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Here is the description of Deutsch-Jozsa problem: Let $f: \{0,1\}^n \mapsto \{0,1\}$ be a function promised to be either constant or balanced ('balanced' means that $f$ outputs as many 0's as 1's).

I need to show that a probabilistic classical algorithm making two evaluations of $f$ can with probability at least 2/3 correctly determine whether $f$ is constant or balanced.

There is also a hint: Your guess does not need to be a deterministic function of the results of the two queries. Your result should not assume any particular a priori probabilities of having a constant or balanced function.

I'm a bit lost here. My thinking is that if the two evaluations are the different, then $f$ is definitely balanced. Otherwise, $f$ could be either constant or balanced. But the chance of success would be depending on the probability of $f$ being constant, which is against the given hint.

How should I approach this problem?

1 Answers 1

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Algorithm:

  1. Independently and uniformly sample two queries from $\{0, 1\}^n$, denoted as $q_1$ and $q_2$.

  2. Evaluate $f(q_1)$ and $f(q_2)$. If $f(q_1) \neq f(q_2)$, output $\color{blue}{\mathsf{balanced}}$; otherwise, output $\color{blue}{\mathsf{balanced}}$ with probability $\frac{1}{3}$ and $\color{red}{\mathsf{constant}}$ with probability $\frac{2}{3}$.

Analysis:

We have

\begin{align} &P(\text{output }\color{blue}{\mathsf{balanced}} \mid f \ \color{blue}{\mathsf{balanced}}) \\[5pt] =\ &P(\text{output }\color{blue}{\mathsf{balanced}}, f(q_1) \neq f(q_2) \mid f \ \color{blue}{\mathsf{balanced}}) \\ &\quad + P(\text{output }\color{blue}{\mathsf{balanced}}, f(q_1) = f(q_2) \mid f \ \color{blue}{\mathsf{balanced}}) \\[5pt] =\ &P(\text{output }\color{blue}{\mathsf{balanced}}\mid f(q_1) \neq f(q_2), f \ \color{blue}{\mathsf{balanced}})\cdot P(f(q_1) \neq f(q_2) \mid f \ \color{blue}{\mathsf{balanced}}) \\ &\quad +P(\text{output }\color{blue}{\mathsf{balanced}}\mid f(q_1) = f(q_2), f \ \color{blue}{\mathsf{balanced}})\cdot P(f(q_1) = f(q_2) \mid f \ \color{blue}{\mathsf{balanced}}) \\[5pt] =\ &1\cdot \frac{1}{2} + \frac{1}{3} \cdot \frac{1}{2} = \frac{2}{3} \end{align} and \begin{align} P(\text{output }\color{red}{\mathsf{constant}} \mid f\ \color{red}{\mathsf{constant}}) = \frac{2}{3} \end{align} Therefore, \begin{align} &P(\text{output is correct}) = P(\text{output }\color{blue}{\mathsf{balanced}}, f \ \color{blue}{\mathsf{balanced}}) + P(\text{output }\color{red}{\mathsf{constant}}, f\ \color{red}{\mathsf{constant}}) \\ =\ &P(\text{output }\color{blue}{\mathsf{balanced}} \mid f \ \color{blue}{\mathsf{balanced}})\cdot P(f\ \color{blue}{\mathsf{balanced}}) \\ &\quad + P(\text{output }\color{red}{\mathsf{constant}}\mid f\ \color{red}{\mathsf{constant}})\cdot P(f\ \color{red}{\mathsf{constant}}) \\ =\ &\frac{2}{3}\cdot P(f\ \color{blue}{\mathsf{balanced}}) + \frac{2}{3} \cdot P(f\ \color{red}{\mathsf{constant}}) = \frac{2}{3} \end{align}