the friendship theorem dr. john s. caughman portland state university
DESCRIPTION
The Friendship Theorem Dr. John S. Caughman Portland State University. Public Service Announcement. “Freshman’s Dream”. ( a+b ) p = a p +b p …mod p …when a, b are integers …and p is prime. Freshman’s Dream Generalizes!. (a 1 +a 2 +…+a n ) p =a 1 p +a 2 p +…+ a n p …mod p - PowerPoint PPT PresentationTRANSCRIPT
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The Friendship Theorem
Dr. John S. CaughmanPortland State University
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Public Service Announcement
(a+b)p=ap+bp …mod p
…when a, b are integers…and p is prime.
“Freshman’s Dream”
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Freshman’s Dream Generalizes!
(a1+a2+…+an)p=a1p +a2
p +…+anp
…mod p…when a, b are integers
…and p is prime.
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Freshman’s Dream Generalizes
(a1+a2+…+an)p = a1p +a2
p +…+anp
tr(A) p = tr(Ap) (mod p)
a4000*a300**a20***a1
A =
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Freshman’s Dream Generalizes!
tr(A) p = tr(Ap) (mod p)
****************
A =
tr(A p) = tr((L+U)p) = tr(Lp +Up) = tr(Lp)+tr(Up)=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 ExampleNancy
John Mark
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Cheap Example of a GraphNancy
John Mark
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What a Graph IS:Nancy
John Mark
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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
John Mark
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 PrincipleIf 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 resultsIf a graph with n vertices has > n2/4 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 TheoryIf 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
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2
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1
2 3
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2 3
4
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4
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4
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2
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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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Related Fact – losing edges
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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 Nx = set of neighbors of x
Let Ny = set of neighbors of y
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x y
How to prove it:
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x y
Nx
How to prove it:
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x y
Ny
How to prove it:
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For each u in Nx define:f(u) = common neighbor of u and y.
x y
How to prove it:
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Pick u1 in Nx.
x y
u1How to prove it:
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f(u1) = common neighbor of u1 and y.
x y
u1
f(u1)
How to prove it:
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x y
How to prove it:
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x yu2
Pick u2 in Nx.
How to prove it:
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x yu2
f(u2)
f(u2) = common neighbor of u2 and y.
How to prove it:
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x y
How to prove it:
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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 Nx to Ny.
x y u*f(u)= f(u*) u
How to prove it:
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So f is one-to-one from Nx to Ny.
x y u*f(u)= f(u*) u
So it can’t be true that |Nx| > |Ny|.
How to prove it:
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So f is one-to-one from Nx to Ny.
x y u*f(u)= f(u*) u
So |Nx| = |Ny|.
How to prove it:
So it can’t be true that |Nx| > |Ny|.
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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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How to prove it:
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…
( )n2
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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…
( ) 1n2
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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 =n2
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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
n2
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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)n2
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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) n2
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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
n2
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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
n2
= 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 PlanEach pair has 1 in common
If x,y not
neighbors,
|Nx|=|Ny|
Some x has n-1
neighbors
|Nx|= k for
all x, and
n =k(k-1)+1
Some Linear
Algebra
Either
Or
?
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Adjacency MatrixCall vertices v1, v2, …, vn. Let A = n x n matrix where: Aij = 1, if vi, vj are neighbors, Aij = 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
v4
v1
v2
v3 0101101101011110
A =
A2 =0101101101011110
0101101101011110
=2121131221211213
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Adjacency Matrix(A2)
ij = # common neighbors of vi, vj
So…….. for our graphs…..
A2 = (k-1) I + J. (J = all 1’s matrix)
(A2) ij = 1 if i, j different, and
(A2) ij = k if i = j.
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Adjacency MatrixA2 = (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 A2 = J, and A J = J. (mod p)Therefore, Ai = J for all i > 1. (mod p)
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Adjacency MatrixAi = J for all i > 1 (mod p)
But tr Ap = (tr A)p (mod p)
So, modulo p, we get: 1 = n = tr J = tr Ap = (tr A)p = 0.
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Putting it all togetherEach pair has 1 in common
If x,y not
neighbors,
|Nx|=|Ny|
Some x has n-1
neighbors
|Nx|= k for
all x and
n =k(k-1)+1
0=1
Either
Or
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Moral:
To make progress in almostany field of math, find a way to sneak linear algebra into it
!
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THANK YOU !