Tuza's Triangle Packing–Covering Conjecture

OPENMajorConjectureProposed 1981 · Standard version

Canonical statement

For a finite simple graph GG, let ν3(G)\nu_3(G) be the maximum number of pairwise edge-disjoint triangles and let τ3(G)\tau_3(G) be the minimum number of edges meeting every triangle. Then
τ3(G)2ν3(G). \tau_3(G)\leq2\nu_3(G).
View source LaTeX
For a finite simple graph \(G\), let \(\nu_3(G)\) be the
maximum number of pairwise edge-disjoint triangles and let \(\tau_3(G)\)
be the minimum number of edges meeting every triangle. Then
\[
  \tau_3(G)\leq2\nu_3(G).
\]

For a finite simple graph GG, write ν3(G)\nu_3(G) for the maximum number of pairwise edge-disjoint triangles and τ3(G)\tau_3(G) for the minimum number of edges meeting every triangle. Removing the three edges of each triangle in a maximum packing destroys all triangles, so τ3(G)3ν3(G)\tau_3(G)\le 3\nu_3(G) trivially, while any cover must use at least one edge from each packed triangle, giving τ3(G)ν3(G)\tau_3(G)\ge\nu_3(G). Tuza conjectured in 1981 that the true worst-case ratio is 22: every graph satisfies τ3(G)2ν3(G)\tau_3(G)\le 2\nu_3(G) [Tuza1981Triangles]. The factor 22 would be sharp, as small complete graphs show.

Haxell proved a general bound with a constant strictly below the trivial factor 33 [Haxell1999Tuza], and by now the fractional relaxation of the problem, asymptotic versions, and many special graph classes are well understood. Recent work continues to extend the verified territory, for instance to random geometric graphs, where almost-perfect triangle packings are constructed [BennettEtAl2026Tuza].

Despite this progress the constant 22 has not been established in full generality, and the conjecture remains open: a proof must handle arbitrary graphs, not only the structured or random classes treated so far.

The boxed statement is the canonical open formulation — not a stronger variant or a related research program. The status reflects the catalog's last review; do your own literature search before investing serious effort.