New quantum lower bound method with applications to

New quantum lower bound method, with applications to direct product theorems Andris Ambainis, U. Waterloo Robert Spalek, CWI, Amsterdam Ronald de Wolf, CWI, Amsterdam

Query model i 0 n n 0 1 0. . . 0 x 1 x 2 x 3 x. N i xi Input x 1, …, x. N accessed by queries. Complexity = the number of queries.

Grover's search 0 1 0. . . 0 x 1 x 2 x 3 x. N n n n Is there i such that xi=1? Queries: ask i, get xi. Classically, N queries required. Quantum: O( N) queries [Grover, 1996]. Speeds up any search problem.

Quantum counting [Boyer et al. , 1998] 0 1 0. . . 0 x 1 x 2 x 3 x. N n Is the fraction of i: xi=1 more than ½+ or less than ½- ? n Classical: queries. n Quantum: queries.

Element distinctness 3 1 17. . . 5 x 1 x 2 x 3 x. N n Are there i, j such that i j but xi=xj? Classically: N queries. n Quantum: O(N 2/3). n

Lower bounds n Search requires N) queries [Bennett et al. , 1997]. Counting: 1/ ) [Nayak, Wu, 1999]. n Element distinctness: (N 2/3) [Shi, n 2002].

Lower bound methods n n Adversary: analyze algorithm, prove it is incorrect on some input. Polynomials: describe algorithm by low degree polynomial.

Which method? Problem Adversary Polynomials Search Yes Counting Yes Element dist. ? Yes

Limits of adversary method n n Certificate for f on input (x 1, x 2, …, x. N): set of variables xi which determine f(x 1, x 2, …, x. N). Search: is there i: xi=1? 0 1 0. . . 0 x 1 x 2 x 3 x. N

Limits of adversary method n n Certificate for f on input (x 1, x 2, …, x. N): set of variables xi which determine f(x 1, x 2, …, x. N). Search: is there i: xi=1? 0 0 0. . . 0 x 1 x 2 x 3 x. N

Certificate complexity n Cx(f): the size of the smallest certificate for f on the input x. § Search: C 0=N, C 1=1.

Limits of adversary method Theorem [Spalek, Szegedy, 2004] n Any quantum adversary lower bound is at most

Example: element distinctness 3 1 17. . . 5 x 1 x 2 x 3 x. N n n Are there i, j: xi= xj? 1 -certificate: {i, j}, xi= xj. Adversary bound: Actual complexity: O(N 2/3).

Example: triangle finding n Graph G, specified by N 2 variables xij: xij=1, if there is edge between i and j. Does G contain a triangle? 1 -certificate: {ij, jk, ik}, xij= xik= xjk=1. Adversary lower bound: at most n The best algorithm: O(N 1. 3) [MSS 03]. n n n

Previous adversary method

Quantum query model U 0 n n Q U 1 … Q UT Fixed starting state. U 0, U 1, …, UT – independent of x 1, x 2, …, x. N. Q – queries. Measuring final state gives the result.

Queries n Basis states for algorithm’s workspace: |i, z , i {1, 2, …, N}. Query transformation: n Example: n n n |i, z , if xi=0; |i, z -|i, z , if xi=1;

Adversary framework Quantum algorithm A |x 1 x 2 … x. N Two registers: HA, HI. n. Query Q: n

Example: Grover search n Start state: | start | 0 , n End state

Density matrices n Measure HA, look at density matrix of HI

Density matrices n n n Sum of off-diagonal entries. N(N-1) entries. Sum for starting state: Sum for end state: 0. Query changes the sym by at most 2 N. ( N) queries needed.

Limits of this approach n n ( end)x, y measures the possibility of distinguishing x from y. If every ( end)x, y small, we can, given x, y: f(x) f(y), distinguish x from y.

Limits of this approach n It might be that: Every x can be Adversary method fails distinguished from every y; n There is no measurement that quantum algorithm distinguishes all x from all y. n f(x)=0 f(y)=1

New method

K-fold search 0 1 0. . . 0 x 1 x 2 x 3 x. N n n n K items i: xi=1, find all of them. O( NK) queries: O( N/K) for each item. This is optimal.

Direct product theorem n n n Theorem [KSW 04] Solving K-fold search with success probability c-K, c>1 requires NK queries. Easy to prove for success probability c. Difficult for probability c-K. Why is this useful? ?

