Quantum partial search for uneven distribution of multiple target items

Abstract

Quantum partial search algorithm is an approximate search. It aims to find a target block (which has the target items). It runs a little faster than full Grover search. In this paper, we consider quantum partial search algorithm for multiple target items unevenly distributed in a database (target blocks have different number of target items). The algorithm we describe can locate one of the target blocks. Efficiency of the algorithm is measured by number of queries to the oracle. We optimize the algorithm in order to improve efficiency. By perturbation method, we find that the algorithm runs the fastest when target items are evenly distributed in database.

Appendix: The proof of equation (66)

The optimal value of $\alpha$ and $\eta$ (31) for even distribution of target items is denoted as $\eta _0$ and $\alpha _0$ (41). Their values can be rewritten as

$$\begin{aligned}&\tan \left( 2\eta _0\sqrt{\frac{z}{K}}\right) =\frac{\sqrt{3\beta -4\beta ^2}}{1-2\beta },\quad \quad \cos \left( 2\alpha _0\sqrt{{{\bar{\tau }}}}\right) =\frac{1-2\beta }{2(1-\beta )},\nonumber \\&\sin \left( 2\alpha _0\sqrt{{{\bar{\tau }}}}\right) =\frac{\sqrt{3-4\beta }}{2(1-\beta )} \end{aligned}$$

(75)

with $\beta =t/K$. On the other hand, the approximate Eqs. (49) and (61) can be reformulated as

$$\begin{aligned} \sum _{i=1}^t\sqrt{\tau _i}\sin \left( 2\alpha _K\sqrt{\tau _i}\right)&=t\sqrt{{{\bar{\tau }}}}\sin \left( 2\alpha _0\sqrt{{{\bar{\tau }}}}\right) +P\delta ^2(\tau ); \end{aligned}$$

(76)

$$\begin{aligned} \sum _{i=1}^t \cos \left( 2\alpha _K\sqrt{\tau _i}\right)&=t\cos \left( 2\alpha _0\sqrt{{{\bar{\tau }}}}\right) +Q\delta ^2(\tau ) \end{aligned}$$

(77)

The coefficient P is found as

$$\begin{aligned} P= & {} -\frac{1}{2} \sin \left( 2\alpha _0\sqrt{{{\bar{\tau }}}}\right) \frac{t\alpha ^2_0}{\sqrt{{{\bar{\tau }}}}}+\frac{1}{4} \cos \left( 2\alpha _0\sqrt{{{\bar{\tau }}}}\right) \frac{t\alpha _0}{{{\bar{\tau }}}}\nonumber \\&\quad -\,\frac{1}{8}\sin \left( 2\alpha _0\sqrt{{{\bar{\tau }}}}\right) \frac{t}{{{\bar{\tau }}}\sqrt{{{\bar{\tau }}}}} +2t{{\bar{\tau }}}\cos \left( 2\alpha _0\sqrt{{{\bar{\tau }}}}\right) \frac{\Delta \alpha _K}{\delta ^2(\tau )} \end{aligned}$$

(78)

Substituting $\Delta \alpha _K$ (63) and the optimal value $\alpha _0$ (75) into above relation, after some algebra, coefficient P equals to

$$\begin{aligned} P=-\frac{(1-\beta )}{\sqrt{3-4\beta }}\frac{t\alpha _0^2}{\sqrt{{{\bar{\tau }}}}} -\frac{(1-2\beta )(1+\beta )}{4(1-\beta )^2}\frac{t\alpha _0}{{{\bar{\tau }}}} +\frac{\sqrt{3-4\beta }}{16(1-\beta )}\frac{t}{{{\bar{\tau }}}\sqrt{{{\bar{\tau }}}}} \end{aligned}$$

(79)

Similar, the coefficient Q in (77) has the expression

$$\begin{aligned} Q=\frac{\sqrt{3-4\beta }}{2(1-\beta )^2}\frac{t\alpha _0}{{{\bar{\tau }}}\sqrt{{{\bar{\tau }}}}} \end{aligned}$$

