Math-1766 · mathematics · Combinatorics

Math-1766

An order-statistic form of Shen's deficit conjecture

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Statement

Definitions and mathematical statement

Let $G$ be a finite simple loopless directed graph on $n$ vertices, with at most one arc from one vertex to another, and suppose every vertex has positive out-degree. Write $d^+(v)=|\{w:v\to w\}|$. Its directed girth $g$ is the least length of a directed cycle with distinct vertices; such a cycle exists because the finite graph has no vertex of out-degree zero. Opposite arcs are allowed, in which case $g$ may equal two.

For any positive integer $r$, define the total deficit below out-degree $r$ by \[ t(G,r)=\sum_{v\in V(G)}\max\{0,r-d^+(v)\}. \] Vertices of degree at least $r$ contribute zero, so surplus degree does not cancel another vertex's deficit. Shen's conjecture asks whether every such graph and every $r\geq1$ satisfy \[ n\geq r(g-1)+1-t(G,r). \] Equivalently, the total deficit should obey $t(G,r)\geq r(g-1)+1-n$ whenever the right side is positive. When the minimum out-degree is at least $r$, the deficit vanishes and the assertion recovers the familiar degree--girth inequality. Here the question is whether the same obstruction persists with precisely one unit of allowed vertex-count loss for each missing outgoing neighbor. No regularity condition is assumed, and there is no restriction on in-degrees. Counting distinct outgoing neighbors, rather than arbitrary parallel arcs, is essential to the stated numerical bound.

Short English statement

Can the degree--girth bound be extended by subtracting exactly the total amount by which vertex out-degrees fall below a chosen threshold?

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