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Paths and Polynomials

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ASPAK2014

The document discusses algorithms and techniques for analyzing graphs and polynomials. It describes how an adjacency matrix A can be used to count walks and common neighbors in a graph. It also explains how to construct a polynomial that is equal to zero if and only if a graph G contains a path of length k, by creating terms for all walks in G and exploiting the property that a+a=0 in finite fields of characteristic two. This allows evaluating whether the polynomial is zero to determine if G contains the desired path.

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Paths & Polynomials
Paths and Polynomials
Paths and Polynomials
Paths and Polynomials
Paths and Polynomials
Paths and Polynomials
Paths and Polynomials
Paths and Polynomials
Paths and Polynomials
Paths and Polynomials
Paths and Polynomials
Paths and Polynomials
Paths and Polynomials
Paths and Polynomials
Paths and Polynomials
Paths and Polynomials
Paths and Polynomials
Paths and Polynomials
Paths and Polynomials
Paths and Polynomials
Paths and Polynomials
Hamiltonian Path has an algorithm that spends O(2n) time and consumes O(2n) space.
Paths and Polynomials
Walks to the rescue…
Walks to the rescue…
Walks to the rescue…
Paths and Polynomials
Adjacency Matrix A[i,j] = 1 iff (i,j) is an edge in G.
Adjacency Matrix A[i,j] = 1 iff (i,j) is an edge in G.
Adjacency Matrix A[i,j] = 1 iff (i,j) is an edge in G.
Adjacency Matrix A[i,j] = 1 iff (i,j) is an edge in G.
Adjacency Matrix A[i,j] = 1 iff (i,j) is an edge in G. What is A2[i,j]?
Adjacency Matrix A[i,j] = 1 iff (i,j) is an edge in G. What is A2[i,j]?
i j Adjacency Matrix A[i,j] = 1 iff (i,j) is an edge in G. What is A2[i,j]?
i j Adjacency Matrix A[i,j] = 1 iff (i,j) is an edge in G. What is A2[i,j]?
i j Adjacency Matrix A[i,j] = 1 iff (i,j) is an edge in G. What is A2[i,j]?
i j Adjacency Matrix A[i,j] = 1 iff (i,j) is an edge in G. What is A2[i,j]?
i j Adjacency Matrix A[i,j] = 1 iff (i,j) is an edge in G. What is A2[i,j]?
i j Adjacency Matrix A[i,j] = 1 iff (i,j) is an edge in G. What is A2[i,j]?
What is A2[i,j] counts the number of common neighbors of i and j.
What is A2[i,j] counts the number of common neighbors of i and j. What is Ak[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] counts the number of walks of length k between i and j.
We know how to count walks!
We know how to count walks! Suppose we now want to count paths.
We know how to count walks! Suppose we now want to count paths. Let U be the collection of all Hamiltonian Walks.
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… n X= Ai i=1 Let U be the collection of all Hamiltonian Walks. Let Ai be the set of walks of length n passing through i.
Consider… n X=U Ai i=1 Let U be the collection of all Hamiltonian Walks. Let Ai be the set of walks of length n passing through i.
Consider… n X=U Ai i=1 Let U be the collection of all Hamiltonian Walks. Let Ai be the set of walks of length n passing through i.
Consider… n X=U Ai i=1 Let U be the collection of all Hamiltonian Walks. Let Ai be the set of walks of length n passing through i. The set of walks that avoid i.
Consider… n X=U Ai i=1 |Z| |X| = |U| ( 1) Z [n] Ai i Z Let U be the collection of all Hamiltonian Walks. Let Ai be the set of walks of length n passing through i. The set of walks that avoid i.
Consider… n X=U Ai i=1 |Z| |X| = |U| ( 1) Z [n] Ai i Z Let U be the collection of all Hamiltonian Walks. Let Ai be the set of walks of length n passing through i. The set of walks that avoid i.
