Graph Isomorphism as Hsp

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Outline

Group theory review

The hidden subgroup problem

Classes of HSP

An example of Abelian HSP

GI as HSP

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Group Theory Review

Definition

A group G is a finite or infinite set of elements with a binary

operation a.k.a. group operation ◦. Any group has following four

fundamental properties. Closure: If A, B ∈ G , then A ◦ B ∈ G .

Associativity: If A, B , C ∈ G , (A ◦ B ) ◦ C = A ◦ (B ◦ C ).

Identity: There is an identity element e such that

∀A ∈ G : e ◦ A = A ◦ e = A. Inverse: There must be an inverse of each element.

Therefore, for each element A ∈ G , the set contains anelement B = A−1 such that A ◦ A−1 = A−1 ◦ A = e .

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continued...

Definition

A subgroup H ⊂ G satisfies the four group requirements.

The order, |H |, of any subgroup, H of a group G must be adivisor of |G |.

Let H with elements hi be the a subgroup of a group G . If anelement x ∈ G , ∈ H , then x ◦ hi for i = 1, 2, . . . constitute theleft coset of the subgroup H with respect to x .

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The Hidden Subgroup Problem

Definition

Find the subgroup H of periods of a function f : G → S under the promise that f is strictly periodic, that is, for all x , y ∈ G ,

f (x ) = f (y ) ⇐⇒ y = xh for some h ∈ H .

In other words 1,Given: G : group, S : set, f : G → S via an oracle.Promise: Subgroup H ≤ G such that f is constant on the leftcosets of H and distinct on different cosets.Task: Find the hidden subgroup H by querying f .

1Taken from Graph isomorphism, the hidden subgroup problem and identifying quantum states by Pranab Sen

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Classes of HSP

Abelian hidden subgroup problem Integer factoring Simon’s problem

Non-abelian hidden subgroup problem Dihedral hidden subgroup problem

Lattice problems

Symmetric hidden subgroup problem Graph isomorphism problems

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Simon’s Problem

Definition

Suppose 2 we are given a function f : {0, 1}n → {0, 1}m, withm ≥ n, and we are promised that either f is 1 − to − 1, or there exists a non-trivial s , such that

∀x = x (f (x )) = f (x ) ⇐⇒ x = x ⊕ s , where ⊕ denotes bitwise exclusive-or. We wish to determine which of these conditions holds

for f , and in the second case, to find s .

2Verbatim from Simon’s paper.

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Simon’s Problem as HSP

Definition

Suppose we are given a function f : {0, 1}n → {0, 1}m, with m ≥ n,and we are promised that either f is 1 − to − 1, or there exists a

non-trivial h, such that ∀x = x

(f (x )) = f (x

) ⇐⇒ x

= x ⊕ h,where ⊕ denotes bitwise exclusive-or. We wish to determine whichof these conditions holds for f , and in the second case, to find h.

Given: G = ({0, 1}n , ⊕): group, S = {0, 1}m: set, f : G → S via

an oracle.Promise: Subgroup h ∈ H ≤ G such that f is constant on theleft cosets of H and distinct on different cosets.Task: Find the hidden subgroup H by querying f .

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Example : Simon’s problem for Z/2, n = 3

Subgroups are closed under inverses and the group operation.

G = S = {x |x ∈ {0, 1}∗ , |x | ≤ 3}

= {000, 001, 010, 011, 100, 101, 110, 111} (1)

and let the blackbox be,

f (000) → 111, f (001) → 101

f (010) → 111, f (011) → 101

f (100) → 010, f (101) → 001

f (110) → 010, f (111) → 001 (2)

Find H .

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(We pretend that we haven’t seen this slide!)

Obviously, the non-trivial period is 010.

f (000) → 111, f (001) → 101f (010) = f (000 ⊕ 010) → 111

f (011) = f (001 ⊕ 010) → 101

f (100) → 010, f (101) → 001

f (110) = f (100 ⊕ 010) → 010f (111) = f (101 ⊕ 010) → 001 (3)

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The Promises are Kept!

H = {000, 010}. So, Promise 1, |H | ≤ |G | =⇒ 2


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Example: Graph Isomorphism (GI) Problem

Definition

Let 3 V (G ) be the vertex set of a simple graph and E (G ) its edge set. Then a graph isomorphism from a simple graph G to a simple

graph H is a bijection f : V (G ) → V (H ) such that (u , v ) ∈ E (G ) ⇐⇒ (f (u ) , f (v )) ∈ E (H ).

3Verbatim from (West 2000).

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A Case of Graph Isomorphism

A and B are two undirected graphs with n = 4 vertices labeled1, 2, 3 and 4 4.

They may be described by two 4 × 4 adjacency matricesrespectively.

The ij -th entry of an adjacency matrix is 1 iff the graph has anedge joining vertices i and j . Rest of the entries are zeros.

We assume a graph will always have at most one edge joiningtwo vertices.

