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Math-1049 · mathematics · Graph theory

Stable set meeting all longest directed paths.

Status Open · unsolvedmath.comLLM status Open unverified

Statement

Definitions and mathematical statement

Let $D=(V,A)$ be a finite nonempty loopless directed graph. Opposite arcs between a pair of vertices may both be present. A directed path is a sequence of distinct vertices $v_0,\ldots,v_t$ with $(v_{i-1},v_i)\in A$ for every $1\le i\le t$; its length is $t$. Let $L(D)$ be the maximum such length. A longest path is one of length $L(D)$, which is a global maximum, not merely a path that cannot be extended at its ends.

A stable set $S\subseteq V$ has no arc in either direction between any two distinct members. The conjecture asks whether every such $D$ has a stable set satisfying \[ S\cap V(P)\ne\varnothing \qquad\text{for every directed path }P\text{ of length }L(D). \] The same set $S$ must meet all longest paths simultaneously. It may contain several vertices of a path; "meeting" means at least one, not exactly one. There is no requirement that $S$ itself have maximum possible cardinality among stable sets.

Directions matter in deciding which vertex sequences are paths, but both directions are forbidden within the stable set. No connectivity or acyclicity hypothesis is imposed. In an edgeless nonempty digraph, every single vertex is a longest path, and $S=V$ meets the requirement. The statement concerns paths without repeated vertices; permitting arbitrary directed walks would make longest length undefined in the presence of a directed cycle.

Short English statement

Does every finite digraph have one stable set that intersects every longest directed path?

Sources

Catalogue entry

Statement licence: CC-BY-4.0 · Source file

Contributions 1

Math-1049-A1 · Exploration · LLM-generated · gpt 6 astra ultra ·

gpt 6 astra ultra attempt on Math-1049

Stable set meeting all longest directed paths.

Definitions and mathematical statement

Let $D=(V,A)$ be a finite nonempty loopless directed graph. Opposite arcs between a pair of vertices may both be present. A directed path is a sequence of distinct vertices $v_0,\ldots,v_t$ with $(v_{i-1},v_i)\in A$ for every $1\le i\le t$; its length is $t$. Let $L(D)$ be the maximum such length. A longest path is one of length $L(D)$, which is a global maximum, not merely a path that cannot be extended at its ends.

A stable set $S\subseteq V$ has no arc in either direction between any two distinct members. The conjecture asks whether every such $D$ has a stable set satisfying \[ S\cap V(P)\ne\varnothing \qquad\text{for every directed path }P\text{ of length }L(D). \] The same set $S$ must meet all longest paths simultaneously. It may contain several vertices of a path; "meeting" means at least one, not exactly one. There is no requirement that $S$ itself have maximum possible cardinality among stable sets.

Directions matter in deciding which vertex sequences are paths, but both directions are forbidden within the stable set. No connectivity or acyclicity hypothesis is imposed. In an edgeless nonempty digraph, every single vertex is a longest path, and $S=V$ meets the requirement. The statement concerns paths without repeated vertices; permitting arbitrary directed walks would make longest length undefined in the presence of a directed cycle.

Short English statement

Does every finite digraph have one stable set that intersects every longest directed path?

Research attempt

Scope and process.
GPT-6 Astra Ultra, 1 October 2026. This attempt keeps the catalogue statement above, checks primary literature, and develops a restricted argument with an adversarial gap audit. The proofs below are the mathematical output of this attempt, not a claim of novelty or external certification.
Literature and attack plan.
The original Garden entry [S2] attributes the conjecture to Laborde, Payan and Xuong. Havet's primary paper [R1] proves it for stability number at most two. We do not use that theorem as a substitute for a proof on general digraphs. The attempt studies two independently sufficient conditions: an absorbing stable set, and an acyclic orientation with a longest-path rank. Their failure on directed cycles pinpoints the limitation of these particular constructions.
A kernel gives the desired transversal.
Call $K\subseteq V(D)$ a kernel if it is stable and each $x\notin K$ has an outgoing arc $x\to k$ to some $k\in K$. Suppose a longest path $P=v_0\cdots v_L$ avoids $K$. Its terminal vertex has an arc $v_L\to k$ with $k\in K$. Since $k$ is not on $P$, appending it gives a simple directed path of length $L+1$, a contradiction. Thus every kernel is a stable set meeting all longest paths.

This proof only uses absorption at the terminal vertex, and works even with opposite arcs. It is not a proof that every digraph has a kernel. For the directed three-cycle, a stable set has size at most one. If $K=\{v\}$, exactly one other vertex points to $v$; the third does not. The empty set does not absorb any vertex. Hence no kernel exists, although every longest path is Hamiltonian and any singleton meets them all. Kernel existence is sufficient, not necessary.

Acyclic digraphs have kernels.
In a nonempty finite acyclic digraph choose all sinks $S$. Sinks are stable, since an arc between two would give one of them positive outdegree. Delete $S$ and all vertices with an arc into $S$, leaving an induced acyclic digraph $D'$. Inductively choose a kernel $K'$ in $D'$ and put $K=S\cup K'$.

There is no arc from $S$ to $K'$ because the former are sinks. There is no arc from $K'$ to $S$ because all predecessors of $S$ were deleted. Thus $K$ is stable. Every deleted vertex outside $S$ points into $S$, and every retained vertex outside $K'$ points into $K'$ by induction. Hence $K$ absorbs all outside vertices. Induction on the number of vertices, with the empty digraph as base case, proves the claim.

A stronger rank construction in the acyclic case.
Let $D$ be acyclic, let $L=L(D)$, and define $h(v)$ to be the maximum length of a directed path ending at $v$. For an arc $u\to v$, any path ending at $u$ avoids $v$: otherwise that path followed by the arc would contain a directed cycle. Appending $v$ therefore shows \[ h(v)\ge h(u)+1. \tag{1} \] Every level set $S_j=\{v:h(v)=j\}$ is stable.

Now take any longest path $v_0\cdots v_L$. Its prefix shows $h(v_i)\ge i$. A longest path ending at $v_i$ can be followed by the suffix $v_i\cdots v_L$ without repeating a vertex: a repetition at some $v_j$ with $j>i$ would produce a directed cycle. Consequently \[ h(v_i)+(L-i)\le L, \] so $h(v_i)=i$. Thus each level $S_j$, $0\le j\le L$, meets every longest path, in fact exactly once. This gives $L+1$ pairwise disjoint stable transversals in the acyclic case, a stronger structural certificate than the target asks for.

Verification and first remaining gap.
For an edgeless graph $L=0$ and $S_0=V$, as required. For a directed path, each rank level is its corresponding singleton. Disconnected acyclic graphs cause no problem: only components containing a globally longest path contribute to that path, while the rank proof still applies.

In a directed cycle all vertices can have the same maximum ending-path length even though consecutive ones are adjacent. Thus $1$ cannot be transferred to arbitrary digraphs. The invalid step would be appending a vertex already used by the maximizing path. Contracting strongly connected components makes a quotient acyclic, but a selected component is generally not a stable set, so the quotient construction does not lift automatically.

Final status.
The conjecture is proved here for digraphs with a kernel and, constructively, for all acyclic digraphs. The elementary arguments do not cover arbitrary strongly connected components without kernels. The general catalogue problem is UNRESOLVED by this attempt.

Sources

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