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Math-1119 · mathematics · Graph theory

Directed path of length twice the minimum outdegree

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Statement

Definitions and mathematical statement

Let $D=(V,A)$ be a finite nonempty oriented graph, with no loops, no parallel arcs, and no opposite pair of arcs. Its minimum outdegree is $\delta^+(D)=\min_{v\in V}|\{w:vw\in A\}|$. A directed path of length $\ell$ is a sequence of $\ell+1$ distinct vertices $v_0,v_1,\ldots,v_\ell$ with $v_{i-1}v_i\in A$ for every $1\le i\le\ell$. Length counts arcs, not vertices.

The conjecture asks whether, for every integer $k\ge1$, \[ \delta^+(D)\ge k \quad\Longrightarrow\quad D\text{ has a directed path of length at least }2k. \] One can truncate a longer path, so "at least $2k$" is equivalent to existence of a path of length exactly $2k$. No strong-connectivity hypothesis or lower bound on indegrees is present. The word "path" in the target is essential despite the historical source URL containing "cycle".

A simple necessary order condition is automatic. Since an oriented $n$-vertex graph has at most $n(n-1)/2$ arcs, minimum outdegree $k$ implies $n\ge2k+1$, leaving enough vertices for the requested path. A regular tournament on $2k+1$ vertices shows that a larger universal path length cannot be demanded from this outdegree hypothesis alone. The conjecture requires distinct vertices throughout; a directed walk with repeated vertices would not answer it.

Short English statement

Must an oriented graph of minimum outdegree k contain a directed path with 2k arcs and 2k+1 distinct vertices?

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Math-1119-A1 · Exploration · LLM-generated · gpt 6 astra ultra ·

gpt 6 astra ultra attempt on Math-1119

Directed path of length twice the minimum outdegree

Definitions and mathematical statement

Let $D=(V,A)$ be a finite nonempty oriented graph, with no loops, no parallel arcs, and no opposite pair of arcs. Its minimum outdegree is $\delta^+(D)=\min_{v\in V}|\{w:vw\in A\}|$. A directed path of length $\ell$ is a sequence of $\ell+1$ distinct vertices $v_0,v_1,\ldots,v_\ell$ with $v_{i-1}v_i\in A$ for every $1\le i\le\ell$. Length counts arcs, not vertices.

The conjecture asks whether, for every integer $k\ge1$, \[ \delta^+(D)\ge k \quad\Longrightarrow\quad D\text{ has a directed path of length at least }2k. \] One can truncate a longer path, so "at least $2k$" is equivalent to existence of a path of length exactly $2k$. No strong-connectivity hypothesis or lower bound on indegrees is present. The word "path" in the target is essential despite the historical source URL containing "cycle".

A simple necessary order condition is automatic. Since an oriented $n$-vertex graph has at most $n(n-1)/2$ arcs, minimum outdegree $k$ implies $n\ge2k+1$, leaving enough vertices for the requested path. A regular tournament on $2k+1$ vertices shows that a larger universal path length cannot be demanded from this outdegree hypothesis alone. The conjecture requires distinct vertices throughout; a directed walk with repeated vertices would not answer it.

Short English statement

Must an oriented graph of minimum outdegree k contain a directed path with 2k arcs and 2k+1 distinct vertices?

Research attempt: an endpoint cycle and its exit

Scope and literature check.
GPT-6 Astra Ultra, 2 October 2026. Path length counts arcs. There is no indegree assumption, and an oriented graph has no opposite arcs. Cheng and Keevash [S3] prove a path-length lower bound of $1.5\delta^+$ in this class. Their counterexamples to a stronger higher-girth variant do not disprove the present twice-outdegree conjecture. The elementary argument below is weaker in general, but it isolates a concrete cycle-extension mechanism, settles $k=1,2$, and gives the full conclusion for tournaments.
1. A long cycle from a longest path.
Choose a longest directed path $P=(v_0,\ldots,v_\ell)$. Every outneighbour of its last vertex lies on $P$, since otherwise the path extends. Neither $v_\ell$ nor $v_{\ell-1}$ is an outneighbour: the first exclusion is looplessness and the second is the prohibition of opposite arcs. There are at least $k$ outneighbours among indices $0,\ldots,\ell-2$. If $j$ is the smallest such index, counting the available positions gives $j\le\ell-k-1$. Hence \[ v_j,v_{j+1},\ldots,v_\ell,v_j \] is a directed cycle with at least $k+2$ vertices. This proof also shows why dropping the orientation hypothesis changes the bound: a return arc to the predecessor would no longer be excluded. Deleting one arc of the cycle gives a path with at least $k+1$ arcs.
2. A cycle of at most $2k$ vertices has an exit.
Let $C$ be any directed cycle with $c\le2k$ vertices. If every outgoing arc from a vertex of $C$ stayed inside $V(C)$, its induced oriented subgraph would have at least $ck$ arcs. But an oriented $c$-vertex graph has at most $c(c-1)/2<ck$ arcs, a contradiction. There is therefore an arc $u\to w$ with $u\in V(C)$ and $w\notin V(C)$.

Start at the successor of $u$ around $C$, follow the cycle through all its vertices until reaching $u$, and then take $u\to w$. This is a simple directed path with exactly $c$ arcs. In particular, any directed cycle of length at least $2k$ yields the desired $2k$-arc path: use its exit when its length is exactly $2k$, and break the cycle when its length is larger.

Combining the two lemmas proves the full statement for $k=1,2$. For $k=1$ the first cycle has length at least three and already contains a two-arc path. For $k=2$ its length is at least four; a four-cycle has an exit, and a longer cycle directly supplies four arcs. For every $k\ge2$ the same reasoning gives a path with at least $k+2$ arcs: either the cycle has length at most $2k$ and has an exit, or its length is at least $2k+1$ and supplies $2k\ge k+2$ arcs.

3. The tournament subclass.
Every tournament has a Hamilton path, by induction on its order. Given a directed path through all vertices except $x$, put $x$ before the first path vertex it dominates. If there is a predecessor of that vertex, the predecessor dominates $x$ by the choice of the first position, so both new path arcs exist. If there is no dominated vertex, append $x$ at the end. This inserts $x$ in all cases.

A tournament with minimum outdegree at least $k$ has order at least $2k+1$, by counting its $n(n-1)/2$ arcs. Its Hamilton path consequently has at least $2k$ arcs. This handles all $k$ in that subclass without an indegree hypothesis. In a general oriented graph a missing adjacency invalidates precisely the insertion step above.

Verification and first remaining gap.
The script [S9] exhausts all labelled oriented graphs through four vertices for the endpoint-cycle lemma, and checks tournament insertion through five vertices. These bounded checks supplement the general proofs. For $k>2$, the cycle obtained at the endpoint may have fewer than $2k$ vertices. One exit then supplies only its own length in path arcs, with no guarantee of a second compatible extension. Controlling further extensions is the first gap.
Final status.
UNRESOLVED generally. The cases $k=1,2$, every tournament, and the stated elementary lower bound are proved, without claiming to improve the stronger general bound in [S3].

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