triples are \`(a,b,p_0)\` and \`(b,p_0,p_1)\`. They cannot both be tight, since otherwise the enlarged path together with \`Q\` would two-cover \`K\`. Reversing a non-tight triple gives \`(p_0,b,a)\` or \`(p_1,p_0,b)\`. Prepending \`(b,a)\` gives the second disjunction.
Appending \`(a,b)\` to \`P\` creates exactly the two possible new triples \`(p_{r-1},p_r,a)\` and \`(p_r,a,b)\`; reversing a non-tight one gives \`(a,p_r,p_{r-1})\` or \`(b,a,p_r)\`. Appending \`(b,a)\` gives the fourth disjunction. ∎
## Cover-comparison and matching lemmas
_GN3 document: `t0000008`_
# Cover-comparison and matching lemmas
## 1. A component drop forces a crossing edge
Let `H` be a boundary tournament and let `W subseteq V(H)`. Suppose `H[W]` has path covers
`R=R_1|...|R_c`
and
`T=T_1|...|T_r`
with `r<c`. Then some ordinary edge `xy` of a component of `T` has its endpoints in two different components of `R`.
If the `R`-component containing `x` is nontrivial and `p` is a neighbor of `x` along that component, then exactly one of
`(y,x,p)`, `(p,x,y)`
is tight.
**Proof.** Suppose every ordinary edge of every component of `T` had both endpoints in one component of `R`. Since each `T_i` is connected, every `T_i` would then lie in a single `R_j`. Because the `T_i` cover `W`, every one of the `c` nonempty components of `R` would contain at least one component of `T`. Distinct components of `R` are disjoint, so these components of `T` would be distinct. Hence `r>=c`, a contradiction.
Thus some edge `xy` of `T` crosses two components of `R`. If the component containing `x` is nontrivial, choose a path neighbor `p` of `x` in that component. The vertices `p,x,y` are distinct, and boundary antisymmetry gives exactly one tight member of the displayed reversal pair. ∎
Each singleton component contains no ordered triple, and every two-vertex ordering is a tight path vacuously. Thus the orders of two singleton components alone impose no tightness condition on any ordered triple of distinct vertices.
## 2. A Cartesian clause lemma
Let `I_1,...,I_m` be nonempty finite sets. For each `j` and each `i in I_j`, let `P_{j,i}` be a Boolean statement. Suppose that for every tuple
`(i_1,...,i_m) in I_1 x ... x I_m`
at least one of
`P_{1,i_1},...,P_{m,i_m}`
is true. Then for some `j`, every statement `P_{j,i}` with `i in I_j` is true.
**Proof.** If no coordinate family were entirely true, choose for every `j` an index `i_j` for which `P_{j,i_j}` is false. The resulting tuple would make all `m` statements false, contradicting the hypothesis. ∎
### Boundary-tournament form
Let `H` be a boundary tournament, and let `k>=1` be an integer with `pc(H)>k`. For `j=1,...,m`, let `{alpha_{j,i}:i in I_j}` be finite families of ordered triples of distinct vertices of `H`, and for each `alpha_{j,i}` let `h_{j,i}` be its reverse.
Assume that for every tuple `(i_1,...,i_m)` there exist `k` pairwise vertex-disjoint vertex-simple sequences whose vertex sets partition `V(H)` and such that:
- every consecutive ordered triple of every sequence, other than the listed triples
`h_{1,i_1},...,h_{m,i_m}`,
is tight; and
- each listed triple `h_{j,i_j}` occurs as a consecutive ordered triple of one of the `k` sequences.
Then for some `j`, every triple `alpha_{j,i}`, `i in I_j`, is tight.
**Proof.** Fix a tuple `(i_1,...,i_m)` and the corresponding `k` sequences. If every listed triple `h_{j,i_j}` were tight, then every consecutive triple in every vertex-simple sequence would be tight. The sequences would therefore be `k` tight paths forming a spanning `k`-path cover of `H`, contrary to `pc(H)>k`.
Hence at least one listed triple `h_{j,i_j}` is non-tight. Boundary antisymmetry makes its reverse `alpha_{j,i_j}` tight. Thus for every tuple at least one of the Boolean statements
`P_{j,i_j} := [alpha_{j,i_j} is tight]`
is true. The Cartesian clause lemma gives an index `j` for which every `alpha_{j,i}` is tight. ∎
## 3. A weighted symmetric-difference lemma for two matchings
Let `G=(V,E)` be a finite graph, and let `F` and `J` be matchings in `G`, with
`|F|=|J|+1`.
Let `w:E -> R_{>=0}` be a nonnegative edge weight such that every edge of `J` has weight zero, and suppose
`W=sum_{e in F} w(e)>0`.
