1 / 24

Quantum Lower Bound for the Collision Problem

Quantum Lower Bound for the Collision Problem. I was born at the Big Bang. Cool! We have the same birthday. Scott Aaronson 1/10/2002 quant-ph/0111102. Collision Problem. Given . Promised: (1) X is one-to-one (permutation) or (2) X is two-to-one.

gracie
Télécharger la présentation

Quantum Lower Bound for the Collision Problem

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Quantum Lower Bound for the Collision Problem I was born at the Big Bang. Cool! We have the same birthday. Scott Aaronson 1/10/2002 quant-ph/0111102

  2. Collision Problem • Given • Promised: • (1) X is one-to-one (permutation) or • (2) X is two-to-one • Problem: Decide which w.h.p., using few queries to the xi • Randomized alg: (n)

  3. One-to-One Two-to-One

  4. Result • Any quantum algorithm for the collision problem uses (n1/5) queries • Shi improved to (n1/4) • (n1/3) when |range| >> n • Previously no lower bound better than (1)

  5. Implications • No polytime blackbox algorithms for • graph isomorphism • nonabelian hidden subgroup • breaking cryptographic hash functions

  6. Implications 2. “Dynamical quantum theories” can’t be simulated in BQP, relative to oracle Define joint distribution over values of observable at times t1, t2, etc. (I.e. classical history) Given polytime quantum algorithm and set of “sampling points,” how hard to sample from this distribution?

  7. Prepare and observe 2nd register • If X is 2-1, obtain (|i+|j)/2 with xi=xj How to Find a Collision in O(1) Queries If Your Memory Is Perfect • Sample 3. Hadamard every bit, and sample again 4. Hadamard every bit again (returning to (|i+|j)/2), and sample again Which basis state (|i or |j) were you “in” after Step 2? After Step 4?

  8. Implications 3. |x|f(x) oracles (Kashefi et al. 2001) more powerful than |x|x|f(x) Requires (n1/7) lower bound for set comparison problem: given sequences x1…xn and y1…yn, decide whether {x1,…,xn}={y1,…,yn} or |{x1,…,xn,y1,…,yn}|>1.1n Can improve to (n1/6) using ideas of Shi

  9. By end: Quantum Query Model • State after t queries: : workbits i: index to query z: output • Query: |,i,z |xi,i,z • Arbitrary unitaries that don’t depend on X

  10. Brassard-Høyer-Tapp (1998) (n1/3) quantum alg for collision problem Grover’s algorithm over n2/3 xi’s Do I collide with any of the pink xi’s? n1/3 xi’s, queried classically, sorted for fast lookup

  11. Lower Bound: Main Ideas • P(X)[0,1], even for g-1 inputs X with g>2. Surprisingly strong constraint. • Take uniform dist. over g-1 inputs • P becomes poly in g of deg  2T. Algebraic magic! • Use approximation theory to show T large

  12. Proof: Let t,X,,i,z = amplitude of |,i,z after t queries. t,X,,i,z is poly of degt, by induction. Base case (t=0) trivial. Unitaries can’t increase degree. Query replaces t,X,,i,z by Lemma (follows Beals et al. 1998): Let (xi,h)=1 if xi=h, 0 otherwise. Then P(X) is poly of deg  2T over the (xi,h).

  13. Let Input Distribution • D(g): Uniform distribution over g-1 inputs • Technicality: g might not divide n • But assume for simplicity that it does

  14. Let • Then for some I, Monomials of P(X) • Claim: If T=O(n) then P(g) is a polynomial of degree  2T in g for integers 1gn. • I(X) = product of r variables (xi,h)

  15. So • since Calculating (I,g): #1 • “Range” of I: Y. w=|Y|. • (I,g) = 0 unless YS (“range” of X)

  16. # of g-1 inputs X with range S s.t. I(X)=1: Calculating (I,g): #2 • Given an S containing Y, # of g-1 inputs of size n: n!/(g!)n/g • Let {y1,…,yw} be distinct values in Y • ri = # of times yi appears in Y • r1 + … + rw = r

  17. Polynomial in g of degree w + (r-w) = r  2T Becomes ~polynomial(g)

  18. Markov’s Inequality Let P(x) be a poly with b1P(x)b2 for all a1xa2 and |dP(x*)/dx|c for some a1x*a2. Then Large derivative Short Long

  19. Lower Bound • 0  P(g)  1 for all 0  g  n • P(1)  1/10 and P(2)  9/10 • So dP/dg  4/5 somewhere • (n1/4) lower bound would follow if g always divided n

  20. Acceptance prob. close to bivariate polynomial in g,N for all g|N s.t. How to Handle n mod g  0: Sketch • Choose N slightly larger than n such that g divides N • Choose g-1 function on {1,…,N} u.a.r, then subfunction of size n

  21. Lower bound obtained when G=n2/5: (continued) • Restrict g’s range to [1,G]; then (g,N) points with g|N are plentiful, so P is bounded • P has large derivative somewhere in either the g or N directions

  22. Large derivative between 1-1 and 2-1 Lots of points at which g|N so P is bounded

  23. Shi’s Improvement to (n1/4) • Choose Nn s.t. g divides N, instead of Nn • If basis state | queries an undefined xi, | “drops out of the universe” • Result: Final state vector has norm in [0,1] Still OK! • P(g,N) is exactly polynomial in (g,N); so g’s range need not be restricted to [1,n2/5]

  24. Uses Paturi’s inequality: • if 0p(x)1 for 0xn and p’()=(1) Shi’s Improvement to (n1/3) • For functions with range {1,…,3n/2}

More Related