A 15/31 Counterexample Family to the Albertson-Berman Conjecture

Abstract: For a graph $G$, let $a(G)$ be the maximum number of vertices in an induced forest. The Albertson-Berman conjecture, posed in 1979, asserts that every $n$-vertex planar graph satisfies $a(G)\ge n/2$. Borodin's bound $a(G)\ge 2n/5$ remains the general lower bound toward this problem. We disprove the conjecture with an explicit 31-vertex plane triangulation $T$ satisfying $a(T)=15$. Moreover, for every integer $k\ge2$, we construct a simple planar graph $M_k$ with $|V(M_k)|=31k$ and $a(M_k)=15k$, so that $a(M_k)/|V(M_k)|=15/31<1/2$. Every member of the family has minimum degree five. The construction starts from a $31$-vertex seed obtained by substituting a $14$-vertex two-terminal gadget into a pentagonal bipyramid, and then uses annular joins along facial triangles to preserve the exact ratio. The resulting graphs are sphere triangulations, and hence maximal planar.

• The paper constructs an infinite family of simple maximal planar graphs with 31k vertices and maximum induced forests of exactly 15k vertices, yielding the counterexample ratio 15/31<1/2.
• A 14-vertex triangulated gadget creates a one-vertex terminal penalty, and a transfer law converts this local obstruction into a global bound through a sparsely decorated pentagonal bipyramid.
• The resulting graphs have minimum degree five and remain maximal planar, narrowing the optimal induced-forest constant to 2/5≤c≤15/31 while leaving 4-connected cases and smaller counterexamples open.

Overview and main result

For a graph $G$, let $a(G)$ denote the maximum order of an induced forest. The Albertson–Berman conjecture (1979) asserts that every planar graph $G$ satisfies $a(G)\ge |V(G)|/2$. The best general lower bound remains Borodin's $a(G)\ge 2n/5$, derived from acyclic 5-colorings, with stronger bounds known for outerplanar graphs ($2n/3$), triangle-free planar graphs ($(71n+72)/128$), and a related $5n/8$ conjecture of Akiyama–Watanabe for bipartite planar graphs. This paper disproves the conjecture in its original simple-graph setting by constructing, for every $k\ge 2$, a simple planar graph $M_k$ with

$a(G)$0

and minimum degree five. Every member is a sphere triangulation, hence maximal planar. The construction proceeds in three stages: a two-terminal gadget with a one-unit terminal penalty, a general transfer law for substituting the gadget onto decorated edges of a base graph, and an annular amplification that joins copies of a 31-vertex seed while preserving the ratio exactly.

The result is notable not only for refuting a long-standing conjecture but for doing so within the class of maximal planar graphs of minimum degree five — a strong structural setting. It also sharpens the picture relative to Makarov's 2026 multigraph construction with asymptotic ratio $a(G)$1, which does not apply to simple graphs.

The two-terminal gadget

The local mechanism is a 14-vertex plane triangulation $a(G)$2, obtained from the icosahedral graph $a(G)$3 by deleting the edge $a(G)$4, with two distinguished adjacent terminals $a(G)$5. Writing $a(G)$6, the internal forest capacity is defined as

$a(G)$7

and the terminal profile is:

selected terminals	internal capacity
$a(G)$8	6
$a(G)$9 or $G$0	6
$G$1	5

The one-unit drop when both terminals are forced into the forest is the entire engine of the counterexample. The proof of the upper bounds is combinatorial and rests on three edge-density lemmas about the icosahedral graph: every five vertices span at least three edges; every six vertices span at least five edges; and every six-set containing either face $G$2 or $G$3 spans at least six edges. For instance, any seven internal vertices span at least eight edges in $G$4, hence at least seven in $G$5, forcing a cycle; and when both terminals are selected, any six-vertex candidate forest would have to be a tree containing an $G$6–$G$7 path, which the edges $G$8 close to a cycle. Matching lower bounds are given by explicit induced trees.

