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The Friendship Theorem Dr. John S. Caughman Portland State University

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Public Service Announcement (a+b) p =a p +b p …mod p …when a, b are integers …and p is prime. “Freshman’s Dream”

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Freshman’s Dream Generalizes! (a 1 +a 2 +…+a n ) p =a 1 p +a 2 p +…+a n p …mod p …when a, b are integers …and p is prime.

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Freshman’s Dream Generalizes (a 1 +a 2 +…+a n ) p = a 1 p +a 2 p +…+a n p tr(A) p = tr(A p ) (mod p) a4a4 000 *a3a3 00 **a2a2 0 ***a1a1 A =

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Freshman’s Dream Generalizes! tr(A) p = tr(A p ) (mod p) **** **** **** **** A = tr(A p ) = tr((L+U) p ) = tr(L p +U p ) = tr(L p )+tr(U p )=0+tr(U) p = tr(A) p Note: tr(UL)=tr(LU) so cross terms combine, and coefficients =0 mod p.

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The Theorem If every pair of people at a party has precisely one common friend, then there must be a person who is everybody's friend.

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Cheap Example Nancy JohnMark

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Cheap Example of a Graph Nancy JohnMark

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What a Graph IS: Nancy JohnMark

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Nancy John Vertices! Mark What a Graph IS:

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Nancy John Edges! Mark What a Graph IS:

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Nancy JohnMark What a Graph IS NOT:

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Nancy Loops! MarkJohn What a Graph IS NOT:

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Nancy Loops! MarkJohn What a Graph IS NOT:

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Nancy John Directed edges! Mark What a Graph IS NOT:

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Nancy John Directed edges! Mark What a Graph IS NOT:

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Nancy John Multi-edges! Mark What a Graph IS NOT:

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Nancy John Multi-edges! Mark What a Graph IS NOT:

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‘Simple’ Graphs… Nancy John Finite Undirected No Loops No Multiple Edges Mark

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Let G be a simple graph with n vertices. The Theorem, Restated If every pair of vertices in G has precisely one common neighbor, then G has a vertex with n-1 neighbors.

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Generally attributed to Erdős (1966). Easily proved using linear algebra. Combinatorial proofs more elusive. The Theorem, Restated

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NOT A TYPICAL “THRESHOLD” RESULT

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Pigeonhole Principle If more than n pigeons are placed into n or fewer holes, then at least one hole will contain more than one pigeon.

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Some threshold results If a graph with n vertices has > n 2 /4 edges, then there must be a set of 3 mutual neighbors. If it has > n(n-2)/2 edges, then there must be a vertex with n-1 neighbors.

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Extremal Graph Theory If this were an extremal problem, we would expect graphs with MORE edges than ours to also satisfy the same conclusion…

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1 23 4

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Of the 15 pairs, 3 have four neighbors in common and 12 have two in common. So ALL pairs have at least one in common. But NO vertex has five neighbors!

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Related Fact – losing edges

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Summary If every pair of vertices in a graph has at least one neighbor in common, it might not be possible to remove edges and produce a subgraph in which every pair has exactly one common neighbor.

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Accolades for Friendship The Friendship Theorem is listed among Abad's “100 Greatest Theorems” The proof is immortalized in Aigner and Ziegler's Proofs from THE BOOK.

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Example 1

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Example 2

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Example 3

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How to prove it: STEP ONE: If x and y are not neighbors, they have the same # of neighbors. Why: Let N x = set of neighbors of x Let N y = set of neighbors of y

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x y How to prove it:

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x y NxNx

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x y NyNy

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For each u in N x define: f(u) = common neighbor of u and y. x y How to prove it:

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Pick u 1 in N x. x y u1u1 How to prove it:

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f(u 1 ) = common neighbor of u 1 and y. x y u1u1 f(u 1 ) How to prove it:

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x y

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x y u2u2 Pick u 2 in N x. How to prove it:

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x y u2u2 f(u 2 ) f(u 2 ) = common neighbor of u 2 and y. How to prove it:

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x y

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x y u* f(u)= f(u*) u How to prove it:

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x y u* f(u)= f(u*) u How to prove it:

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So f is one-to-one from N x to N y. x y u* f(u)= f(u*) u How to prove it:

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So f is one-to-one from N x to N y. x y u* f(u)= f(u*) u So it can’t be true that |N x | > |N y |. How to prove it:

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So f is one-to-one from N x to N y. x y u* f(u)= f(u*) u So |N x | = |N y |. How to prove it: So it can’t be true that |N x | > |N y |.

