Math-1561 · mathematics · Combinatorics

Math-1561

A corrected colored reduction and a Sullivan-2 counterexample normal form

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Statement

Definitions and mathematical statement

Let $D=(V,E)$ be a nonempty finite oriented graph, meaning a simple directed graph with neither loops nor opposite pairs of arcs. Define \[ N^+(v)=\{u:v\to u\},\qquad N^-(v)=\{u:u\to v\}, \qquad d^\pm(v)=|N^\pm(v)|. \] Let $N_2^+(v)$ consist of vertices at directed distance exactly two from $v$: they are endpoints of paths $v\to u\to w$, with $w\ne v$ and $w\notin N^+(v)$. The number of paths is irrelevant; only distinct endpoints are counted. The source's Conjecture 6.4 asks whether some vertex always satisfies \[ |N_2^+(v)|+d^+(v)\ge2d^-(v). \] Thus the number of vertices reachable in one or two steps should, for at least one starting point, be at least twice that point's indegree. The two out-neighbourhoods are disjoint by definition, so their sum counts distinct vertices rather than walks.

All graph sizes and all orientations of all underlying simple graphs are included. In particular, the digraph need not be a tournament, strongly connected, or balanced in its degrees. A vertex of indegree zero satisfies the assertion automatically. The inequality concerns one vertex chosen after the graph is given; an average inequality or an estimate for two-edge walk counts would be a different claim. The indegree on the right is retained exactly as in the workshop, rather than replaced by the outdegree from Seymour's conjecture.

Short English statement

Must an oriented graph contain a vertex whose first and second out-neighbourhoods together are at least twice its indegree?

Sources

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