The Hunt for a Quantum Algorithm for Graph Isomorphism

Cristopher Moore, University of New MexicoAlexander Russell, University of Connecticut

Leonard J. Schulman, Caltech

The Hidden Subgroup Problem

Given a function f(x), find the y such that

for all x.Given a function f on a group G, find the subgroup H consisting of h such that

for all g.

f(x + y) = f(x)

f(gh) = f(g)

The Hidden Subgroup Problem

This captures many quantum algorithms: indeed, most algorithms which give an exponential speedup.

: Simon’s problem : factoring, discrete log (Shor) : Pell’s equation (Hallgren)

What can the non-Abelian HSP do?

Z∗

n

Zn

2

Z

Graph Isomorphism

Define a function f on . If both graphs are rigid, then either f is 1–1 and , or f is 2–1 and for some involution m (of a particular type).

?

H = 1S2n

H = 1, m

Standard Method: Coset States

Start with a uniform superposition,

Measuring f gives a random coset of H:

or, if you prefer, a mixed state:

1√|G|

∑

g∈G

|g〉

|cH〉 =1√|H|

∑

h∈H

|ch〉

ρ =1

|G|

∑

c∈G

|cH〉 〈cH|

The Fourier Transform

We now perform a basis change. In ,

and in ,

Why? Because these are homomorphisms from G to . These form a basis for with many properties (e.g. convolution)

|k〉 =1√n

∑

x

e2πikx/n

Zn

Zn

2

C C[G]

|k〉 =1√2n

∑

x

(−1)k·x

Group Representations

Homomorphisms from groups to matrices:

For instance, consider this three-dimensional representation of .Any representation can be decomposed into a direct sum of irreducible representations.

A5

σ : G → U(V )

Heartbreaking Beauty

Given a “name” and a row and column i, j,

Miraculously, these form an orthogonal basis for :

ρ

C[G]

|σ, i, j〉 =

√dσ

|G|

∑g

σ(g)ij

∑

σ∈ bG

d2

σ = |G|

Given a state in and a group element g, we can apply various group actions:

We can think of as a representation of G under any of these actions.Under (left or right) multiplication, the regular representation contains copies of each .

Group Actions

C[G]

|x〉 → |xg〉 or

∣∣g−1x⟩or

∣∣g−1xg

⟩

C[G]

dσ σ ∈ G

For most group families, the QFT can be carried out efficiently, in polylog(|G|) steps [Beals 1997; Høyer 1997; M., Rockmore, Russell 2004]

Weak sampling: just the nameStrong sampling: name, row and column in a basis of our choice (some bases may be much more informative than others)Intermediate: strong, but with a random basis

Levels of Measurement

σ

σ, i, j

The mixed state over (left) cosets

is left G-invariant, hence block-diagonal.Measuring the irrep name (weak sampling) loses no coherence.Strong sampling is the only thing left to do!

Fourier Sampling is Optimal

ρ =1

|G|

∑

c∈G

|cH〉 〈cH|

For each irrep , we have a projection operator

The probability we observe is

Compare with the Plancherel distribution ( , the completely mixed state)

Projections and Probabilities

πσH =

1

|H|

∑

h∈H

σ(h)

σ

σ

d2σ

|G|H = 1

dσ|H| rkπσH

|G|

If , we have ( )

In , is exponentially small, so the observed distribution is very close to PlancherelWeak sampling fails [Hallgren, Russell, Ta-Shma 2000]

Random basis fails [Grigni, Schulman, Vazirani, Vazirani 2001]

But, strong is stronger for some G... [MRRS 2004]

Weak Sampling Fails

rkπσH =

dσ

2

(1 +

χσ(m)

dσ

)H = 1, m χσ(g) = trσ(g)

χσ(m)/dσSn

But what about a basis of our choice? Given , we observe a basis vector b with probability

Here we have

How much does vary with m?

Now for Strong Sampling

‖πHb‖2

rkπH

σ

‖πHb‖2 =1

2(1 + 〈b, mb〉)

〈b, mb〉

Expectation of an irrep over m’s conjugates is

so

To turn the second moment into a first moment,

Controlling the Variance

Expm

σ(m) =χσ(m)

dσ

Expm〈b, mb〉 =

χσ(m)

dσ

|〈b, mb〉|2 = 〈b ⊗ b∗, m(b ⊗ b

∗)〉

σ

Decompose into irreducibles:

Then

How much of lies in low-dimensional ?