Selected-edge substitution and the transfer law

Given a base graph $G$9 and a set $a(G)\ge |V(G)|/2$0 of "decorated" edges, one substitutes a copy of $a(G)\ge |V(G)|/2$1 onto each edge of $a(G)\ge |V(G)|/2$2, identifying terminals with endpoints. Each substitution adds 12 vertices and 35 edges. Defining

$a(G)\ge |V(G)|/2$3

the transfer law states exactly:

$a(G)\ge |V(G)|/2$4

The upper bound follows by summing the gadget's terminal-conditioned capacities over all copies; the lower bound assembles equality witnesses, using the fact that the union of two forests along a connected subtree is a forest. In the full-edge specialization $a(G)\ge |V(G)|/2$5 this recovers $a(G)\ge |V(G)|/2$6; for $a(G)\ge |V(G)|/2$7 this yields a 76-vertex block with ratio $a(G)\ge |V(G)|/2$8, showing that decorating every base edge is suboptimal. Undecorated base edges impose cycle constraints without paying their 12 internal vertices — the key to improving the ratio below.

The pentagonal bipyramid seed

The base graph is the pentagonal bipyramid $a(G)\ge |V(G)|/2$9 (7 vertices, 15 edges), with only the two disjoint rim edges $a(G)\ge 2n/5$0 and $a(G)\ge 2n/5$1 decorated. A three-case argument shows that if $a(G)\ge 2n/5$2 is a forest then $a(G)\ge 2n/5$3, with equality attained at $a(G)\ge 2n/5$4; hence $a(G)\ge 2n/5$5. The substituted graph $a(G)\ge 2n/5$6 has 31 vertices, 85 edges, and $a(G)\ge 2n/5$7. Adding the two diagonals $a(G)\ge 2n/5$8 and $a(G)\ge 2n/5$9 inside the two quadrilateral faces left by the edge sums produces a maximal planar seed $2n/3$0 with 87 edges ($2n/3$1) and $2n/3$2, witnessed by an explicit 15-vertex induced path. Its degree multiset is $2n/3$3.

A remark extends the analysis to odd bipyramids $2n/3$4 with a maximum rim matching decorated, giving ratios $2n/3$5, minimized at $2n/3$6 — so the pentagonal case is optimal within that family.

Annular amplification

Copies of $2n/3$7 are joined along vertex-disjoint facial triangles ("ports") $2n/3$8 and $2n/3$9 via a triangulated cylinder of six cross-edges forming six triangular faces. Both ports avoid the path witness $(71n+72)/128$0, so no annulus edge connects selected witness vertices across seeds. Three lemmas complete the proof: each join preserves the sphere triangulation topology (yielding $(71n+72)/128$1 edges); the per-seed bound $(71n+72)/128$2 sums to $(71n+72)/128$3 while the copied witnesses give $(71n+72)/128$4; and since the unique degree-four vertex lies in port $(71n+72)/128$5 and gains two cross-neighbours, $(71n+72)/128$6. The full degree histogram has maximum degree nine, and each $(71n+72)/128$7 is 3-connected with separating triangles.

Consequence: the Albertson–Berman conjecture fails already for simple planar graphs of minimum degree five, and the optimal constant $(71n+72)/128$8 satisfies $(71n+72)/128$9.

Verification and limitations

The proof is entirely symbolic; an accompanying script independently reconstructs the seed from the stated data and certifies $5n/8$0 via two exact algorithms (branch-and-bound feedback vertex set and a 0–1 ILP with separated cycle cuts), neither of which uses the transfer law or the displayed witness. The script does not optimize full $5n/8$1 directly: the universal bound over $5n/8$2 relies on the body arguments.

Three limitations are conceded explicitly. First, the interval $5n/8$3 for the optimal constant is wide on both sides. Second, the construction does not address 4-connected planar graphs, since every $5n/8$4 contains the separating triangle $5n/8$5; whether the conjecture fails there remains open. Third, minimality is established only within this construction: 31 vertices is the smallest output of this method, but smaller counterexamples may exist.

Conclusion

This paper refutes the Albertson–Berman conjecture through an explicit infinite family of maximal planar graphs with exact induced-forest ratio $5n/8$6 and minimum degree five. The architecture — a one-unit terminal penalty in a 14-vertex gadget, converted into a global vertex deficit by a clean transfer law over a sparsely decorated odd bipyramid, then amplified without loss by annular joins — is modular and may admit other instantiations. The central open problem is now to determine the true value of the optimal constant $5n/8$7, currently bracketed between Borodin's $5n/8$8 and the $5n/8$9 established here.