Decompose `F triangle J` into its alternating connected components. For such a component `C`, put
`delta(C)=|F intersect C|-|J intersect C|`
and
`omega(C)=sum_{e in F intersect C} w(e)`.
Then exactly one of the following holds:
1. some union `S` of alternating components satisfies
`sum_{C in S} delta(C)=1`
and
`sum_{C in S} omega(C)<W`;
2. there is a unique alternating path `C_*` with `delta(C_*)=1`; it satisfies `omega(C_*)=W`, there is no component with `delta=-1`, and every other alternating component has `delta=0` and weight zero.
**Proof.** Every alternating component of two matchings has `delta in {-1,0,1}`. Moreover
`sum_C delta(C)=|F|-|J|=1`
and
`sum_C omega(C)=W`.
Assume the first conclusion fails. In particular, every component with `delta=1` must have weight at least `W`, because that component alone would otherwise satisfy conclusion 1. Since all component weights are nonnegative, their total is `W`, and `W>0`, there can be at most one component with `delta=1`. The total excess is one, so such a component exists; call it `C_*`. Necessarily `omega(C_*)=W`, and every other component has weight zero.
If some component had `delta=-1`, then with only one component of excess `+1` the sum of all `delta` values could not equal `1`. Thus no such component exists, and every remaining component is balanced. A symmetric-difference component with one more `F`-edge than `J`-edge is an alternating path, so `C_*` is the unique `F`-heavy alternating path. ∎
The positivity hypothesis `W>0` is essential for this formulation: with zero total weight, several zero-weight `F`-heavy components can coexist with `J`-heavy components.
A useful specialization takes `w` to be the indicator of edges crossing a fixed vertex partition. If `J` uses no crossing edge, the second alternative says that one alternating path contains every crossing edge of `F`.
## Cover augmentation lemmas
_GN3 document: `t0000009`_
# Cover augmentation lemmas
## 1. Concatenating two components forces a reversed joining triple
Let `H` be a boundary tournament with `pc(H)>2`, and let
`A|B|C`
be a spanning three-path cover of `H` in which all three components are nontrivial. Write
`A=(a_0,...,a_p)`, `B=(b_0,...,b_q)`
with `p,q>=1`.
If `A` is followed by `B`, the only new consecutive triples are
`(a_{p-1},a_p,b_0)`
and
`(a_p,b_0,b_1)`.
They cannot both be tight. Hence at least one of
`(b_0,a_p,a_{p-1})`, `(b_1,b_0,a_p)`
is tight.
The same conclusion holds for each of the six ordered pairs of distinct components among `A,B,C`.
**Proof.** Every consecutive triple wholly inside `A` or `B` is tight. If both new consecutive triples were tight, the concatenation of `A` and `B` would be a tight path and, together with the untouched third component, would give a spanning two-path cover. Boundary antisymmetry gives the reverse of each non-tight joining triple. ∎
## 2. One cut and two joins in an edge-ordered path cover
Let `G` be an edge-ordered complete graph and let `F` be a spanning cover by `c>=3` vertex-disjoint increasing paths. Orient each path increasingly.
For a vertex `v` of one of the paths, let `L(v)` be its incoming path edge when present and let `L(v)=-infinity` at a source. Let `U(v)` be its outgoing path edge when present and let `U(v)=+infinity` at a terminal.
Let `p->v` be an edge of one component `C`. Let `t` be the terminal vertex of a second component `A`, and let `s` be the source vertex of a third component `B`, with the three components distinct. If
`L(t) < tv < U(v)`
and
`L(p) < ps < U(s)`,
then deleting the path edge `pv` and inserting the edges `tv` and `ps` produces a spanning increasing path cover with `c-1` components.
The statement remains valid when `A` or `B` is a singleton, using the endpoint conventions above.
**Proof.** Write
`A=(a_0,...,t)`,
`C=(c_0,...,p,v,...,c_m)`,
`B=(s,...,b_q)`.
After deleting `pv`, replace the three components by
`(a_0,...,t,v,...,c_m)`
and
`(c_0,...,p,s,...,b_q)`.
The two displayed inequalities are exactly the comparisons needed at the two new joins. All inherited comparisons remain unchanged. The new paths are disjoint, span the same vertices, and replace three old components by two. ∎
## 3. Two cuts and a singleton cross-swap
Let an edge-ordered complete graph have a spanning three-path cover
`A|B|(z)`
with `A,B` nontrivial increasing paths. Orient `A,B` increasingly. Choose path edges
`q->w` in `A`, `p->v` in `B`.
With the same `L,U` notation, suppose
`L(q)<qz<zv<U(v)`
and
`L(p)<pw<U(w)`.
