Circular Pattern Matching with $k$ Mismatches

TL;DR

This paper introduces the first worst-case efficient algorithms for circular pattern matching with up to k mismatches, providing practical bounds and extending recent techniques from the k-mismatch problem.

Contribution

It presents the first non-trivial worst-case upper bounds for the circular pattern matching with k mismatches problem, including two algorithms with improved time complexities.

Findings

An O(nk)-time algorithm for the problem.

An O(n + (n/m)k^4)-time algorithm extending recent techniques.

Improved the main algorithm's complexity from O(n + (n/m)k^5) to O(n + (n/m)k^4).

Abstract

The $k$-mismatch problem consists in computing the Hamming distance between a pattern $P$ of length $m$ and every length-$m$ substring of a text $T$ of length $n$, if this distance is no more than $k$. In many real-world applications, any cyclic rotation of $P$ is a relevant pattern, and thus one is interested in computing the minimal distance of every length-$m$ substring of $T$ and any cyclic rotation of $P$. This is the circular pattern matching with $k$ mismatches ($k$-CPM) problem. A multitude of papers have been devoted to solving this problem but, to the best of our knowledge, only average-case upper bounds are known. In this paper, we present the first non-trivial worst-case upper bounds for the $k$-CPM problem. Specifically, we show an $O(nk)$-time algorithm and an $O(n+\frac{n}{m}\,k^4)$-time algorithm. The latter algorithm applies in an extended way a technique that was very recently developed for the $k$-mismatch problem [Bringmann et al., SODA 2019]. A preliminary version of this work appeared at FCT 2019. In this version we improve the time complexity of the main algorithm from $O(n+\frac{n}{m}\,k^5)$ to $O(n+\frac{n}{m}\,k^4)$.

Equations

$${\mathsf{rot}}_x(S)=VU\text{, where }S=UV\text{ and }|U|=x.$$

$$M(p,x)=\{(i,(i-p+x)\mathrm{mod}m):i\in {\mathbf{W}}_p\}.$$

$${\mathsf{Chain}}_q(I,a)=I\cup (I\oplus q)\cup (I\oplus 2q)\cup \cdots \cup (I\oplus aq).$$

$$t^{\prime }[x,y]=\sum _{i=1}^x\sum _{j=1}^{y}t[i,j].$$

$$Z^{\prime }:=\{z\in Z:{\exists }_{x\in X}z\equiv x(\mathrm{mod}q)\},$$

$$\{z\in Z_j:z\mathrm{mod}q\in X\}.$$

$$\mathsf{Mis}(S,S^{\prime })=\{i=0,\dots ,|S|-1:S[i]\ne S^{\prime }[i]\}.$$

$${\mathsf{Misper}}_k(S,i,j)={\mathsf{LeftMisper}}_k(S,i,j)\cup {\mathsf{RightMisper}}_k(S,i,j)$$

$$I={\mathsf{Misper}}_{k+1}(S,i,j)\text{ and }{I}^{\prime }={\mathsf{Misper}}_{k+1}(S^{\prime },i,j).$$

$$\text{textsc}Pairs-Match(T,I,P,J)=\bigcup _{i\in I,j\in J}\text{textsc}Pair-Match(T,i,P,j).$$

$$L_1=J\cap {\mathbf{W}}_p,L_2=J^{\prime }\cap {\mathbf{W}}_x.$$

Full text

A preliminary version of this work appeared at FCT 2019. In this version we improve the time complexity of the main algorithm from $\mathcal{O}(n+\frac{n}{m}\,{k^{5}})$ to $\mathcal{O}(n+\frac{n}{m}\,{k^{4}})$.

1 Introduction

Pattern matching is a fundamental problem in computer science [1]. It consists in finding all substrings of a text $T$ of length $n$ that match a pattern $P$ of length $m$. In many real-world applications, a measure of similarity is usually introduced allowing for approximate matches between the given pattern and substrings of the text. The most widely-used similarity measure is the Hamming distance between the pattern and all length-$m$ substrings of the text.

Computing the Hamming distance between $P$ and all length-$m$ substrings of $T$ has been investigated for the past 30 years. The first efficient solution requiring $\mathcal{O}(n\sqrt{m\log m})$ time was independently developed by Abrahamson [2] and Kosaraju [3] in 1987. The $k$-mismatch version of the problem asks for finding only the substrings of $T$ that are close to $P$, specifically, at Hamming distance at most $k$. The first efficient solution to this problem running in $\mathcal{O}(nk)$ time was developed in 1986 by Landau and Vishkin [4]. It took almost 15 years for a breakthrough result by Amir et al. improving this to $\mathcal{O}(n\sqrt{k\log k})$ [5]. More recently, there has been a resurgence of interest in the $k$-mismatch problem. Clifford et al. gave an $\mathcal{O}((n/m)(k^{2}\log k)+n\text{polylog}n)$-time algorithm [6], which was subsequently improved further by Gawrychowski and Uznański to $\mathcal{O}((n/m)(m+k\sqrt{m})\text{polylog}n)$ [7]. In [7], the authors have also provided evidence that any further progress in this problem is rather unlikely.

The $k$-mismatch problem has also been considered on compressed representations of the text [8, 9, 10, 11], in the parallel model [12], and in the streaming model [13, 6, 14]. Furthermore, it has been considered in non-standard stringology models, such as the parameterized model [15] and the order-preserving model [16].

