Question

We are given a n-partite graph G. Each partition has n vertices, some of which may be isolated. Let us number the vertices in some $i^{th}$ partition as $V_{i1},V_{i2},...,V_{in}$. Now each non- isolated vertex $V_{ij}$ has at least one neighbor in each of the remaining $n-1$ partitions s.t. for a given numbering of the vertices, the n vertices (vertex $V_{ij}$ and its $n-1$ neighbors) form a permutation on the second index. For e.g., consider a 4-partite graph. Each partition has 4 vertices. Using the numbering as given above, vertex $V_{12}$ has as neighbors vertices $V_{21},V_{34},V_{43}$. Vertex $V_{12}$ can have other neighbors as well. We need to show that the graph G will always contain a n-clique. A stronger claim would be to say that every vertex is part of some n-clique.

Answer 1

False as stated: take large $n$ and for each vertex $V_{ij}$ choose some random permutation and draw the corresponding edges. Now, we have $n^n$ possible cliques to form. Let's look at the probability that $V_{11},\dots, V_{nn}$ is a clique. The chance that $V_{11}$ acquires $k$ fixed neighbors in this subgraph when its edges are drawn is $\frac{(n-1-k)!}{(n-1)!}$. Multiplying by ${n-1\choose k}$, we see that the chance to get $k$ vertices is less than $\frac 1{k!}$. Thus, if $X_1$ is the number of acquired neighbors for $V_{11}$, then $Ee^{X_1}\le e^e$. Thus $Ee^{\sum X_j}\le e^{en}$ by independence. But the clique means that $\sum X_j\ge n(n-1)/2$, so the probability that we have a given clique is at most $e^{en}e^{-\frac{n(n-1)}2}$, which is smaller than $n^{-n}$ by an order of magnitude.

I strongly suspect that you meant something else. Actually, my first impulse when seeing your post was to say "Consider the graph consisting of isolated vertices only" but then I decided that it would be too cheap.