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STEP 1: If x and y are not neighbors, they have the same # of neighbors. STEP 2: Either some x has n-1 neighbors or ALL vertices have same # of neighbors. Why: Assume no vertex has n-1 neighbors. Let A = {x : x has max # of neighbors, k}. B = {y : y has < k neighbors}. How to prove it:

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A = {x : x has max # of neighbors, k}. B = {y : y has < k neighbors}. By Step 1, all in A are neighbors to all in B! Set || Possible Size. A 0, 1, 2, …., n B 0, 1, 2, …., n

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STEP 1: If x and y are not neighbors, they have the same # of neighbors. STEP 2: Either some x has n-1 neighbors or ALL vertices have k neighbors. STEP 3: If all vertices have k neighbors, then n = k (k-1) + 1. How to prove it:

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Why: STEP 3: If all vertices have k neighbors, then n = k (k-1) + 1. Count paths of length 2…

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How to prove it: Why: STEP 3: If all vertices have k neighbors, then n = k (k-1) + 1. Count paths of length 2… ( ) n2n2

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How to prove it: Why: STEP 3: If all vertices have k neighbors, then n = k (k-1) + 1. Count paths of length 2… ( ) 1 n2n2

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How to prove it: Why: STEP 3: If all vertices have k neighbors, then n = k (k-1) + 1. Count paths of length 2… ( ) 1= n2n2

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How to prove it: Why: STEP 3: If all vertices have k neighbors, then n = k (k-1) + 1. Count paths of length 2… ( ) 1= n n2n2

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How to prove it: Why: STEP 3: If all vertices have k neighbors, then n = k (k-1) + 1. Count paths of length 2… ( ) 1= n (k) n2n2

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How to prove it: Why: STEP 3: If all vertices have k neighbors, then n = k (k-1) + 1. Count paths of length 2… ( ) 1= n (k) (k-1) n2n2

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How to prove it: Why: STEP 3: If all vertices have k neighbors, then n = k (k-1) + 1. Count paths of length 2… ( ) 1= n (k) (k-1) ________ 2 n2n2

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How to prove it: STEP 3: If all vertices have k neighbors, then n = k (k-1) + 1. ( ) 1= n (k) (k-1) ________ 2 n2n2 = n (k) (k-1) ________ 2 (n)(n-1) 2 n = k (k-1) + 1

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STEP 1: If x and y are not neighbors, they have the same # of neighbors. STEP 2: Either some x has n-1 neighbors or ALL vertices have k neighbors. STEP 3: If all vertices have k neighbors, then n = k (k-1) + 1. How to prove it:

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The Master Plan Each pair has 1 in common If x,y not neighbors, |N x |=|N y | Some x has n-1 neighbors |N x |= k for all x, and n =k(k-1)+1 Some Linear Algebra Either Or ?

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Adjacency Matrix Call vertices v 1, v 2, …, v n. Let A = n x n matrix where: A ij = 1, if v i, v j are neighbors, A ij = 0, if not. A is called the adjacency matrix of G. Notice that the trace of A is 0. {

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Adjacency Matrix v4v4 v1v1 v2v2 v3v A = A 2 = =

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Adjacency Matrix (A 2 ) ij = # common neighbors of v i, v j So…….. for our graphs….. A 2 = (k-1) I + J. (J = all 1’s matrix) (A 2 ) ij = 1 if i, j different, and (A 2 ) ij = k if i = j.

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Adjacency Matrix A 2 = (k-1) I + J (J = all 1’s matrix) A J = (k) J Now let p be a prime divisor of k-1. Then k = 1 and n = k(k-1)+1 = 1 (mod p) So A 2 = J, and A J = J. (mod p) Therefore, A i = J for all i > 1. (mod p)

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Adjacency Matrix A i = J for all i > 1 (mod p) But tr A p = (tr A) p (mod p) So, modulo p, we get: 1 = n = tr J = tr A p = (tr A) p = 0.

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Putting it all together Each pair has 1 in common If x,y not neighbors, |N x |=|N y | Some x has n-1 neighbors |N x |= k for all x and n =k(k-1)+1 0=1 Either Or

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Moral: To make progress in almost any field of math, find a way to sneak linear algebra into it !

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Related Result 1 Let m > 0 and k > 0 be integers. A graph with at least m points is an (m,k)-graph if any m-tuple of its points has exactly k common adjacent points. The friendship theorem characterizes all (2, 1)-graphs. For m > 2 and k > 0, there is only one (m,k)-graph, namely the complete graph on m+k points.

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Related Result 2 There are other generalizations of finite friendship graphs. Instead of graphs with edges encoding pairs of friends and adjacency matrix satisfying “A 2 -J is diagonal", one could consider incidence matrices of finite geometries: these are associated with bijections between blockset and pointset of these geometries.

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Friendship references Huneke, Craig The friendship theorem. Amer. Math. Monthly 109 (2002), no. 2, Brunat, Josep M. A proof of the friendship theorem by elementary methods. (Catalan) Butl. Soc. Catalana Mat. No. 7 (1992), Hammersley, J. M. The friendship theorem and the love problem. Surveys in combinatorics (Southampton, 1983), , London Math. Soc. Lecture Note Ser., 82, Cambridge Univ. Press, Cambridge, Sudolsk\'y, Marián A generalization of the friendship theorem. Math. Slovaca 28 (1978), no. 1, Skala, H. L. A variation of the friendship theorem. SIAM J. Appl. Math. 23 (1972), Longyear, Judith Q.; Parsons, T. D. The friendship theorem. Nederl. Akad. Wetensch. Proc. Ser. A 75=Indag. Math. 34 (1972), Wilf, Herbert S. The friendship theorem Combinatorial Mathematics and its Applications (Proc. Conf., Oxford, 1969) pp Academic Press, London

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THANK YOU !

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