In the Deutsch–Jozsa problem, we are given a black boxquantum computer known as an oracle that implements some function. The function takes n-digit binary values as input and produces either a 0 or a 1 as output for each such value. We are promised that the function is either constant or balanced. The task then is to determine if is constant or balanced by using the oracle.
Motivation
The Deutsch–Jozsa problem is specifically designed to be easy for a quantum algorithm and hard for any deterministic classical algorithm. The motivation is to show a black box problem that can be solved efficiently by a quantum computer with no error, whereas a deterministic classical computer would need a large number of queries to the black box to solve the problem. More formally, it yields an oracle relative to which EQP, the class of problems that can be solved exactly in polynomial time on a quantum computer, and P are different. Since the problem is easy to solve on a probabilistic classical computer, it does not yield an oracle separation with BPP, the class of problems that can be solved with bounded error in polynomial time on a probabilistic classical computer. Simon's problem is an example of a problem that yields an oracle separation between BQP and BPP.
Classical solution
For a conventional deterministic algorithm where n is number of bits, evaluations of will be required in the worst case. To prove that is constant, just over half the set of inputs must be evaluated and their outputs found to be identical. The best case occurs where the function is balanced and the first two output values that happen to be selected are different. For a conventional randomized algorithm, a constant evaluations of the function suffices to produce the correct answer with a high probability. However, evaluations are still required if we want an answer that is always correct. The Deutsch-Jozsa quantum algorithm produces an answer that is always correct with a single evaluation of.
History
The Deutsch–Jozsa Algorithm generalizes earlier work by David Deutsch, which provided a solution for the simple case where.
Specifically we were given a boolean function whose input is 1 bit, and asked if it is constant. The algorithm as Deutsch had originally proposed it was not, in fact, deterministic. The algorithm was successful with a probability of one half. In 1992, Deutsch and Jozsa produced a deterministic algorithm which was generalized to a function which takes bits for its input. Unlike Deutsch's Algorithm, this algorithm required two function evaluations instead of only one. Further improvements to the Deutsch–Jozsa algorithm were made by Cleve et al., resulting in an algorithm that is both deterministic and requires only a single query of. This algorithm is still referred to as Deutsch–Jozsa algorithm in honour of the groundbreaking techniques they employed.
Algorithm
For the Deutsch–Jozsa algorithm to work, the oracle computing from has to be a quantum oracle which doesn't decohere. It also mustn't leave any copy of lying around at the end of the oracle call. The algorithm begins with the bit state. That is, the first n bits are each in the state and the final bit is. A Hadamard transform is applied to each bit to obtain the state We have the function implemented as a quantum oracle. The oracle maps the state to, where is addition modulo 2. Applying the quantum oracle gives For each is either 0 or 1. Testing these two possibilities, we see the above state is equal to At this point the last qubit may be ignored and therefore below is remained: We apply a Hadamard transform to each qubit to obtain where is the sum of the bitwise product. Finally we examine the probability of measuring, which evaluates to 1 if is constant and 0 if is balanced. In other words, the final measurement will be if is constant and will yield some other states if is balanced.
Deutsch's algorithm
Deutsch's algorithm is a special case of the general Deutsch–Jozsa algorithm. We need to check the condition. It is equivalent to check , if zero, then is constant, otherwise is not constant. We begin with the two-qubit state and apply a Hadamard transform to each qubit. This yields We are given a quantum implementation of the function that maps to. Applying this function to our current state we obtain We ignore the last bit and the global phase and therefore have the state Applying a Hadamard transform to this state we have if and only if we measure a zero and if and only if we measure a one. So with certainty we know whether is constant or balanced.
