Transcript Quantum Lower Bounds The Polynomial and Adversary Methods

Quantum Lower Bounds
The Polynomial and Adversary
Methods
Scott Aaronson
September 14, 2001
Prelim Exam Talk
Motivation
• Quantum computing: model of computation
based on our best-confirmed physical theory
• To understand quantum computing, must know
limitations as well as capabilities
• E.g., can QC’s decide NP in polynomial time?
– Smart money says no, but proving it implies PNP
• Popular alternative: study restricted models
Talk Overview
• Intro to Quantum Model
• Polynomial Method (Beals et al. 1998)
– N lower bound for search
– Part of D(f)1/6 bound for total functions
• Adversary Method (Ambainis 2000)
– Density matrices and entanglement
– N lower bound for search (again)
The Quantum Model
• State of computer: superposition over binary
strings
• To each string Y, associate complex amplitude Y
• Y |Y|2 = 1
• On measuring, see Y with probability |Y|2
• Dirac ket notation: State written | = Y Y |Y
• Each |Y is called a basis state
Unitary Evolution
• Quantum state changes by multiplying amplitude
vector with unitary matrix: |(t+1)= U|(t)
• U is unitary iff U-1=U†, † conjugate transpose
(Linear transformation that preserves norm=1)
• Example:
1/2 -1/2
1/2 1/2
(|0+ |1)/2 = |1
• Circuit model: U must be efficiently computable
Black-box model: No such restriction
Query Model
• Algorithm state is i,z,ai,z,a|i,z,a
(i: index to query
z: workspace
a: answer bit)
• Input: X=x1…xn  {0,1}n
• Query replaces each |i,z,a by (-1)x[i]|i,z,a
• Algorithm alternates unitaries and queries:
U0  O1  U1  …  UT-1  OT  UT
• Ui are arbitrary, but independent of input
• By end, i,z|i,z,f(X)|2  2/3 for every X
Lower Bounds by Polynomials
Beals, Buhrman, Cleve, Mosca, de Wolf, FOCS 1998
Key Idea
• Let Q2(f) = minimum no. of queries used by quantum
alg that evaluates f:{0,1}n{0,1} w.p.  2/3 for all
X=x1…xn  {0,1}n
• If quantum algorithm makes T queries, acceptance
probability is degree-2T polynomial over input bits
• Implies Q2(f)  ~deg(f)/2, where ~deg(f) = min
degree of polynomial p s.t. |p(X)–f(X)|  1/3 for all X
• Show ~deg(f) is large for function f of interest
Lemma: Q2(f)  ~deg(f)/2.
Proof: After T queries, amplitude i,z,a of basis state
|i,z,a is a complex-valued multilinear polynomial of
degree  T over x1,…,xn. By induction.
• Base case: Before any queries, i,z,a is degree-0
polynomial.
• Query: Replaces each i,z,a by (1-2xi)i,z,a. Increases
degree by  1. Since xi{0,1}, can replace xi xi by xi.
• Unitary: Replaces each i,z,a by linear combination of
i’,z’,a’. So degree doesn’t increase.
Separating real and imaginary parts, i,z|i,z,f(X)|2 is a realvalued multilinear polynomial of degree  2T.
Lemma (Minsky, Papert 1968): If p: RnR is a multilinear
polynomial, there’s a polynomial q: RR s.t.
(1) deg(q)  deg(p),
(2) q(|X|) = psym(X) = (1/n!) S(n)p((X))
(|X|: Hamming weight of X
S(n): Symmetric group)
Proof: Let d = deg(psym)  deg(p). Let Vj = sum of all
products of j distinct variables. Since psym is symmetrical,
psym(X) = a0 + a1V1 + … + adVd
for some aiR. Vj assumes value
choose(|X|,j) = |X|(|X|-1)(|X|-2)…(|X|-j+1)/j!
on X, which is a polynomial of degree j of |X|. So construct
q(|X|) of degree d accordingly.
Approximate Degree of OR
Theorem (Ehlich, Zeller 1964; Rivlin, Cheney 1966): Let
p: RR be a polynomial s.t. b1  p(i)  b2 for every integer
0  i  n and |dp(x)/dx|  c for some real 0  x  n. Then
deg(p)  [cn / (c + b2 – b1)].
Corollary (Nisan, Szegedy 1994): ~deg(ORn) = (n).
Proof: Let r: RR be symmetrization of approximating
polynomial for ORn. Then 0  r(i)  1 for every integer
0  i  n, and dr(x)/dx  1/3 for some x  [0,1] because
r(0)  1/3 and r(1)  2/3. So
deg(r)  [n/3 / (1/3 + 1 – 0)].
Definitions: C(f) and bs(f)
For total Boolean function f and input X:
XB = X with variables in set B flipped
Certificate complexity CX(f) = Minimum size of set A s.t.
f(X) = f(XB) for all B disjoint from A
C(f) = maxX CX(f)
Block sensitivity bsX(f) = Maximum number of disjoint sets
B s.t. f(X)  f(XB)
bs(f) = maxX bsX(f)
Immediate: bs(f)  C(f)  D(f), D(f) deterministic query
complexity
Bound for Total Boolean Functions
Theorem (Beals et al.): D(f) = O(Q2(f)6) for all total f.