Consider… n X=U Ai i=1 |Z| |X| = |U| ( 1) Z [n] Ai i Z Let U be the collection of all Hamiltonian Walks. Let Ai be the set of walks of length n passing through i. The set of walks that avoid i. The set of walks that avoid X.
Consider… n X=U Ai i=1 |Z| |X| = |U| ( 1) Z [n] Ai i Z Let U be the collection of all Hamiltonian Walks. Let Ai be the set of walks of length n passing through i. The set of walks that avoid i. 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 polynomials are equal, finding a perfect matching, primality testing, and so on.
Basic Idea (x1 + 3x2 x3 )(3x1 + x4 1) · · · (x7 x2 ) Simplifying is expensive, but evaluating is cheap. 0?
Basic Idea (x1 + 3x2 x3 )(3x1 + x4 1) · · · (x7 x2 ) Simplifying is expensive, but evaluating is cheap. The probability that a random assignment corresponds to a root is low. 0?
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.
Paths and Polynomials
x[v1,2 ]x[v2,3 ] · ·2,3 ] · k· x[vk 1,k]y[v2 ] · · · 2 ] · · ·]y[vk ] x[v1,2 ]x[v · x[v· 1,k ]y[v1 1 ]y[v y[vk Edges Vertices
x[v1,2 ]x[v2,3 ] · ·2,3 ] · k· x[vk 1,k]y[v2 ] · · · 2 ] · · ·]y[vk ] x[v1,2 ]x[v · x[v· 1,k ]y[v1 1 ]y[v y[vk Edges Vertices We want terms corresponding to walks to cancel out,
x[v1,2 ]x[v2,3 ] · ·2,3 ] · k· x[vk 1,k]y[v2 ] · · · 2 ] · · ·]y[vk ] x[v1,2 ]x[v · x[v· 1,k ]y[v1 1 ]y[v y[vk Edges Vertices We want terms corresponding to walks to cancel out, somehow.
The Trick
Paths and Polynomials
Evaluate these polynomials over finite fields of characteristic two.
Evaluate these polynomials over finite fields of characteristic two. a+a=0
x[v1,2 ]x[v2,3 ] · ·2,3 ] · k· x[vk 1,k]y[v2 ] · · · 2 ] · · ·]y[vk ] x[v1,2 ]x[v · x[v· 1,k ]y[v1 1 ]y[v y[vk Edges Vertices
x[v1,2 ]x[v2,3 ] · ·2,3 ] · k· x[vk 1,k]y[v2 ] · · · 2 ] · · ·]y[vk ] x[v1,2 ]x[v · x[v· 1,k ]y[v1 1 ]y[v y[vk Edges 1,2 ]x[v2,3 ] · · · x[vk 1,k ]y[v1 ]y[v2 ] · · · y[vk ] Vertices
x[v1,2 ]x[v2,3 ] · ·2,3 ] · k· x[vk 1,k]y[v2 ] · · · 2 ] · · ·]y[vk ] x[v1,2 ]x[v · x[v· 1,k ]y[v1 1 ]y[v y[vk Edges 1,2 ]x[v2,3 ] · · · x[vk 1,k ]y[v1 ]y[v2 ] · · · y[vk ] Vertices
x[v1,2 ]x[v2,3 ] · ·2,3 ] · k· x[vk 1,k]y[v2 ] · · · 2 ] · · ·]y[vk ] x[v1,2 ]x[v · x[v· 1,k ]y[v1 1 ]y[v y[vk Edges 1,2 ]x[v2,3 ] · · · x[vk 1,k ]y[v1 ]y[v2 ] · · · y[vk ] Vertices
x[v1,2 ]x[v2,3 ] · ·2,3 ] · k· x[vk 1,k]y[v2 ] · · · 2 ] · · ·]y[vk ] x[v1,2 ]x[v · x[v· 1,k ]y[v1 1 ]y[v y[vk Edges 1,2 ]x[v2,3 ] · · · x[vk 1,k ]y[v1 ]y[v2 ] · · · y[vk ] 1,2 ]x[v2,3 ] · · · x[vk 1,k ]y[v1 ]y[v2 ] · · · y[vk ] 1,2 ]x[v2,3 ] · · · x[vk 1,k ]y[v1 ]y[v2 ] · · · y[vk ] 1,2 ]x[v2,3 ] · · · x[vk 1,k ]y[v1 ]y[v2 ] · · · y[vk ] 1,2 ]x[v2,3 ] · · · x[vk 1,k ]y[v1 ]y[v2 ] · · · y[vk ] Vertices
