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ecoist
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 Barely compatible gatherings   « on: Jun 6th, 2008, 9:06pm » Quote Modify

A barely compatible gathering of people, BCP, is a set of people with the following properties.

1) Every person in the set is friends with the same number, k, of people in the set.
2) If person a is not friends with person b, then a and b have exactly one friend in common in the set.  If a and b are friends, then a and b have no friend in common in the set.

Show that there are only a finite number of integers n such that there exists a barely compatible gathering of n people.

Examples.
n=2.  Two friends.

n=5.  People equal the vertices of a pentagon.  Two vertices are friends if they are adjacent.

n=10.  People equal the ten 2-subsets of {1,...,5}.  Two 2-subsets are friends if they are disjoint.  (The Petersen graph)
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Obob
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 Re: Barely compatible gatherings   « Reply #1 on: Jun 7th, 2008, 3:16pm » Quote Modify

In graph-theoretic terms, there are only finitely many regular graphs of girth => 5 and diameter <= 2.
 « Last Edit: Jun 7th, 2008, 3:26pm by Obob » IP Logged
Aryabhatta
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 Re: Barely compatible gatherings   « Reply #2 on: Jun 8th, 2008, 11:21am » Quote Modify

Here is an attempt

Let A be the adjacency matrix of the graph.

Then we can easily see that

A^2  + A = (k-1)I + J, where J is the all ones matrix.

now if s is an eigenvalue of A with eigenvector x, then we have that

[s^2 + s - (k-1)] x = Jx.

Thus we have that either s = k (in which case x is [1 1 ... 1]) or 2s = -1 +- sqrt(4k-3)  (roots of s^2 + s - (k-1) = 0)

We can also show that the multiplicity of the eigenvalue k is 1. (true for any regular graph of degree k)

So since trace of A is zero, for some integers P and Q we must have that

P (-1 + sqrt(4k-3)) + Q(-1 - sqrt(4k-3)) + k = 0.

i.e
-k + P + Q = (P-Q) sqrt(4k-3)

We also have the P+Q = n-1 = k^2 (the eigenvalue corresponding to [1 1 ... 1] of A^2 is n-1, and must be square of corresponding for A, which is k)

So P-Q = (-k + k^2)/sqrt(4k-3)

Thus we must have that 4k-3 = t^2 for some integer t.

and that k(k-1)/t is an integer.

Thus we have that

(t^2+3)(t^2-1)/16t is an integer.

Thus we have that (t^4 + 2t^2 -3)/t is also an integer. (Mutiply by 16 and expand the numberator)

Thus t divides 3.

Hence there are only a finite number of such k.

 « Last Edit: Jun 8th, 2008, 11:35am by Aryabhatta » IP Logged
Eigenray
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 Re: Barely compatible gatherings   « Reply #3 on: Jun 9th, 2008, 1:27am » Quote Modify

Very close...but you're missing a factor of 2 in the trace equation.

Here is a similar problem:

There are n>3 people such that any two of them have exactly one common friend.  Show that there is one person who is friends with everybody else.
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ecoist
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 Re: Barely compatible gatherings   « Reply #4 on: Jun 9th, 2008, 2:50pm » Quote Modify

If I'm not mistaken, eigenray, your problem is known as the Friendship Theorem.  I learned of this problem in the form: if there are n people with the property that any two of them have exactly one friend in common, then Dale Carnegie is there.  Eons ago Dale Carnegie wrote a book entitled "How to Win Friends and Inflluence People".  About a year ago I saw online a purely combinatorial proof of the Friendship Theorem.
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