All Quantum Adversaries Are Equivalent Robert Špalek joint work with Mario Szegedy
Quantum query complexity � Want to compute Boolean function f � Input queried by oracle calls O x | i , b , z � = | i , b ⊕ x i , z � Allow arbitrary unitary operations between � Length of computation t is the number of oracle calls Final state | ϕ t x � = U t O x U t − 1 . . . U 1 O x U 0 | 0 � Measure the leftmost qubit | q x � of | ϕ t x � to get the outcome ⇒ Pr [ q x = f ( x )] ≥ 2 Bounded-error ⇐ 3 � quantum query complexity Q 2 ( f ) is the minimal length of computation of a bounded-error algorithm 2
Adversary lower bounds � [Bennett, Bernstein, Brassard & Vazirani, 1997] Hybrid method • computation starts at a fixed state | ϕ 0 x � = | ϕ 0 y � • inner product � ϕ k x | ϕ k y � changes little after one query • output states | ϕ t x � and | ϕ t y � almost orthogonal if f ( x ) � = f ( y ) = ⇒ number of queries must be big � [Ambainis, 2000] Quantum adversary • examine average over many input pairs 3
Example lower bound for parity [Ambainis, 2000] Unweighted quantum adversary � Let A = f − 1 ( 0 ) and B = f − 1 ( 1 ) . Pick R ⊆ A × B . � Compute m = min x ∈ A |{ y : ( x , y ) ∈ R }| , m ′ = min y ∈ B |{ x : ( x , y ) ∈ R }| , ℓ = max x ∈ A , i ∈ [ n ] |{ y : ( x , y ) ∈ R & x i � = y i }| , ℓ ′ = max y ∈ B , i ∈ [ n ] |{ x : ( x , y ) ∈ R & x i � = y i }| . � mm ′ � Then Q 2 ( f ) = Ω ( ℓℓ ′ ) For parity: � R = { ( x , y ) : | x | = n 2 , | y | = n 2 + 1, | y − x | = 1 } 2 , m ′ = n 2 + 1 , ℓ = ℓ ′ = 1 . Hence Q 2 ( parity ) = Ω ( n ) � m = n 4
Weighted adversary lower bounds � [Høyer, Neerbek & Shi, 2001] • used spectral norm of weighted adversary matrix • specialized for binary search and sorting � [Barnum, Saks & Szegedy, 2003] Spectral method • general bound in terms of spectral norms • one weighted adversary matrix � [Ambainis, 2003] Weighted quantum adversary • weight scheme: n + 1 adversary matrices 5
Dual adversary lower bounds � [Laplante & Magniez, 2003] Kolmogorov complexity bound • general lower bound in terms of K ( x | y ) [conditional prefix-free Kolmogorov complexity K ( x | y ) is the length of the shortest program P taken from a prefix-free set such that P ( y ) = x ] • subsumes all known adversary bounds � [Laplante & Magniez, 2003] “MiniMax” bound • combinatorial version of the Kolmogorov complexity bound 6
Our results � Equality of bounds: [BSS03] • spectral [Ambainis, 2003] • weighted primal [Zhang, 2004] • “strong” weighted [LM03] � • Kolmogorov dual [LM03] • MiniMax � Limitations of the method: � � • min ( C 0 ( f ) n , C 1 ( f ) n ) for partial f � • C 0 ( f ) C 1 ( f ) for total f Some of them were known for some of the methods. 7
Inclusion of adversary lower bounds hybrid unweighted primal dual [we] HNS01 [LM03] weighted [LM03] spectral strong weighted Kolmogorov MiniMax [we] 8
Primal versus dual bounds λ ( Γ ) � [BSS03] Spectral Adversary SA ( f ) = max max i λ ( Γ i ) Γ Γ ≥ 0 symmetric with Γ [ x , y ] = 0 when f ( x ) = f ( y ) Γ i [ x , y ] = Γ [ x , y ] when x i � = y i , otherwise 0 λ ( Γ ) spectral norm of Γ 1 � [LM03] MiniMax MM ( f ) = min max � p x x , y p x ( i ) p y ( i ) ∑ i : x i � = y i f ( x ) � = f ( y ) p x probability distribution on n bits � [our paper] SA ( f ) = MM ( f ) • follows from duality in semidefinite programming • two non-trivial transformations needed 9
Reduce MiniMax to spectral 1/2 � � f ( x ) � = f ( y ) ∑ 1. MM ( f ) = 1/ µ max = 1 max min p x ( i ) p y ( i ) p x x , y i : x i � = y i � 2. Define R i [ x , y ] = p x ( i ) p y ( i ) and rewrite it as maximize µ subject to ∀ i : R i is non-negative symmetric rank-1 , ∑ i R i ◦ I = I , ∑ i R i ◦ D i ≥ µ F . 3. Relax into ∀ i : R i � 0 . The best solution actually is rank-1. 10
Reduce MiniMax to spectral 2/2 4. By duality of semidefinite programming, µ max = µ min minimize µ = Tr ∆ maximize µ subject to ∆ is diagonal subject to ( ∀ i ) R i � 0, ⇐ ⇒ Z ≥ 0 ∑ i R i ◦ I = I , Z · F = 1 ∑ i R i ◦ D i ≥ µ F . ( ∀ i ) ∆ − Z ◦ D i � 0 5. (Simplified) With a little calculation, w.l.o.g. ∆ = I and maximize Z · F subject to Z ≥ 0 ( ∀ i ) I − Z ◦ D i � 0 which is exactly the spectral bound . 11
