# regular graph construction

Is there a construction of (2b-2)-regular graph with 4b-3 or 4b-4 vertices, such that no two vertices share more than (b-1) vertices??

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If you can do it with $4b-5$ vertices, then I think you would have solved the Hadamard conjecture. –  Brendan McKay Aug 26 '12 at 16:42

Partial answer: If 4b-3 is a prime power, then the Paley graph of 4b-3 vertices will have this property, see http://en.wikipedia.org/wiki/Paley_graph.

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