4This is a special case of the exposition from (Jozsa 2000)

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Isomorphic Graphs

0 1 1 1

1 0 1 01 1 0 11 0 1 0

0 1 1 1

1 0 0 11 0 0 11 1 1 0

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Permutations

P 4 is the group of all permutations of n = 4 symbols 1, 2, 3and 4.

A and B are isomorphic if ∃Π ∈ P 4 operates on A to produceB .

Such a permutation is,

1 2 3 41 2 4 3

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Symmetry Group of a Graph

Pick two permutations, Π1 = 1 2 3 4

2 1 3 4 and

Π2 =

1 2 3 41 4 3 2

. Π1 ◦ Π2 and Π2 ◦ Π1, operating on A,

separately, produces two very different graph. So, symmetric groupsare non-abelian i.e. group operations are non-commutative.

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Non-abelianness of Graph Permutation

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The Question

Can we detect the isomorphism of any two graph defined on nvertices in poly (n) steps?

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A Bigger Space

C is the disjoint union of A and B .

V (C ) = 2 × n = 2 × 4 = 8.

The vertices are labeled as 1, . . . , n, n + 1, . . . 2n = 1, . . . 8 .

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Symmetry Group of C

The permutation group on 2n vertices is P 2n, in our case, P 8.

For our example, any permutation will be a 8 × 2 matrix.

A symmetric group, K , on the new graph C will be a subgroupof P 8 × P 8 i.e. K ⊂ P 8 × P 8.

Any symmetric permutation on C (because of the disjointunion)

case 1: either separately permutes the vertices {1, 2, 3, 4} and

{5, 6, 7, 8}, case 2: or swaps these two sets of vertices.

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Subgroups in P 2n

Let, H = P 4 × P 4 .

The elements of H are ordered pairs of vertices of A and B .

The group operation is as usual composition of permutation.

If case 1 is true, K ⊂ H .

Let σ be the special permutation i.e. involution. So, σ = σ−1

and σ2 = e .

So, we could define a group on 2n = 8 vertices as {{e , σ} , ◦}.

Any nontrivial group operation on H , σH , will just swap thevertices in the ordered pairs.

So, if case 2 is true, K ⊂ σH .

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Verification

Claim 1:If A and B are isomorphic exactly half of themembers of K are in H and half are in σH .

Claim 2:If A and B are not isomorphic then K ⊂ H

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Claim 1: K 2

⊂ H , K 2

⊂ σH

Let A and B be the two graphs we previously seen.

K is the symmetric group over A ∪ B .

K contains all the ordered pairs of elements from P 4 and P 4

respectively. By definition, all such ordered pairs are in H .

Being a symmetric group, K also has the previously mentionedordered pairs in reverse order.

The involution σ can generate each of those reversed pairs byoperating on individual elements of H respectively.

So, these two sets of ordered pairs are equal in number andconstitutes K .

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Claim 2: K ⊂ H

Redefine A and B .

The symmetric group, K is defined over A ∪ B = C . The vertices will be relabeled to 1 . . . 8.

Every element of K separately permutes {1, 2, 3, 4} and{5, 6, 7, 8}. So, K ⊂ H .

Any swap operation σ on K will destroy the symmetry. So,K ⊂ σH .

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The Ambient Group, G

Let G = H ∪ σH . H is a subgroup.

σ is a coset of H .

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GI as HSP

Definition

Find the symmetric hidden subgroup G on the disjoint union of two

graphs, A and B , each of n vertices, in the permutation group P 2nby querying an oracle, f : P 2n → P 2n.

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Break

Thank you!

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Example: Graph Isomorphism (GI) Problem

Definition

Let 5

V (G ) be the vertex set of a simple graph and E (G ) its edge set. Then a graph isomorphism from a simple graph G to a simple

graph H is a bijection f : V (G ) → V (H ) such that (u , v ) ∈ E (G ) ⇐⇒ (f (u ) , f (v )) ∈ E (H ).

5Verbatim from (West 2000).

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A Case of Graph Isomorphism

A and B are two undirected graphs with n = 4 vertices labeled1, 2, 3 and 4 6.

They may be described by two 4 × 4 adjacency matricesrespectively.

The ij -th entry of an adjacency matrix is 1 iff the graph has anedge joining vertices i and j . Rest of the entries are zeros.

We assume a graph will always have at most one edge joiningtwo vertices.

6This is a special case of the exposition from (Jozsa 2000)

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Isomorphic Graphs

0 1 01 0 10 1 0

0 1 11 0 01 1 0

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The group

P 2n × P 2n is the permutation group of 2n symbols. The first(second) n symbols correspond to the vertices of G 1 (G 2). For ourexample, we consider S 6 hence it is non-abelian.

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The sub group

K ≤ P 2n × P 2n is the set of all permutations of the vertex setwhich leaves the edge set the same. So, K is a symmetry group.

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Promise 1

K ≤ P 2n × P 2n

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Questions and Answers

Thank you!

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Graph Isomorphism as Hsp

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