The Friendship Theorem - Portland State University

Report
The Friendship
Theorem
Dr. John S. Caughman
Portland State University
Public Service Announcement
“Freshman’s Dream”
(a+b)p=ap+bp
…mod p
…when a, b are integers
…and p is prime.
Freshman’s Dream Generalizes!
(a1+a2+…+an)p=a1p +a2p +…+anp
…mod p
…when a, b are integers
…and p is prime.
Freshman’s Dream Generalizes
A=
a1
0
0
0
*
a2
0
0
*
*
a3
0
*
*
*
a4
(a1+a2+…+an)p = a1p +a2p +…+anp
tr(A) p = tr(Ap) (mod p)
Freshman’s Dream Generalizes!
A=
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
tr(A) p = tr(Ap) (mod p)
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.
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.
Cheap Example
Nancy
John
Mark
Cheap Example of a Graph
Nancy
John
Mark
What a Graph IS:
Nancy
John
Mark
What a Graph IS:
Nancy
John
Vertices!
Mark
What a Graph IS:
Nancy
John
Edges!
Mark
What a Graph IS NOT:
Nancy
John
Mark
What a Graph IS NOT:
Nancy
John
Loops!
Mark
What a Graph IS NOT:
Nancy
John
Loops!
Mark
What a Graph IS NOT:
Nancy
John
Directed edges!
Mark
What a Graph IS NOT:
Nancy
John
Directed edges!
Mark
What a Graph IS NOT:
Nancy
John
Multi-edges!
Mark
What a Graph IS NOT:
Nancy
John
Multi-edges!
Mark
‘Simple’ Graphs…
Nancy
John
• Finite
• Undirected
• No Loops
• No Multiple Edges
Mark
The Theorem, Restated
Let G be a simple graph with n vertices.
If every pair of vertices in G has precisely one
common neighbor, then G has a vertex with n-1
neighbors.
The Theorem, Restated
Generally attributed to Erdős (1966).
Easily proved using linear algebra.
Combinatorial proofs more elusive.
NOT A TYPICAL
“THRESHOLD” RESULT
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.
Some threshold results
If 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.
Extremal Graph Theory
If this were an extremal problem, we
would expect graphs with MORE edges
than ours to also satisfy the same
conclusion…
1
1
2
1
2
3
1
4
2
3
4
4
4
2
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!
Related Fact – losing edges
Related Fact – losing edges
Related Fact – losing edges
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.
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.
Example 1
Example 2
Example 3
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
How to prove it:
x
y
How to prove it:
y
x
Nx
How to prove it:
y
x
Ny
How to prove it:
x
y
For each u in Nx define:
f(u) = common neighbor of u and y.
How to proveu it:
1
x
Pick u1 in Nx.
y
How to proveu it:
1
f(u1)
x
y
f(u1) = common neighbor of u1 and y.
How to prove it:
x
y
How to prove it:
x
u2
Pick u2 in Nx.
y
How to prove it:
f(u2)
x
u2
y
f(u2) = common neighbor of u2 and y.
How to prove it:
x
y
How to prove it:
u
x
u*
f(u)= f(u*)
y
How to prove it:
u
x
u*
f(u)= f(u*)
y
How to prove it:
u
x
u*
f(u)= f(u*)
y
So f is one-to-one from Nx to Ny.
How to prove it:
u
x
u*
f(u)= f(u*)
y
So f is one-to-one from Nx to Ny.
So it can’t be true that |Nx| > |Ny|.
How to prove it:
u
x
u*
f(u)= f(u*)
y
So f is one-to-one from Nx to Ny.
So it can’t be true that |Nx| > |Ny|.
So
|Nx| = |Ny|.
How to prove it:
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}.
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
.
How to prove it:
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:
STEP 3: If all vertices have k neighbors, then
n = k (k-1) + 1.
Why:
Count paths of length 2…
How to prove it:
STEP 3: If all vertices have k neighbors, then
n = k (k-1) + 1.
Why:
Count paths of length 2…
(2 )
n
How to prove it:
STEP 3: If all vertices have k neighbors, then
n = k (k-1) + 1.
Why:
Count paths of length 2…
(2 )
n
1
How to prove it:
STEP 3: If all vertices have k neighbors, then
n = k (k-1) + 1.
Why:
Count paths of length 2…
(2 )
n
1 =
How to prove it:
STEP 3: If all vertices have k neighbors, then
n = k (k-1) + 1.
Why:
Count paths of length 2…
(2 )
n
1 = n
How to prove it:
STEP 3: If all vertices have k neighbors, then
n = k (k-1) + 1.
Why:
Count paths of length 2…
(2 )
n
1 = n (k)
How to prove it:
STEP 3: If all vertices have k neighbors, then
n = k (k-1) + 1.
Why:
Count paths of length 2…
(2 )
n
1 = n (k) (k-1)
How to prove it:
STEP 3: If all vertices have k neighbors, then
n = k (k-1) + 1.
Why:
Count paths of length 2…
(2 )
n
(k) (k-1)
1 = n________
2
How to prove it:
STEP 3: If all vertices have k neighbors, then
n = k (k-1) + 1.
( )
n
(k) (k-1)
1 = n________
2
2
(k) (k-1)
(n)(n-1) = n________
2
2
n = k (k-1) + 1
How to prove it:
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.
The Master Plan
If x,y not
Each pair
has 1 in
common
Some x has
n-1
neighbors
neighbors,
|Nx|=|Ny|
Either
Or
Some
Linear
Algebra
?
|Nx|= k for
all x, and
n =k(k-1)+1
Adjacency Matrix
Call 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.
Adjacency Matrix
v1
v4
v2
0
1
1
1
A=
v3
A2 =
0
1
1
1
1
0
1
0
1
1
0
1
1
0
1
0
0
1
1
1
1
0
1
0
1
1
0
1
1
0
1
0
1
0
1
0
=
1
1
0
1
1
0
1
0
3
1
2
1
1
2
1
2
2
1
3
1
1
2
1
2
Adjacency Matrix
(A2) ij = # common neighbors of vi, vj
So…….. for our graphs…..
(A2) ij = 1
(A2) ij = k
if i, j different, and
if i = j.
A2 = (k-1) I + J.
(J = all 1’s matrix)
Adjacency Matrix
A2 = (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)
Adjacency Matrix
Ai = 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.
Putting it all together
If x,y not
Each pair
has 1 in
common
Some x has
n-1
neighbors
neighbors,
0=1
|Nx|=|Ny|
Either
Or
|Nx|= k for
all x and
n =k(k-1)+1
Moral:
To make progress in almost
any field of math, find a way to
sneak linear algebra into it !
THANK YOU !

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