Grover's diffusion on subset of input space

Question

Is it possible to run grover's diffusion step on a subset of the possible input space?

By this, I mean is it possible to do the diffusion process with a state space isn't in a total superposition of all states.

Let's say I have a 2 qubit system in a equal superposition of states $\left|00\right>$, $\left|01\right>$ and $\left|10\right>$, but have exactly a 0 probability of measuring a $\left|11\right>$. Also, let's say I have a function $f$ that only outputs a 1 for the input 00. Is there a modification to Grover's where I can do the diffusion only on the $\left|00\right>$, $\left|01\right>$ and $\left|10\right>$ states and leave the $\left|11\right>$ state unchanged?

If that is possible, does it speedup the diffusion? Say I have a state with $n$ qubits but $2^x$ non-zero states where $x < n$, does this modified grover's algorithm still run in $\mathcal O(2^{n/2})$, or can it run in $\mathcal O(2^{x/2})$?

Answer 1

This can work. There's no reliance on powers of two or anything like that in the basic conception of the algorithm.

If $S$ is a subset of computational basis states with $N$ elements and you have a superposition: $$\left |\phi\right> = \frac{1}{\sqrt{N}}\sum_{x \in S}\left|x\right>$$ then basically all you need to do is change the classic Grover diffusion operator with the complete $n$-qubit uniform superposition $\left|\psi\right>$: $$2\left|\psi\right>\left<\psi\right| - I$$

into:

$$2\left|\phi\right>\left<\phi\right| - I$$

A common way of composing the Grover diffusion operator for an $n$-qubit system without fancy gates is to make it out of two Hadamards and some implementation of the diagonal matrix $2\left|0\right>\left<0\right| - I$ via:

$$H^{\otimes n}(2 \left|0\right>\left<0\right| - I)H^{\otimes n}$$.

To get our new diffusion operator, we just need to replace the left Hadamard gate with a gate $B$ and the right Hadamard with $B^{\dagger}$ (if they are not the same) where $B\left|0\right> = \left|\phi\right>$

From a fully general perspective, if determining whether something is in $S$ is itself difficult, implementing $B$ and having $\left|\phi\right>$ in the first place would void any advantage gained by subsetting.

Assuming no difficulty with $B$, it will run in $\mathcal{O}(\sqrt{N})$ with your new $N$. The ideal stopping point will also still be $\approx \frac{\pi}{4}\sqrt{N}$ iterations.