Assume $f(x)$ is an $n$ bits to $m$ bits function and we want to use Grover's search algorithm to find $x$ such that $f(x) = a$, where $a$ is some $m$-bit predetermined value.

When using the algorithm, can we separate the calls to $f(x)$ and $f(x)=a$? That is, instead of querying $F(x) \equiv (f(x) = a)$ around $2^{n/2}$ times (interleaved with amplification), is it possible to query $f(x)$ once at the begining and then repeat $2^{n/2}$ times $f(x)=a$ (interleaved with amplification) ?

  • $\begingroup$ It feels like you'd be applying the diffusion operator to an $(n+m)$-qubit input rather than merely an $n$-qubit input? You might need $2^{(n+m)/2}$ such repetitions? I'm not sure. $\endgroup$ – Mark S May 16 at 11:52
  • $\begingroup$ Don't see why. The diffusion is applied to the input $n$ bits (just like in the database search). $\endgroup$ – mike_dole_z3 May 16 at 12:37
  • $\begingroup$ It seems like you want to create a composite function, first $f(x)$ and then $F(x)$. Initially you'll create a superposition $\vert x\rangle\vert 000\cdots\rangle\vert 0\rangle$, then $\vert x\rangle\vert f(x)\rangle\vert 0\rangle$, and then $\vert x \rangle\vert f(x)\rangle\vert F(x)\rangle$. How would you rotate? Conditioned on $F(x)=1$? $\endgroup$ – Mark S May 16 at 13:48
  • $\begingroup$ Yes, conditioned on $F(x)=1$. Also, keep in mind that $|x\rangle|f(x)\rangle$ is entangled. $\endgroup$ – mike_dole_z3 May 16 at 13:54
  • $\begingroup$ But if you diffuse just on $\vert x\rangle$, $\vert f(x)\rangle$ gets in the way to stop the interference from the diffusion operator? $\endgroup$ – Mark S May 16 at 22:40

It sounds like you want a state given to you as $\vert x\rangle\vert f(x)\rangle$, and you wish to only conditionally rotate when $f(x)=a$, diffuse over $x$, conditionally rotate when $f(x)=a$, diffuse over $x$, ... conditionally rotate when $f(x)=a$, diffuse over $x$, and measure.

If I understand your question, then I think the issue is in the "diffuse over $x$."

Because you can't simply diffuse over the mixed state $\vert x\rangle$, but you must diffuse over the pure state $\vert x\rangle\vert f(x)\rangle$.

If $\vert f(x)\rangle$ has not been uncomputed back to $\vert 0\rangle$, then the diffusion will not lead to constructive/destructive interference in the first register. Even if you are diffusing only over the first register $\vert x\rangle$, the second register $\vert f(x)\rangle$ gets in the way.

  • $\begingroup$ Thanks. Can you please provide reference for "you can't simply diffuse over the mixed state". $\endgroup$ – mike_dole_z3 May 28 at 0:35

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