Albertson–Berman induced-forest conjecture

Every planar graph GG contains a vertex subset S⊆V(G)S\subseteq V(G) such that the induced subgraph G[S]G[S] is a forest and ∣S∣≥∣V(G)∣2|S|\geq \frac{|V(G)|}{2}.

References

Progress summary

Refreshed
Claimed solved

Two August 2026 preprints claim the conjecture is false, with a stronger counterexample than the first, but neither claim has independent verification.

The Albertson–Berman conjecture, posed in 1979, asserts that every planar graph contains an induced forest on at least half its vertices. August 2026 work claims this statement is false.

Known results

  • Borodin’s acyclic 55-colorability gives a(G)≥25na(G)\geq \frac{2}{5}n for planar graphs.
  • The conjecture holds for planar graphs of girth at least 55.
  • The conjecture would imply an independent set of size at least n4\frac{n}{4} without using the Four Color Theorem.

August 2026 claimed disproofs and refinement

One preprint constructs a 3131-vertex plane triangulation with a(G)=15a(G)=15, and an infinite family with ratio 1531<12\frac{15}{31}<\frac{1}{2}. Another claims a 3939-vertex example with ratio 1939\frac{19}{39}. A newer report claims to determine the minimum counterexample order, increase connectivity, and improve the ratio further; these claims remain unverified, and one reported construction is explicitly incomplete.

Current status (as of August 2026): the conjecture has claimed counterexamples, including a reported ratio of 1531\frac{15}{31}, but no independent verification is recorded and the exact post-disproof structural conclusions remain unconfirmed.

Sources

Solutions 0

No solutions have been posted yet.