In many real-world applications, such as in bioinformatics [17, 18, 19, 20] or in image processing [21, 22, 23, 24], any cyclic shift (rotation) of $P$ is a relevant pattern, and thus one is interested in computing the minimal distance of every length-$m$ substring of $T$ and any cyclic rotation of $P$, if this distance is no more than $k$. This is the circular pattern matching with $k$ mismatches ($k$-CPM) problem. A multitude of papers [25, 26, 27, 28, 29, 30] have thus been devoted to solving the $k$-CPM problem but, to the best of our knowledge, only average-case upper bounds are known; i.e. in these works the assumption is that text $T$ is uniformly random. The main result states that, after preprocessing pattern $P$, the average-case optimal search time of $\mathcal{O}(n\frac{k+\log m}{m})$ [31] can be achieved for certain values of the error ratio $k/m$ (see [29, 25] for more details on the preprocessing costs). Note that the exact (no mismatches allowed) version of the CPM problem can be solved as fast as exact pattern matching; namely, in $\mathcal{O}(n)$ time [32].

In this paper, we draw our motivation from (i) the importance of the $k$-CPM problem in real-world applications and (ii) the fact that no (non-trivial) worst-case upper bounds are known. Trivial here refers to running the fastest-known algorithm for the $k$-mismatch problem [7] separately for each of the $m$ rotations of $P$. This yields an $\mathcal{O}(n(m+k\sqrt{m})\text{polylog}n)$-time algorithm for the $k$-CPM problem. This is clearly unsatisfactory: it is a simple exercise to design an $\mathcal{O}(nm)$-time or an $\mathcal{O}(nk^{2})$-time algorithm. In an effort to tackle this unpleasant situation, we present two much more efficient algorithms: a simple $\mathcal{O}(nk)$-time algorithm and an $\mathcal{O}(n+\frac{n}{m}\,{k^{4}})$-time algorithm. Our second algorithm applies in an extended way a technique that was developed very recently for $k$-mismatch pattern matching in grammar compressed strings by Bringmann et al. [9]. We also show that both of our algorithms can be implemented in $\mathcal{O}(m)$ space.

A preliminary version of this work was published as [33].

Our approach

We first consider a simple version of the problem (called Anchor-Match) in which we are given a position in $T$ (an anchor) which belongs to potential $k$-mismatch circular occurrences of $P$. A simple $\mathcal{O}(k)$-time algorithm is given (after linear-time preprocessing) to compute all relevant occurrences. By considering separately each position in $T$ as an anchor we obtain an $\mathcal{O}(nk)$-time algorithm. The concept of an anchor is extended to the so-called matching pairs: when we know a pair of positions, one in $P$ and the other in $T$, that are aligned. Then comes the idea of a sample $\mathcal{S}$, which is a fragment of $P$ of length $\Theta(m/k)$ which supposedly exactly matches a corresponding fragment in $T$. We choose $\mathcal{O}(k)$ samples and work for each of them and for windows of $T$ of size $2m$. As it is typical in many versions of pattern matching, our solution is split into periodic and non-periodic cases. If $\mathcal{S}$ is non-periodic the sample occurs only $\mathcal{O}(k)$ times in a window and each occurrence gives a matching pair (and consequently two possible anchors). Then we perform Anchor-Match for each such anchor. The hard part is the case when $\mathcal{S}$ is periodic. Here we compute all exact occurrences of $\mathcal{S}$ and obtain $\mathcal{O}(k)$ groups of occurrences, each one being an arithmetic progression. Now each group is processed using the approach “few matches or almost periodicity” of Bringmann et al. [9]. In the latter case periodicity is approximate allowing up to $k$ mismatches. Finally, we are able to decrease the exponent of $k$ by one in the complexity using a marking trick.

2 Preliminaries

Let $S=S[0]S[1]\cdots S[n-1]$ be a string of length $|S|=n$ over an integer alphabet $\Sigma$. The elements of $\Sigma$ are called letters. For two positions $i$ and $j$ on $S$, we denote by $S[i\mathinner{.\,.}j]=S[i]\cdots S[j]$ the fragment of $S$ that starts at position $i$ and ends at position $j$ (it equals the empty string $\varepsilon$ if $j<i$). A prefix of $S$ is a fragment that starts at position [math], i.e. of the form $S[0\mathinner{.\,.}j]$, and a suffix is a fragment that ends at position $n-1$, i.e. of the form $S[i\mathinner{.\,.}n-1]$. For an integer $p$, we define the $p$th power of $S$, denoted by $S^{p}$, as the string obtained from concatenating $p$ copies of $S$. $S^{\infty}$ denotes the string obtained by concatenating infinitely many copies. If $S$ and $S^{\prime}$ are two strings of the same length, then by $S=_{k}S^{\prime}$ we denote the fact that $S$ and $S^{\prime}$ have at most $k$ mismatches, that is, that the Hamming distance between $S$ and $S^{\prime}$ does not exceed $k$.

We say that a string $S$ has period $q$ if $S[i]=S[i+q]$ for all $i=0,\ldots,|S|-q-1$. String $S$ is periodic if it has a period $q$ such that $2q\leq|S|$. We denote the smallest period of $S$ by $\textsf{per}(S)$. Fine and Wilf’s periodicity lemma [34] asserts that if a string of length $n$ has periods $p$ and $q$ and $n\geq p+q-1$, then the string has a period $\gcd(p,q)$.