Then deleting `qw,pv` and adding `qz,zv,pw` gives a spanning cover by two increasing paths.
**Proof.** Let `A_q` be the prefix of `A` ending at `q`, `A_w` the suffix beginning at `w`, and similarly let `B_p,B_v` be the prefix and suffix of `B` determined by `p->v`. The two new paths are
`A_q,z,B_v`
and
`B_p,A_w`.
They are disjoint and span all vertices. The first displayed chain gives every new comparison in the first path, and the second gives the single new comparison in the second path. ∎
## 4. Repeated singleton transfers force an extreme triangle edge
Let `G` be an edge-ordered complete graph with no spanning cover by two increasing paths. Let `x,y` be distinct vertices.
Let `T=(v,t_1,...,t_r)`, `r>=0`, be an increasing path disjoint from an increasing path `A` and from `{x,y}`. Suppose
`A | (y,v,t_1,...,t_r) | (x)`
and
`A | (x,v,t_1,...,t_r) | (y)`
are spanning three-path covers of `G`. Then
`vx<xy` and `vy<xy`.
Thus `xy` is the largest edge of the triangle `{x,y,v}`.
Dually, let `S=(s_0,...,s_r,w)`, `r>=0`, be an increasing path disjoint from an increasing path `A` and from `{x,y}`. If
`A | (s_0,...,s_r,w,y) | (x)`
and
`A | (s_0,...,s_r,w,x) | (y)`
are spanning three-path covers of `G`, then
`xy<wx` and `xy<wy`,
so `xy` is the smallest edge of `{x,y,w}`.
If both pairs of covers exist for the same pair `{x,y}`, with the displayed vertices `v` and `w`, then
`vx,vy < xy < wx,wy`,
and both
`(v,x,y,w)`, `(v,y,x,w)`
are increasing paths.
**Proof.** In the first pair of covers, if `xy<yv`, then `(x,y,v,t_1,...,t_r)` is increasing and together with `A` gives a spanning two-path cover. Hence `yv<xy`. The other cover similarly gives `xv<xy`.
For the second pair of covers, if `wy<xy`, then `(s_0,...,s_r,w,y,x)` is increasing and together with `A` gives a spanning two-path cover. Hence `xy<wy`. If `wx<xy`, then `(s_0,...,s_r,w,x,y)` is increasing and together with `A` gives a spanning two-path cover. Hence `xy<wx`.
Combining the two sets of inequalities gives the final two increasing four-vertex paths. ∎
## 5. Extreme two-vertex-component barriers in an edge-ordered three-cover
Let `G` be an edge-ordered complete graph with no spanning cover by two increasing paths. For a named increasing path component `A`, write `L_A(v)` and `U_A(v)` for the incoming and outgoing path edges at `v`, with the same endpoint conventions `-infinity,+infinity` as in Section 2.
### Globally smallest two-vertex component
Suppose
`(x,y)|A|B`
is a spanning increasing three-cover and `xy` is the globally smallest edge. Let `p->v` be a path edge of `A`, and let `s` be the source of `B`. If
`L_A(p)<ps<U_B(s)`,
then
`xv>=U_A(v)` and `yv>=U_A(v)`.
### Globally largest two-vertex component
Suppose instead
`(X,Y)|A|B`
is a spanning increasing three-cover and `XY` is the globally largest edge. Let `p->v` be a path edge of `A`, and let `t` be the terminal of `B`. If
`L_B(t)<tv<U_A(v)`,
then
`pX<=L_A(p)` and `pY<=L_A(p)`.
Here the inequalities use the formal endpoint values `-infinity,+infinity`; for example a conclusion `e>=+infinity` means that the stated antecedent cannot occur when the relevant vertex is terminal.
**Proof.** For the minimum-edge case, orient the two-vertex component as `(x,y)`. Since `xy` is globally smallest, `xy<yv`. If also `yv<U_A(v)`, then Section 2 applies with terminal `y` of the two-vertex component, cut edge `p->v`, and source `s`, producing a spanning two-path cover. Thus `yv>=U_A(v)`. Reverse the order to `(y,x)` and repeat to obtain `xv>=U_A(v)`.
For the maximum-edge case, orient the two-vertex component as `(X,Y)`. Since `pX<XY`, if also `L_A(p)<pX`, Section 2 applies using source `X` of that component and terminal `t` of `B`, again producing a two-path cover. Thus `pX<=L_A(p)`. Reverse the order to `(Y,X)` and repeat for `pY`. ∎
## 6. Cutting a component around a three-vertex component
Let `H` be a boundary tournament with `pc(H)>2`. Suppose
`(a,s,c)|R|Q`
is a spanning three-path cover, where
`R=(u_0,...,u_k)`, `k>=1`.
