Math-1129 · mathematics · Graph theory

Math-1129

Triangle-packing vs triangle edge-transversal.

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Statement

Definitions and mathematical statement

Let $G=(V,E)$ be a finite simple undirected graph. A triangle is the three-edge subgraph on three pairwise adjacent vertices. A triangle packing is a family of triangles with pairwise disjoint edge sets; sharing vertices is permitted. Write $\nu_{\triangle}(G)$ for the largest number of triangles in such a family.

A triangle edge-transversal is a set $F\subseteq E$ meeting the edge set of every triangle. Equivalently, deleting $F$ leaves a triangle-free graph. Write $\tau_{\triangle}(G)$ for the minimum size of a triangle edge-transversal. Both parameters are zero when the graph has no triangle.

Tuza's conjecture asks whether \[ \tau_{\triangle}(G)\le2\nu_{\triangle}(G) \] for every finite simple graph $G$. Equivalently, if no packing has more than $k$ triangles, some set of at most $2k$ edges destroys every triangle. "At most" is the appropriate convention: graphs with fewer than $2k$ edges need not admit a set of exactly that size.

The elementary inequalities are $\nu_{\triangle}(G)\le\tau_{\triangle}(G)\le3\nu_{\triangle}(G)$. For the upper bound, the union of the edges of a maximum packing meets every triangle, since otherwise the packing could be enlarged. The conjecture improves that factor three to two. For $K_4$, the packing number is one and the transversal number is two, so the proposed factor cannot be reduced. Both quantities are integral; fractional assignments of weights to edges or triangles are different relaxations of the problem.

Short English statement

Can all triangles be destroyed using at most twice as many edges as the maximum number of edge-disjoint triangles?

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