Graphs where every two vertices have odd number of mutual neighbours There was a rather cute question last week about graphs where every pair of distinct vertices has an odd number of mutual neighbours.
The question was to show that such a graph must have an odd number of vertices, and it can be accomplished with a nice algebraic graph theory argument.
But let's up the ante a bit: can we actually characterize the graphs with this property? 
Here are some examples in the family:


*

*complete graphs of odd order

*anything obtained by gluing together a bunch of odd complete graphs at a single vertex

*a graph of the form A - B - C where A and C have the "even" version of this property (every pair of vertices have even number common neighbours) B is an odd complete graph, and A is completely joined to B, B completely joined to C.


Is this the lot?
 A: Take a Steiner triple system on $v$ points.  Let $X$ be the graph with the $v(v-1)/6$ triples
as its vertices, two triples adjacent if the have exactly one point in common.  We need
$v\equiv1,3$ modulo 6.  Then two adjacent triples have exactly $(v+3)/2$ common neighbours, and two disjoint triples have exactly 9 common neighbours.  If we take $v\equiv3,7$ modulo
12 we get examples.
Of course I am just constructing strongly regular graphs with $\lambda$ and $\mu$ odd.
The are strongly regular graphs with this property besides the ones listed, for example generalized quadrangles with $s$ and $t$ even.  Further examples appear in Andries Brouwer's on-line tables (http://www.win.tue.nl/~aeb/graphs/srg/srgtab.html), or Gordon Royle's
(http://units.maths.uwa.edu.au/~gordon/remote/srgs/).  
This suggest that a classification might be difficult.
A: Maybe I'm missing something, but I'm not sure that the third condition actually generates what I'll call odd graphs.  For example, if I let $A$ be the graph consisting of a single vertex and $B$ be the graph consisting of two isolated vertices, then clearly both $A$ and $B$ are even.  However, if I form the $A-B-C$ construction with this choice of $A$ and $B$ I get a graph with an even number of vertices, which can't be odd by the proof of the cute question.  
We can fix this by further insisting that each vertex of $A$ and $C$ have odd degree.  I'll call such graphs oddly even.  Note that a disjoint union of two oddly even graphs is still oddly even, so it really isn't necessary to have a $A-B-C$ construction, we only need a $A-B$ construction.  Furthermore, it is not necessary that $B$ is an odd clique; $B$ can in fact be any odd graph.
We thus have the following theorem.
Theorem.  Let $A$ be an oddly even graph and $B$ be an odd graph.  Then the graph $A-B$ formed by taking $A$ and $B$ and adding all edges between $A$ and $B$ is odd.  
So, I guess the answer to the question is no.
