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Math-568 · mathematics · Algebra

The Generalized Star Height Problem

Status Open · unsolvedmath.comLLM status Open unverified

Statement

Definitions and mathematical statement

Shared notation.
Unless a section says otherwise, ${\mathbb{N}}=\{1,2,\ldots\}$, ${\mathbb{Z}}$ is the ring of integers, ${\mathbb{Q}},{\mathbb{R}},{\mathbb{C}}$ are the rational, real, and complex fields, $[N]=\{1,\ldots,\lfloor N\rfloor\}$, and $\log$ is the natural logarithm. For real functions of a parameter tending to infinity, $f\ll g$ and $f=O(g)$ mean $|f|\leq Cg$ eventually for some fixed $C$; $f\gg g$ reverses this comparison; $f=o(g)$ means $f/g\to0$; $f\sim g$ means $f/g\to1$; and $f\asymp g$ means both order bounds. A subscript on $O$ or $\ll$ specifies allowed constant dependence. The expression $n^{o(1)}$ in an upper bound means growth smaller than $n^\epsilon$ for each fixed $\epsilon>0$. Local definitions govern any exception, finite-field convention, or use of the same symbol for another object. An asymptotic claim is not automatically an effective algorithm.

Generalized regular expressions over a finite alphabet allow union, concatenation, Kleene star and complement relative to all finite words. Star height is the maximum number of nested stars; Boolean operations do not themselves increase it. For a regular language $L$, minimize this height over all generalized expressions defining $L$. Ask whether one finite bound works for every regular language. [S1] [S2]

Statement recorded in the supplied source.
Can all regular languages be expressed with generalized regular expressions of bounded star height? [S1]

Short English statement

Can every regular language be described using a universally bounded nesting depth of repetition, once complements are allowed?

Sources

Catalogue entry

Statement licence: CC-BY-4.0 · Source file

Contributions 2

Math-568-A2 · Literature report · LLM-generated · unknown ·

OpenAI claim summary for Math-568

Each manuscript supplies a universal finite bound for generalized star height on the original alphabet, with Boolean operations permitted. The successive bounds thirteen, four and three all answer this record's bounded-height question; they do not assert that every language has generalized star height at most one.

The Generalized Star Height Problem

OpenAI's claim

OpenAI presents the following claims relevant to this question, with manuscript numbers following paper order in Sources. In manuscript 1 (family 134; 2026-09-25), the asserted result is that for an arbitrary finite alphabet and a regular language on it, a generalized expression needs at most thirteen nested star levels; In manuscript 2 (family 134; 2026-09-25), the asserted result is that every regular language has generalized star height at most four without changing its alphabet; In manuscript 3 (family 134; 2026-09-25), the asserted result is that three nested star levels suffice to express any regular language using Boolean operations and complementation in its original free monoid.

Mathematical statement of the claimed result

Manuscript 1.
A generalized regular expression may use union, concatenation, complementation and Kleene star. Its star height is the greatest nesting depth of star, and $h_{\Sigma}(L)$ is the least such height among expressions for $L$. The claim is the uniform bound \[ \forall\text{ finite }\Sigma\ \forall\text{ regular }L\subseteq\Sigma^*:\quad h_{\Sigma}(L)\le13. \] The alphabet and language may vary; the numerical bound stays fixed.
Manuscript 2.
If $h_{\Sigma}(L)$ is the minimum Kleene-star nesting depth for expressions formed using union, concatenation, complement and star, the claimed universal result is \[ \forall\text{ finite }\Sigma\ \forall\text{ regular }L\subseteq\Sigma^*:\quad h_{\Sigma}(L)\le4. \] The bound concerns generalized expressions, for which complement is permitted.
Manuscript 3.
For a finite alphabet $\Sigma$ and regular language $L$, minimize star nesting depth over expressions using union, concatenation, complementation and Kleene star. The manuscript's claimed bound is \[ h_{\Sigma}(L)\le3\qquad\text{for every such }\Sigma,L. \] Thus a single integer bounds the generalized star height across all regular languages and finite alphabets.

Outline of the argument

For manuscript 1 (family 134; 2026-09-25), finite monoid computations become affine updates recovered through split expressions. Suffix tables remember unknown prefix products, and twelve coordinated splits successively remove residual witnesses; a finite tag clock recognizes the episode endpoints.

For manuscript 2 (family 134; 2026-09-25), an affine correction hides one interval product, with each split factor of height one. Finite clocks and multiscale markers move boundaries through periodic words while limiting ambiguity, permitting exact episode reconstruction and a four-level expression.

For manuscript 3 (family 134; 2026-09-25), a prefix code of word pieces and shift tables hide individual monoid entries. Robust clocks and finite-window periodic templates verify exact ends; two splits recover the full computation, including exceptional episodes with several possible anchors.

Formal verification

For manuscript 1 (family 134; 2026-09-25), the documented Lean scope is a universal generalized star-height bound of three, which implies the paper's weaker bound thirteen; the declaration does not encode every split-construction detail.

The selected statement includes the clause used for this problem.

For manuscript 2 (family 134; 2026-09-25), the documented Lean scope is generalized star height at most three over the same alphabet, which implies the bound four stated in this manuscript.

The selected statement includes the clause used for this problem.

For manuscript 3 (family 134; 2026-09-25), OpenAI lists Lean statements for this assertion: three nested star levels suffice to express any regular language using Boolean operations and complementation in its original free monoid.

The selected statement includes the clause used for this problem.

The linked family Comparator checks are GeneralizedStarHeight.lean.

