Meshulam's game explained

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.^{[1] [2]}

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^[3] proved that, for every graph G:

η_H(I(G))\geq\Psi(G)

where is the homological connectivity of plus 2.

Examples

1. If G is the empty graph, then Ψ(G) = 0, since no explosions are needed.
2. 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.
3. 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.
4. If G has an independence domination number of at least k,

, then . 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 .

1. • Note: this also implies that

, where is the line graph of G, and 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).

1. • Similarly, when H is an r-partite hypergraph,

.^[4]

1. If G is the complete bipartite graph K_n,n, and L(G) is its line graph, then

.^{[5] [6]} 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 explosions are required to remove all vertices.

1. • Note: this result was generalized later: if F is any subgraph of K_n,n, then

.

Proof for the case 1

To illustrate the connection between Meshulam's game and connectivity, we prove it in the special case in which

, which is the smallest possible value of . We prove that, in this case, , i.e., NON can always destroy the entire graph using at most one explosion. means that is not connected. This means that there are two subsets of vertices, X and Y, where no edge in connects any vertex of X to any vertex of Y. But is the independence complex of G; so in G, every vertex of X is connected to every vertex of Y. Regardless of how CON plays, he must at some step select an edge between a vertex of X and a vertex of Y. NON can explode this edge and destroy the entire graph.

In general, the proof works only one way, that is, there may be graphs for which

.

See also

• Hall-type theorems for hypergraphs

Notes and References

1. Aharoni. Ron. Berger. Eli. Ziv. Ran. 2007-05-01. Independent systems of representatives in weighted graphs. Combinatorica. 27. 3. 253–267. 10.1007/s00493-007-2086-y. 43510417. 0209-9683.
2. Aharoni. Ron. Berger. Eli. Kotlar. Dani. Ziv. Ran. 2017-01-04. On a conjecture of Stein. Abhandlungen aus dem Mathematischen Seminar der Universität Hamburg. 87. 2. 203–211. 10.1007/s12188-016-0160-3. 119139740. 0025-5858.
3. Meshulam. Roy. 2003-05-01. Domination numbers and homology. Journal of Combinatorial Theory, Series A. en. 102. 2. 321–330. 10.1016/S0097-3165(03)00045-1. 0097-3165. free.
4. Haxell. Penny. Narins. Lothar. Szabó. Tibor. 2018-08-01. Extremal hypergraphs for Ryser's Conjecture. Journal of Combinatorial Theory, Series A. en. 158. 492–547. 10.1016/j.jcta.2018.04.004. 0097-3165.
5. Björner. A.. Lovász. L.. Vrećica. S. T.. Živaljević. R. T.. 1994. Chessboard Complexes and Matching Complexes. Journal of the London Mathematical Society. en. 49. 1. 25–39. 10.1112/jlms/49.1.25. 1469-7750.
6. Shareshian. John. Wachs. Michelle L.. 2009-10-01. Top homology of hypergraph matching complexes, p-cycle complexes and Quillen complexes of symmetric groups. Journal of Algebra. en. 322. 7. 2253–2271. 0808.3114. 10.1016/j.jalgebra.2008.11.042. 5259429. 0021-8693.