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Quantum Certificate Complexity

Quantum Certificate Complexity. Scott Aaronson UC Berkeley.

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Quantum Certificate Complexity

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  1. Quantum Certificate Complexity Scott Aaronson UC Berkeley

  2. 0-1-NPC - #L - #L/poly - #P - #W[t] - +EXP - +L - +L/poly - +P - +SAC1 - A0PP - AC - AC0 - AC0[m] - ACC0 - AH - AL – AlgP/poly - AM - AM intersect coAM - AmpMP - AP - AP - APP - APP - APX - AVBPP - AvE - AvP - AW[P] - AWPP - AW[SAT] - AW[*] - AW[t] - βP - BH - BPE - BPEE - BPHSPACE(f(n)) - BPL - BPPKT - BPP-OBDD - BPPpath - BPQP - BPSPACE(f(n)) - BPTIME(f(n)) - BQNC - BQNP - BQP-OBDD - BQP/log - BQP/qlog - BQTIME(f(n)) - k-BWBP - C=L - C=P - CFL - CLOG - CH - Check - CkP - CNP - coAM - coC=P - cofrIP - Coh - coMA - coModkP - compIP - compNP -coNE - coNEXP - coNL - coNP - coNP/poly - coRE - coRNC - coRP - coUCC - CP - CSIZE(f(n)) - CSL - CZK - D#P - Δ2P - δ-BPP - δ-RP - DET - DisNP - DistNP - DP - DSPACE(f(n)) - DTIME(f(n)) - Dyn-FO - Dyn-ThC0 - E - EE - EEE - EESPACE - EEXP - EH - ELEMENTARY - ELkP - EPTAS - k-EQBP - EQP - EQTIME(f(n)) - ESPACE - EXP - EXP/poly - EXPSPACE - Few - FewP - FNL - FNL/poly - FNP - FO(t(n)) - FOLL – FPNP[log] - FPR - FPRAS - FPT - FPTnu - FPTsu - FPTAS - FQMA - frIP - F-TAPE(f(n)) - F-TIME(f(n)) - GapL - GapP - GC(s(n),C) - GPCD(r(n),q(n)) - G[t] - HkP - HVSZK - IC[log,poly] - IP - L - LIN - LkP - LOGCFL - LogFew - LogFewNL - LOGNP - LOGSNP - L/poly - LWPP - MA - MA’ - MAC0 - MA-E - MA-EXP - mAL - MaxNP - MaxPB - MaxSNP - MaxSNP0 - mcoNL - MinPB - MIP - MIPEXP - (Mk)P - mL - mNC1 - mNL - mNP - ModkL - ModkP - ModP - ModZkL - mP - MP - MPC - mP/poly - mTC0 - NC - NC0 - NC1 - NC2 - NE - NEE - NEEE - NEEXP - NEXP - NIQSZK - NISZK - NISZKh - NL - NLIN - NLOG - NL/poly - NPC - NPC - NPI - NP intersect coNP - (NP intersect coNP)/poly - NPMV - NPMV-sel - NPMVt - NPMVt-sel - NPO - NPOPB - NP/poly - (NP,P-samplable) - NPR - NPSPACE - NPSV - NPSV-sel - NPSVt - NPSVt-sel - NQP - NSPACE(f(n)) - NT - NTIME(f(n)) - OCQ - OptP - P#P - P#P[1] - PBP - k-PBP - PC - PCD(r(n),q(n)) - P-close - PCP(r(n),q(n)) - PEXP - PF - PFCHK(t(n)) - Φ2P - PhP - Π2P - PK - PKC - PL - PL1 - PLinfinity - PLF - PLL - P/log - PNP - PNP[k] - PNP[log] - P-OBDD - PODN - polyL - PP - PPA - PPAD - PPADS - P/poly - PPP - PQUERY - PR - PR - PrHSPACE(f(n)) - PromiseBPP - PromiseBQP - PromiseP - PromiseRP - PrSPACE(f(n)) - P-Sel - PSK - PT1 - PTAPE - PTAS - PT/WK(f(n),g(n)) - PZK - QAC0 - QAC0[m] - QACC0 - QAM - QCFL - QH - QIP - QIP(2) - QMA+ - QMA(2) - QMAlog - QMAM - QMIP - QMIPle - QMIPne - QNC0 - QNCf0 - QNC1 - QP - QPSPACE - QSZK - R - RE - REG - RevSPACE(f(n)) - RHL - RL - RNC - RPP - RSPACE(f(n)) - S2P - SAC - SAC0 - SAC1 - SBP - SC - SEH - SFk - Σ2P - SKC - SL - SLICEWISE PSPACE - SNP - SO-E - SP - span-P - SPARSE - SPL - SPP - SUBEXP - symP - SZK - SZKh - TALLY - TC0 - Θ2P - TREE-REGULAR - UCC - UL - UL/poly - UP - US - VNCk - VNPk - VPk - VQPk - W[1] - WAPP - W[P] - WPP - W[SAT] - W[*] - W[t] - W*[t] - XP - XPuniform - YACC - ZPE - ZPP SHAMELESS PLUG http://www.cs.berkeley.edu/~aaronson/zoo.html

