Quantum algorithms are at most polynomially faster for any symmetric function Andris Ambainis University of Latvia

When does a problem have exponential speedups? What properties are necessary/sufficient?

Simon’s problem Black box F:{0, 1} n  {0, 1} n. Hidden y  {0, 1} n. F(x) = F(z) if and only if z = x  y. Goal: find y. Classical: O(2 n/2 ) evaluations of f. Quantum: O(n) evaluations of f.

Decision version of Simon’s problem Black box F:{0, 1} n  {0, 1} n. One of two cases: –F(x) = F(z) if and only if z = x  y. –F(x) are all distinct. Determine which of two cases we have. Classical: O(2 n/2 ) evaluations of f. Quantum: O(n) evaluations of f.

Collision problem Black box F:{0, 1} n  {0, 1} n. One of two cases: –For each x, there is unique z≠x: F(x) = F(z). –F(x) are all distinct. Determine which of two cases we have. Classical: O(2 n/2 ) evaluations of f. Quantum: O(2 n/3 ) evaluations of f.

Simon’s problem Simon’s problem: –F(x) = F(z) if and only if z = x  y. –F(x) are all distinct. Which of those two? Collision problem: –For each x, there is unique z≠x: F(x) = F(z). –F(x) are all distinct. Which of those two? Permuting F(x) transforms Simon’s into the collision problem.

Folk conjecture Conjecture If G – symmetric (e.g., the collision problem), then the quantum complexity of G at most polynomially smaller than the classical complexity of G.

Terminology F:{1,..., N}  {1,..., M}. Property G(F)=G(F(1),..., F(N)). Define x i = F(i). G(x 1, x 2,..., x N )– function of N variables x 1, x 2,..., x N  {1,..., M}. Evaluations F(i) = queries about x i.

Two types of symmetries Function G is symmetric if G(x 1,..., X N ) = G(xx G(x 1,..., X N ) = G(x  (1),..., x  (N) ) = G(  (x 1 ),...,  (x N )). Q  i  (i) x  (i)  Q ixixi  (x i ) Q ixixi

Main result Theorem If G has both types of symmetries, R 2 (G)=O*(Q 2 11 (G)), where –R 2 (G) is the number of queries to compute G by a probabilistic algorithm; –Q 2 (G) is the number of queries to compute G by a probabilistic algorithm. * some log factors are omitted.

Previous results [Beals et al., 1998] If f(x 1,..., x N ),x 1,..., x N  {0, 1} is total (defined for all x 1,..., x N ), then D(f)=O(Q 2 6 (f)). Incomparable to our result: –[Beals et al.]: total, possibly non-symmetric. –Our result: symmetric, possibly not total.

Proof of main result Theorem R 2 (G)=O*(Q 2 11 (G)). For simplicity, prove R 2 (G)=O*(Q 2 14 (G)). 1. Probabilistic strategy with T 2 queries; 2. If the probabilistic strategy fails, we deduce a quantum lower bound of Ω(T 1/7 ).

Input types Since G(x 1,..., x N ) is symmetric, it only depends on:... the number of i:x i =1 the number of i:x i =2 the number of i:x i =3 y1y1 y2y2 y3y3... Input type

Random sampling Input type (y 1,..., y M ). Query x i for T 2 randomly chosen indices i. Estimate y 1,..., y M based on results. With high probability, the estimates y’ i are within O(N/T) of the actual y i.

Two cases Estimates (y’ 1,... y’ M ). Assume all y’ i are within O(N/T) of true y i. Case 1 This assumption uniquely determines G(y 1,..., y M ). Case 2 We have G(y 1,..., y M )=0, G(y 1 ’’,..., y M ’’)=1, both consistent with the assumption.

Distinguishing problem y1y1 y2y2 y3y3... y’’ 1 y’’ 2 y’’ 3... y i and y’’ i differ by at most O(N/T). G=0G=1 Claim Distinguishing between these two types requires Ω(T 1/7 ) queries.

Case 1 y1y1 y2y2 y3y3... y’’ 1 y’’ 2 y’’ 3... G=0G=1 Total difference: If D small, the types are hard to distinguish.

Case 1 Theorem Ifis O(N/ε), then Ω(1/√ ε) queries are needed to distinguish between y i and y’’ i. Proof: quantum adversary [A, 2000].

Case 2 y1y1 y2y2 y3y3... y’’ 1 y’’ 2 y’’ 3... G=0G=1 ε. more than N/ ε. Each difference at most O(N/T). ε different rows. At least T/ε different rows.

Simplifying assumption y1y1 y2y2 y3y3... y’’ 1 y’’ 2 y’’ 3... G=0G=1. Every two rows differ by 0 or k. Lemma Under this assumption, Ω(m 1/5 ) queries are needed, m – number of. Different rows.

Intermediate input type y1y1 y2y2 y3y3... y’’ 1 y’’ 2 y’’ 3... G=0 G=1 z1z1 z2z2 z3z Require Ω(m 1/5 ) queries to distinguish. Require Ω(m 1/5 ) queries to distinguish.

SImplifying.... y1y1 y2y2 y3y3... G=0 z1z1 z2z2 z3z For simplicity, remove all equal rows.

Simplifying... ymym... zmzm 0 0 y1y1 z1z1 Distinguishing these requires Ω(m 1/5 ) queries!

Set equality Variables x 1,..., x m, x’ 1,..., x’ m. Promise: {x 1,..., x m } and {x’ 1,..., x’ m } are either equal or disjoint. Task: distinguish the two cases. Theorem [Midrijānis, 2004] Set equality requires Theorem [Midrijānis, 2004] Set equality requires Ω(m 1/5 ) queries.

Set equality ≤ Distinguishing ymym... zmzm 0 0 y1y1 z1z1 Replicate:  x 1,... x m - z 1,..., z m times;  x’ 1,... x’ m - k times each. Are {x 1,... X m } and {x’ 1,... x’ m } equal or disjoint?

Putting it all together If m ≥ T 5/7, then the lower bound of Ω(m 1/5 ) = Ω(T 1/7 ) Ω(m 1/5 ) = Ω(T 1/7 ) follows from set equality. Otherwise, each row differs by O(N/T). The total difference at most By the other argument, Ω(T 1/7 ) queries needed.

More information? A. Ambainis, arxiv preprint 0902.xxxx.

Open problems 1. Improve the exponent 11 in R 2 (G)=O*(Q 2 11 (G)). 2. Prove better lower bound for set equality (the best algorithm is O(m 1/3 ). 3. Prove R 2 (G)=O(Q 2 c (G)) for functions G(x 1,..., x N ) that are only symmetric w. r. t. permuting x 1,..., x N.

SImplifying the problem... Y Y’’ Y1Y1 Y2Y2 Y3Y3