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

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Presentation on theme: "The Friendship Theorem Dr. John S. Caughman Portland State University."— Presentation transcript:

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

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

3 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.

4 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 =

5 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.

6 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.

7 Cheap Example Nancy JohnMark

8 Cheap Example of a Graph Nancy JohnMark

9 What a Graph IS: Nancy JohnMark

10 Nancy John Vertices! Mark What a Graph IS:

11 Nancy John Edges! Mark What a Graph IS:

12 Nancy JohnMark What a Graph IS NOT:

13 Nancy Loops! MarkJohn What a Graph IS NOT:

14 Nancy Loops! MarkJohn What a Graph IS NOT:

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

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

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

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

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

20 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.

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

22 NOT A TYPICAL “THRESHOLD” RESULT

23 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.

24 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.

25 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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31 1 2

32 1 23

33 1 23 4

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37 2

38 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!

39 Related Fact – losing edges

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42 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.

43 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.

44 Example 1

45 Example 2

46 Example 3

47 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

48 x y How to prove it:

49 x y NxNx

50 x y NyNy

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

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

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

54 x y

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

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

57 x y

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

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

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

61 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:

62 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 |.

63 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:

64 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

65 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:

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

67 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

68 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

69 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

70 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

71 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

72 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

73 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

74 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

75 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:

76 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 ?

77 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. {

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

79 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.

80 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)

81 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.

82 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

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

84 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.

85 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.

86 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

87 THANK YOU !


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