For a string $S$, by $\mathsf{rot}_{x}(S)$ for $0\leq x<|S|$, we denote the string that is obtained from $S$ by moving the prefix of $S$ of length $x$ to its suffix. We call the string $\mathsf{rot}_{x}(S)$ (or its representation $x$) a rotation of $S$. More formally, we have

[TABLE]

2.1 Anatomy of Circular Occurrences

In what follows, we denote by $m$ the length of the pattern $P$ and by $n$ the length of the text $T$. We say that $P$ has a $k$-mismatch circular occurrence (in short $k$-occurrence) in $T$ at position $p$ if $T[p\mathinner{.\,.}p+m-1]=_{k}\mathsf{rot}_{x}(P)$ for some rotation $x$. In this case, the position $x$ in the pattern is called the split point of the pattern and $p+(m-x)\bmod m$ 111 The modulo operation is used to handle the trivial rotation with $x=0$. is called the anchor in the text. In other words, if $P=UV$ and its rotation $VU$ occurs in $T$, then the first position of $V$ in $P$ is the split point of this occurrence, and the first position of $U$ in $T$ is the anchor of this occurrence (see Fig. 1).

The main problem in scope can now be stated as follows.

$k$-CPM Problem

Input: Text $T$ of length $n$, pattern $P$ of length $m$, and positive integer $k$.

Output: All positions of $T$ that contain a $k$-occurrence of $P$.

For an integer $z$, let us denote $\mathbf{W}_{z}=[z\mathinner{.\,.}z+m-1]$ (window of size $m$). Intuitively, this window corresponds to a length-$m$ fragment of the text $T$. For a $k$-occurrence at position $p$ of $T$ with rotation $x$, we introduce a set of pairs of positions in the fragment of the text and the corresponding positions from the original (unrotated) pattern $P$:

[TABLE]

The pairs $(i,j)\in M(p,x)$ are called matching pairs of an occurrence $p$ with rotation $x$. In particular, $(p+((m-x)\bmod m),0)\in M(p,x)$. An example is provided in Fig. 2.

3 An $\mathcal{O}(nk)$-time Algorithm

We first introduce an auxiliary problem in which one wants to compute all $k$-occurrences of $P$ in $T$ with a given anchor $\mathbf{a}$. This problem describes the core computational task in our first solution.

Anchor-Match Problem

Input: Text $T$ of length $n$, pattern $P$ of length $m$, positive integer $k$, and position $\mathbf{a}$.

Output: All $k$-occurrences $p$ of $P$ in $T$ with anchor $\mathbf{a}$, represented as a collection of $\mathcal{O}(k)$ intervals.

For a string $X$ let us denote by $X_{(i)}$ and $X^{(i)}$ its fragments of length $m$ starting at position $i$ and ending at position $i$, respectively. Moreover, for a binary string $X$, by $||X||$ we denote the arithmetic sum of characters in $X$. We define the following auxiliary problem.

Light-Fragments Problem

Input: Positive integers $m$, $k$ and a string $V$ of length $n$ over alphabet $\{0,1\}$ containing $\mathcal{O}(k)$ non-zero characters. The string $V$ is specified by its positions with non-zero characters (sorted increasingly).

Output: The set $A\;=\;\{\,i\ :\ ||V_{(i)}||\,\leq\,k\,\}$ represented as a collection of $\mathcal{O}(k)$ intervals.

Lemma 1**.**

The Light-Fragments problem can be solved in $\mathcal{O}(k)$ time.

Proof.

Let $I$ be the set of positions of $V$ with non-zero characters; $|I|=\mathcal{O}(k)$ by definition. We define a piecewise constant function $f$ on integers such that $f(x)=|I\cap[0,x]|$. Let $g(x)=f(x+m-1)-f(x-1)$. Then $g(x)=|I\cap[x,x+m-1]|$. Function $f$ has $\mathcal{O}(k)$ pieces, so $g$ has $\mathcal{O}(k)$ pieces as well and both can be computed in $\mathcal{O}(k)$ time since $I$ is sorted. In the end we report the pieces where $g$ has value at most $k$. ∎

Let us recall a standard algorithmic tool for preprocessing text $T$. We denote the length of the longest common prefix (resp. suffix) of two strings $U$ and $V$ by $\mathsf{lcp}(U,V)$ (resp. $\mathsf{lcs}(U,V)$). There is an $\mathcal{O}(n)$-sized data structure answering such queries over suffixes (resp. prefixes) of $T$ in $\mathcal{O}(1)$ time after $\mathcal{O}(n)$-time preprocessing. It consists of the suffix array of $T$ and a data structure for answering range minimum queries; see [1]. Using the kangaroo method [4, 12], these queries can handle mismatches; after an $\mathcal{O}(n)$-time preprocessing of $T$, longest common prefix (resp. suffix) queries with up to $k$ mismatches can be answered in $\mathcal{O}(k)$ time.

Lemma 2**.**

After $\mathcal{O}(n)$-time preprocessing, the answer to Anchor-Match problem, represented as a union of $\mathcal{O}(k)$ intervals, can be computed in $\mathcal{O}(k)$ time.

Proof.

In the preprocessing we prepare a data structure for $\mathsf{lcp}$ queries in $P\#T$, for a special character $\#$ that occurs neither in $P$ nor in $T$.

Consider now a query for an anchor $\mathbf{a}$ over $T$. Let $L=T^{(\mathbf{a}-1)}$ and $R=T_{(\mathbf{a})}$. We define a binary string $L^{\prime}$ such that $L^{\prime}[i]=1$ if and only if $L[i]\neq P[i]$. We define $R^{\prime}$ analogously. Let $L^{\prime\prime}$ be the longest suffix of $L^{\prime}$ such that $||L^{\prime\prime}||\leq k$. Let $R^{\prime\prime}$ be the longest prefix of $R^{\prime}$ such that $||R^{\prime\prime}||\leq k$.

