Albertson–Berman conjecture on induced forests in planar graphs

For every simple planar graph GG with n=V(G)n=|V(G)| vertices, if a(G)a(G) denotes the maximum number of vertices in an induced forest of GG, then a(G)n2a(G)\geq \frac{n}{2}.

Progress summary

Solved

A 2026 preprint claims the conjecture is false by exhibiting a 39-vertex planar graph whose largest induced forest has fewer than half the vertices, but this has not been independently verified.

The conjecture, open since 1979, asserts that every simple planar graph on nn vertices contains an induced forest with at least n/2n/2 vertices. Two 2026 sources now claim explicit counterexamples, so the assertion is no longer plausibly an untouched open problem, but neither claim has independent confirmation.

Known results

  • Borodin’s acyclic 55-colorability theorem gives a(G)2n/5a(G)\geq 2n/5.
  • Disjoint copies of K4K_4 show that the conjectured n/2n/2 bound would be tight.
  • A 2016 paper records the statement as a challenging conjecture and the 2n/52n/5 bound as the best general lower bound.

August 2026 claimed counterexamples

An arXiv preprint claims a fully proved simple planar graph G39G_{39} with a(G39)=19a(G_{39})=19, giving 19/39<1/219/39<1/2, while calling the manuscript unfinished and unpolished. A separate Zenodo record claims 33-connected maximal planar graphs MkM_k with 31k31k vertices and a(Mk)=15ka(M_k)=15k, and supplies a Python verifier; neither claim has independent verification or a reported retraction.

Current status (as of August 2026): The n/2n/2 conjecture remains unverified and is now contradicted by two claimed explicit constructions; independent checking is needed, and the best valid lower bound and optimal constant remain open.

Sources
Sources & referencesView supporting material

Primary source

arXiv

Solutions 0

No solutions have been posted yet.