Then at least one of the following holds:
1. `(s,c,u_0)` is not tight;
2. `(u_k,a,s)` is not tight;
3. `k>=2` and both `(c,u_0,u_1)` and `(u_{k-1},u_k,a)` are not tight.
**Proof.** Delete an edge `u_i u_{i+1}` of `R` and consider the vertex sequence
`(u_{i+1},...,u_k,a,s,c,u_0,...,u_i)`.
Together with `Q`, this would be a spanning two-path cover if every new consecutive triple were tight.
When `k=1`, deleting the sole edge leaves only the two new triples `(u_k,a,s)` and `(s,c,u_0)`, so at least one is non-tight.
Assume `k>=2` and both of those triples are tight. Deleting the first edge of `R` leaves only one additional new triple, `(u_{k-1},u_k,a)`, so it must be non-tight. Deleting the last edge leaves only `(c,u_0,u_1)`, which must also be non-tight. ∎
Boundary antisymmetry supplies the corresponding reversed tight triple whenever one of the displayed joining triples is non-tight.
## 7. Joining two components relative to a fixed two-cover
Let `G` be a boundary tournament. Let
`J=P|Q|R`
be a spanning three-path cover and let `F` be a spanning exact two-path cover of the same vertex set.
Choose an ordered pair of distinct components, say
`P=(p_0,...,p_k)`, `Q=(q_0,...,q_l)`,
and concatenate them using the ordinary edge `p_k q_0`. The only new consecutive triples that can occur are
`(p_{k-1},p_k,q_0)` when `k>=1`,
and
`(p_k,q_0,q_1)` when `l>=1`.
Hence either one of the displayed triples is non-tight, in which case its reverse is a tight triple on three distinct vertices, or the concatenation gives a spanning exact two-path cover.
Moreover, among distinct ordinary edges joining endpoints of two components of `J` and used to concatenate those components, at most one can produce the same ordinary path forest as `F`. Consequently there is a concatenation for which either a new consecutive triple is non-tight and supplies its tight reverse, or the resulting exact two-cover has ordinary path forest different from that of `F`.
**Proof.** Every consecutive triple wholly inside `P,Q,R` is inherited. If both existing new triples are tight, the concatenation of `P` and `Q`, together with `R`, is an exact two-cover.
Every successful concatenation adds exactly one ordinary edge to the ordinary path forest of `J`. If such a concatenation has the same ordinary forest as `F`, then the forest of `J` is contained in that of `F` and the added joining edge is the unique edge of `F` not already in `J`. Thus at most one distinct joining edge can reconstruct the forest of `F`.
The three unordered pairs of components of `J` supply three distinct ordinary endpoint-joining edges, since the components are pairwise vertex-disjoint. Choose one different from the possible unique edge that reconstructs `F`, orient the corresponding pair of components in either concatenation order, and apply the first assertion. ∎
## Path insertion and endpoint replacement lemmas
_GN3 document: `t0000010`_
# Path insertion and endpoint replacement lemmas
## 1. Two-sided endpoint replacement forces reversed order
Let `H` be a boundary tournament. Let
`Q=(q_0,q_1,...,q_r)`, `r>=2`,
be a Hamilton tight path of `H[X]` on a vertex set `X subseteq V(H)`, and let `y in V(H)-X`. Suppose `H[X union {y}]` is non-Hamiltonian. Assume nevertheless that there is
- a Hamilton tight path `L` on `(X-{q_0}) union {y}`, and
- a Hamilton tight path `R` on `(X-{q_r}) union {y}`.
Then at least one of the three pairs
`(Q,L)`, `(Q,R)`, `(L,R)`
has two common vertices occurring in different relative orders.
Consequently `PATH_FORESTS_AND_INTERSECTIONS.md` Section 4 applies to at least one of these pairs and yields a reversed common edge, a tight triple reversing an ordered edge at an intersection, or a vertex-simple tight cycle.
**Proof.** Suppose every pair has its common vertices in the same relative order.
The common vertices of `Q` and `L` are `q_1,...,q_r`, so `L` is obtained by inserting `y` into that displayed order. If `y` were not in one of the first two positions, then `L` would begin with `(q_1,q_2)`, and prepending `q_0` would give a Hamilton tight path on `X union {y}`. Hence
`L=(y,q_1,...,q_r)`
or
`L=(q_1,y,q_2,...,q_r)`.
Similarly, comparison of `Q` with `R` shows that
`R=(q_0,...,q_{r-1},y)`
or
`R=(q_0,...,q_{r-2},y,q_{r-1})`.
Compare these four possible pairs. The first form of `L` and the first form of `R` put `y` and `q_1` in opposite orders. The first form of `L` with the second form of `R` does the same when `r>=3`; when `r=2` it gives the tight Hamilton path
`(q_0,y,q_1,q_2)`.
