TITLE: Construction of a graph
QUESTION [5 upvotes]: To construct a specific kind of undirected graph $G=(V,E)$, which $|V|=n>2$. For convenience, label the vertices with $v_1,v_2,\dots ,v_n\in V$, and $(v_i,v_j)\in E$ means there is a edge between vertices $v_i,v_j$.
And the graph has the following property:


$(v_1,v_2)\in E$
For $i\geq 3$, $(v_i,v_1)\in E\Rightarrow (v_i,v_2)\notin E$,and $(v_i,v_2)\in E\Rightarrow (v_i,v_1)\notin E$
For $i\neq j,(v_i,v_j) \notin E \Rightarrow \exists v_{k_1},v_{k_2}(k_1\neq k_2)$ which $(v_{k_1},v_i),(v_{k_2},v_i),(v_{k_1},v_j),(v_{k_2},v_j)\in E$ and $\forall l\neq k_1,k_2, (v_l,v_i) \notin E$ or $(v_l,v_j)\notin E$


If $n$ is even, for instance, $n=12$, we can construct the graph like this:

But I couldn't construct one when $n$ is odd, for instance, when $n=5$, etc.

If $n\geq 3$ is an odd number, is it possible to construct a graph meet the above-mentioned property?

REPLY [3 votes]: Notice that for every $n>3$ there should be a $C_4$ with no diagonal. For $n=5$, it is easy to see then that no such graphs exist. 
For $n=7$, the Mycielski construction over $C_3$ fits. In other words, mark the vertices of an equilateral triangle with side length 2 and the midpoints of its sides; conect each two vertices at distance 1. Now add one more vertex which is connected to the three vertices of the initial triangle.