# Reduction from Clique to something else

Given $G(V,E)$ and $k$. Is there a clique with size $k$?

Given set $X = \{x_1,x_2,\dots,x_n \}$, and collection $A = \{A_1,A_2,...,A_n\}$ of sub-sets of $X$ and $k$. Are there are $k$ different elements in $A$, that every two of them has intersection that isn't empty? For example $A_1 = \{ x_1,x_2,x_3 \}$, $A_2 = \{x_4,x_5,x_1 \}$ so their intersection is $x_1$ and its valid, but it should be applied for every two $A_i,A_j$ in $A$.

Prove: That unnamed problem (which surely has a name) is NP-Hard.

The reduction is..

$X=E$, and every $A_i$ in $A$ is composed of edges that are connected to vertex $v_i$.

I can't understand the reduction, can someone explain why it works please?

$X = E$, so it remains to define the collection $A$. For each vertex $v_i$ in $V$ create $A_i$ as following: for each edge $x_t$ incident to the vertex $v_i$, add $x_t$ to $A_i$. Thus the edge $x_t=(v_i, v_j) \in E \Leftrightarrow A_i \cap A_j = \{x_t\}$, and every pair $A_i, A_j$ has at most one common element. So, if there are $k$ different elements in $A$ such that every two $A_i, A_j$ of them has non-empty intersection which is exactly the edge between vertices $v_i$ and $v_j$, then these $k$ vertices are pairwise connected meaning that they form a clique.