The second form of `L` with the first form of `R` puts `y` and `q_2` in opposite orders when `r>=3`; when `r=2` it gives
`(q_0,q_1,y,q_2)`.
Finally, the second form of `L` with the second form of `R` puts `y` and `q_2` in opposite orders when `r>=4`; for `r=2`, the common vertices `q_1,y` occur in opposite orders; and for `r=3` the two paths combine to
`(q_0,q_1,y,q_2,q_3)`.
Thus either some pair reverses the order of common vertices or `X union {y}` is Hamiltonian. The latter is excluded. ∎
## 2. Barrier gaps for a noninsertable vertex in an increasing path
Let `G` be an edge-ordered complete graph and let
`Q=(q_0,...,q_m)`, `m>=1`,
be an increasing path. Let `x` be a vertex outside `Q`. For `0<=i<=m-1`, call the gap between `q_i,q_{i+1}`
- **left-feasible** if `i=0` or `q_{i-1}q_i < q_i x`;
- **right-feasible** if `i=m-1` or `xq_{i+1} < q_{i+1}q_{i+2}`.
Assume insertion of `x` between `q_i,q_{i+1}` never gives an increasing path.
Then some gap is both left- and right-feasible, and at every such gap
`xq_{i+1} < xq_i`.
Let `beta(x)` be the least right-feasible gap. Then `beta(x)` is also left-feasible and
`xq_{beta(x)+1} < xq_{beta(x)}`.
Moreover, if `x,y` are two such noninsertable vertices and `beta(x)<beta(y)`, then, with `i=beta(x)`,
`xq_{i+1} < q_{i+1}q_{i+2} < yq_{i+1}`.
Define the boundary tournament `H_G` on `V(G)` by declaring
`(u,v,w)` tight exactly when `uv<vw`.
Then the displayed inequalities imply that
`(x,q_{i+1},y)`
is tight in `H_G`.
**Proof.** Put `e_i=q_iq_{i+1}`. Let `L_0` be true and, for `i>=1`, let `L_i` mean `e_{i-1}<q_i x`. Let `R_{m-1}` be true and, for `i<=m-2`, let `R_i` mean `xq_{i+1}<e_{i+1}`.
If no gap satisfied both `L_i` and `R_i`, then `L_0` would force `R_0` to fail. Whenever `L_i` holds and `R_i` fails, we have
`e_i<e_{i+1}<xq_{i+1}`,
so `L_{i+1}` holds. Induction gives `L_{m-1}`, contradicting the automatic truth of `R_{m-1}`. Hence some gap is left- and right-feasible.
At such a gap all comparisons needed for insertion are correct except possibly the comparison of the two edges `q_i x` and `xq_{i+1}`. Since insertion fails, totality of the edge order gives
`xq_{i+1}<q_i x`.
For `beta(x)`, every earlier `R_j` fails. The same propagation from `L_0` gives `L_{beta(x)}`, so the displayed inequality follows.
Finally let `beta(x)=i<beta(y)`. Right-feasibility of `x` gives
`xq_{i+1}<e_{i+1}`,
while minimality of `beta(y)` makes `R_i(y)` false, hence
`e_{i+1}<yq_{i+1}`.
By the definition of `H_G`, the comparison `xq_{i+1}<q_{i+1}y` is exactly the tightness of `(x,q_{i+1},y)`. ∎
## 3. Local obstruction when every insertion position fails
Let `H` be a boundary tournament. Let
`B=(b_1,...,b_m)`, `m>=2`,
be a tight path in `H`, and let `x in V(H)-V(B)`. In the comparison digraph `Gamma(H)` put
`e_i={b_i,b_{i+1}}` for `1<=i<=m-1`,
`f_i={x,b_i}` for `1<=i<=m`.
Assume that inserting `x` in every one of the `m+1` positions of the displayed order of `B` fails to give a tight path. Then for some `1<=t<=m-1`, one of the following occurs.
1. `2<=t<=m-1` and
`f_t -> e_{t-1} -> e_t -> f_t`.
2. The following arcs hold:
`e_t -> f_t`,
`f_{t+1} -> f_t`,
together with `e_{t-1}->f_t` when `t>1`, and `f_{t+1}->e_{t+1}` when `t<m-1`. When `t=m-1`, the failed final insertion additionally gives
`f_m -> e_{m-1}`.
Every displayed arc in either alternative involves only `x` together with at most four consecutive vertices of `B`.
**Proof.** Failure of the insertion at the left end gives `e_1->f_1`, while failure at the right end gives `f_m->e_{m-1}`.