Using the kangaroo method [4, 12], the strings $L^{\prime\prime}$, $R^{\prime\prime}$ can be constructed in $\mathcal{O}(k)$ time. The Anchor-Match problem now reduces to the Light-Fragments problem for string $V=L^{\prime\prime}R^{\prime\prime}$. ∎

For an illustration of Lemma 2 inspect Fig. 3.

Proposition 1**.**

$k$-CPM can be solved in $\mathcal{O}(nk)$ time and $\mathcal{O}(n)$ space.

Proof.

We invoke the algorithm of Lemma 2 for all $\mathbf{a}\in[0\mathinner{.\,.}n-1]$ and obtain $\mathcal{O}(nk)$ intervals of $k$-occurrences of $P$ in $T$. Instead of storing all the intervals, we count how many intervals start and end at each position of $T$. We can then compute the union of the intervals by processing these counts from left to right. ∎

4 Algorithmic tools

In this section we introduce further algorithmic tools to get to our second solution.

4.1 Internal Queries in a Text

Let $T$ be a string of length $n$ called text. An Internal Pattern Matching (IPM) query, for two given fragments $F$ and $G$ of the text, such that $|G|\leq 2|F|$, computes the set of all occurrences of $F$ in $G$. If there are more than two occurrences, they form an arithmetic sequence with difference $\textsf{per}(F)$. A data structure for IPM queries in $T$ can be constructed in $\mathcal{O}(n)$ time and answers queries in $\mathcal{O}(1)$ time (see [35] and [36, Theorem 1.1.4]). It can be used to compute all occurrences of a given fragment $F$ of length $p$ in $T$, expressed as a union of $\mathcal{O}(n/p)$ pairwise disjoint arithmetic sequences with difference $\textsf{per}(F)$, in $\mathcal{O}(n/p)$ time.

4.2 Simple Geometry of Arithmetic Sequences of Intervals

We next present algorithms that will be used in subsequent proofs for handling regular sets of intervals.

For an interval $I$ and an integer $r$, let $I\oplus r\;=\;\{\,i+r\>:\;i\in I\,\}$. We define

[TABLE]

This set is further called an interval chain (with difference $q$). Note that it can be represented in $\mathcal{O}(1)$ space using four integers: $a$, $q$, and the endpoints of $I$. An illustration of an interval chain representing the output for a problem defined in the next subsection can be found in Fig. 4.

For a given value of $q$, let us fit the integers from $[1\mathinner{.\,.}n]$ into the cells of a grid of width $q$ so that the first row consists of numbers 1 through $q$, the second of numbers $q+1$ to $2q$, etc. Let us call this grid $\mathcal{G}_{q}$. A chain $\mathsf{Chain}_{q}$ can be conveniently represented in the grid $\mathcal{G}_{q}$ using the following lemma from [37].

Lemma 3** ([37]).**

The set $\mathsf{Chain}_{q}(I,a)$ is a union of $\mathcal{O}(1)$ orthogonal rectangles in $\mathcal{G}_{q}$. The coordinates of the rectangles can be computed in $\mathcal{O}(1)$ time.

Lemma 4 can be used to compute a union of interval chains.

Lemma 4**.**

Given $c$ interval chains, all of which have difference $q$ and are subsets of $[0\mathinner{.\,.}n]$, the union of these chains, expressed as a subset of $[0\mathinner{.\,.}n]$, can be computed in $\mathcal{O}(n+c)$ time.

Proof.

By Lemma 3, the problem reduces to computing the union of $\mathcal{O}(c)$ rectangles on a grid of total size $n$. Let $t$ be a 2D array of the same shape as $\mathcal{G}_{q}$, initially set to zeroes. For a rectangle with opposite corners $(x_{1},y_{1})$ and $(x_{2},y_{2})$, with $x_{1}\leq x_{2}$ and $y_{1}\leq y_{2}$, we increment $t[x_{1},y_{1}]$, decrement $t[x_{2}+1,y_{1}]$ and $t[x_{1},y_{2}+1]$, and increment $t[x_{2}+1,y_{2}+1]$ (provided that the respective cells are within the array). This takes $\mathcal{O}(c)$ time. We then compute prefix sums of $t$, which are defined as

[TABLE]

Such values can be computed in time proportional to the size of the grid, i.e., in $\mathcal{O}(n)$ time. Finally, we note that $(x,y)$ is contained in $t^{\prime}[x,y]$ rectangles, concluding the proof. ∎

Remark 1**.**

The proof of Lemma 4 is essentially based on an idea that was used, for example, for reducing the decision version of range stabbing queries in 2D to weighted range counting queries in 2D (cf. [38]).

We will also use the following auxiliary lemma.

Lemma 5**.**

Let $X$ and $Z$ be intervals and $q$ be a positive integer. The set

[TABLE]

represented as a disjoint sum of at most three interval chains, each with difference $q$, can be computed in $\mathcal{O}(1)$ time.

Proof.

If $|X|\geq q$, then $Z^{\prime}=Z$ is an interval and thus an interval chain. If $|X|<q$, then $Z^{\prime}$ can be divided into disjoint intervals of length smaller than or equal to $|X|$. The intervals from the second until the penultimate one (if any such exist), have length $|X|$. Hence, they can be represented as a single chain, as the first element of each such interval is equal $\bmod\;q$ to the first element of $X$. The two remaining intervals can be treated as chains as well. ∎

4.3 The Aligned-Light-Sum Problem

We define the following abstract problem that resembles the Light-Fragments problem from Section 3.

