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Quantum Computing MAS 725 Hartmut Klauck NTU 9.4.2012 Today Lower bounds in the black box model polynomial method adversary method The black box model Input x1,…,xn with query access We want to compute some function f(x1,…,xn) Examples: Deutsch‘s problem: x1,x2 are the values g(0) and g(1) for a function g, we want to know if g is balanced ) Compute parity of x1,x2 Grover: Find i mit xi=1, consider f=OR The black box model We measure the number of queries an algorithm does in the worst case, cost of an algorithm Not time, space etc. The query complexity of a function f is the minimal cost of an algorithm computing f Types of algorithms Deterministic algorithms: classical, no error, the complexity is denoted D(f) Randomized algorithms: allow error probability 1/3, over the randomness of the algorithm, complexity R(f) Quantum algorithms: We count the number of quantum queries. With error 1/3: Q(f) Withour error: QE(f) Examples We know some bounds already: QE(XOR2)=1, but R(XOR2)=2 QE(XORn)· d n/2e Q(OR)=O(n1/2) R(OR)=(n) Q(OR)=(n1/2) How can we show quantum lower bounds in general? How much better can quantum algorithms be compared to randomized algorithms? Examples We showed the lower bound for OR with an adversary argument Are there other more general techniques? For certain problems (Simon, Period Finding) we have seen exponential speedups What is the largest speedup for a total Boolean function? What is the hardest function? Consider ID: ID(x)=x How many queries are needed to compute ID? R(ID)=n We get no information on a position we do not query Formally: apply the Yao principle: fix random string, then a deterministic algo with n-1 queries must have error 1/2. Q(ID)· 1/2 n/2+O(n ) Ansatz: We use the sign oracle (-1)xi |ii Formula for Hadamard transform: Another Hadamard maps back to |xi So suffices to generate this superposition Doing this exactly needs n queries Number of queries corresponds to |y| Use only y with|y|· n/2+O(n1/2) Algorithm Generate uniform superposition over all y with |y|· n/2+O(n1/2) On |yi query all bits xi with yi=1 Apply Hadamard Success probability? 99% of all strings have Hamming weight at most n/2+O(n1/2) Hence resulting superposition is close to the desired one and we will have small error For example error 5% with n/2+n1/2 queries Conclusion n/2· Q(ID)· n/2+n1/2 ...if we can show Q(XOR)¸ n/2 The polynomial method A quantum black box algorithm is a sequence of unitaries and query unitaries We construct a low degree polynomial from such an algorithm that represents the computed function Then we analyze the minimum degree for such polynomials The polynomial method We claim that the amplitudes of the final state of a T query algorithm can be written as polynomials of degree T Proof by induction: T=0: amplitudes do not depend on the input, i.e. are constants T! T+1: i(x) is given by a degree T polynomial Next we apply a unitary that does not depend on x The new ®i(x) is a linear combination of degree T polynomials, degree unchanged The query transformation: state |ii|ai|ki maps to |ii|a© xii|ki The new iak(x) is xi¢ i, a©1, k + (1-xi)¢ i, a, k Degree is no more than T+1 The polynomial method The acceptance probability on input x is the sum of squred amplitudes Hence can be written as a polynomial of degree 2T We may replace xik by xi The result is multilinear Conclusion Given a quantum algorithm with T queries that computed f exactly we get a multilinear polynomial of degree 2T that satisfies p(x)=f(x) for all x. If the quantum algorithm has error 1/, then there is a polynomial with p(x)2[0,1/3] for f(x)=0 and p(x)2[2/3,1] for f(x)=1 Now we have to consider the degree of polynomials representing Boolean functions Exact quantum algorithms For every total Boolean function f there is a unique multilinear polynomial that represents f exactly deg(f) denotes the degree of this polynomial QE(f)¸ deg(f)/2 Example: XOR, the polynomial is Degree is n, and QE(XOR) =n/2 Multilinear polynomials Claim: For every Boolean function there is a unique multilinear polynomial, which represents f exactly, i.e. f(x)= p(x) for all Boolean x. Proof: Assume f(x)=p(x)=q(x) for alle Boolean x, yet pq Then p-q is a multilinear polynomial for g(x)=0, and p-q is not the zero polynomial Take a minimum degree monomial m in p-q with nonzero coefficient a0. Let z be the string string, that