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Problem packetResearch packetR818

R818Sourced evidence

One-way liveness forces at least h(h+1)/4 deterministic states

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Authored summary

For the alphabet of all \(h\) by \(h\) Boolean matrices, an \(h\)-state 1NFA recognizes strings with nonzero Boolean product, and every equivalent 2DFA has at least \(h(h+1)/4\) states.

The record cites sources for its explanation.

Recorded status: reported

Recorded scope: one-way liveness over the full alphabet of h by h Boolean matrices for every integer h at least 1

Complete recorded scope and conditions
{
  "kind": "family",
  "statement": "one-way liveness over the full alphabet of h by h Boolean matrices for every integer h at least 1",
  "family": "the one-way-liveness languages owl_h"
}

Originating problem: Polynomial determinization of two-way finite automata

Recorded relationships: The fixed-alphabet determinization question remains open

Authored record and scope
Authored title
One-way liveness forces at least h(h+1)/4 deterministic states
Record type
claim
Stored status
reported
Evidence grade
sourced
Recorded scope data
{ "kind": "family", "statement": "one-way liveness over the full alphabet of h by h Boolean matrices for every integer h at least 1", "family": "the one-way-liveness languages owl_h" }
Linked research record IDs
R815

2Authored explanation

Let \(\Sigma_h=\{0,1\}^{h\times h}\). A word \(A_1\cdots A_m\) is live when the Boolean product \(A_1\cdots A_m\) has a nonzero entry. Equivalently, the layered graph described by the matrices has a path joining its first and last columns. An \(h\)-state 1NFA follows a path by storing its current row index.

Adeogun and Kapoutsis construct \(1+\binom{h+1}{2}\) smooth connectivity properties. Each successive pair is separated by liveness and satisfies the suffix-of-choice condition in their main lemma. The two exit-size sequences of any solving 2DFA must decrease strictly a total of at least \(\binom{h+1}{2}\) times. One sequence therefore has at least half that many strict decreases, which yields \[ |Q|\ge \frac12\binom{h+1}{2}=\frac{h(h+1)}4. \] The alphabet size is \(|\Sigma_h|=2^{h^2}\), so this statement by itself uses an alphabet that grows with \(h\).

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Replay material: source only

3Evidence

Replay package: source only

A verification source is cited. This record has no executable replay attached.

Verification source: arxiv.org ↗, Adeogun and Kapoutsis, arXiv:2602.24279v2, Sections 2.2 and 4.1-4.4, especially Theorem 1 on p. 15

4What was measured

5How it connects

Recorded for

Machine-readable record

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  "slug": "twnfa-claim-one-way-liveness-quadratic-bound",
  "type": "claim",
  "title": "One-way liveness forces at least h(h+1)/4 deterministic states",
  "summary": "For the alphabet of all \\(h\\) by \\(h\\) Boolean matrices, an \\(h\\)-state 1NFA recognizes strings with nonzero Boolean product, and every equivalent 2DFA has at least \\(h(h+1)/4\\) states.",
  "relevance": "For Polynomial determinization of two-way finite automata, record twnfa-claim-one-way-liveness-quadratic-bound (“One-way liveness forces at least h(h+1)/4 deterministic states”) records a bound, answer, status fact, or structural consequence. The record states: For the alphabet of all \\(h\\) by \\(h\\) Boolean matrices, an \\(h\\)-state 1NFA recognizes strings with nonzero Boolean product, and every equivalent 2DFA has at least \\(h(h+1)/4\\) states.",
  "relevance_source": "recorded",
  "body": "Let \\(\\Sigma_h=\\{0,1\\}^{h\\times h}\\). A word \\(A_1\\cdots A_m\\) is live when the Boolean product \\(A_1\\cdots A_m\\) has a nonzero entry. Equivalently, the layered graph described by the matrices has a path joining its first and last columns. An \\(h\\)-state 1NFA follows a path by storing its current row index.\n\nAdeogun and Kapoutsis construct \\(1+\\binom{h+1}{2}\\) smooth connectivity properties. Each successive pair is separated by liveness and satisfies the suffix-of-choice condition in their main lemma. The two exit-size sequences of any solving 2DFA must decrease strictly a total of at least \\(\\binom{h+1}{2}\\) times. One sequence therefore has at least half that many strict decreases, which yields\n\\[\n|Q|\\ge \\frac12\\binom{h+1}{2}=\\frac{h(h+1)}4.\n\\]\nThe alphabet size is \\(|\\Sigma_h|=2^{h^2}\\), so this statement by itself uses an alphabet that grows with \\(h\\).",
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    "family": "the one-way-liveness languages owl_h"
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      "slug": "R815",
      "title": "The fixed-alphabet determinization question remains open",
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    {
      "slug": "R817",
      "title": "Two singleton contexts isolate any Boolean-matrix entry",
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    {
      "slug": "R816",
      "title": "The one-way-liveness bound transfers to a fixed binary alphabet",
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      "slug": "R813",
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      "slug": "two-way-nfa-polynomial-determinization",
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7Provenance

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