(80)

With the help of approximation Eqs. (76) and (77), the right-hand side of Eq. (64) in first order of $\delta ^2(\tau )$ becomes

$$\begin{aligned} \text {RHS of }(64)= & {} \frac{2\sqrt{K} \left( t\sqrt{{{\bar{\tau }}}}\sin \left( 2\alpha _0\sqrt{{{\bar{\tau }}}}\right) +P\delta ^2(\tau ) \right) }{\sqrt{z} \left( 2t\cos \left( 2\alpha _0\sqrt{{{\bar{\tau }}}}\right) +2Q\delta ^2(\tau )+K-2t\right) }\nonumber \\= & {} \frac{2t\sqrt{K{{\bar{\tau }}}}\sin \left( 2\alpha _0\sqrt{{{\bar{\tau }}}}\right) }{\sqrt{z}\left( 2t\cos \left( 2\alpha _0\sqrt{{{\bar{\tau }}}}\right) +K-2t\right) } +\frac{2\sqrt{K}P\delta ^2(\tau )}{\sqrt{z}\left( 2t\cos \left( 2\alpha _0\sqrt{{{\bar{\tau }}}}\right) +K-2t\right) }\nonumber \\&-\,\frac{4t\sqrt{K{{\bar{\tau }}}}\sin \left( 2\alpha _0\sqrt{{{\bar{\tau }}}}\right) Q \delta ^2(\tau )}{\sqrt{z}\left( 2t\cos \left( 2\alpha _0\sqrt{{{\bar{\tau }}}}\right) +K-2t\right) ^2} \end{aligned}$$

(81)

Combining above result with left-hand side of Eq. (64), and using the identity (65), we have the result

$$\begin{aligned}&\left( 1+\tan ^2\left( 2\eta _0\sqrt{\frac{z}{K}}\right) \right) \Delta \eta _K\nonumber \\&\quad = \frac{KP\delta ^2(\tau )}{z\left( 2t\cos \left( 2\alpha _0\sqrt{{{\bar{\tau }}}}\right) +K-2t\right) } -\frac{2tK\sqrt{{{\bar{\tau }}}}\sin \left( 2\alpha _0\sqrt{{{\bar{\tau }}}}\right) A\delta ^2(\tau )}{z\left( 2t\cos \left( 2\alpha _0\sqrt{{{\bar{\tau }}}}\right) +K-2t\right) ^2} \end{aligned}$$

(82)

Note that $\eta _0$ and $\alpha _0$ have the expressions (75). Then above equation gives

$$\begin{aligned} \Delta \eta _K =\left( (1-2\beta )\frac{P}{t{{\bar{\tau }}}}-\beta \sqrt{3-4\beta }\frac{Q}{t\sqrt{{{\bar{\tau }}}}}\right) \delta ^2(\tau ) \end{aligned}$$

(83)

Last, substituting coefficient P (79) and Q (80) into above equation, after some algebra, we solve $\Delta \eta _K$ in first order of $\delta ^2(\tau )$ explicitly:

$$\begin{aligned} \Delta \eta _{K}= & {} -\left( \frac{(1-\beta )(1-2\beta )}{\sqrt{3-4\beta }}\frac{\alpha _0^2}{{{\bar{\tau }}}\sqrt{{{\bar{\tau }}}}} +\frac{(4\beta ^3-8\beta ^2+3\beta +1)}{4(1-\beta )^2}\frac{\alpha _0}{{{\bar{\tau }}}^2} \right. \nonumber \\&\quad \left. +\,\frac{(1-2\beta )\sqrt{3-4\beta }}{16(1-\beta )}\frac{1}{{{\bar{\tau }}}^2\sqrt{{{\bar{\tau }}}}} \right) \delta ^2(\tau ) \end{aligned}$$

(84)