α x x 0 α x x f(x) α x x α x ( 1) f(x) x f(x) x f(x) α x = α x x 2

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1 Quadratic speedup for unstructured search - Grover s Al- CS 94- gorithm /8/07 Spring 007 Lecture Unstructured Search Here s the problem: You are given an efficient boolean function f : {1,,} {0,1}, and are promised that for eactly one a {1,,}, f(a) = 1 Think of this as a table of size, where eactly one element has value 1, and all the others are 0 How long does it take to determine the a such that f(a) = 1? Since we assume f can be computed classically in polynomial time, we can also compute it in superposition: α 0 α f() As we saw before, we can use circuit for f to put information about f() in the phase by effecting the transformation: Here is another way of creating this phase state: α α ( 1) f() ( ) 0 1 α ( ) f() f() α = α ( ) f() f() ( ) 0 1 = α ( 1) f() ow, we might as well assume f is a black bo or oracle All we need to do is design an algorithm that finds a : f(a) = Black-Bo Techniques A black-bo algorithm attempts to determine a without actually eamining the structure of f, but it is allowed to evaluate f on any input or superposition of inputs As we saw before, any black-bo quantum algorithm must make Ω( n/ ) queries to f In compleity theory, black-bo techniques such as simulation and diagonalization are one way to determine whether or not two compleity classes are equal Since these techniques don t look inside the actual machine, they would not be able to tell if you gave your machine access to an oracle f, which is called relativization Compleity classes can be defined relative to oracles: for eample, P f is the class of problems that can be solved in polynomial time if you have CS 94-, Spring 007, Lecture

2 a v v θ φ θ θ + φ u = 1 =1 e Figure 01: To rotate v by θ to v, we reflect around e and then reflect around ψ 0 access to the oracle f ow if there is an oracle f relative to which two compleity classes are equal, and an oracle g relative to which they are not, then these black-bo techniques cannot be used to determine whether or not the two classes are equal In the case of P and P, f P f = P f, and g P g P g, so no black-bo technique can answer the P =? P question There are statements in compleity theory that don t relative, such as IP = PSPACE and MIP = EXP However, these rely on a single technique using the principle of local checkability, and we don t know how to apply it to either P =? P or P? BQP So we have no idea how to resolve these open questions 0 Grover s algorithm Grover s algorithm finds a in O( ) steps Consider the dimensional Hilbert space spanned by 1,, We wish to find a Consider the uniform superposition u = 1 Then a u = 1/ = cos θ a,u Consider the two dimensional subspace spanned by a and u Let e be the state orthogonal to a in this subspace Let θ be the angle between u and e Then sin θ = cos( π θ) = cos θ a,u = 1/ and therefore θ 1/ See Figure 01 for an illustration of these vectors a is the target, so we want to increase θ But how? One way to rotate a vector is to make two reflections In particular, we can rotate a vector v by θ by reflecting about e and then reflecting about u This transformation is also illustrated in Figure 01 Each step of our algorithm is a rotation by θ (we discuss the implementation below) This means that we need π/ θ iterations for the algorithm to complete We know that θ 1, so we need O( ) iterations for the algorithm to complete In the end, we get very close to a, and then with high probability, a measurement of the state yields a How do you implement the two reflections? CS 94-, Spring 007, Lecture 11 0-

3 1 Reflection about e is easy We can reflect about the hyperplane orthogonal to a by flipping the phase of the component in the direction of a, ie carry out the transformation α α ( 1) f() For the reflection about u, let us first to determine how to reflect about the basis state 0 We can do so by flipping the phase of all components ecept 0, ie carry out the transformation α α0 0 + α 0 Since u = H n 0, we can flip around u by first applying the Hadamard transform H n, then reflecting about 0, and finally applying the Hadamard again to switch back from the Hadamard basis 03 Another approach Let s look at the search algorithm differently, with all superpositions The rotation about u, D, is an inversion about the mean : 1 For = n, D can be decomposed and rewritten as: D = H H = H I H 0 0 = H H I / / / / / / = I / / / / 1 / / / / 1 / = / / / 1 Observe that D is epressed as the product of three unitary matrices (two Hadamard matrices separated by a conditional phase shift matri) Therefore, D is also unitary Regarding the implementation, both the Hadamard and the conditional phase shift transforms can be efficiently realized within O(n) gates CS 94-, Spring 007, Lecture

