Quantum Computing MAS 725 Hartmut Klauck NTU TexPoint fonts used in EMF. Read the TexPoint manual before you delete this box.: A A A A

Today Lower bounds in the black box model polynomial method adversary method

The black box model Input x 1,…,x n with query access We want to compute some function f(x 1,…,x n ) Examples: Deutsch‘s problem: x 1,x 2 are the values g(0) and g(1) for a function g, we want to know if g is balanced ) Compute parity of x 1,x 2 Grover: Find i mit x i =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: Q E (f)

Examples We know some bounds already: Q E (XOR 2 )=1, but R(XOR 2 )=2 Q E (XOR n ) · d n/2 e Q(OR)=O(n 1/2 ) R(OR)=  (n) Q(OR)=  (n 1/2 ) 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) · n/2+O(n 1/2 ) Ansatz: We use the sign oracle (-1) x i |i i Formula for Hadamard transform: Another Hadamard maps back to |x i 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(n 1/2 )

Algorithm Generate uniform superposition over all y with |y| · n/2+O(n 1/2 ) On |y i query all bits x i with y i =1 Apply Hadamard Success probability? 99% of all strings have Hamming weight at most n/2+O(n 1/2 ) Hence resulting superposition is close to the desired one and we will have small error For example error 5% with n/2+n 1/2 queries

Conclusion n/2 · Q(ID) · n/2+n 1/2...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 |i i |a i |k i maps to |i i |a © x i i |k i The new  iak (x) is x i ¢  i, a © 1, k + (1-x i ) ¢  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 x i k by x i 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 Q E (f) ¸ deg(f)/2 Example: XOR, the polynomial is Degree is n, and Q E (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 p  q 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 a  0. Let z be the string string, that contains all x i 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 Q E (OR) ¸ n/2 But actually Q E (OR)=n

Q E (OR) We consider the amplitudes of the final state of an optimal algorithm for OR (no error), as polynomials of degree T Basis states |0y i are rejecting. B is the set of those For i in B we have p i (x)=0 when x is not 0 n There is a j in B with p j (0 n )  0 Consider the real part q(x) on 1-p j (x)/p j (0 n ) Then: deg (q) · T = Q E (OR) and q(0 n )=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 Q E (f) ¸ (log n)/2 –o(log n) And Q E (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: x 1 /3 + x 2 /3 + 1/3, adeg(OR 2 )=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 f 0,…,f n Symmetrization turns a multilinear polynomial for f into a univariate polynomial of the same degree p(k) 2 [0,1/3] for f k =0 and p(k) 2 [2/3,1] for f k =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 k 2 {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 p sym (x)= Lemma: If p is a multilinear polynom of degree d, then there is a univariate degree d polynomial with q(|x|) = p sym (x) for all x Proof: Let V j denote the sum of all products of j variables Then p sym can be written as  i=0  d  b i V i Value of V j on x is The sum is

The approximation degree of OR But what about OR? We know already Q(OR)=n 1/2 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: 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) 2/3 Hence p’(x) ¸ 1/3 for some x 2 [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: x 1 =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 0 n needs a certificate of size n

An observation There are 1-certificates and 0-certificates x i =1 is 1-cert for OR x 1 =0,…, x n =0 is 0-cert for OR For all f: 1-cert and 0-cert need to share at least one variable

Certificate and adversary

Adversary bound Example: OR on 0 n 10 n-1 … 0 n-1 1 p x : (1…1) (1 0….) (01 0…) (0…01) Rescale by 1/n 1/2 n 1/2 Adv(OR) · n 1/2

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

Generalized adversary bound This bound is asymptotically equal to Q(f)