Proof overview:
1. bs(f) = O(Q2(f)2). Follows easily from n lower
bound for ORn.
2. C(f)  bs(f)2. Proved on next slide.
3. D(f)  C(f) bs(f). Proof omitted. Idea: To evaluate f,
repeatedly query a 1-certificate consistent with
everything queried so far. Need to repeat at most
bs(f) times.
Lemma (Nisan 1991): C(f)  bs(f)2.
Proof: Let X  {0,1}n be input, B1,…,Bb be disjoint minimal
blocks s.t. b = bsX(f)  bs(f). Claim: C = iBi  {0,1},
variables set according to X, is a certificate for X of size
 bs(f)2.
1. If C were not a certificate, let X’ be input that agrees with C
s.t. f(X’)  f(X). Let X’ = XB. Then B is a sensitive block
for X disjoint from iBi, contradiction.
2. For each 1  i  b, |Bi|  bs(f). For if we flip a Bi-variable
in XB[i], function value must flip from f(XB[i]) to f(X),
otherwise Bi wouldn’t be minimal. So every singleton in Bi
is a sensitive block for f on XB[i]. Hence size of C is  bs(f)
bs(f).
(Is lemma tight? Open problem!)
Quantum Adversary Method
Ambainis, STOC’2000, to appear in JCSS
• Key Idea
– Give algorithm superposition of inputs
– Consider (I=inputs, A=algorithm) as bipartite
quantum state.
– Initially I and A are unentangled. By end of
computation, they must be highly entangled.
– Upper-bound how much entanglement can
increase via a single query.
– How? Density matrices.
Density Matrices
• Mixed state: distribution over quantum states
I.e., one part of composite state
(Not mixed: pure)
• Non-unique decomposition into pure states:
|0 w.p. ½, |1 w.p. ½
= (|0+|1)/2 w.p. ½, (|0-|1)/2 w.p. ½
• Density matrix:  = ipi|ii|
where || has (i,j) entry i*j
•  represents all measurable information
Entanglement
• Quantum state is entangled if not a mixture
of product states
(States for which measuring one subsystem reveals
nothing about other subsystems)
• Examples:
½(|00+|01+|10+|11): unentangled
|00 w.p. ½, |11 w.p. ½: unentangled
(|00+|11)/2: entangled (EPR pair)
Plan of Attack
• Input: (1/|S|) XS|X
• t = i,z,a pt,i,z,a|t,i,z,at,i,z,a| after t queries
• Initially: input and algorithm unentangled  0
is pure state  (0)XY = 1/|S| for all X,Y
• By end: highly entangled  T highly mixed 
|(T)XY|1/(3|S|) (say) for all X,Y with f(X)f(Y)
• Goal: Upper-bound
At = X,Y:f(X)f(Y) (|(t-1)XY|-|(t)XY|)
Lemma: For all X,Y with f(X)f(Y),
|(0)XY|-|(T)XY| = (1/|S|).
Proof: Let i,z,a i,z,a|i,z,a, i,z,a i,z,a|i,z,a be final
algorithm states on X and Y respectively. Then
(T)XY = (1/|S|) i,z,a i,z,a* i,z,a
 (1/|S|) [i,z,a|i,z,a|2] [i,z,a|i,z,a|2]
(by Cauchy-Schwarz)
 (1/|S|) { [i,z|i,z,0|2] [i,z|i,z,0|2]
+ [i,z|i,z,1|2] [i,z|i,z,1|2] }
 (2/|S|)[(1- )], where  is error prob.
Theorem: Q2(ORn) = (n).
Proof: Let S contain all X{0,1}n of Hamming wt 1.
A0=n-1 and ATn/2 (say); we show At-1-At = O(n).
At-1-At  X,Y:f(X)f(Y) |()XY-(’)XY| (=t-1, ’=t)
 i,z,a pi,z,a X,Y:f(X)f(Y) |(i,z,a)XY-(’i,z,a)XY|.
Now (i,z,a)XY = i,z,a,X*i,z,a,Y, since i,z,a is a pure state.
A query maps i,z,a,X to (-1)x[i]i,z,a,X, (i,z,a)XY to
(1)x[i]+y[i](i,z,a)XY. And for all X,Y, x[i]y[i] for only two
values of i, so only four rows/columns change. So
At-1-At  8 maxY X |i,z,a,X*i,z,a,Y|
 8 X |i,z,a,X| = O(n) by Cauchy-Schwarz.
Game-Tree Search
• For some problems, adversary method yields
better bound than polynomial method
• I.e. AND of n OR’s of n vars each
• Upper bound: recursive Grover, O(n log n)
• bs(f) = O(Q2(f)2) yields only Q2(f) = (4n)
• Adversary method: Q2(f) = (n)
• Idea: each XS can be changed in n places
to produce Y s.t. f(X)f(Y), YS.