If some `t<=m-2` satisfies `f_{t+1}->e_{t+1}`, choose the least such `t`; otherwise put `t=m-1`. For `t=1`, the left-end failure gives `e_t->f_t`. For `t>1`, minimality says `f_t->e_t` does not hold, so boundary antisymmetry gives `e_t->f_t`.
If `t>1` and `f_t->e_{t-1}`, then tightness of `B` gives `e_{t-1}->e_t`, producing alternative 1. Otherwise `e_{t-1}->f_t` whenever `t>1`.
Suppose first that `t<m-1`. By construction `f_{t+1}->e_{t+1}`. If `f_t->f_{t+1}`, then all comparisons needed to insert `x` between `b_t,b_{t+1}` would point forward:
`e_{t-1}->f_t->f_{t+1}->e_{t+1}`,
with the first comparison omitted when `t=1`. This contradicts failed insertion, so `f_{t+1}->f_t`.
If `t=m-1`, the same argument uses the arc `f_m->e_{m-1}` supplied by failure of the right-end insertion. When the predecessor comparison exists and does not form alternative 1, failed middle insertion forces `f_m->f_{m-1}`. The case `m=2` is the same with no predecessor edge. Thus alternative 2 holds. ∎
## 4. Opposite extensions of one ordered pair concatenate
Let `a,b,x,y` be distinct vertices of a boundary tournament.
If
`(x,a,b)` and `(a,b,y)`
are tight, then
`(x,a,b,y)`
is a tight path. Likewise, if
`(y,a,b)` and `(a,b,x)`
are tight, then
`(y,a,b,x)`
is a tight path.
Two left extensions of `(a,b)`, or two right extensions of `(a,b)`, do not by themselves imply a four-vertex path; the useful automatic conclusion is the opposite-end concatenation above.
## Johnson-graph density bounds
_GN3 document: `t0000011`_
# Johnson-graph density bounds
Let `J(r,k)` be the Johnson graph whose vertices are the `k`-subsets of an `r`-element set, with two vertices adjacent exactly when their intersection has order `k-1`.
## 1. A quadratic density bound from local `(k+1)`-set occupancy
Let `B` be a family of `k`-subsets of an `r`-element set, where `r>=k+1`. Write
`b=|B|`, `M=binom(r,k)`, `p=b/M`.
Suppose every `(k+1)`-subset contains at most `t` members of `B`. Then
`(r-k+1)p^2-p <= t(t-1)(r-k)/(k(k+1))`,
and therefore
`p <= [1+sqrt(1+4t(t-1)(r-k)(r-k+1)/(k(k+1)))]/[2(r-k+1)]`.
If `t>=2` and equality holds throughout, then every `(k+1)`-subset contains exactly `t` members of `B`, and every `(k-1)`-subset is contained in the same number of members of `B`.
**Proof.** For each `(k+1)`-set `U`, put
`y_U=|B intersect binom(U,k)|`.
Every edge of `J(r,k)[B]` has a unique union of order `k+1`, so
`e(B)=sum_U binom(y_U,2) <= binom(r,k+1) binom(t,2)`.
For each `(k-1)`-set `S`, put
`x_S=|{F in B:S subset F}|`.
Every Johnson edge has a unique intersection of order `k-1`, hence
`e(B)=sum_S binom(x_S,2)=(sum_S x_S^2-kb)/2`,
because `sum_S x_S=kb`. By Cauchy-Schwarz,
`sum_S x_S^2 >= k^2b^2/binom(r,k-1)`.
Combining the lower and upper bounds for `e(B)` and using
`binom(r,k-1)=binom(r,k)k/(r-k+1)`,
`binom(r,k+1)=binom(r,k)(r-k)/(k+1)`,
and `b=p binom(r,k)` gives the displayed quadratic inequality. Solving it for the positive root gives the density bound.
If equality holds and `t>=2`, then equality is required both in the pointwise estimate `binom(y_U,2)<=binom(t,2)` and in Cauchy-Schwarz. Thus every `y_U=t` and all `x_S` are equal. ∎
## 2. A degree bound from the same local occupancy hypothesis
Under the same hypotheses,
`p <= [k+(t-1)(r-k)]/[k(r-k+1)]`.
More precisely, every vertex of `J(r,k)[B]` has degree at most
`(t-1)(r-k)`,
while the average degree of `J(r,k)[B]` is at least
`k(r-k+1)p-k`.
If equality holds, then every member `F of B` has exactly `t-1` neighbors of `B` inside each `(k+1)`-set containing `F`, and every `(k-1)`-subset lies in the same number of members of `B`.
