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I do not know how to approach an apparent variant of the friends and strangers problem:

"There are 251 students in a class, where every student lists exactly 168 other students who they can work with. For any two students in the class, if student A puts student B on his or her list, then student B will also have student A on his or her list. Show that there must be some group of 4 students who are all willing to work with one another."

Is it correct to take the 251 students as the pigeons, and 168 as the holes? So ⌈251/168⌉ = 2, which is the minimum number of students who choose someone who did not choose them?

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    Your final question, "is [2] the minimum number of students who choose someone who did not choose them?", is at odds with the quoted exercise: "For any two students in the class, if student A puts student B on his or her list, then student B will also have student A on his or her list." So I think you mistook the point of the exercise. – 2017-02-15

1 Answers 1

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The exercise describes a simple undirected graph $G$ on $251$ vertices which is regular of degree $168$. That is every vertex (student) has $168$ edges (connections to other students with whom they can work). Does $G$ contain a $4$-clique, $K_4$?

This situation is immediately solved by Turán's Thm., which gives an upper bound on the number of edges that an $n$-vertex graph can have and remain $K_{r+1}$-free.

That upper bound is:

$$ \left\lfloor \left( 1 - \frac{1}{r} \right) \cdot \frac{n^2}{2} \right\rfloor $$

For the immediate case we take $n=251$ and $r=3$ (since we are concerned with avoiding $(r+1)$-cliques), and we get at most $21000$ edges.

But counting the edges in $G$, we find it has $251\cdot 168/2 = 21084$ edges. So $G$ is not free of $4$-cliques.

With a little more effort we can show the data of this exercise imply a stronger conclusion:

Each vertex of $G$ belongs to at least one $4$-clique.

To see this, fix any vertex $u\in G$ and consider the deleted neighborhood $H$ of $u$, the $168$ other vertices of $G$ adjacent to $u$. Since $G\setminus \{u\}$ has $250$ vertices, there are $250-168=82$ vertices in $G$ but not in $H\cup \{u\}$.

Since each $v\in H$ has $167$ neighbors in $G\setminus \{u\}$, such $v$ must have at least $167-82 = 85$ neighbors in $H$. Now we apply the special case of Turán's Thm. when $r=2$, namely Mantel's Thm., which says:

The maximum number of edges in an $n$-vertex triangle-free graph is $\lfloor n^2/4 \rfloor$.

Putting $n=168$ here gives an upper bound of $168^2/4 = 7056$ edges if $H$ is triangle-free. But since each $v\in H$ has (at least) $85$ neighbors there, $H$ must have at least $85\cdot 168/2 = 7140$ edges.

It follows that $H$ contains a triangle, and together with the first vertex $u$ it forms a $4$-clique in $G$.

Readers with even a mild interest in the proof of Turán's Thm. (1941) are encouraged to have a look at Martin Aigner's exposition, Turán's Graph Theorem (1995). A half dozen proofs are surveyed there, starting with a half-page treatment of the special case used above, Mantel's Thm. (1906).