Quantum Pattern Matching Fast on Average

Abstract

The d-dimensional pattern matching problem is to find an occurrence of a pattern of length $m \times \dots \times m$ within a text of length $n \times \dots \times n$, with $n \ge m$. This task models various problems in text and image processing, among other application areas. This work describes a quantum algorithm which solves the pattern matching problem for random patterns and texts in time $\widetilde{O}((n/m)^{d/2} 2^{O(d^{3/2}\sqrt{\log m})})$. For large m this is super-polynomially faster than the best possible classical algorithm, which requires time $\widetilde{\Omega }( n^{d/2} + (n/m)^d)$. The algorithm is based on the use of a quantum subroutine for finding hidden shifts in d dimensions, which is a variant of algorithms proposed by Kuperberg.

Appendix: Proof of Claim 14

In this appendix, we analyse the algorithm of Sect. 5 for identifying hidden shifts. Let $L_i = |\mathcal {L}_i|$ denote the number of states in the list at the start of the i’th stage. Our first task is to find a lower bound on $L_{i+1}$ in terms of $L_i$ which holds with high probability. For ease of analysis, we consider a different, “relaxed” process where any state is allowed to be combined with any other state, rather than being restricted to the same bin; and where we assume there are always an even number of states. We then relate the relaxed process to the real process followed by the algorithm by bounding the number of states that the real process would need to discard, given its restriction to only combining states in the same bin.

Let $L_i^{(j)}$ denote the number of states we would have in the list $\mathcal {L}_i$ after j steps of the i’th stage, based on following the relaxed process; note that $L_i^{(0)} = L_i$. Also let $S_i^{(j)}$ denote the number of successful pairings in the j’th step of the i’th stage (again allowing any pair of states to be combined). Finally let $t_i$ denote the number of steps taken in the i’th stage (i.e. until the total number of states is at most $n^2$). Then

$$\begin{aligned} L_i^{(j)} = \frac{L_i^{(j-1)}}{2} - S_i^{(j)},\;\;\;\; L_{i+1} \ge \sum _{j=1}^{t_i} S_i^{(j)} - t_i 2^{db_i}, \end{aligned}$$

(1)

where the second inequality allows for the fact that at each step we may need to discard at most $2^{db_i}$ states in the real process, as compared with the relaxed process. The probability that each combination operation applied to a pair succeeds is independent and equal to 1 / 2. Using a Chernoff bound, we have

$$\begin{aligned} \Pr [|S_i^{(j)} - L_i^{(j-1)}/4| \ge \ln n \sqrt{L_i^{(j-1)}/4} ] \le 2 e^{-(\ln n)^2 / 2} = 2n^{-(\ln n)/2}. \end{aligned}$$

(2)

As there will be ${{\mathrm{poly}}}(n)$ steps in total throughout the algorithm, this quantity is small enough that we can take a union bound over all steps and assume that this event never happens. Making this assumption, we have

$$\begin{aligned} L_i^{(j+1)} \ge \frac{L_i^{(j)}}{2} - \frac{L_i^{(j)}}{4} - \frac{\ln n}{2} \sqrt{L_i^{(j)}} = \frac{L_i^{(j)}}{4}\left( 1 - \frac{2\ln n}{\sqrt{L_i^{(j)}}} \right) . \end{aligned}$$

(3)

As we have $L_i^{(j)} \ge n^2$ for all steps j, then

$$\begin{aligned} L_i^{(j+1)} \ge \frac{L_i^{(j)}}{4}\left( 1 - \frac{2\ln n}{n} \right) , \end{aligned}$$

implying

$$\begin{aligned} L_i^{(j)} \ge L_i \left( \frac{1 - (2\ln n)/n}{4} \right) ^j. \end{aligned}$$

(4)

By (1), (2) and (3),

$$\begin{aligned} L_{i+1}\ge & {} \sum _{j=1}^{t_i} S_i^{(j)} - t_i 2^{db_i} \ge \sum _{j=1}^{t_i} \frac{L_i^{(j-1)}}{4}\left( 1 - \frac{2\ln n}{\sqrt{L_i^{(j-1)}}}\right) \\&- t_i 2^{db_i} \ge \frac{1}{4} \sum _{j=1}^{t_i} L_i^{(j-1)} \left( 1 - \frac{2\ln n}{n}\right) - t_i 2^{db_i} \end{aligned}$$

and so by (4), writing $\xi = (2\ln n)/n$,

$$\begin{aligned} L_{i+1}\ge & {} \frac{L_i}{4} (1-\xi ) \sum _{j=0}^{t_i-1} \left( \frac{1 - \xi }{4} \right) ^j - t_i 2^{db_i} = \frac{L_i}{4} (1 - \xi ) \frac{1 - ((1-\xi )/4)^{t_i} }{1 - (1-\xi )/4} - t_i 2^{db_i} \\= & {} L_i\frac{(1 - \xi ) (1 - ((1-\xi )/4)^{t_i}) }{3 + \xi } - t_i 2^{db_i}. \end{aligned}$$

