In graph theory, Meshulam's game is a game used to explain a theorem of Roy Meshulam related to the homological connectivity of the independence complex of a graph, which is the smallest index k such that all reduced homological groups up to and including k are trivial. The formulation of this theorem as a game is due to Aharoni, Berger and Ziv.
Description The game-board is a graph G. It is a zero-sum game for two players, CON and NON. CON wants to show that I(G), the independence complex of G, has a high connectivity; NON wants to prove the opposite. At his turn, CON chooses an edge e from the remaining graph. NON then chooses one of two options:
Disconnection – remove the edge e from the graph. Explosion – remove both endpoints of e, together with all their neighbors and the edges incident to them. The score of CON is defined as follows:
If at some point the remaining graph has an isolated vertex, the score is infinity; Otherwise, at some point the remaining graph contains no vertex; in that case the score is the number of explosions. For every given graph G, the game value on G (i.e., the score of CON when both sides play optimally) is denoted by Ψ(G).
Game value and homological connectivity Meshulam proved that, for every graph G: η H ( I ( G ) ) ≥ Ψ ( G ) {\displaystyle \eta _{H}(I(G))\geq \Psi (G)} where η H ( I ( G ) ) {\displaystyle \eta _{H}(I(G))} is the homological connectivity of I ( G ) {\displaystyle I(G)} plus 2.
Examples If G is the empty graph, then Ψ(G) = 0, since no explosions are needed. If G has k connected components, then Ψ(G) ≥ k. Regardless of the order in which CON offers edges, each explosion made by NON destroys vertices in a single component, so NON needs at least k explosions to destroy all vertices. If G is a union of k vertex-disjoint cliques, each of which contains at least two vertices, then Ψ(G) = k, since each explosion completely destroys a single clique. If G has an independence domination number of at least k, i γ ( G ) ≥ k {\displaystyle i\gamma (G)\geq k} , then Ψ ( G ) ≥ k {\displaystyle \Psi (G)\geq k} . Proof: Let A be an independent set with domination number at least k. CON starts by offering all edges (a,b) where a is in A. If NON disconnects all such edges, then the vertices of A remain isolated so CON's score is infinity. If NON explodes such an edge, then the explosion removes from A only the vertices that are adjacent by b (the explosion at a does not destroy vertices of A, since A is an independent set). Therefore, the remaining vertices of A require at least k-1 vertices to dominate, so the domination number of A decreased by at most 1. Therefore, NON needs at least k explosions to destroy all vertices of A. This proves that Ψ ( G ) ≥ i γ ( G ) {\displaystyle \Psi (G)\geq i\gamma (G)} . Note: this also implies that Ψ ( L ( G ) ) ≥ ν ( G ) / 2 {\displaystyle \Psi (L(G))\geq \nu (G)/2} , where L ( G ) {\displaystyle L(G)} is the line graph of G, and ν ( G ) {\displaystyle \nu (G)} is the size of the largest matching in G. This is because the matchings in G are the independent sets in L(G). Each edge in G is a vertex in L(G), and it dominates at most two edges in the matching (= vertices in the independent set). Similarly, when H is an r-partite hypergraph, Ψ ( L ( H ) ) ≥ ν ( H ) / r {\displaystyle \Psi (L(H))\geq \nu (H)/r} . If G is the complete bipartite graph Kn,n, and L(G) is its line graph, then Ψ ( L ( G ) ) ≥ ⌊ 2 n / 3 ⌋ {\displaystyle \Psi (L(G))\geq \lfloor 2n/3\rfloor } . Proof: L(G) can be seen as an n-by-n array of cells, where each row is a vertex on one side, each column is a vertex on the other side, and each cell is an edge. In the graph L(G), each cell is a vertex, and each edge is a pair of two cells in the same column or the same row. CON starts by offering two cells in the same row; if NON explodes them, then CON offers two cells in the same column; if NON explodes them again, then the two explosions together destroy 3 rows and 3 columns. Therefore, at least ⌊ 2 n / 3 ⌋ {\displaystyle \lfloor 2n/3\rfloor } explosions are required to remove all vertices. Note: this result was generalized later: if F is any subgraph of Kn,n, then Ψ ( L ( G ) ) ≥ | F | n − n 3 − 1 2 {\displaystyle \Psi (L(G))\geq {\frac {|F|}{n}}-{\frac {n}{3}}-{\frac {1}{2}}} .
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