Aligned-Light-Sum Problem

Input: Positive integers $m$, $k$, $q$ and strings $U,V$ over alphabet $\{0,1\}$, each containing $\mathcal{O}(k)$ non-zero characters. The strings are specified by their positions with non-zero characters.

Output: The set $A\;=\;\{\,i\ :\ (\exists\,j)\;||U_{(i)}||\,+\,||V_{(j)}||\,\leq\,k\ \land\ j\equiv i\pmod{q}\}$.

Lemma 6**.**

The Aligned-Light-Sum problem can be solved in $\mathcal{O}(k^{2})$ time with the output represented as a collection of $\mathcal{O}(k^{2})$ interval chains, each with difference $q$.

Proof.

Let $I$ and $I^{\prime}$ be the positions with non-zero characters in $U^{\prime}$ and $V^{\prime}$, respectively. We partition the set $\{0,\ldots,|U^{\prime}|-m\}$ into intervals such that for all $j$ in an interval, the set $\mathbf{W}_{j}\cap I$ is the same. For this, we use a sliding window approach. We generate events corresponding to $x$ and $x-m+1$ for all $x\in I$ and sort them. When $j$ crosses an event, the set $\mathbf{W}_{j}\cap I$ changes. Thus we obtain a partition of $\{0,\ldots,|U^{\prime}|-m\}$ into intervals $Z_{1},\ldots,Z_{n_{1}}$. We obtain a similar partition of $\{0,\ldots,|V^{\prime}|-m\}$ into intervals $Z^{\prime}_{1},\ldots,Z^{\prime}_{n_{2}}$. We have $n_{1},n_{2}=\mathcal{O}(k)$.

Let us now fix $Z_{j}$ and $Z^{\prime}_{j^{\prime}}$. First we check if the condition on the total number of non-zero characters is satisfied for any $z\in Z_{j}$ and $z^{\prime}\in Z^{\prime}_{j^{\prime}}$. If so, we compute the set $Z^{\prime}_{j^{\prime}}\bmod q=\{z^{\prime}\bmod q\,:\,z^{\prime}\in Z^{\prime}_{j^{\prime}}\}$. It is a single circular interval and can be computed in constant time. The required result is

[TABLE]

By Lemma 5, this set can be represented as a union of three chains, each with difference $q$ and, as such, can be computed in $\mathcal{O}(1)$ time. The conclusion follows. ∎

5 An $\mathcal{O}(n+{k^{5}})$-time Algorithm for Short Texts

In this section we proceed by assuming that $m\leq n\leq 2m$ and aim at an $\mathcal{O}(n+k^{5})$-time algorithm. In the next sections we remove this assumption and reduce the exponent of $k$ to 4.

A (deterministic) sample is a short fragment $\mathcal{S}$ of the pattern $P$. An occurrence in the text without any mismatch is called exact. We introduce a problem of Sample-Match that consists in finding all $k$-occurrences of $P$ in $T$ such that $\mathcal{S}$ matches exactly a fragment of length $|\mathcal{S}|$ in $T$.

We split the pattern $P$ into $2k+3$ fragments of length $\left\lfloor\frac{m}{2k+3}\right\rfloor$ or $\left\lceil\frac{m}{2k+3}\right\rceil$ each. In any $k$-occurrence of $P$ in $T$ at least $k+2$ of those fragments will occur exactly in $T$ (up to $k$ fragments may occur with one mismatch and at most one fragment will contain the split point).

Remark 2**.**

We force at least $k+2$ fragments (instead of just one) to match exactly for two reasons: (1) in order to have more than a half of them match exactly, which will guarantee that the interval chains that are obtained from applications of the Aligned-Light-Match problem have the same difference and thus can be unioned using Lemma 4 (see the proof of Proposition 2); and (2) for the marking trick in the next section.

Let us henceforth fix a sample $\mathcal{S}$ as one of these fragments, let $p_{\mathcal{S}}$ be its starting position in $P$, and let $m_{\mathcal{S}}=|\mathcal{S}|$. We assume that the split point $x$ in $P$ is to the right of $\mathcal{S}$, i.e., that $x\geq p_{\mathcal{S}}+m_{\mathcal{S}}$. The opposite case—that $x<p_{\mathcal{S}}$—can be handled analogously.

5.1 Matching Non-Periodic Samples

Let us assume that $\mathcal{S}$ is non-periodic. Further let $j$ denote the starting position of $\mathcal{S}$ in $P$ and $i$ denote a starting position of an occurrence of $\mathcal{S}$ in $T$. We introduce a problem in which, intuitively, we compute all $k$-occurrences of $P$ in $T$ which align $T[i]$ with $P[j]$.

Pair-Match Problem

Input: Text $T$ of length $n$, pattern $P$ of length $m$, positive integer $k$, and two integers $i\in[0\mathinner{.\,.}n-1]$ and $j\in[0\mathinner{.\,.}m-1]$.

Output: The set $A(i,j)$ of all positions in $T$ where we have a $k$-mismatch occurrence of $\mathsf{rot}_{x}(P)$ for some $x$ such that $(i,j)$ is a matching pair.

We show how to solve the Sample-Match problem for a non-periodic sample $\mathcal{S}$ in $\mathcal{O}(k^{2})$ time in two steps. First, we show how to solve the Pair-Match problem for a given $(i,j)$ in $\mathcal{O}(k)$ time. We then apply this solution for each occurrence $i$ of $\mathcal{S}$ in $T$ by noticing that we can have $\mathcal{O}(k)$ such occurrences in total.

