Quantum Lower Bounds The Polynomial and Adversary Methods Scott Aaronson September 14, 2001 Prelim Exam Talk.

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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 QCs 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: Circuit model: U must be efficiently computable Black-box model: No such restriction 1/ 2 -1/ 2 1/ 2 ( |0 + |1 )/ 2 = |1

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=x 1 …x n {0,1} n Query replaces each |i,z,a by (-1) x[i] |i,z,a Algorithm alternates unitaries and queries: U 0 O 1 U 1 … U T-1 O T U T U i 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 Q 2 (f) = minimum no. of queries used by quantum alg that evaluates f:{0,1} n {0,1} w.p. 2/3 for all X=x 1 …x n {0,1} n If quantum algorithm makes T queries, acceptance probability is degree-2T polynomial over input bits Implies Q 2 (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: Q 2 (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 x 1,…,x n. By induction. Base case: Before any queries, i,z,a is degree-0 polynomial. Query: Replaces each i,z,a by (1-2x i ) i,z,a. Increases degree by 1. Since x i {0,1}, can replace x i x i by x i. Unitary: Replaces each i,z,a by linear combination of i,z,a. So degree doesnt increase. Separating real and imaginary parts, i,z | i,z,f(X) | 2 is a real- valued multilinear polynomial of degree 2T.

Lemma (Minsky, Papert 1968): If p: R n R is a multilinear polynomial, theres a polynomial q: R R s.t. (1) deg(q) deg(p), (2) q(|X|) = p sym (X) = (1/n!) S(n) p( (X)) (|X|: Hamming weight of XS(n): Symmetric group) Proof: Let d = deg(p sym ) deg(p). Let V j = sum of all products of j distinct variables. Since p sym is symmetrical, p sym (X) = a 0 + a 1 V 1 + … + a d V d for some a i R. V j 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. b 1 p(i) b 2 for every integer 0 i n and |dp(x)/dx| c for some real 0 x n. Then deg(p) [cn / (c + b 2 – b 1 )]. Corollary (Nisan, Szegedy 1994): ~deg(OR n ) = ( n). Proof: Let r: R R be symmetrization of approximating polynomial for OR n. 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: X B = X with variables in set B flipped Certificate complexity C X (f) = Minimum size of set A s.t. f(X) = f(X B ) for all B disjoint from A C(f) = max X C X (f) Block sensitivity bs X (f) = Maximum number of disjoint sets B s.t. f(X) f(X B ) bs(f) = max X bs X (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(Q 2 (f) 6 ) for all total f. Proof overview: 1.bs(f) = O(Q 2 (f) 2 ). Follows easily from n lower bound for OR n. 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, B 1,…,B b be disjoint minimal blocks s.t. b = bs X (f) bs(f). Claim: C = i B i {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 = X B. Then B is a sensitive block for X disjoint from i B i, contradiction. 2.For each 1 i b, |B i | bs(f). For if we flip a B i -variable in X B[i], function value must flip from f(X B[i] ) to f(X), otherwise B i wouldnt be minimal. So every singleton in B i is a sensitive block for f on X B[i]. Hence size of C is bs(f) bs(f). (Is lemma tight? Open problem!)

Quantum Adversary Method Ambainis, STOC2000, 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: = i p i | 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 p t,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 A t = 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: Q 2 (OR n ) = ( n). Proof: Let S contain all X {0,1} n of Hamming wt 1. A 0 =n-1 and A T n/2 (say); we show A t-1 -A t = O( n). A t-1 -A t X,Y:f(X) f(Y) |( ) XY -() XY | ( = t-1, = t ) i,z,a p i,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 A t-1 -A t 8 max Y 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 ORs of n vars each Upper bound: recursive Grover, O( n log n) bs(f) = O(Q 2 (f) 2 ) yields only Q 2 (f) = ( 4 n) Adversary method: Q 2 (f) = ( n) Idea: each X S can be changed in n places to produce Y s.t. f(X) f(Y), Y S.

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