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Quantum Certificate Complexity Scott Aaronson UC Berkeley

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0-1-NP C - #L - #L/poly - #P - #W[t] - +EXP - +L - +L/poly - +P - +SAC 1 - A 0 PP - AC - AC 0 - AC 0 [m] - ACC 0 - 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 - BP H SPACE(f(n)) - BPL - BPP KT - BPP-OBDD - BPP path - 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 - C k P - CNP - coAM - coC=P - cofrIP - Coh - coMA - coMod k P - compIP - compNP -coNE - coNEXP - coNL - coNP - coNP/poly - coRE - coRNC - coRP - coUCC - CP - CSIZE(f(n)) - CSL - CZK - D#P - Δ 2 P - δ-BPP - δ-RP - DET - DisNP - DistNP - DP - DSPACE(f(n)) - DTIME(f(n)) - Dyn-FO - Dyn-ThC 0 - E - EE - EEE - EESPACE - EEXP - EH - ELEMENTARY - EL k P - EPTAS - k- EQBP - EQP - EQTIME(f(n)) - ESPACE - EXP - EXP/poly - EXPSPACE - Few - FewP - FNL - FNL/poly - FNP - FO(t(n)) - FOLL – FP NP[log] - FPR - FPRAS - FPT - FPT nu - FPT su - FPTAS - FQMA - frIP - F-TAPE(f(n)) - F- TIME(f(n)) - GapL - GapP - GC(s(n),C) - GPCD(r(n),q(n)) - G[t] - H k P - HVSZK - IC[log,poly] - IP - L - LIN - L k P - LOGCFL - LogFew - LogFewNL - LOGNP - LOGSNP - L/poly - LWPP - MA - MA - MAC 0 - MA-E - MA-EXP - mAL - MaxNP - MaxPB - MaxSNP - MaxSNP 0 - mcoNL - MinPB - MIP - MIP EXP - (M k )P - mL - mNC 1 - mNL - mNP - Mod k L - Mod k P - ModP - ModZ k L - mP - MP - MPC - mP/poly - mTC 0 - NC - NC 0 - NC 1 - NC 2 - NE - NEE - NEEE - NEEXP - NEXP - NIQSZK - NISZK - NISZK h - NL - NLIN - NLOG - NL/poly - NPC - NP C - NPI - NP intersect coNP - (NP intersect coNP)/poly - NPMV - NPMV-sel - NPMV t - NPMV t -sel - NPO - NPOPB - NP/poly - (NP,P-samplable) - NP R - NPSPACE - NPSV - NPSV-sel - NPSV t - NPSV t -sel - NQP - NSPACE(f(n)) - NT - NTIME(f(n)) - OCQ - OptP - P #P - P #P[1] - PBP - k-PBP - P C - PCD(r(n),q(n)) - P-close - PCP(r(n),q(n)) - PEXP - PF - PFCHK(t(n)) - Φ 2 P - PhP - Π 2 P - P K - PKC - PL - PL 1 - PL infinity - PLF - PLL - P/log - P NP - P NP[k] - P NP[log] - P-OBDD - PODN - polyL - PP - PPA - PPAD - PPADS - P/poly - P PP - PQUERY - PR - P R - Pr H SPACE(f(n)) - PromiseBPP - PromiseBQP - PromiseP - PromiseRP - PrSPACE(f(n)) - P-Sel - PSK - PT 1 - PTAPE - PTAS - PT/WK(f(n),g(n)) - PZK - QAC 0 - QAC 0 [m] - QACC 0 - QAM - QCFL - QH - QIP - QIP(2) - QMA+ - QMA(2) - QMA log - QMAM - QMIP - QMIP le - QMIP ne - QNC 0 - QNC f 0 - QNC 1 - QP - QPSPACE - QSZK - R - RE - REG - RevSPACE(f(n)) - R H L - RL - RNC - RPP - RSPACE(f(n)) - S 2 P - SAC - SAC 0 - SAC 1 - SBP - SC - SEH - SF k - Σ 2 P - SKC - SL - SLICEWISE PSPACE - SNP - SO-E - SP - span-P - SPARSE - SPL - SPP - SUBEXP - symP - SZK - SZK h - TALLY - TC 0 - Θ 2 P - TREE-REGULAR - UCC - UL - UL/poly - UP - US - VNC k - VNP k - VP k - VQP k - W[1] - WAPP - W[P] - WPP - W[SAT] - W[*] - W[t] - W * [t] - XP - XP uniform - YACC - ZPE - ZPP SHAMELESS PLUG http://www.cs.berkeley.edu/~aaronson/zoo.html

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Overview Most of whats 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 Ill discuss today: Well need both big quantum lower bound methods (adversary method and polynomial method)

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f:{0,1} n {0,1} is a total Boolean function D(f)(deterministic query complexity) R 0 (f)(zero-error randomized) R 2 (f)(bounded-error randomized) Q 2 (f)(bounded-error quantum) Q 0 (f)(zero-error quantum) Q E (f)(exact quantum) Background

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Example

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Certificate Complexity C X (f) = min # of queries needed to distinguish X from every Y s.t. f(Y) f(X) Block Sensitivity bs X (f) = max # of disjoint blocks B {x 1,…,x n } s.t. flipping B changes f(X) Example: For f=MAJ(x 1,x 2,x 3,x 4,x 5 ), letting X=11110, 11110 11110 C X (MAJ)=3bs X (MAJ)=2 C(f) and bs(f)

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Randomized Certificate Complexity RC X (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(x 1,…,x n ), letting X=00…0, RC X (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)

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Let D 0,D 1 be distributions over f -1 (0), f -1 (1) s.t. D 0 looks locally similar to every 1-input, and D 1 looks locally similar to every 0-input: Then Ambainis Adversary Method (special case)

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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, x i y i w.p. O(1/RC(f)) Adversary method then yields For upper bound, use weighted Grover

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g g Example where C(f) = (QC(f) 2.205 )

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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(Q 2 (f) 2 Q 0 (f) 2 ) (de Wolf), D(f) = O(Q 2 (f) 6 ) (Beals et al.)

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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 monomialsi.e. maxonomials of x 1 x 2 – x 2 + 2x 3 are x 1 x 2, 2x 3 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

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Lemma: O(ndeg(f) log n) iterations suffice w.h.p. Proof: Let S be current set of monomials, and Initially (S) n ndeg(f) ndeg(f)! Were 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

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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 theres a square frame of size n 1/3 n 1/3, f(X)=0 otherwise

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Open Problems Is If so we get ~

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