Paths and Polynomials

Paths & Polynomials

Hamiltonian Path has an algorithm thatspends O(2n) time andconsumes O(2n) space.

Walks to the rescue…

Adjacency Matrix A[i,j] = 1 iff (i,j) is an edge in G.


What is A2[i,j]?


What is A2[i,j]?

i

j

What is A2[i,j] counts the number of common neighbors of i and j.


What is Ak[i,j]?


What is Ak[i,j]?

What is Ak[i,j] counts the number of walks of length kbetween i and j.

We know how to count walks!


Suppose we now want to count paths.



Let U be the collection of all Hamiltonian Walks.



Let Ai be the set of walks of length n passing through i.




Consider…

X =n�

i=1

Ai



Consider…

X = U �n�

i=1

Ai



Consider…

X = U �n�

i=1

Ai

The set of walks that avoid i.



Consider…

X = U �n�

i=1

Ai


|X| = |U| ��

Z�[n]

(�1)|Z|�

i�Z

Ai



Consider…

X = U �n�

i=1

Ai


|X| = |U| ��

Z�[n]

(�1)|Z|�

i�Z

Ai

The set of walks that avoid X.

Polynomial Identity Testing

Input: A polynomial p(x). Question: Is p(x) identically zero?

(x1 + 3x2 � x3)(3x1 + x4 � 1) · · · (x7 � x2) � 0?

Polynomial Identity Testing

Input: A polynomial p(x). Question: Is p(x) identically zero?

(x1 + 3x2 � x3)(3x1 + x4 � 1) · · · (x7 � x2) � 0?

Captures several problems, like checking if two polynomialsare equal, finding a perfect matching, primality testing,

and so on.

Basic Idea

(x1 + 3x2 � x3)(3x1 + x4 � 1) · · · (x7 � x2) � 0?

Simplifying is expensive, but evaluating is cheap.

Basic Idea

(x1 + 3x2 � x3)(3x1 + x4 � 1) · · · (x7 � x2) � 0?

Simplifying is expensive, but evaluating is cheap.

The probability that a random assignmentcorresponds to a root is low.

Given a (directed) graph G and a number k… !!!

Let’s try to construct a polynomial that is zero !

if and only if !

G has a path of length k.

x[v1,2]x[v2,3] · · · x[vk�1,k]y[v1]y[v2] · · · y[vk]

Edges


Vertices


Edges


Vertices

We want terms corresponding to walks to cancel out,


Edges


Vertices

We want terms corresponding to walks to cancel out,

somehow.

The Trick

Evaluate these polynomials over finite fieldsof characteristic two.

Evaluate these polynomials over finite fieldsof characteristic two.

a + a = 0


Edges


Vertices


Edges


Vertices

x[v1,2]x[v2,3] · · · x[vk�1,k]y[v1]y[v2] · · · y[vk]x[v1,2]x[v2,3] · · · x[vk�1,k]y[v1]y[v2] · · · y[vk]x[v1,2]x[v2,3] · · · x[vk�1,k]y[v1]y[v2] · · · y[vk]x[v1,2]x[v2,3] · · · x[vk�1,k]y[v1]y[v2] · · · y[vk]x[v1,2]x[v2,3] · · · x[vk�1,k]y[v1]y[v2] · · · y[vk]


Edges


Vertices






For a walk W, we will dump all these terms into the formula:


�

�:[k]�[k]

x[v1,2]x[v2,3] · · · x[vk�1,k]y[v1, �(1)]y[v2, �(2)] · · · y[vk, �(k)]


�

�:[k]�[k]


Sum this over all walks W:


�

�:[k]�[k]


Sum this over all walks W:

�

W:=v1,...,vk

�(W, �)

This polynomial captures exactly the paths in G,

i.e, it is identically zero precisely when G has no paths of length k.

�

W:=v1,...,vk

�(W, �)

�

W:=v1,...,vk

�(W, �)

How do we evaluate this polynomial now?

Inclusion ExclusionRELOADED

Paths and Polynomials

Technology

Transcript of Paths and Polynomials