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# Quantum Measurements for Hidden Subgroup Problems with Optimal Sample Complexityself.__wrap_n=self.__wrap_n||(self.CSS&&CSS.supports("text-wrap","balance")?1:2);self.__wrap_b=(e,t,r)=>{let n=(r=r||document.querySelector([data-br="${e}"])).parentElement,a=e=>r.style.maxWidth=e+"px";r.style.maxWidth="";let s=n.clientWidth,i=n.clientHeight,l=s/2-.25,o=s+.5,u;if(s){for(a(l),l=Math.max(r.scrollWidth,l);l+1<o;)a(u=Math.round((l+o)/2)),n.clientHeight===i?o=u:l=u;a(o*t+s*(1-t))}r.__wrap_o||"undefined"!=typeof ResizeObserver&&(r.__wrap_o=new ResizeObserver(()=>{self.__wrap_b(0,+r.dataset.brr,r)})).observe(n)};self.__wrap_n!=1&&self.__wrap_b(":R12quuultfautta:",1) One of the central issues in the hidden subgroup problem is to bound the sample complexity, i.e., the number of identical samples of coset states sufficient and necessary to solve the problem. In this paper, we present general bounds for the sample complexity of the identification and decision versions of the hidden subgroup problem. As a consequence of the bounds, we show that the sample complexity for both of the decision and identification versions is $\Theta(\log|\HH|/\log p)$ for a candidate set $\HH$ of hidden subgroups in the case that the candidate subgroups have the same prime order $p$, which implies that the decision version is at least as hard as the identification version in this case. In particular, it does so for the important instances such as the dihedral and the symmetric hidden subgroup problems. Moreover, the upper bound of the identification is attained by the pretty good measurement, which shows that the pretty good measurements can identify any hidden subgroup of an arbitrary group with at most $O(\log|\HH|)$ samples. Simplify Updated on September 24, 2006 Copy BibTeX Edited 2 times Loading... Summary There is no AI-powered summary yet, because we do not have a budget to generate summaries for all articles. 1. Buy subscription We will thank you for helping thousands of people to save their time at the top of the generated summary. If you buy our subscription, you will be able to summarize multiple articles. Pay$8
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