Lemma 7**.**

After $\mathcal{O}(n)$-time preprocessing, the Pair-Match problem can be solved in $\mathcal{O}(k)$ time, where the output is represented as a union of $\mathcal{O}(k)$ intervals.

Proof.

Recall that the Anchor-Match problem returns all $k$-occurrences of $P$ in $T$ with a given anchor. The Pair-Match problem can be essentially reduced to the Anchor-Match problem, since for a given matching pair of characters in $P$ and $T$, there are at most two ways of choosing the anchor depending on the relation between $j$ and a split point: these are $i-j$ and $i+|P|-j$ (see Fig. 5). Clearly, we choose $i-j$ as an anchor only if $i-j\geq 0$ and $i+|P|-j$ only if $i+|P|-j<|T|$. We then have to take the intersection of the answer with $[i-m+1\mathinner{.\,.}i]$ to ensure that the $k$-occurrence contains position $i$. ∎

Lemma 8**.**

After $\mathcal{O}(n)$-time preprocessing, the Sample-Match problem for a non-periodic sample $\mathcal{S}$ can be solved in $\mathcal{O}(k^{2})$ time and outputs a union of $\mathcal{O}(k^{2})$ intervals of occurrences.

Proof.

Since $\mathcal{S}$ is non-periodic, it has $\mathcal{O}(k)$ occurrences in $T$, which can be computed in $\mathcal{O}(k)$ time after an $\mathcal{O}(n)$-time preprocessing using IPM queries [35, 36] in $P\#T$. Let $j$ be the starting position of $\mathcal{S}$ in $P$ and $i$ be a starting position of an occurrence of $\mathcal{S}$ in $T$. For each of the $\mathcal{O}(k)$ such pairs $(i,j)$, the computation reduces to the Pair-Match problem for $i$ and $j$. The statement follows by Lemma 7. ∎

5.2 Matching Periodic Samples

Let us assume that $\mathcal{S}$ is periodic, i.e., it has a period $q$ with $2q\leq|\mathcal{S}|$. A fragment of a string $S$ containing an inclusion-maximal arithmetic sequence of occurrences of $\mathcal{S}$ in $S$ with difference $q$ is called here an $\mathcal{S}$-run. If $\mathcal{S}$ matches a fragment in the text, then the match belongs to an $\mathcal{S}$-run. For example, the underlined fragment of $S=\texttt{bb\text@underline{ababab}aa}$ is an $\mathcal{S}$-run for $\mathcal{S}=\texttt{abab}$.

Lemma 9**.**

If $\mathcal{S}$ is periodic, the number of $\mathcal{S}$-runs in the text is $\mathcal{O}(k)$ and they can all be computed in $\mathcal{O}(k)$ time after $\mathcal{O}(n)$-time preprocessing.

Proof.

We construct the data structure for IPM queries on $P\#T$. This allows us to compute the set of all occurrences of $\mathcal{S}$ in $T$ as a collection of $\mathcal{O}(k)$ arithmetic sequences with difference $\textsf{per}(\mathcal{S})$. We then check for every two consecutive sequences if they can be joined together. This takes $\mathcal{O}(k)$ time and results in $\mathcal{O}(k)$ $\mathcal{S}$-runs. ∎

For two equal-length strings $S$ and $S^{\prime}$, we denote the set of their mismatches by

[TABLE]

We say that position $a$ in $S$ is a misperiod with respect to the fragment $S[i\mathinner{.\,.}j]$ if $S[a]\neq S[b]$ where $b$ is the unique position such that $b\in[i\mathinner{.\,.}j]$ and $(j-i+1)\mid(b-a)$.

We define the set $\mathsf{LeftMisper}_{k}(S,i,j)$ as the set of $k$ maximal misperiods that are smaller than $i$ and $\mathsf{RightMisper}_{k}(S,i,j)$ as the set of $k$ minimal misperiods that are greater than $j$. Each of the sets can have less than $k$ elements if the corresponding misperiods do not exist. We further define

[TABLE]

and $\mathsf{Misper}(S,i,j)=\bigcup_{k=0}^{\infty}\mathsf{Misper}_{k}(S,i,j)$.

The following lemma captures a combinatorial property behind the new technique of Bringmann et al. [9]. The intuition is shown in Fig. 6.

Lemma 10**.**

Assume that $S=_{k}S^{\prime}$ and that $S[i\mathinner{.\,.}j]=S^{\prime}[i\mathinner{.\,.}j]$. Let

[TABLE]

If $I\cap I^{\prime}=\emptyset$, then $\mathsf{Mis}(S,S^{\prime})=I\cup I^{\prime}$, $I=\mathsf{Misper}(S,i,j)$, and $I^{\prime}=\mathsf{Misper}(S^{\prime},i,j)$.

Proof.

