Talk:Strongly regular graph

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Potentially inaccurate statement in examples

I am not an expert on graph theory, so I hesitate to update the article itself. However, I believe the categorical statement:

The strongly regular graphs with λ=0 are triangle free. The seven listed above are the only known ones.

is incorrect. If I understand the subject correctly, every complete bipartite graph on an even number of vertexes is also strongly regular, with signature srg(2 k, k, 0, k), and bipartite graphs with signature srg(2 k, k, 0, p), with p < k, are easily constructed. — Preceding unsigned comment added by 98.26.29.11 (talk) 12:45, 22 May 2012 (UTC)[reply]

I need to retract the construction remark - the only strongly regular bipartite graphs are complete.

Proof of $(v-k-1)\mu =k(k-\lambda -1)$

I was unable to find a reference to cite for this formula. However, the proof is pretty simple:

Consider a strongly regular graph G with parameters (v,k,λ,μ) and vertex set V. Let n(x) be the set of vertexes in V that are adjacent to the vertex x in G, and let p(x) be the set of vertexes in V that are neither the vertex x nor in n(x). Then, consider a particular vertex q. Every vertex t in n(q) forms an edge with λ vertexes in n(q). Further, t forms an edge with q. This leaves k-λ-1 edges that have one vertex in n(q) and one vertex in p(q). Taken over all vertexes adjacent to q, this gives k (k-λ-1).

Next, consider the set of vertexes p(q). For every vertex t in p(q), q forms an edge with μ vertexes in n(t). By the partitioning of V, there are v-k-1 vertexes in p(q). Thus, the vertexes in p(q) form (v-k-1) μ edges with the vertexes in n(q). But this is the same set of edges that were counted in the preceding paragraph. Hence, (v-k-1) μ = k (k-λ-1). — Preceding unsigned comment added by 66.194.221.34 (talk) 16:48, 4 June 2012 (UTC)[reply]

Potentially inaccurate statement in examples

Proof of ( v − k − 1 ) μ = k ( k − λ − 1 ) {\displaystyle (v-k-1)\mu =k(k-\lambda -1)}

Proof of $(v-k-1)\mu =k(k-\lambda -1)$