The Hunt for a Quantum Algorithm for Graph Isomorphism Cristopher Moore, University of New Mexico Alexander Russell, University of Connecticut Leonard J. Schulman, Caltech
The Hidden Subgroup Problem Given a function f(x) , find the y such that f ( x + y ) = f ( x ) for all x. Given a function f on a group G , find the subgroup H consisting of h such that f ( gh ) = f ( g ) for all g .
The Hidden Subgroup Problem This captures many quantum algorithms: indeed, most algorithms which give an exponential speedup. : Simon’s problem Z n 2 : factoring, discrete log (Shor) Z ∗ n : Pell’s equation (Hallgren) Z What can the non-Abelian HSP do?
Graph Isomorphism ? Define a function f on . If both graphs S 2 n are rigid, then either f is 1–1 and , H = { 1 } or f is 2–1 and for some H = { 1 , m } involution m (of a particular type).
Standard Method: Coset States 1 � Start with a uniform superposition, | g � � | G | g ∈ G Measuring f gives a random coset of H : 1 � | cH � = | ch � � | H | h ∈ H or, if you prefer, a mixed state: 1 � | cH � � cH | ρ = | G | c ∈ G
The Fourier Transform We now perform a basis change. In , Z n 1 � e 2 π ikx/n √ n | k � = x and in , Z n 2 1 � ( − 1) k · x √ | k � = 2 n x Why? Because these are homomorphisms from G to . These form a basis for C [ G ] C with many properties (e.g. convolution)
Group Representations Homomorphisms from groups to matrices: σ : G → U ( V ) For instance, consider this three-dimensional representation of . A 5 Any representation can be decomposed into a direct sum of irreducible representations.
Heartbreaking Beauty Given a “name” and a ρ row and column i, j , � d σ � | σ , i, j � = σ ( g ) ij | G | g Miraculously, these form an orthogonal basis for : C [ G ] � d 2 σ = | G | σ ∈ b G
Group Actions Given a state in and a group element g , C [ G ] we can apply various group actions: � � g − 1 x � g − 1 xg � � � | x � → | xg � or or We can think of as a representation of G C [ G ] under any of these actions . Under (left or right) multiplication, the regular σ ∈ � representation contains copies of each . G d σ
Levels of Measurement 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 name σ Strong sampling : name, row and column σ , i, j in a basis of our choice (some bases may be much more informative than others) Intermediate: strong, but with a random basis
Fourier Sampling is Optimal The mixed state over (left) cosets 1 � | cH � � cH | ρ = | G | c ∈ G 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!
Projections and Probabilities For each irrep , we have a projection operator σ 1 � H = σ ( h ) π σ | H | h ∈ H d σ | H | rk π σ H The probability we observe is σ | G | d 2 Compare with the Plancherel distribution σ | G | ( , the completely mixed state) H = { 1 }
Weak Sampling Fails χ σ ( g ) = tr σ ( g ) If , we have ( ) H = { 1 , m } � � 1 + χ σ ( m ) H = d σ rk π σ 2 d σ χ σ ( m ) /d σ In , is exponentially small, so the S n observed distribution is very close to Plancherel Weak sampling fails [Hallgren, Russell, Ta-Shma 2000] Random basis fails [Grigni, Schulman, Vazirani, Vazirani 2001] But, strong is stronger for some G... [MRRS 2004]
Now for Strong Sampling But what about a basis of our choice? Given , σ we observe a basis vector b with probability � π H b � 2 rk π H � π H b � 2 = 1 Here we have 2(1 + � b , m b � ) How much does vary with m ? � b , m b �
Controlling the Variance Expectation of an irrep over m ’s conjugates is σ Exp m σ ( m ) = χ σ ( m ) d σ Exp m � b , m b � = χ σ ( m ) so d σ To turn the second moment into a first moment, | � b , m b � | 2 = � b ⊗ b ∗ , m ( b ⊗ b ∗ ) �
Controlling the Variance Decompose into irreducibles: σ ⊗ σ ∗ σ ⊗ σ ∗ ∼ � a τ τ = τ ∈ b G Then Var m � π H b � 2 ≤ 1 χ τ ( m ) 2 � � � � Π σ ⊗ σ ∗ ( b ⊗ b ∗ ) � � 4 τ d τ � τ ∈ b G How much of lies in low-dimensional ? b ⊗ b ∗ τ
Strong Sampling Fails Using simple counting arguments, we show that almost all of lies in high-dimensional b ⊗ b ∗ σ ⊗ σ ∗ subspaces of . τ Since is exponentially small, the χ τ ( m ) /d τ observed distribution on for any basis is b exponentially close to uniform. No subexponential set of experiments on coset states can solve Graph Isomorphism. [M., Russell, Schulman 2005]
Entangled Measurements For any group, there exists a measurement on the tensor product of coset states ρ ⊗ · · · ⊗ ρ � �� � k with [Ettinger, Høyer, Knill 1999] k = poly(log | G | ) What can we prove about entangled measurements?