Application: sorting n Theorem [KSW 04] A quantum algorithm for sorting x 1, x 2, …, x. N with S qubits of workspace must use queries.

Proof n n n Divide algorithm into stages: first K items sorted, next K items sorted, … Suffices to show each stage requires ( NK) queries. Each stage reduces to K-fold search.

Proof n n At the beginning of ith stage, we get S qubits from the previous stage. Theorem K-fold search requires ( NK) queries, even if we allow K/C qubits of advice.

Proof n n n Theorem K-fold search requires ( NK) queries, even if we allow K/C qubits of advice. Proof Replace advice by completely mixed state. Success probability p with advice => Success probability p 2 -K/C, no advice.

Direct product theorem n n n Theorem Solving K-fold search with success probability c-K, c>1 requires NK queries. [KSW 04]: proof by polynomials method. This talk: (new) adversary method.

Proof sketch n n “Know-0”, “Know-1”, …, “Know-k” states. Describe quantum state as

Proof n Adversary framework Quantum algorithm A n |x 1 x 2 … x. N Start state for input:

Proof n State of HI if we know n Subspace Tj spanned by all

Proof n n T 0 T 1 …. TK n T 0 T 1 … TK. T 0 – starting state. TK – entire HI. Tj – “know at-most j” subspace

Proof n T 0 T 1 … TK Sj=Tj (Tj-1).

Proof n T 0 S 1 … Sj=Tj (Tj-1). SK Sj is “know-j” subspace.

Proof n n | - state of algorithm including the input register |x 1 … x. N. | j belongs to HA Sj. Probability of “know-j”:

Proof n n Start state: p 0=1, p 1=…=p. K=0. Change in one query: After NK queries, p. K/2+1, …, p. K are exponentially small. Success probability exponentially small.

Threshold functions 0 1 0. . . 0 x 1 x 2 x 3 x. N n n n F(x 1, x 2, …, x. N)=1 if xi=1 for at least t values i {1, 2, …, N}. F(x 1, x 2, …, x. N)=0 if xi=1 for at most t-1 values i {1, 2, …, N}. Query complexity: ( Nt).

Threshold functions 0 1 0. . . 0 x 1 x 2 x 3 x. N n n n F(x 1, x 2, …, x. N)=1 if xi=1 for at least t values i {1, 2, …, N}. F(x 1, x 2, …, x. N)=0 if xi=1 for at most t-1 values i {1, 2, …, N}. Query complexity: ( Nt).

Threshold functions n n n K instances of threshold function. (K Nt) queries. Theorem Solving all K instances with probability at most c-K requires K Nt queries.

Proof Algorithm n n … K input registers. Each input register initially , | 0 , | 1 - uniform over |x 1 … x. N with t-1 and t values i: xi=1.

Proof n n n For each instance, states “solved”, “know-0”, “know-1”, … “know-(t-1)”. For K instances, vector of K states. Progress of a state: n n n “solved” – progress t/2. “know-t/2”, … “know-(t-1)” – progress t/2. “know-j”, j

Proof n If progress of final state less than t. K/4, the probability of getting all K correct answers is c-K. Decompose current state n Potential function n

Proof n n n Start state: P( )=1. For pj, j t. K/4 to be more than c-K, One query increases P( ) by at most a factor of

Proof n F(x 1, x 2, …, x. N)=0, “know-j”: n F(x 1, x 2, …, x. N)=1, “know-j”:

Proof n Starting state: n “Solved”: n “Know-j”

Application: testing linear inequalities n n aij known, xi, bj accessed by queries. Which inequalities are true?

Our result n n Memory limited to S (qu)bits. Classically: (N 2/S) queries. Quantum: (N 3/2 t 1/2/S 1/2) queries. Lower bound follows from threshold function lower bound.

Conclusion n New quantum lower bound method, by eigenspace analysis. Direct product theorems for K-fold search and threshold functions. Consequences for time-space tradeoffs.

More details n n A. Ambainis. A new quantum lower bound method, with application to direct product theorem for search, quant-ph/0508200. A. Ambainis, R. Spalek, R. de Wolf, Quantum direct product theorems for symmetric functions and time-space tradeoffs , quant-ph/0511200.

Open problems n AND OR OR n x 1 x 2 x 3 x 4 n AND-OR tree: best lower bound O( N), N – number of variables. Algorithm: O(N. 753). Adversary lower bound for element distinctness?