Let $J=\mathsf{Misper}(S,i,j)$ and $J^{\prime}=\mathsf{Misper}(S^{\prime},i,j)$. We first observe that $I\cup I^{\prime}\subseteq\mathsf{Mis}(S,S^{\prime})$ since $I\cap I^{\prime}=\emptyset$. Then, $S=_{k}S^{\prime}$ implies that $|\mathsf{Mis}(S,S^{\prime})|\leq k$ and hence $|I|\leq k$ and $|I^{\prime}|\leq k$, which in turn implies that $I=J$ and $I^{\prime}=J^{\prime}$. The observation that $\mathsf{Mis}(S,S^{\prime})\subseteq J\cup J^{\prime}$ concludes the proof. ∎

A string $S$ is $k$-periodic w.r.t. an occurrence $i$ of $Q$ if $|\mathsf{Misper}(S,i,i+|Q|-1)|\leq k$. In this case, $|Q|$ is called the $k$-period. In particular, in the conclusion of the above lemma $S$ and $S^{\prime}$ are $|I|$-periodic and $|I^{\prime}|$-periodic, respectively, w.r.t. $Q=S[i\mathinner{.\,.}j]=S^{\prime}[i\mathinner{.\,.}j]$. This notion forms the basis of the following auxiliary problem in which we search for $k$-occurrences in which the rotation of the pattern and the fragment of the text are $k$-periodic for the same period $Q$.

Let $U$ and $V$ be two strings and $J$ and $J^{\prime}$ be sets containing positions in $U$ and $V$, respectively. We say that length-$m$ fragments $U[p\mathinner{.\,.}p+m-1]$ and $V[x\mathinner{.\,.}x+m-1]$ are $(J,J^{\prime})$-disjoint if the sets $(\mathbf{W}_{p}\cap J)\ominus p$ and $(\mathbf{W}_{x}\cap J^{\prime})\ominus x$ are disjoint. For example, if $J=\{2,4,11,15,16,17\}$, $J^{\prime}=\{5,6,15,18,19\}$, and $m=12$, then $U[3\mathinner{.\,.}14]$ and $V[6\mathinner{.\,.}17]$ are $(J,J^{\prime})$-disjoint for:

$U=$     ab$\bullet$ a$\bullet$b abc ab$\bullet$ abc \bullet$$\bullet$$\bullet

$V=$ abc ab$\bullet$ $\bullet$bc abc abc $\bullet$bc \bullet$$\bulletc

Let us introduce an auxiliary problem that is obtained in the case that is shown in the conclusion of the above lemma (i.e., misperiods in the rotation of $P$ and the corresponding fragment of $T$ are not aligned); see also Fig. 7.

Periodic-Periodic-Match Problem

Input: A string $U$ which is $2k$-periodic w.r.t. an exact occurrence $i$ of a length-$q$ string $Q$ and a string $V$ which is $2k$-periodic w.r.t. an exact occurrence $i^{\prime}$ of the same string $Q$ such that $m\leq|U|,|V|\leq 2m$ and

$J=\mathsf{Misper}(U,i,i+q-1),\ \ J^{\prime}=\mathsf{Misper}(V,i^{\prime},i^{\prime}+q-1)$.

(The strings $U$ and $V$ are not stored explicitly.)

Output: The set of positions $p$ in $U$ for which there exists a $(J,J^{\prime})$-disjoint $k$-occurrence $U[p\mathinner{.\,.}p+m-1]$ of $V[x\mathinner{.\,.}x+m-1]$ for $x$ such that

$i-p\equiv i^{\prime}-x\pmod{q}$.

Intuitively, the modulo condition on the output of the Periodic-Periodic-Match problem corresponds to the fact that the approximate periodicity is aligned.

In the Periodic-Periodic-Match problem we search for $k$-occurrences in which none of the misperiods in $J$ and $J^{\prime}$ are aligned. In this case each of the misperiods accounts for one mismatch in the $k$-occurrence. In the lemma below we reduce the Periodic-Periodic-Match problem to the Aligned-Light-Sum problem, in which we only require that the total number of misperiods in an occurrence is at most $k$. This way all the $(J,J^{\prime})$-disjoint $k$-occurrences can be found. Also additional occurrences where two misperiods are aligned can be reported, but they are still valid $k$-occurrences (actually, $k^{\prime}$-occurrences for some $k^{\prime}<k$).

Lemma 11**.**

We can compute in $\mathcal{O}(k^{2})$ time a set of $k$-occurrences of $P$ in $T$ represented as $\mathcal{O}(k^{2})$ interval chains, each with difference $q$, that is a superset of the solution to the Periodic-Periodic-Match problem.

Proof.

In the Periodic-Periodic-Match problem the modulo condition forces the exact occurrences of the approximate period to match. Hence, it guarantees that all the positions except the misperiod positions match. Now the Aligned-Light-Sum problem highlights these positions inside the string.

Claim**.**

Periodic-Periodic-Match can be reduced in $\mathcal{O}(k)$ time to the Aligned-Light-Sum problem so that we obtain a superset of the desired result. The potential extra positions do not satisfy only the $(J,J^{\prime})$-disjointness condition.

Proof.

Let the parameters $m$, $k$ and $q$ remain unchanged. We create strings $U^{\prime}$ and $V^{\prime}$ of length $|U|$ and $|V|$, respectively, with positions with non-zero characters in the sets $I$ and $I^{\prime}=J^{\prime}$, respectively. Then we prepend $U^{\prime}$ with $z=(i^{\prime}-i)\bmod q$ zeros. Let $A$ be the solution to the Aligned-Light-Sum problem for $U^{\prime}$ and $V^{\prime}$. Then $(A\ominus z)\cap\mathbb{Z}_{\geq 0}$ is a superset of the solution to Periodic-Periodic-Match; the elements of the set that correspond to matches where non-zero elements of the strings $U^{\prime}$, $V^{\prime}$ were aligned do not satisfy the disjointness condition. ∎

Now the thesis follows from Lemma 6. ∎

Let us further define

[TABLE]

Let $A$ be a set of positions in a string $S$ and $m$ be a positive integer. We then denote $A\bmod m=\{a\bmod m\,:\,a\in A\}$ and by $\mathsf{frag}_{A}(S)$ we denote the fragment $S[\min(A)\mathinner{.\,.}\max(A)]$. We provide pseudocode for an algorithm that computes all $k$-occurrences of $P$ such that $\mathcal{S}$ matches a fragment of a given $\mathcal{S}$-run (see Algorithm 1); inspect also Fig. 8.

