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Iranian Mathematical Olympiad

Combinatorics Difficulty 6.4 National Olympiad Prove it Iran

a) Prove that every graph with 2n2n vertices and n+2n+2 edges has an independent set of size nn.

b) Find the number of graphs with 2n2n vertices and n+3n+3 edges having no independent set of size nn.

Solution

a) Let AA be a maximal independent subset of vertices and BB be its complement. Assume to the contrary that AA has at most n−1n-1 elements and so BB consists of at least n+1n+1 vertices. Since AA is maximal, there should be an edge from every vertex in BB to AA, so there are at least n+1n+1 edges between AA and BB. This means that there are at most 1 edge in BB let's say between uu and vv. Now B−{u}B - \{u\} is an independent subset of vertices with at least nn elements, which is a contradiction.

b) We prove that the only answer is the union of two disjoint triangles and n−3n-3 edges between n−3n-3 disjoint pair of vertices. Let AA be a maximal independent subset of vertices, by assumption it has at most n−1n-1 vertices and so its complement BB has at least n+1n+1 vertices. Similar to the previous part, we can conclude that there are at least n+1n+1 edges connecting vertices of BB to AA and there are at least 2 edges connecting vertices in BB. As the total number of edges is assumed to be n+3n+3, we deduce that AA should have exactly n−1n-1 vertices and there are exactly 2 edges in BB. Also, every element of BB is connected to exactly one element of AA.

If two elements of BB, say uu and vv, are connected to some vertex w∈Aw \in A, then uvuv must be an edge, because otherwise, (A−{w})∪{u,v}(A - \{w\}) \cup \{u, v\} form an independent set of size nn which is a contradiction. This observation together with the fact that there are two edges in BB implies that vertices in AA can not be connected to more than two vertices in BB.

Since ∣A∣=n−1|A| = n-1 and ∣B∣=n+1|B| = n+1, there should be two vertices x1,x2∈Ax_1, x_2 \in A and two disjoint pairs {y1,z1},{y2,z2}\{y_1, z_1\}, \{y_2, z_2\} in BB such that xix_i is adjacent to yi,ziy_i, z_i (i=1,2i=1,2) and so yi,ziy_i, z_i are connected. Also, there is a bijection between A−{x1,x2},B−{y1,z1,y2,z2}A - \{x_1, x_2\}, B - \{y_1, z_1, y_2, z_2\} such that the corresponding vertices are connected. This gives n+3n+3 edges in the graph and since there are exactly n+3n+3 edges, we conclude that the graph is the union of two triangles x1y1z1,x2y2z2x_1y_1z_1, x_2y_2z_2 and n−3n-3 disjoint edges as desired.

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