contains all xi in m as ones, and contains no other ones m(z)=1, and p(z)=a, contradiction Another example Polynomial for OR: Also has degree n Hence QE(OR)¸ n/2 But actually QE(OR)=n QE(OR) We consider the amplitudes of the final state of an optimal algorithm for OR (no error), as polynomials of degree T Basis states |0yi are rejecting. B is the set of those For i in B we have pi(x)=0 when x is not 0n There is a j in B with pj(0n) 0 Consider the real part q(x) on 1-pj(x)/pj(0n) Then: deg (q) · T = QE(OR) and q(0n)=0 and q(x)=1 for other x, hence deg(q)¸ deg(OR) = n Some facts about polynomials If a Boolean f depends on n variables, we have deg(f)¸ log n - loglog n All symmetric, nonconstant f have degree n-o(n) Hence QE(f)¸ (log n)/2 –o(log n) And QE(f)¸ n/2-o(n) for symmetric nonconstant f Approximating polynomials Given a quantum algorithm with T queries that we get a multilinear polynomial of degree 2T with p(x)2[0,1/3] for f(x)=0 and p(x)2[2/3,1] for f(x)=1 adeg(f) denotes the minimal degree of such a polynomial Then: Q(f)¸ adeg(f)/2 Example OR on 2 bits: x1/3 + x2/3 + 1/3, adeg(OR2)=1 Symmetrization Let f denote a symmetric function, i.e, f is constant on all x with |x|=k, i.e., we have function values f0,…,fn Symmetrization turns a multilinear polynomial for f into a univariate polynomial of the same degree p(k)2[0,1/3] for fk=0 and p(k)2[2/3,1] for fk=1 XOR We get a polynomial such that p(k)2[0,1/3] for even k and p(k)2[2/3,1] for odd k k2{0,…,n} Clearly p(k)-1/2 has n roots! And degree n Q(XOR)=n/2 Symmetrization Let be a permutation of [1,…,n], p polynomial in n var. Set psym(x)= Lemma: If p is a multilinear polynom of degree d, then there is a univariate degree d polynomial with q(|x|) = psym(x) for all x Proof: Let Vj denote the sum of all products of j variables Then psym can be written as i=0...d bi Vi Value of Vj on x is The sum is The approximation degree of OR But what about OR? We know already Q(OR)=n1/2 But adeg(OR) could be smaller Symmetrization: p(0) 2[0,1/3] p(1),…,p(n)2[2/3,1] What is the minimal degree of such a polynomial? A result from approximation theory Theorem: Let p be a polynomial with 0· p(x)· 1 for all integers 0· x· n such that |p’(x)|¸ c for some real 0· x· n Then deg(p)¸ (cn/(c+1))1/2 But: p(0)<1/3 and p(1)>2/3 Hence p’(x)¸ 1/3 for some x2 [0,1] adeg(OR)¸ (n/4)1/2 We recover the lower bound for search etc. Some more facts For a total Boolean function f we have D(f)=O(adeg(f)6) Hence also D(f)=O(Q(f)6) This is clearly only true for total functions The best speedup that is known (Grover) is only quadratic Polynomial method is very useful for functions with a lot of symmetry, example Element Distinctness The adversary method This method actually characterizes Q(f) [in its strongest form] Leads to a characterization as a semidefinite program Original idea is to bound the progress achieved by one query in distinguishing pairs of inputs Certificate complexity A certificate for x is a set of positions and values that fixes the function value for all x that are consistent with them Example: x1=1 fixes OR XOR has no certificate of length <n C(f) is the max over all x of the min certificate for x C(XOR)=n C(OR)=n 0n needs a certificate of size n An observation There are 1-certificates and 0-certificates xi=1 is 1-cert for OR x1=0,…, xn=0 is 0-cert for OR For all f: 1-cert and 0-cert need to share at least one variable Certificate and adversary C( f ) = m in px 2 f 0;1gn m ax x X px ( j ) j such t hat X px ( j ) py ( j ) ¸ 1 if f ( x) 6 = f ( y) j :x j 6 = yj A dv( f ) = m inn m ax px 2 R x X px ( j ) 2 j such t hat X j :x j 6 = yj px ( j ) py ( j ) ¸ 1 if f ( x) 6 = f ( y) Adversary bound A dv( f ) = m inn m ax jjpx jj 2 px 2 R x such t hat X px ( j ) py ( j ) ¸ 1 if f ( x) 6 = f ( y) j :x j 6 = yj Example: OR on 0n 10n-1 … 0n-11 px : (1…1) (1 0….) (01 0…) (0…01) Rescale by 1/n1/2 n1/2 Adv(OR)· n1/2 Adversary bound Need to show two things: Q(f)=(Adv(f)) How to prove lower bounds on Adv(f) How to prove lower bounds Adversary bound as stated is a minimization problem, so we take the dual A dv( f ) = m ax ¡ 2RD£ D jj¡ jj such t hat ¡ ( x; y) ¸ 0 ¡ ( x; y) = 0 if f ( x) = f ( y) 8j jj¡ ± X x;y:x j 6 = yj jxi hyj jj · 1 Generalized adversary bound A dv § ( f ) = m ax ¡ 2 R D £ nD jj¡ jj such t hat ¡ ( x; y) = 0 if f ( x) = f ( y) 8j jj¡ ± X jxi hyjjj · 1 x;y:x j 6 = yj This bound is asymptotically equal to Q(f)