Assume $f(x)$ is n-bit to n-bit function. Let $F(x)$ be defined as $T$ iterations of $f(x)$, i.e. $F(x) = f^T(x)$.

Quantum algorithm relies on $F(x)$; it calls it $R$ times. What is the best query complexity of the algorithm in terms of calls to $f(x)$:

  1. Can we do better than $R \cdot T$ queries while maintaining negligible quantum memory complexity?

  2. Can we do better than $R \cdot T$ queries with additional quantum memory? If so, then how much? Can we do less than $T$?


1 Answer 1


That really depends on the function $f$ and the size of $R\cdot T$. Generically, I don't think that you can expect improvements over $R\cdot T$, but improvements are possible in some special cases.

For example, with the function $f$, there's a similar question in classical, and there are instances where speedups are possible, such as modular exponentiation: $f(x)=x\text{ mod }N$. There are better ways of calculating $x^a\text{ mod }N$ than just calculating $F^a(x)$, but it's specific to that function.

If you want really large $R\cdot T$, or non-integer values, there are quantum algorithms based on phase estimation. You want to have a look at this paper.

  • $\begingroup$ thanks. how about 2. ? $\endgroup$ Jun 3, 2019 at 3:47
  • $\begingroup$ @la_guesso34 I am not aware of any results in that direction. It is easy to imagine time improvements via parallelism, but probably using the same number of queries. $\endgroup$
    – DaftWullie
    Jun 3, 2019 at 5:37
  • $\begingroup$ The logic is as follows: Unitary operator describing $f$ is simply a matrix M. Then $f^T$ is $M^T$. So, to apply $F$ we just need to apply $\tilde{M} = M^T$. $\endgroup$ Jun 4, 2019 at 2:38

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