**Proof.** Fix `F in B`. Every Johnson neighbor `G of F` has a unique union `U=F union G` of order `k+1`. There are `r-k` such supersets `U` of `F`, and each contains at most `t-1` further members of `B`. Hence
`deg_B(F)<= (t-1)(r-k)`.
Thus
`2e(B)<=b(t-1)(r-k)`.
Using the intersection counts `x_S` from the previous proof,
`2e(B)=sum_S x_S(x_S-1)=sum_S x_S^2-kb`.
Cauchy-Schwarz gives
`2e(B)>=k^2b^2/binom(r,k-1)-kb`.
Comparing the two inequalities and substituting
`binom(r,k-1)=binom(r,k)k/(r-k+1)`, `b=p binom(r,k)`
yields
`k(r-k+1)p-k <= (t-1)(r-k)`,
which is equivalent to the claimed bound. Equality forces equality in every pointwise degree bound and in Cauchy-Schwarz, giving the stated regularity conditions. ∎
For `k=2,t=2` the second bound gives `p<=r/[2(r-1)]`. For `k=5,t=2` it gives `p<=r/[5(r-4)]`.
## 3. Complement-pair cut bound in `J(10,5)`
Let `Omega` be a ten-element set and let `F⊆binom(Omega,5)` contain no complementary pair. Put
`bar(F)={Omega-A : A in F}`.
In the Johnson graph `J(10,5)`, whose vertices are the five-subsets of `Omega` and whose adjacent vertices meet in four elements, let `e(F,bar(F))` denote the number of Johnson edges with one endpoint in `F` and one endpoint in `bar(F)`. Then
`e(F,bar(F)) <= 15|F|`.
**Proof.** For every four-set `C⊆Omega`, let
`K_C={C∪{v} : v in Omega-C}`.
This is a clique of order six in `J(10,5)`, and every Johnson edge belongs to exactly one such clique, namely the clique indexed by the intersection of its endpoints. Put
`r_C=|F∩K_C|`, `s_C=|bar(F)∩K_C|`.
Since `F` contains no complementary pair, `F` and `bar(F)` are disjoint. Hence `r_C+s_C<=6`, and therefore
`r_C s_C <= (r_C+s_C)^2/4 <= (3/2)(r_C+s_C)`.
Summing over all four-sets counts every edge from `F` to `bar(F)` exactly once. Each five-set contains exactly five four-subsets, so
`sum_C r_C=5|F|`, `sum_C s_C=5|bar(F)|=5|F|`.
Consequently
`e(F,bar(F)) <= (3/2)(10|F|)=15|F|`. ∎
## 4. Hamilton-five density hierarchy in boundary tournaments
Let `H` be a boundary tournament, let `W⊆V(H)` have order `r>=6`, and fix `S⊆W` of order `s∈{0,1,2,3}`. Let `h_5(W;S)` be the number of five-subsets `F` such that
`S⊆F⊆W`
and `H[F]` is Hamiltonian. Then the fraction of five-subsets containing `S` that are non-Hamiltonian is at most
`(r-s)/[(5-s)(r-4)]`.
Equivalently,
`h_5(W;S) >= [1-(r-s)/((5-s)(r-4))] binom(r-s,5-s)`.
For `s=0`, this gives Hamilton-five density at least
`4(r-5)/[5(r-4)]`.
For fixed vertex, pair, and triple, the corresponding asymptotic guaranteed densities tend respectively to `3/4`, `2/3`, and `1/2`. For every `r>10`, these bounds strictly improve the elementary fixed-subset density bounds obtained by direct double counting from the four-of-six theorem.
**Proof.** Let `B_S` be the family of sets
`F-S`
where `S⊆F⊆W`, `|F|=5`, and `H[F]` is non-Hamiltonian. Then `B_S` is `k=(5-s)`-uniform on the `N=r-s` vertices of `W-S`.
Every `(k+1)=(6-s)`-subset `U⊆W-S` corresponds to the six-set `S∪U`. By `SMALL_ORDER_HAMILTONICITY.md` Section 6, at least four of the six five-subsets of `S∪U` are Hamiltonian. Hence at most two members of `B_S` lie inside `U`.
Apply Section 2 with `t=2`. The bad-set density satisfies
`p_bad <= [k+(N-k)]/[k(N-k+1)] = N/[k(N-k+1)]`.
Substituting `N=r-s`, `k=5-s`, and `N-k+1=r-4` gives
`p_bad <= (r-s)/[(5-s)(r-4)]`.
Complementing inside the family of all five-subsets containing `S` gives the displayed Hamiltonian density. ∎
## Complement-free Johnson bounds on five-subsets of a ten-set
_GN3 document: `t0000012`_
# Complement-free Johnson bounds on five-subsets of a ten-set
## Theorem. Spectral bound for intersection-one pairs
Let `Omega` be a ten-element set. Let `G` be the graph whose vertices are the `252` five-subsets of `Omega`, with two distinct five-sets adjacent exactly when they meet in one element.
