A counter-example to Baranyai's combinatorial characterisation for 3-rigidity

Recently Baranyai described a necessary combinatorial characterisation of graph rigidity for dimension 3. In this short note we provide a counter-example to the converse of the condition. Additionally, we provide an alternative proof to the Baranyai’s necessary condition.

💡 Research Summary

The paper revisits a recent claim by Tamás Baranyai (2026) concerning a combinatorial characterisation of generic rigidity for three‑dimensional graphs. Baranyai’s theorem states that if a graph G = (V,E) is minimally 3‑rigid (i.e., it is 3‑rigid but ceases to be so after deleting any edge) and contains at least four vertices, then for every edge e ∈ E there exists a partition (S₁,S₂,S₃) of the edge set satisfying three conditions: (i) |S_i| = |V| − i for i = 1,2,3; (ii) e ∈ S₁; (iii) the three derived graphs (V, S₁ ∪ S₂), (V, S₁ ∪ S₃)/e, and (V, S₂ ∪ S₃ ∪ e)/e are each minimally 2‑rigid. Baranyai also asserted that the converse holds, i.e., that these combinatorial conditions are sufficient for a graph to be minimally 3‑rigid.

The authors of the present note provide a clear counter‑example to this converse. They use the classic “double‑banana” graph: two copies of K₅ with one edge removed (each copy is (2,3)-tight) are joined at a pair of vertices that form a 2‑vertex separating set. The whole graph is (3,6)-tight (|E| = 3|V| − 6) and, as the authors demonstrate, for every edge e one can explicitly construct a partition (S₁,S₂,S₃) meeting Baranyai’s three conditions. Nevertheless, the graph is not 3‑rigid because the separating pair acts as a hinge, allowing one “banana” to rotate relative to the other while preserving all edge lengths. This shows that Baranyai’s necessary condition is indeed necessary, but the claimed sufficiency is false.

Beyond the counter‑example, the paper supplies an alternative proof of the original necessary condition that relies solely on linear algebra, avoiding the more involved geometric arguments used by Baranyai. The proof proceeds through three lemmas:

1. Lemma 4.1 is a standard Laplace‑expansion result: for any n × n matrix X and a set C of k columns, there exists a set R of k rows such that both the submatrix X_{R,C} and its complementary submatrix X_{R̄, C̄} are invertible when X is invertible.

2. Lemma 4.2 shows that for a minimally 3‑rigid graph G and a spanning tree T containing an edge e, one can partition the remaining edges E \ T into two subsets R₁ and R₂ of sizes |V| − 2 and |V| − 3 respectively, so that the graphs (V, T ∪ R₁) and (V, R₂ ∪ T)/e are both minimally 2‑rigid. The proof selects a generic realisation p of G, aligns two vertices so that the vector between them lies on the x‑axis, and then constructs a square matrix X from the rigidity matrix by deleting six carefully chosen columns. By showing X is invertible (using a kernel argument based on skew‑symmetric matrices) and applying Lemma 4.1, the required partition is extracted.

3. Lemma 4.3 refines the construction by guaranteeing a spanning tree F that contains the chosen edge e and such that the graph obtained by contracting e after adding back the edges outside F is minimally 2‑rigid. This involves adding a dummy parallel edge, forming a matrix Z, and again using Laplace expansion to locate a suitable set of rows S that yields the desired tree and contraction properties.

With these lemmas, the authors reconstruct Baranyai’s partition: set S₁ = F, S₂ = R₁, S₃ = R₂. The three conditions (i)–(iii) follow directly from the lemmas, thereby providing a concise, purely algebraic proof of the necessary part of Baranyai’s theorem.

In the final section, the authors discuss implications for higher dimensions. They recall that coning a graph (adding a new vertex adjacent to all existing vertices) preserves minimal rigidity when the dimension is increased by one. Consequently, the double‑banana counter‑example can be “coned” to produce counter‑examples for any dimension d ≥ 3. Motivated by this, they propose Conjecture 5.1: a graph with at least four vertices is (3,6)-tight if and only if for every edge e there exists a partition (S₁,S₂,S₃) satisfying the same size constraints and such that the three derived graphs are (2,3)-tight. This conjecture separates the purely counting condition (3,6)-tightness from genuine 3‑rigidity, suggesting a refined combinatorial characterisation that may hold even though the original sufficiency claim does not.

Overall, the paper makes two significant contributions: it definitively disproves the converse of Baranyai’s combinatorial condition by exhibiting a concrete counter‑example, and it offers a clean linear‑algebraic proof of the original necessary condition, thereby deepening our understanding of the subtle relationship between combinatorial sparsity counts and geometric rigidity in three dimensions.