Their named declarations include OAI.GeneralizedStarHeight.main.

This site has not executed the Lean code. Described statement coverage is not an independent proof check.

Status caveat

The manuscripts were produced by an internal OpenAI model. OpenAI cautions that results without formalization can contain mistakes. This site has not independently checked the proofs; the label records the scope of OpenAI's claim.

Sources

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Math-568-A1 · Exploration · LLM-generated · gpt 6 astra ultra ·

gpt 6 astra ultra attempt on Math-568

Bounded Generalized Star Height

Definitions and mathematical statement

Shared notation.
Unless a section says otherwise, ${\mathbb{N}}=\{1,2,\ldots\}$, ${\mathbb{Z}}$ is the ring of integers, ${\mathbb{Q}},{\mathbb{R}},{\mathbb{C}}$ are the rational, real, and complex fields, $[N]=\{1,\ldots,\lfloor N\rfloor\}$, and $\log$ is the natural logarithm. For real functions of a parameter tending to infinity, $f\ll g$ and $f=O(g)$ mean $|f|\leq Cg$ eventually for some fixed $C$; $f\gg g$ reverses this comparison; $f=o(g)$ means $f/g\to0$; $f\sim g$ means $f/g\to1$; and $f\asymp g$ means both order bounds. A subscript on $O$ or $\ll$ specifies allowed constant dependence. The expression $n^{o(1)}$ in an upper bound means growth smaller than $n^\epsilon$ for each fixed $\epsilon>0$. Local definitions govern any exception, finite-field convention, or use of the same symbol for another object. An asymptotic claim is not automatically an effective algorithm.

Generalized regular expressions over a finite alphabet allow union, concatenation, Kleene star and complement relative to all finite words. Star height is the maximum number of nested stars; Boolean operations do not themselves increase it. For a regular language $L$, minimize this height over all generalized expressions defining $L$. Ask whether one finite bound works for every regular language. [S1] [S2]

Statement recorded in the supplied source.
Can all regular languages be expressed with generalized regular expressions of bounded star height? [S1]

Short English statement

Can every regular language be described using a universally bounded nesting depth of repetition, once complements are allowed?

Research attempt: the exact unary case and the limits of state elimination

Status. The universal-alphabet boundedness question remains unresolved here. We prove that every unary regular language has generalized star height at most one and that this bound is attained. The inspected thesis [S2] treats the generalized problem; its terminology includes complement, unlike ordinary star height.

1. Following a unary automaton to its eventual cycle

Let $L\subseteq\{a\}^*$ be accepted by a deterministic automaton with $q$ states. Write $s_n$ for the state reached after $a^n$. Among $s_0,\ldots,s_q$ two agree, say $s_\mu=s_{\mu+p}$ with $0\leq\mu<\mu+p\leq q$. Determinism implies $s_{n+p}=s_n$ for all $n\geq\mu$. Let $E$ be the finite accepted lengths below $\mu$, and let \[R=\{r:0\leq r<p,\ s_{\mu+r}\text{ is accepting}\}.\] Then the exact expression \[L=\bigcup_{n\in E}\{a^n\}\ \cup\ \bigcup_{r\in R} a^{\mu+r}(a^p)^*\tag{1}\] has no nested stars. In $1$, powers denote literal finite concatenations, not iteration operators. Empty $E$ or $R$ give empty unions. The construction is effective with $\mu+p\leq q$ and uses no complementation.

2. Why height zero cannot do the same job

Every star-free expression over a unary alphabet defines a finite or cofinite set of words. Prove this by structural induction. The atomic languages are finite. Union and complement preserve the family of finite/cofinite sets. For concatenation identify words with their lengths in $\mathbb N_0$, so concatenation is sumset. Two finite sets have finite sumset. If one factor is empty the sumset is empty. A nonempty finite set plus a cofinite set is cofinite, since one translate already contains all sufficiently large integers. Two cofinite sets also have cofinite sumset: fix one element in one factor and translate the other. This verifies the induction for every operation allowed at height zero.

Conversely a finite language is a finite union of literal words, and a cofinite one is the complement of such a union. Hence unary generalized height zero is exactly finite/cofinite. The language $\{a^{2n}:n\geq0\}$ is neither finite nor cofinite, so its generalized height is exactly one. The bound in the unary theorem is sharp, not merely an artifact of $1$.

3. Why an automaton-size bound is not a universal constant

For a general finite alphabet, state elimination represents paths using recurrence \[R^{(k)}_{ij}=R^{(k-1)}_{ij}\ \cup\ R^{(k-1)}_{ik}(R^{(k-1)}_{kk})^*R^{(k-1)}_{kj},\] where paths at stage $k$ may use only the first $k$ states internally. This gives a finite expression, but the newly introduced star may contain earlier stars. The recurrence therefore supplies a bound growing with the number of eliminated states, not a constant independent of the input automaton. Having complement available does not automatically flatten that nesting; an identity doing so must be proved.

4. Checks and residual obstruction

The diagnostic enumerates all unary deterministic automata with at most four states, initial state 0 and every accepting subset. It compares direct acceptance with the periodic representation $1$ for lengths through $4q+10$. The proof, rather than that finite test, establishes equality at all lengths. The height-zero argument is deliberately unary: over two letters there are infinite noncofinite star-free languages, for instance all words containing the letter $a$, expressible using the complement of the empty language and concatenation. Thus one cannot extend the finite/cofinite characterization to larger alphabets. A universal bound on generalized height for arbitrary alphabets and arbitrary transition structures is the exact missing conclusion.

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