x[v1,2 ]x[v2,3 ] · ·2,3 ] · k· x[vk 1,k]y[v2 ] · · · 2 ] · · ·]y[vk ] x[v1,2 ]x[v · x[v· 1,k ]y[v1 1 ]y[v y[vk Edges 1,2 ]x[v2,3 ] · · · x[vk 1,k ]y[v1 ]y[v2 ] · · · y[vk ] 1,2 ]x[v2,3 ] · · · x[vk 1,k ]y[v1 ]y[v2 ] · · · y[vk ] 1,2 ]x[v2,3 ] · · · x[vk 1,k ]y[v1 ]y[v2 ] · · · y[vk ] 1,2 ]x[v2,3 ] · · · x[vk 1,k ]y[v1 ]y[v2 ] · · · y[vk ] 1,2 ]x[v2,3 ] · · · x[vk 1,k ]y[v1 ]y[v2 ] · · · y[vk ] Vertices
Paths and Polynomials
For a walk W, we will dump all these terms into the formula:
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)]
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)] Sum this over all walks W:
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)] Sum this over all walks W: (W, ) W:=v1 ,...,vk
Paths and Polynomials
This polynomial captures exactly the paths in G, i.e, it is identically zero precisely when G has no paths of length k.
(W, ) W:=v1 ,...,vk
(W, ) W:=v1 ,...,vk How do we evaluate this polynomial now?
(W, ) W:=v1 ,...,vk How do we evaluate this polynomial now?
Paths and Polynomials
Inclusion Exclusion RELOADED

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Paths and Polynomials

  • 22. Hamiltonian Path has an algorithm that spends O(2n) time and consumes O(2n) space.
  • 24. Walks to the rescue…
  • 25. Walks to the rescue…
  • 26. Walks to the rescue…
  • 28. Adjacency Matrix A[i,j] = 1 iff (i,j) is an edge in G.
  • 29. Adjacency Matrix A[i,j] = 1 iff (i,j) is an edge in G.
  • 30. Adjacency Matrix A[i,j] = 1 iff (i,j) is an edge in G.
  • 31. Adjacency Matrix A[i,j] = 1 iff (i,j) is an edge in G.
  • 32. Adjacency Matrix A[i,j] = 1 iff (i,j) is an edge in G. What is A2[i,j]?
  • 33. Adjacency Matrix A[i,j] = 1 iff (i,j) is an edge in G. What is A2[i,j]?
  • 34. i j Adjacency Matrix A[i,j] = 1 iff (i,j) is an edge in G. What is A2[i,j]?
  • 35. i j Adjacency Matrix A[i,j] = 1 iff (i,j) is an edge in G. What is A2[i,j]?
  • 36. i j Adjacency Matrix A[i,j] = 1 iff (i,j) is an edge in G. What is A2[i,j]?
  • 37. i j Adjacency Matrix A[i,j] = 1 iff (i,j) is an edge in G. What is A2[i,j]?
  • 38. i j Adjacency Matrix A[i,j] = 1 iff (i,j) is an edge in G. What is A2[i,j]?
  • 39. i j Adjacency Matrix A[i,j] = 1 iff (i,j) is an edge in G. What is A2[i,j]?
  • 40. What is A2[i,j] counts the number of common neighbors of i and j.
  • 41. What is A2[i,j] counts the number of common neighbors of i and j. What is Ak[i,j]?