Bounds on Multiregister Sampling Weak sample each register, observing σ = σ 1 ⊗ · · · ⊗ σ k Given a subset I of the k registers, decompose that part of the tensor product: � � a I σ i ∼ τ τ = i ∈ I τ ∈ b G This group action multiplies these registers by g and leaves the others fixed.
Bounds on Multiregister Sampling Second moment: analogous to one register, consider . Given subsets I and J, define σ ⊗ σ ∗ χ τ ( m ) � 2 � E I,J ( b ) = � Π I,J � � ( b ⊗ b ∗ ) τ d τ τ ∈ b G For an arbitrary entangled basis, [M., Russell 2005] Var m � Π H b � 2 ≤ 1 � E I,J ( b ) 4 k I,J ⊆ [ k ]: I,J � = ∅
Bounds on Multiregister Sampling With some additional work, this general bound can be used to show that registers Ω ( n log n ) are necessary for [Hallgren, Rötteler, Sen; M., Russell] S n But what form might this measurement take? Note that each subset of the registers contributes some information...
Subset Sum and the Dihedral Group The HSP in the dihedral group reduces to D n random cases of Subset Sum [Regev 2002] 2 O ( √ log n ) Leads to a -time and -register algorithm [Kuperberg 2003] Subset Sum gives the optimal multiregister measurement [Bacon, Childs, van Dam 2005]
More Abstractly... If , there is a missing harmonic : H = { 1 , m } � π ( h ) = 0 h ∈ H Weak sampling gives random two-dimensional irreps ; think of these as integers . ± j σ j σ j ⊗ σ k ∼ Tensor products: = σ j + k ⊕ σ j − k σ 0 ∼ Find subset that gives . ⊕ π =
Subsets in General Suppose H has a missing harmonic . τ W I For each subset I , consider the subspace τ resulting from applying the group action to I . (In , this flips the integers j in this subset.) D n If the hidden subgroup is a conjugate of H , W I then the state is perpendicular to for all I . τ C [ G k ] How much of does this leave? What W I fraction is spanned by the ? τ
Independent Subspaces Say that two subspaces V, W of a space U are independent if, just as for random vectors in U , Exp v ∈ V � Π W v � 2 = dim W W dim U or equivalently V tr Π V Π W = tr Π V tr Π W dim U dim U dim U Being in V or W are “independent events.”
Each Subset Contributes I � = J W I W J For , and are independent. τ τ W τ = span I W I Therefore, is large: τ dim W τ 1 dim C [ G k ] ≥ 1 − 1 + 2 k / | G | If , probability of “some subset k ≥ log 2 | G | being in ” is if the hidden subgroup is ≥ 1 / 2 τ trivial, but is zero if it is a conjugate of H . [M., Russell 2005]
Find An Informative Subset! C [ G k ] Divide into subspaces; for each one, find a subset I for a large fraction of the completely W I σ 0 ∼ mixed state is in : e.g. in . D n ⊕ π = τ “Pretty Good Measurement” (i.e., Subset Sum for ) is optimal for Gel’fand pairs... [MR 2005] D n ...but it is not optimal for [Childs] . What is? S n And, is it related to Subset Something?
The Hunt Continues vs. Beauty and Truth The Adversary
Acknowledgments
Recommend
More recommend
