Math-1569 · mathematics · Combinatorics

Math-1569

Local average outdegree and short directed cycles

Status Partially solved · unsolvedmath.comLLM status Open unverified

Top Open Problems Not reported cataloguedPartially solvedOpen

Statement

Definitions and mathematical statement

Let $D=(V,E)$ be a finite simple loopless digraph on $n$ vertices, with at least one outgoing arc at every vertex. Opposite arcs are allowed. Write $d^+(v)$ for the number of arcs leaving $v$. The directed girth $g(D)$ is the length of its shortest directed cycle; such a cycle exists because following outgoing arcs in a finite graph must eventually revisit a vertex. For a positive integer $r$, assume the local degree-sum condition \[ d^+(u)+d^+(v)\ge2r \qquad\text{for every arc }u\to v. \] Shen's conjecture asks whether these assumptions imply \[ g(D)\le\left\lceil\frac nr\right\rceil. \] The degree sum is imposed on endpoints of every directed edge, not on all pairs of vertices and not merely on average over the graph. A vertex may individually have outdegree smaller than $r$, provided its incident outgoing-edge endpoint sums meet the requirement. Consequently this is a local average-degree extension of the usual minimum-outdegree short-cycle question.

The assertion is uniform over $n,r$ and all such digraphs; $r$ is not the number of vertices or an average degree computed afterwards. The minimum-outdegree-one assumption is retained independently of the degree-sum condition, excluding sinks that the latter might not constrain. Degrees count neighbours in a simple digraph; parallel arcs would artificially enlarge them without reducing girth.

Short English statement

Does a lower bound on the sum of the endpoint outdegrees of each arc force the same short-cycle bound as a minimum-outdegree condition?

Sources

Catalogue entry

Statement licence: CC-BY-4.0 · Source file

Contributions 0

No attempts yet. Significant partial results and substantial reductions are welcome. Contribute an attempt.