  • 42. 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] counts the number of walks of length k between i and j.
  • 43. We know how to count walks!
  • 44. We know how to count walks! Suppose we now want to count paths.
  • 45. We know how to count walks! Suppose we now want to count paths. Let U be the collection of all Hamiltonian Walks.
  • 46. 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.
  • 47. Consider… n X= Ai i=1 Let U be the collection of all Hamiltonian Walks. Let Ai be the set of walks of length n passing through i.
  • 48. Consider… n X=U Ai i=1 Let U be the collection of all Hamiltonian Walks. Let Ai be the set of walks of length n passing through i.
  • 49. Consider… n X=U Ai i=1 Let U be the collection of all Hamiltonian Walks. Let Ai be the set of walks of length n passing through i.
  • 50. Consider… n X=U Ai i=1 Let U be the collection of all Hamiltonian Walks. Let Ai be the set of walks of length n passing through i. The set of walks that avoid i.
  • 51. Consider… n X=U Ai i=1 |Z| |X| = |U| ( 1) Z [n] Ai i Z Let U be the collection of all Hamiltonian Walks. Let Ai be the set of walks of length n passing through i. The set of walks that avoid i.
  • 52. Consider… n X=U Ai i=1 |Z| |X| = |U| ( 1) Z [n] Ai i Z Let U be the collection of all Hamiltonian Walks. Let Ai be the set of walks of length n passing through i. The set of walks that avoid i.
  • 53. Consider… n X=U Ai i=1 |Z| |X| = |U| ( 1) Z [n] Ai i Z Let U be the collection of all Hamiltonian Walks. Let Ai be the set of walks of length n passing through i. The set of walks that avoid i. The set of walks that avoid X.
  • 54. Consider… n X=U Ai i=1 |Z| |X| = |U| ( 1) Z [n] Ai i Z Let U be the collection of all Hamiltonian Walks. Let Ai be the set of walks of length n passing through i. The set of walks that avoid i. The set of walks that avoid X.
  • 55. Polynomial Identity Testing Input: A polynomial p(x). Question: Is p(x) identically zero? (x1 + 3x2 x3 )(3x1 + x4 1) · · · (x7 x2 ) 0?
  • 56. 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 polynomials are equal, finding a perfect matching, primality testing, and so on.
  • 57. Basic Idea (x1 + 3x2 x3 )(3x1 + x4 1) · · · (x7 x2 ) Simplifying is expensive, but evaluating is cheap. 0?
  • 58. Basic Idea (x1 + 3x2 x3 )(3x1 + x4 1) · · · (x7 x2 ) Simplifying is expensive, but evaluating is cheap. The probability that a random assignment corresponds to a root is low. 0?
  • 59. 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.
  • 61. x[v1,2 ]x[v2,3 ] · ·2,3 ] · k· x[vk 1,k]y[v2 ] · · · 2 ] · · ·]y[vk ] x[v1,2 ]x[v · x[v· 1,k ]y[v1 1 ]y[v y[vk Edges Vertices
  • 62. x[v1,2 ]x[v2,3 ] · ·2,3 ] · k· x[vk 1,k]y[v2 ] · · · 2 ] · · ·]y[vk ] x[v1,2 ]x[v · x[v· 1,k ]y[v1 1 ]y[v y[vk Edges Vertices We want terms corresponding to walks to cancel out,
  • 63. x[v1,2 ]x[v2,3 ] · ·2,3 ] · k· x[vk 1,k]y[v2 ] · · · 2 ] · · ·]y[vk ] x[v1,2 ]x[v · x[v· 1,k ]y[v1 1 ]y[v y[vk Edges Vertices We want terms corresponding to walks to cancel out, somehow.
  • 66. Evaluate these polynomials over finite fields of characteristic two.
