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I know that the f/oracles in Simons algorithm is 2-1, and that $f(x)=f(y)\iff y=x\oplus s$. My question is if we can have gave different oracles/functions, that has the same codomain/output for a given s. And if so how many such oracles can we have ? And is there a way to compute how many?

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  • $\begingroup$ I'm not sure to understand your question, are you asking, for a given $f$ satisfying Simon's promise, how many unitaries $U_f$ are there such that $U_f|x,y\rangle=|x,y\oplus f(x)\rangle$? $\endgroup$
    – Tristan Nemoz
    Apr 2 at 12:20
  • $\begingroup$ No, i was wondering how many f's can we have that satisfies the condition in Simon algorithm, for a specific s, $\endgroup$ Apr 2 at 12:30

1 Answer 1

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Let $f:\{0, 1\}^n\to\{0,1\}^m$ satisfying Simon's promise for a given $s$. We want to count how such many $f$ are there.

Let us assume $s\neq 0$ for now. Let $S_0$ be a subset of $\{0, 1\}^n$ such that $\left|S_0\right|=2^{n-1}$ and such that for all $x,y\in S_0$, $x\neq y\oplus s$. Basically, if we know $f$ on $S_0$, we completely know $f$ on $\{0, 1\}^n$. Let us denote $x_1,\cdot,x_{2^{n-1}}$ the elements of $S_0$.

  • We first choose $f\left(x_1\right)$, for which we have $2^m$ choices.
  • We then choose $f\left(x_2\right)$, for which we have $2^m-1$ choices.
  • ...
  • We then choose $f\left(x_{2^{n-1}}\right)$, for which we have $2^m-2^{n-1}+1$ choices.

All in all, this results in $\prod\limits_{i=0}^{2^n-1}\left(2^m-i\right)=\frac{2^m!}{\left(2^m-2^{n-1}\right)!}$ functions.

If $s=0$, a similar reasoning yields $\frac{2^m!}{\left(2^m-2^n\right)!}$ functions.

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  • $\begingroup$ Thank you, just to dobbel check. The reason $|S_{0}|=2^{n-1}$ is because f is 2 to 1, so we only map to half of the possible states? $\endgroup$ Apr 2 at 14:39
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    $\begingroup$ And is probably should be $x\neq y\oplus s$ and $s\neq y\oplus s$ $\endgroup$ Apr 2 at 14:46
  • $\begingroup$ @PinkElephants Yes, once you hav chosen $f(0)$ for instance, the value of $f(s)$ is forced. So, once you have chosen a value for half the possible inputs, the rest of them is forced. Nice catch for your second comment! $\endgroup$
    – Tristan Nemoz
    Apr 2 at 14:53

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