4 Consider D operating on a vector α to generate another vector β : D α 1 α = α 1 α, where = 1 Here, is the mean amplitude, so the epression describes a reflection of about the mean Thus, the operation D can be considered an inversion about the mean with respect to i=1 The quantum search algorithm iteratively improves the probability of measuring a solution Here s how: 1 Start state is ψ 0 = u = 1 Invert the phase of a using f 3 Then invert about the mean using D 4 Repeat steps and 3 O( ) times, so in each iteration α a increases by This process is illustrated in Figure 0 Suppose we just want to find a with probability 1 Until this point, the rest of the basis vectors will have amplitude at 1 least In each iteration of the algorithm, α a increases by at least = Eventually, α a = 1 The number of iterations to get to this α a is 04 More applications Grover s algorithm is often called a database search algorithm However, in order to perform the reflection about e, α α ( 1) f(), it is necessary to be able to query f in superposition Thus, it is more accurate to refer to this as unstructured search instead of database search But there are a number of applications of unstructured search: 1 Find the minimum in O( ) steps Eercise Approimately count elements, or generate random ones 3 O( 1/3 ) algorithm for the collision problem 4 Speed up the test for matri multiplication In this problem we are given three matrices, A, B, and C, and are told that the product of the first two equals the third We wish to verify that this is indeed true An efficient (randomized) way of doing this is picking a random array r, and checking to see whether Cr = ABr = A(Br) Classically, we can do the check in O(n ) time, but using a similar approach to Grover s algorithm we can speed it up to O(n 175 ) time CS 94-, Spring 007, Lecture

5 (a) k i (b) k i (c) k i Figure 0: The first three steps of Grover s algorithm We start with a uniform superposition of all basis vectors in (a) In (b), we have used the function f to invert the phase of α k After running the diffusion operator D, we amplify α k while decreasing all other amplitudes CS 94-, Spring 007, Lecture

6 041 The Collision Problem Given a two-to-one function f : X Y, we want to find a collision, ie 1 and such that f( 1 ) = f( ) A probabilistic classical algorithm to do so is as follows: 1 Pick a random subset K X of cardinality O( ), where = X Compute the pair (, f()) for each K, and sort them by the second element 3 Scan the sorted list for a collision Due to the birthday parado, this algorithm will succeed with probability at least 1/, if K > 118 The algorithm queries f O( ) times, and it runs in time O( log ) A faster quantum algorithm to find a collision is as follows: 1 Pick a random subset K X of cardinality O( 3 ), where = X Compute the pair (, f()) for each K, and sort them by the second element 3 Scan the sorted list for a collision If there is one, then output it and quit 4 Construct the function g : X {0,1}, where g() = 1 if and only if y K f() = f(y) This function can be evaluated in O(log 3 ) by computing f() and then running binary search on the list above for an element equal to (y, f(y)) 5 Run Grover s search over the set X\K for the function g Since the elements in K do not collide with one another, they each collide with one element in X\K, so there are K elements in X\K such that g() = 1 As a result, the number of queries to g in this search is in O( X\K / K ) = O( / 3 ) = O( 3 ) The search will output an such that g() = 1, and we can then use binary search to scan the sorted list for the (y, f(y)) such that f() = f(y) This algorithm queries f O( 3 ) times, and it runs in time O( 3 log 3 ) In the quantum algorithm above, we used the fact that if there are t marked items, then Grover s algorithm runs in O( /t) steps We leave the proof of this as an eercise for the reader CS 94-, Spring 007, Lecture

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