  3. Overview • Most of what’s known about quantum computing can be cast in the query complexity model • Despite its simplicity, open problems abound • We make progress on some of these by studying randomized certificate complexity RC(f) and quantum certificate complexity QC(f) • Main results I’ll discuss today: • We’ll need both big quantum lower bound methods (adversary method and polynomial method)

  4. Background • f:{0,1}n{0,1} is a total Boolean function • D(f) (deterministic query complexity) •  R0(f) (zero-error randomized) •  R2(f) (bounded-error randomized) • Q2(f) (bounded-error quantum)  Q0(f) (zero-error quantum) •  QE(f) (exact quantum)

  5. Example

  6. Certificate Complexity • CX(f) = min # of queries needed to distinguish X from every Y s.t. f(Y)f(X) • Block Sensitivity • bsX(f) = max # of disjoint blocks B{x1,…,xn} s.t. flipping B changes f(X) • Example: For f=MAJ(x1,x2,x3,x4,x5), letting X=11110, • 11110 11110 • CX(MAJ)=3 bsX(MAJ)=2 C(f) and bs(f)

  7. Randomized Certificate Complexity • RCX(f) = min # of randomized queries needed to distinguish X from any Y s.t. f(Y)f(X) with ½ prob. • Quantum Certificate Complexity QC(f) • Example: For f=MAJ(x1,…,xn), letting X=00…0,RCX(MAJ) = 1 • Observations: Anything a prover might provide a verifier besides X, the verifier can compute for itself • One-sided and two-sided error are equivalent • Different notions of nondeterministic quantum query complexity: Watrous 2000, de Wolf 2002 RC(f) and QC(f)

  8. Ambainis’ Adversary Method(special case) Let D0,D1 be distributions over f-1(0), f-1(1) s.t. D0 looks “locally similar” to every 1-input, and D1 looks “locally similar” to every 0-input: Then

  9. Claim: • Any randomized certificate for input X can be made nonadaptive with constant blowup • By minimax theorem, exists distribution over {Y:f(Y)f(X)} s.t. for all i, xiyi w.p. O(1/RC(f)) • Adversary method then yields • For upper bound, use “weighted Grover”

  10. g g Example where C(f) = (QC(f)2.205)

  11. New Quantum/Classical Relation For total f, where ndeg(f) = min degree of poly p s.t. p(X)0  f(X)=1 Previous: D(f) = O(Q2(f)2Q0(f)2) (de Wolf), D(f) = O(Q2(f)6) (Beals et al.)

  12. Idea (follows Buhrman-de Wolf): Let p be s.t. p(X)0  f(X)=1 Maxonomials of p are monomials not dominated by other monomials—i.e. maxonomials of x1x2 – x2 + 2x3 are x1x2, 2x3 Nisan-Smolensky: For every 0-input X and maxonomial M of p, X has a sensitive block whose variables are all in M Consequence: Randomized 0-certificate must intersect each maxonomial w.p.  ½ Randomized algorithm: Keep querying a randomized 0-certificate, until either one no longer exists or p=0

  13. Lemma: O(ndeg(f) log n) iterations suffice w.h.p. Proof: Let S be current set of monomials, and Initially (S)  nndeg(f) ndeg(f)! We’re done when (S)=0 Claim: Each iteration decreases (S) by expected amount (S)/4e Reason:  1/e of (S) is concentrated on maxonomials, each of which decreases in degree w.p.  ½

  14. Local Proofs • When faced with a hard problem, analyze limitations of known techniques (Baker-Gill-Solovay, Razborov-Rudich) • Is • I claim that a ‘yes’ answer would require “global analysis” of Boolean functions • Given nn lattice of bits X, let f(X)=1 if there’s a square ‘frame’ of size n1/3n1/3, f(X)=0 otherwise

  15. Open Problems ~ • Is • Is If so we get

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