You are in a strange party where everyone knows exactly 22 other person and any two who don’t know each other have exactly 6 mutual friends. How many are there in the party?

Source 40- Strange Party

My research:

I know that the adjacency matrix $M$ is symmetrical and if $M[i,j] = 0$ then i-th and j-th rows have 6 times 1 at the same index position. I am thinking about encoding problem into SAT formulae and trying to brute force SAT solver with different number of literals (elements of upper triangle matrix of adjacency matrix).

I have read some articles about Moore graphs and it seems that the upper limit for the vertices is $22^2 + 1 = 485$.

Do you have any hints to the solution?

  • 1
    $\begingroup$ What is a party? I didn't think that people can be in groups of more than 6 people, let alone where they know 22. $\endgroup$
    – Asaf Karagila
    Nov 2 '20 at 21:07
  • $\begingroup$ There's a trivial answer of 23 people (everyone knows everyone, so the 6 mutual friends rule doesn't get used). I'm guessing that's not what you're looking for though. $\endgroup$
    – Dave
    Nov 2 '20 at 21:25

This is not a solution. This just demonstrate that $ n \leq 100$.

Take the typical graph theoretic representation.
Let friend be represented by a red edge, stranger be represented by a blue edge.
We have $d_r (v_i) = 22, d_b (v_i) = n - 23$, so total number of red edges is $11n$ and total number of blue edges is $ \frac{n(n-23)}{2}$.

We count the number of triples of red-red-blue triangles.
For each blue edge, the condition gives us exactly 6 red-red-blue triangles.
For each red-red emanating from a vertex, it could give us a red-red-red or red-red-blue triangle.

Hence, $ 6 \times \frac{ n ( n-23) } { 2 } \leq n \times { 22 \choose 2 } \Rightarrow n \leq 100$.


A partial answer: the possible numbers of vertices are $23, 40, 100$, and some unknown subset of $\{51,52, \dots, 99\}$.

For a vertex to have degree $22$, we need at least $23$ vertices. This is possible: take the clique $K_{23}$. The second condition is satisfied trivially.

If we want the second condition to be satisfied nontrivially, then there must be non-adjacent vertices $v$ and $w$. This requires $6$ more vertices in $N(v) \cap N(w)$, $16$ more in $N(v) \setminus N(w)$, and $16$ more in $N(w) \setminus N(v)$, for at least $40$ total.

$40$ vertices is possible: consider the graph with vertex set $\{v_1, \dots, v_{20}, w_1, \dots, w_{20}\}$, with a complete graph on the first $20$ and the last $20$ vertices. Additionally, add edges $v_i w_i$, $v_i w_{i+1}$, and $v_i w_{i-1}$ (with arithmetic modulo $20$). Two non-adjacent vertices $v_i$ and $w_j$ have $\{w_{i-1}, w_i, w_{i+1}, v_{j-1}, v_j, v_{j+1}\}$ as common neighbors.

If there are more than $40$ vertices, then in the lower bound above there must be at least one vertex $u$ not adjacent to $v$ or $w$. Let's say that $|N(u) \cap N(v) \cap N(w)| = k$. Then there are $6-k$ vertices in each pairwise intersection of $N(u), N(v), N(w)$ leaving $22 - k - 2(6-k) = k+10$ vertices adjacent to only one of $u,v,w$. That's a total of at least $51$ vertices.

I don't know if this is achievable, but there's an upper bound. On the high end, the Higman–Sims graph has $100$ vertices, and is an example because it is strongly regular. Every vertex has degree $22$, every two non-adjacent vertices have $6$ common neighbors, and for a bonus every two adjacent vertices have no common neighbors. As the other answer argues, this is the largest possible number of vertices.

  • $\begingroup$ The first was clever. I'm not seeing your second example though... $\endgroup$
    – Mike
    Nov 2 '20 at 18:45
  • $\begingroup$ I see it now. Thanks! $\endgroup$
    – Mike
    Nov 2 '20 at 18:49
  • $\begingroup$ I also found the 40 vertices example, but wasn't able to adapt it to > 40. $\endgroup$
    – Calvin Lin
    Nov 2 '20 at 19:25
  • $\begingroup$ @CalvinLin $41, \dots, 50$ are also impossible, and I have serious doubts about $51$. $\endgroup$ Nov 2 '20 at 19:38

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