Editing Deutsch–Jozsa algorithm (section)

== Classical solution ==
For a conventional [[deterministic]] algorithm where <math>n</math> is the number of bits, <math>2^{n-1} + 1</math> evaluations of <math>f</math> will be required in the worst case. To prove that <math>f</math> is constant, just over half the set of inputs must be evaluated and their outputs found to be identical (because the function is guaranteed to be either balanced or constant, not somewhere in between). The best case occurs where the function is balanced and the first two output values are different. For a conventional [[randomized algorithm]], a constant <math>k</math> evaluations of the function suffices to produce the correct answer with a high probability (failing with probability <math>\epsilon \leq 1/2^{k}</math> with <math>k \geq 1</math>). However, <math>k=2^{n-1} + 1</math> evaluations are still required if we want an answer that has no possibility of error. The Deutsch-Jozsa quantum algorithm produces an answer that is always correct with a single evaluation of <math>f</math>.