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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

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Quantum Lower Bounds The Polynomial and Adversary Methods

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

  2. 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

  3. 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)

  4. 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

  5. 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 (|0+ |1)/2 = |1

  6. 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

  7. Lower Bounds by PolynomialsBeals, 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

  8. 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 real-valued multilinear polynomial of degree  2T.

  9. Lemma (Minsky, Papert 1968): If p: RnR is a multilinear polynomial, there’s a polynomial q: RR s.t. • deg(q)  deg(p), • 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.

  10. 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)].

  11. 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

  12. Bound for Total Boolean Functions • Theorem (Beals et al.): D(f) = O(Q2(f)6) for all total f. • Proof overview: • bs(f) = O(Q2(f)2). Follows easily from n lower bound for ORn. • C(f)  bs(f)2. Proved on next slide. • 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.

  13. 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. • 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. • 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!)

  14. Quantum Adversary MethodAmbainis, 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.

  15. 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

  16. 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)

  17. 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|)

  18. 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.

  19. 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,api,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.

  20. 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.

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