I recently came across the proposition that for a database containing $N$ elements with $m<\sqrt{N}$ marked elements, applying Grover's algorithm with any $T$ $(0<T<\sqrt{N}-1)$ iterations (that is, $T$ is uniformly chosen from $0$ to $\sqrt N-1$) would give you a marked state with probability of at least $\frac{1}{8}$.

However I don't quite understand how to prove this. I know that, given $m$ marked elements in $N$ total elements, the probability that you would measure a marked state at the $T^{th}$ iteration is $$ \sin ^2\bigg((T+1/2)\sqrt\frac{m}{N}\bigg)$$

Suppose you have $T=1$. Then you get the probability as $ \sin^2\bigg(1.5\sqrt\frac{m}{N}\bigg)$, and since you have $ \min(m)=1$, as $N$ increases indefinitely, you can make the value of this function as close to 0 as you want.

Is there something I'm missing?


1 Answer 1


I suspect you're misinterpreting the statement a little. The claim will not be that for every possible value of $T$ the probability is at least 1/8. Instead, the success probability, averaged across all possible values of $T$, is at least 1/8. This means that you're trying to assess something like $$ \frac{1}{\sqrt{N}-2}\sum_{T=1}^{\sqrt{N}-2}\sin^2((2T+1)\theta) $$ where $\sin\theta=\sqrt{\frac{M}{N}}$. So, sure, there are some small terms in that sum, but there are also some large terms, and they balance out.

I tried to quickly (i.e. not particularly carefully, hence I won't reproduce what I did here) to bound this, and came up with 3/8 rather than the claimed 1/8.

  • $\begingroup$ I actually got to a stage where I have an expression of the form $\frac{1}{2}-\frac{sin(4\sqrt{N} \theta)}{4\sqrt{N} sin(2\theta)}$ so having the second term there be $\frac{3}{8}$ would get me a solution (using Lemma 2, section 4 of Boyer's paper: arxiv.org/abs/quant-ph/9605034) so would you mind giving me some pointers? $\endgroup$
    – requiemman
    Oct 25, 2023 at 12:01
  • $\begingroup$ That looks like where I got to. I said that $\sin(4\sqrt{N}\theta)\leq 1$, and you can substitute $\sin(2\theta)=2\sqrt{M(N-M)}/N$. This is smallest for $M=1$, which leaves the second term being 1/8, I think, so the overall conclusion seems to be a bound of 3/8. $\endgroup$
    – DaftWullie
    Oct 25, 2023 at 12:43

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