Let `F` be a complement-free family of `m` five-subsets, meaning that `A in F` implies `Omega-A notin F`. Then the average degree of the induced graph `G[F]` is at most
`7 + 18m/252`.
In particular, `m<=126`, so the average degree is at most `16`.
**Proof.** Let `J` be the adjacency matrix of the ordinary Johnson graph `J(10,5)`, where two five-sets are adjacent when they meet in four elements. For `0<=j<=5`, let `U_j` denote the usual `j`th Johnson eigenspace. The eigenvalues of `J` on
`U_0,U_1,U_2,U_3,U_4,U_5`
are respectively
`25,15,7,1,-3,-5`.
The distance-four adjacency matrix of `J(10,5)` is exactly the adjacency matrix `A` of `G`, since Johnson distance four means intersection size one. The standard Johnson one-swap recurrence gives
`A = p_4(J)`
with
`p_4(t)=(t^4-32t^3+166t^2+864t-1575)/576`.
Evaluating this polynomial at the six Johnson eigenvalues gives the eigenvalues of `A`:
`25,-15,7,-1,-3,5`.
Thus `G` is `25`-regular, and every eigenvalue on the orthogonal complement of the constant vectors is at most `7`.
Let `chi` be the characteristic vector of `F` and write
`chi=(m/252)1+v`,
where `v` is orthogonal to `1`. Then
`||v||^2=m-m^2/252`.
By the Rayleigh bound,
`2e(G[F]) = chi^T A chi`
`<=25m^2/252 + 7(m-m^2/252)`.
Dividing by `m` gives average degree at most
`7+18m/252`.
Finally, the `252` five-subsets split into `126` complementary pairs, and a complement-free family contains at most one member of each pair. Hence `m<=126`, giving the bound `16`. ∎
## Ternary path-system insertion theorem
_GN3 document: `t0000013`_
# Ternary path-system insertion theorem
Let `V` be a finite set and let `E` be a set of ordered triples of distinct vertices. Call a vertex sequence **tight** when every consecutive ordered triple belongs to `E`.
Assume the following two properties.
1. For every three distinct vertices `u,v,w`, at least one of `(u,v,w)` and `(w,u,v)` belongs to `E`.
2. Reversal symmetry holds: `(u,v,w) in E` if and only if `(w,v,u) in E`.
## Theorem. Every exterior vertex has two insertion positions
Let
`P=(v_0,...,v_{k-1})`
be a tight path and let `x notin V(P)`. Then at least two of the `k+1` positions obtained by inserting `x` into the displayed order of `P` give a tight path.
Consequently every `n`-vertex system satisfying the two assumptions has at least
`2^(n-1)`
Hamilton tight paths.
This bound is sharp.
**Proof.** Fix an adjacent pair `v_i,v_{i+1}` of `P`. On the three-set `{x,v_i,v_{i+1}}`, reversal symmetry means that tightness depends only on which vertex is in the middle. Let
`q_y=1`
when the two reverse orders having middle vertex `y` are tight. Applying the first axiom to the three cyclic choices gives
`q_x or q_{v_i}`,
`q_{v_i} or q_{v_{i+1}}`,
`q_{v_{i+1}} or q_x`.
Hence at most one of the three values `q_x,q_{v_i},q_{v_{i+1}}` is zero.
The adjacent pair `v_i,v_{i+1}` can therefore forbid at most one insertion position: immediately before `v_i`, between `v_i,v_{i+1}`, or immediately after `v_{i+1}`, according to which middle vertex is the unique forbidden one. Every new triple created by inserting `x` is associated with one adjacent pair of `P`. Thus the `k-1` adjacent pairs can forbid at most `k-1` of the `k+1` insertion positions. At least two positions remain tight.
For the counting statement, fix a vertex `x`. Every Hamilton tight path on `V-{x}` has at least two insertions of `x`, and deleting `x` from a resulting path recovers both the parent path and the insertion position. Hence, if `N(V)` denotes the number of Hamilton tight paths,
`N(V)>=2N(V-{x})`.
Starting from `N=1` on one vertex gives `N(V)>=2^(n-1)`.
For sharpness, fix a total order on `V` and declare `(a,b,c)` tight exactly when `b` is not the largest of `{a,b,c}`. The two axioms hold. A Hamilton ordering is tight exactly when it has no interior local maximum. Such an ordering decreases to the minimum element and then increases. It is determined uniquely by choosing which of the other `n-1` vertices lie to the left of the minimum, so there are exactly `2^(n-1)` Hamilton tight paths. ∎