Tight bounds on spectral norm � [Mathias, 1990] λ ( Γ ) ≤ max r x ( M ) c y ( N ) x , y Γ [ x , y ] > 0 • Γ [ x , y ] = M [ x , y ] · N [ x , y ] symmetric, M , N ≥ 0 r x ( M ) the x -th row norm, c y ( N ) the y -th column norm • The bound is tight, i.e. there always exist M , N s.t. equality is reached. [our paper] We add conditioning on Γ [ x , y ] > 0 , which was not there � On the other hand, λ ( Γ ) ≥ δ T Γ δ for every | δ | = 1 � [our paper] (Strong) weighted adversary is the spectral adversary with bounds on λ ( Γ ) and λ ( Γ i ) expanded using the inequalities above. 12
Spectral versus (strong) weighted adversary λ ( Γ ) [BSS03] Spectral Adversary SA ( f ) = max max i λ ( Γ i ) Γ [Amb03, Zha04] Strong Weighted Adversary w like Γ , w i ≥ 0 with w i [ x , y ] = 0 when f ( x ) = f ( y ) or x i = y i and w i [ x , y ] w i [ y , x ] ≥ w [ x , y ] 2 for x i � = y i ∑ y ∗ w [ x , y ∗ ] ∑ x ∗ w [ y , x ∗ ] � SWA ( f ) = max min ∑ y ∗ w i [ x , y ∗ ] ∑ x ∗ w i [ y , x ∗ ] w , w i x , y , i w [ x , y ] > 0, xi � = yi � Γ → w : w [ x , y ] : = Γ [ x , y ] δ [ x ] δ [ y ] for δ = principal eigen-vector of Γ w [ x , y ] wt ( x ) wt ( y ) for wt ( x ) = ∑ y ∗ w [ x , y ∗ ] √ � w → Γ : Γ [ x , y ] : = 13
Limitation of all adversary methods Easy to prove in the dual formulation! Let f be total. � � f ( x ) � = f ( y ) ∑ � MM ( f ) = 1 max min p x ( i ) p y ( i ) p x x , y i : x i � = y i � Let C f ( x ) be some minimal certificate for f ( x ) . Define p x ( i ) = 1/ |C f ( x ) | if i ∈ C f ( x ) , otherwise 0. � For every f ( x ) � = f ( y ) , there is j ∈ C f ( x ) ∩ C f ( y ) with x j � = y j 1 1 � � ∑ p x ( i ) p y ( i ) ≥ p x ( j ) p y ( j ) = ≥ � � C 0 ( f ) C 1 ( f ) C f ( x ) C f ( y ) i : x i � = y i � Hence MM ( f ) ≤ C 0 ( f ) C 1 ( f ) . 14
Consequences of the limitation Cannot prove good lower bounds on problems with small certificates: � element distinctness: C 0 = 2 , C 1 = n , hence limited by O ( √ n ) tight bound Θ ( n 2/3 ) proved by the polynomial method [AS04] � triangle finding: C 0 = n 2 , C 1 = 3 , hence limited by O ( n ) � verification of matrix multiplication: C 0 = 2 n , C 1 = n 2 , limited by O ( n 3/2 ) � binary And-Or trees: C 0 = C 1 = √ n , hence limited by O ( √ n ) The complexities of the last 3 problems are open. 15
Conclusion � Linear algebraic proof of equivalence of: • spectral • weighted • strong weighted � Using semidefinite programming, equivalence with MiniMax � With [LM03] , Kolmogorov bound also fits there � Simple proof of limitations of all bounds 16
Proof of spectral adversary � Decompose the quantum state | ϕ x � = ∑ i | i �| ϕ x , i � . Then � ϕ x | ϕ y � = ∑ i � ϕ x , i | ϕ y , i � . � After one query | ϕ ′ x � = ∑ i ( − 1 ) x i | i �| ϕ x , i � . Then � ϕ ′ x | ϕ ′ y � = ∑ i ( − 1 ) x i + y i � ϕ x , i | ϕ y , i � . Hence � ϕ ′ x | ϕ ′ y � − � ϕ x | ϕ y � = 2 ∑ i : x i � = y i � ϕ x , i | ϕ y , i � . � Define progress function Ψ t = ∑ x , y Γ [ x , y ] δ x δ y · � ϕ t x | ϕ t y � , where δ is the principial eigen-vector of Γ with | δ | = 1 . � Ψ 0 = ∑ x , y Γ [ x , y ] δ x δ y · 1 = λ ( Γ ) , Ψ T is constant times smaller. λ ( Γ ) But Ψ t + 1 − Ψ t ≤ max i λ ( Γ i ) , hence T ≥ max i λ ( Γ i ) . 17
Recall Ψ t = ∑ x , y Γ [ x , y ] δ x δ y · � ϕ t x | ϕ t y � and � ϕ t + 1 | ϕ t + 1 � − � ϕ t x | ϕ t y � = 2 ∑ i : x i � = y i � ϕ x , i | ϕ y , i � . x y Define column vector a i [ x ] = δ x | ϕ x , i | Ψ t + 1 − Ψ t 2 ∑ x , y ∑ Γ [ x , y ] δ x δ y � ϕ x , i | ϕ y , i � = i : x i � = y i 2 ∑ x , y ∑ ≤ Γ i [ x , y ] δ x δ y · | ϕ x , i | · | ϕ y , i | i a T λ ( Γ i ) | a i | 2 2 ∑ i Γ i a i ≤ 2 ∑ = i i | a i | 2 = 2 max δ 2 x | ϕ x , i | 2 λ ( Γ i ) ∑ λ ( Γ i ) ∑ i ∑ ≤ 2 max i i x i | ϕ x , i | 2 = 2 max δ 2 λ ( Γ i ) ∑ x ∑ = λ ( Γ i ) 2 max i i x i
Recommend
More recommend
