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Let D denote a $1-(n, κ, m)$ design where two distinct blocks have at most one point in common (i.e. D is a partial linear space). Then the block graph $\Gamma(D)$ has the blocks of D as vertices and two vertices are adjacent whenever the blocks intersect.and if D is a transversal design $TD_1(κ, m)$

is $\Gamma(D)$ a strongly regular graph with parameter $(m^2,k*(m-1),k^2-3k+m,k(k-1))$?

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What is your background, masoud? What do you already know about this problem? It will help people who can answer judge what they should write. Putting in some relevant definitions helps. – David Roberts Apr 17 '12 at 4:15
Also, please provide context. This has the appearance of a homework problem. – Chris Godsil Apr 17 '12 at 13:13
no, I see it in the a article, but i can't understand why it's right – mj125 Apr 17 '12 at 15:28
Definition: Let $k ≥ 2$ and $n ≥ 1$. A transversal design $TD(k, n)$ is a triple $(X, G, B)$ such that the following properties are satisfied: 1. $X$ is a set of kn elements called points, 2. $G$ is a partition of X into k subsets of size n called groups, 3. $B$ is a set of k-subsets of X called blocks, 4. any group and any block contain exactly one common point, and 5. every pair of points from distinct groups is contained in exactly one block. – mj125 Apr 17 '12 at 15:35
Think of each block as a vector of length $k$, where the $i$-th entry is the element of the $i$-th group that lies in the block. Since there are exactly $n$ blocks that contain a given point, the number of neighbors of a vertex in the block graph is $k(m-1)$. Given two blocks $\alpha$ and $\beta$ with no points in common, for each two distinct groups there is exactly one block that agrees with $\alpha$ on the first group and with $\beta$ on the second. So these two blocks have exactly $k(k-1)$ common neighbors. You can arrive at the $k^2-3k+m$ similarly. – Chris Godsil Apr 17 '12 at 17:38

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