Controlling the Variance

σ ⊗ σ∗

σ ⊗ σ∗ ∼

=

⊕

τ∈ bG

aττ

b ⊗ b∗

τ

Varm ‖πHb‖2 ≤1

4

∑

τ∈ bG

χτ (m)

dτ

∥∥∥Πσ⊗σ∗

τ (b ⊗ b∗)

∥∥∥2

Using simple counting arguments, we show that almost all of lies in high-dimensional subspaces of .

Since is exponentially small, the observed distribution on for any basis is exponentially close to uniform.

No subexponential set of experiments on coset states can solve Graph Isomorphism. [M., Russell, Schulman 2005]

Strong Sampling Fails

τ

b ⊗ b∗

σ ⊗ σ∗

χτ (m)/dτ

b

For any group, there exists a measurement on the tensor product of coset states

with [Ettinger, Høyer, Knill 1999]

What can we prove about entangled measurements?

Entangled Measurements

k = poly(log |G|)

ρ ⊗ · · ·⊗ ρ︸ ︷︷ ︸

k

Weak sample each register, observing

Given a subset I of the k registers, decompose that part of the tensor product:

This group action multiplies these registers by g and leaves the others fixed.

Bounds on Multiregister Sampling

σ = σ1 ⊗ · · ·⊗ σk

⊗

i∈I

σi∼=

⊕

τ∈ bG

aIττ

Second moment: analogous to one register, consider . Given subsets I and J, define

For an arbitrary entangled basis, [M., Russell 2005]

Bounds on Multiregister Sampling

σ ⊗ σ∗

EI,J(b) =∑

τ∈ bG

χτ (m)

dτ

∥∥ΠI,Jτ (b ⊗ b

∗)∥∥2

Varm ‖ΠHb‖2 ≤1

4k

∑

I,J⊆[k]:I,J "=∅

EI,J(b)

With some additional work, this general bound can be used to show that registers are necessary for [Hallgren, Rötteler, Sen; M., Russell]

But what form might this measurement take?

Note that each subset of the registers contributes some information...

Bounds on Multiregister Sampling

Ω(n log n)Sn

The HSP in the dihedral group reduces to random cases of Subset Sum [Regev 2002]

Leads to a -time and -register algorithm [Kuperberg 2003]

Subset Sum gives the optimal multiregister measurement [Bacon, Childs, van Dam 2005]

Subset Sum and the Dihedral Group

Dn

2O(

√log n)

If , there is a missing harmonic:

Weak sampling gives random two-dimensional irreps ; think of these as integers .

Tensor products:

Find subset that gives .

More Abstractly...

H = 1, m∑

h∈H

π(h) = 0

σj ±j

σj ⊗ σk∼= σj+k ⊕ σj−k

σ0∼= ⊕ π

Suppose H has a missing harmonic .

For each subset I, consider the subspace resulting from applying the group action to I. (In , this flips the integers j in this subset.)

If the hidden subgroup is a conjugate of H, then the state is perpendicular to for all I.

How much of does this leave? What fraction is spanned by the ?

Subsets in General

WIτ

WIτ

C[Gk]W

Iτ

τ

Dn

Say that two subspaces V, W of a space U are independent if, just as for random vectors in U,

or equivalently

Being in V or W are “independent events.”

Independent Subspaces

Expv∈V ‖ΠW v‖2=

dimW

dimU

trΠV ΠW

dimU=

trΠV

dimU

tr ΠW

dimU

VW

For , and are independent.

Therefore, is large:

If , probability of “some subset being in ” is if the hidden subgroup is trivial, but is zero if it is a conjugate of H. [M., Russell 2005]

Each Subset Contributes

I != J WIτ W

Jτ

dimWτ

dim C[Gk]≥ 1 −

1

1 + 2k/|G|

Wτ = spanIWIτ

k ≥ log2 |G|≥ 1/2τ

Divide into subspaces; for each one, find a subset I for a large fraction of the completely mixed state is in : e.g. in .

“Pretty Good Measurement” (i.e., Subset Sum for ) is optimal for Gel’fand pairs... [MR 2005]

...but it is not optimal for [Childs]. What is? And, is it related to Subset Something?

Find An Informative Subset!

σ0∼= ⊕ π Dn

Sn

Dn

C[Gk]

WIτ

The Hunt Continues

The AdversaryBeauty and Truth

vs.

Acknowledgments