Quantum Certificate Complexity
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Transcript Quantum Certificate Complexity
Quantum Certificate Complexity
Scott Aaronson
UC Berkeley
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 - kEQBP - 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)) - FTIME(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
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:
QC f
RC f , R0 f O Q2 f Q0 f log n
2
• We’ll need both big quantum lower bound methods
(adversary method and polynomial method)
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)
Example
f OR x1,
D OR n
R0 OR n
1 2
R2 OR
n
1
, xn
QE OR n
Q0 OR n
Q2 OR
n
Certificate Complexity C f
C f
max
X
X
CX(f) = min # of queries needed to distinguish X
from every Y s.t. f(Y)f(X)
Block Sensitivity bs f
bs f
max
X
X
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
X
RC
f
max
RC
X
f
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
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:
X f 1 0 , i 1,..., n
Y f 1 1 , i 1,..., n
Then
1
Q2 f
.
Pr xi yi
Y D1
Pr xi yi .
X D0
Claim: QC f
RC f
• 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, xiyi w.p. O(1/RC(f))
• Adversary method then yields
RC f
• For upper bound, use “weighted Grover”
Example where C(f) = (QC(f)2.205)
g
g
k x1
x29
12
0 if k
g k 1 if k 13,14,15,16
0 if k
17
New Quantum/Classical Relation
For total f,
R0 f O RC f ndeg f log n
O Q2 f Q0 f log n
2
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.)
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
Lemma: O(ndeg(f) log n) iterations suffice w.h.p.
Proof: Let S be current set of monomials, and
S
deg M !
M S
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
Local Proofs
• When faced with a hard problem, analyze
limitations of known techniques (Baker-GillSolovay, Razborov-Rudich)
2
• Is RC f o bs f ?
• I claim that a ‘yes’ answer would require “global
analysis” of Boolean functions
• Given nn lattice of bits X, let f(X)=1 if there’s a
square ‘frame’ of size n1/3n1/3, f(X)=0 otherwise
bs
0n
RC
1/ 3
f
n
0n
f n
2/ 3
Open Problems
~
• Is deg f
RC f ?
• Is R0 f O RC f
If so we get
2
?
R f OR f
R0 f O Q2 f ,
4
2
0
2