A rough lower bound that follows from (3) for large enough n is that $L_i^{(j+1)} \ge L_i^{(j)}/8$; and as $L_i^{(j+1)} \le L_i^{(j)}/2$ always, we have $1/8 \le L_i^{(j+1)}/L_i^{(j)} \le 1/2$. Using $L_i^{(0)} = L_i$ and $L_i^{(t_i)} \le n^2$ we obtain $\log _2 (L_i / n^2) \le t_i \le 3\log _2 (L_i / n^2)$. So

$$\begin{aligned} ((1-\xi )/4)^{t_i} \le 4^{-t_i} \le n^4 / L_i^2. \end{aligned}$$

Assuming that $L_i \ge n^3$ for all $1 \le i \le S$, we have

$$\begin{aligned} \frac{(1 - \xi ) (1 - ((1-\xi )/4)^{t_i}) }{3+\xi } \ge \frac{(1- (2\ln n) / n)(1-1/n^2)}{3+ (2\ln n) / n} = \frac{1- O((\log n)/n)}{3}. \end{aligned}$$

Further assume that $L_i = 2^{O(\sqrt{n})}$ for all i, implying $t_i = O(\sqrt{n})$. Then

$$\begin{aligned} L_{i+1} \ge \frac{(1- O((\log n)/n))L_i}{3} - O(\sqrt{n}) 2^{db_i}. \end{aligned}$$

We now need to determine how large $L_1$ needs to be such that $L_{S+1}$ is still quite large. Working backwards, we can take

$$\begin{aligned} L_{i-1} \le 3(1+O((\log n)/n))L_i + O(\sqrt{n})2^{db_{i-1}}. \end{aligned}$$

Thus

$$\begin{aligned} L_1\le & {} \left( 3(1+O((\log n)/n)) \right) ^S L_{S+1} + \left( 3(1+O((\log n)/n)) \right) ^{S-1} O(\sqrt{n})2^{db_S} \\&+ \dots + O(\sqrt{n})2^{db_1}. \end{aligned}$$

Assuming that $L_{S+1} = O(1)$, and $S=O(\sqrt{n})$, we have

$$\begin{aligned} L_1 = O\left( \sqrt{n} \sum _{i=1}^S 3^i 2^{db_i}\right) \end{aligned}$$

We now need to pick values $S,\, b_i$ for the number of stages and the number of bits zeroed at each stage, such that $\sum _{i=1}^S b_i = n-1$, to minimise $\sum _{i=1}^S 3^i 2^{d b_i}$. We choose these values to make all the above terms equal to $2^{c \sqrt{n}}$ for some fixed c, i.e. $b_i = (c\sqrt{n} - (\log _2 3)i)/d$ (for simplicity ignoring the fact that $b_i$ has to be rounded to an integer). Relaxing to the constraint $\sum _i b_i = n$ for simplicity, we obtain $S c \sqrt{n} - (\log _2 3)S(S+1)/2 = dn$. Hence $c = d\sqrt{n}/S + (\log _2 3)(S+1)/(2\sqrt{n})$. Minimising this over S, we get that the minimum is found at $S = \sqrt{2 \log _3 2} \sqrt{dn}$, giving

$$\begin{aligned} c = \frac{\sqrt{d}}{\sqrt{2 \log _3 2}} + \frac{\log _2 3(\sqrt{2 \log _3 2} \sqrt{dn} + 1)}{2 \sqrt{n}} = \sqrt{(2\log _2 3)d} + O(1/\sqrt{n}). \end{aligned}$$

Thus

$$\begin{aligned} L_1 = O(\sqrt{n}\,S \, 2^{(\sqrt{2(\log _2 3)d}+O(1/\sqrt{n}))\sqrt{n}}) = O(n 2^{\sqrt{(2\log _2 3)dn}}) = O(n 2^{1.781\dots \sqrt{dn}}) \end{aligned}$$

as claimed.