  • 67. Evaluate these polynomials over finite fields of characteristic two. a+a=0
  • 68. x[v1,2 ]x[v2,3 ] · ·2,3 ] · k· x[vk 1,k]y[v2 ] · · · 2 ] · · ·]y[vk ] x[v1,2 ]x[v · x[v· 1,k ]y[v1 1 ]y[v y[vk Edges Vertices
  • 69. x[v1,2 ]x[v2,3 ] · ·2,3 ] · k· x[vk 1,k]y[v2 ] · · · 2 ] · · ·]y[vk ] x[v1,2 ]x[v · x[v· 1,k ]y[v1 1 ]y[v y[vk Edges 1,2 ]x[v2,3 ] · · · x[vk 1,k ]y[v1 ]y[v2 ] · · · y[vk ] Vertices
  • 70. x[v1,2 ]x[v2,3 ] · ·2,3 ] · k· x[vk 1,k]y[v2 ] · · · 2 ] · · ·]y[vk ] x[v1,2 ]x[v · x[v· 1,k ]y[v1 1 ]y[v y[vk Edges 1,2 ]x[v2,3 ] · · · x[vk 1,k ]y[v1 ]y[v2 ] · · · y[vk ] Vertices
  • 71. x[v1,2 ]x[v2,3 ] · ·2,3 ] · k· x[vk 1,k]y[v2 ] · · · 2 ] · · ·]y[vk ] x[v1,2 ]x[v · x[v· 1,k ]y[v1 1 ]y[v y[vk Edges 1,2 ]x[v2,3 ] · · · x[vk 1,k ]y[v1 ]y[v2 ] · · · y[vk ] Vertices
  • 72. x[v1,2 ]x[v2,3 ] · ·2,3 ] · k· x[vk 1,k]y[v2 ] · · · 2 ] · · ·]y[vk ] x[v1,2 ]x[v · x[v· 1,k ]y[v1 1 ]y[v y[vk Edges 1,2 ]x[v2,3 ] · · · x[vk 1,k ]y[v1 ]y[v2 ] · · · y[vk ] 1,2 ]x[v2,3 ] · · · x[vk 1,k ]y[v1 ]y[v2 ] · · · y[vk ] 1,2 ]x[v2,3 ] · · · x[vk 1,k ]y[v1 ]y[v2 ] · · · y[vk ] 1,2 ]x[v2,3 ] · · · x[vk 1,k ]y[v1 ]y[v2 ] · · · y[vk ] 1,2 ]x[v2,3 ] · · · x[vk 1,k ]y[v1 ]y[v2 ] · · · y[vk ] Vertices
  • 73. x[v1,2 ]x[v2,3 ] · ·2,3 ] · k· x[vk 1,k]y[v2 ] · · · 2 ] · · ·]y[vk ] x[v1,2 ]x[v · x[v· 1,k ]y[v1 1 ]y[v y[vk Edges 1,2 ]x[v2,3 ] · · · x[vk 1,k ]y[v1 ]y[v2 ] · · · y[vk ] 1,2 ]x[v2,3 ] · · · x[vk 1,k ]y[v1 ]y[v2 ] · · · y[vk ] 1,2 ]x[v2,3 ] · · · x[vk 1,k ]y[v1 ]y[v2 ] · · · y[vk ] 1,2 ]x[v2,3 ] · · · x[vk 1,k ]y[v1 ]y[v2 ] · · · y[vk ] 1,2 ]x[v2,3 ] · · · x[vk 1,k ]y[v1 ]y[v2 ] · · · y[vk ] Vertices
  • 75. For a walk W, we will dump all these terms into the formula:
  • 76. 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)]
  • 77. 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)] Sum this over all walks W:
  • 78. 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)] Sum this over all walks W: (W, ) W:=v1 ,...,vk
  • 80. This polynomial captures exactly the paths in G, i.e, it is identically zero precisely when G has no paths of length k.
  • 82. (W, ) W:=v1 ,...,vk How do we evaluate this polynomial now?
  • 83. (W, ) W:=v1 ,...,vk How do we evaluate this polynomial now?