Lemma 12**.**

After $\mathcal{O}(n)$-time preprocessing, algorithm Run-Sample-Matching works in $\mathcal{O}(k^{3})$ time and returns a compact representation that consists of $\mathcal{O}(k^{3})$ intervals and $\mathcal{O}(k^{2})$ interval chains, each with difference $q$. Moreover, if there is at least one interval chain, then some rotation of the pattern $P$ is $k$-periodic with a $k$-period $\textsf{per}(\mathcal{S})$.

Proof.

See Algorithm 1. The sets $J$ and $J^{\prime}$ can be computed in $\mathcal{O}(k)$ time:

Claim**.**

If $S$ is a string of length $n$, then the sets $\mathsf{RightMisper}_{k}(S,i,j)$ and $\mathsf{LeftMisper}_{k}(S,i,j)$ can be computed in $\mathcal{O}(k)$ time after $\mathcal{O}(n)$-time preprocessing.

Proof.

For $\mathsf{RightMisper}_{k}(S,i,j)$, we use the kangaroo method [4, 12] to compute the longest common prefix with at most $k$ mismatches of $S[j+1\mathinner{.\,.}n-1]$ and $U^{\infty}$ for $U=S[i\mathinner{.\,.}j]$. The value $\mathsf{lcp}(X^{\infty},Y)$ for a fragment $X$ and a suffix $Y$ of a string $S$, occurring at positions $a$ and $b$, respectively, can be computed in constant time as follows. If $\mathsf{lcp}(S[a\mathinner{.\,.}n-1],S[b\mathinner{.\,.}n-1])<|X|$ then we are done. Otherwise the answer is given by $|X|+\mathsf{lcp}(S[b\mathinner{.\,.}n-1],S[b+|X|\mathinner{.\,.}n-1])$. The computations for $\mathsf{LeftMisper}_{k}(S,i,j)$ are symmetric. ∎

The $\mathcal{O}(k^{3})$ and $\mathcal{O}(k^{2})$ time complexities of computing $X$ and $Y$ follow from Lemmas 7 and 11, respectively (after $\mathcal{O}(n)$-time preprocessing). The sets $X$ and $Y$ consist of $\mathcal{O}(k^{3})$ intervals and $\mathcal{O}(k^{2})$ interval chains, each with difference $q$.

As for the “moreover” statement, by Lemma 10, if any occurrence $q$ is reported in the Periodic-Periodic-Match problem, then it implies the existence of $x$ such that $V[x\mathinner{.\,.}x+m-1]$ is $k$-periodic with a $k$-period $q$. However, $V[x\mathinner{.\,.}x+m-1]$ is a rotation of the pattern $P$. This concludes the proof. ∎

The correctness of the algorithm follows from Lemma 10, as shown in the lemma below.

Lemma 13**.**

Assume $n\leq 2m$. Let $\mathcal{S}$ be a periodic sample in $P$ with smallest period $q$ and $R$ be an $\mathcal{S}$-run in $T$. Let $X$ and $Y$ be defined as in the pseudocode of Run-Sample-Matching. Then $X\cup Y$ is a set of $k$-occurrences of $P$ in $T$ which is a superset of the solution to Sample-Match for $\mathcal{S}$ in $R$.

Proof.

Both Pair-Match and Periodic-Periodic-Match problems return positions of $k$-occurrences of $P$ in $T$. Assume that $\mathsf{rot}_{x}(P)$, for $x\geq p_{\mathcal{S}}+m_{\mathcal{S}}$, has a $k$-mismatch occurrence in $T$ at position $p$ such that the designated fragment $\mathcal{S}$ matches a fragment of $R$ exactly. Hence, it suffices to show that $p\in X\cup Y$.

Let $J=\mathsf{Misper}_{k+1}(T,s,s+q-1)$ and $J^{\prime}=\mathsf{Misper}_{k+1}(P^{2},m+p_{\mathcal{S}},m+p_{\mathcal{S}}+q-1)$. We define $L_{1}$ and $L_{2}$ as the subsets of $J$ and $J^{\prime}$, respectively, that are relevant for this $k$-occurrence, i.e.,

[TABLE]

Further let $L^{\prime}_{2}=L_{2}\bmod m$. If any $i\in L_{1}$ and $j\in L^{\prime}_{2}$ are a matching pair for this $k$-occurrence, then it will be found in the Pairs-Match problem, i.e. $p\in X$. Let us henceforth consider the opposite case.

Let $S=T[p\mathinner{.\,.}p+m-1]$, $S^{\prime}=\mathsf{rot}_{x}(P)$, and $i=m-x+p_{\mathcal{S}}$ be the starting position of $Q=\mathcal{S}[0\mathinner{.\,.}q-1]$ in both strings. Further let $I=L_{1}\ominus p$ and $I^{\prime}=L_{2}\ominus x$. We have that $I\cap I^{\prime}=\emptyset$ by our assumption that misperiods do not align. We make the following claim.

Claim**.**

$\mathsf{Mis}(S,S^{\prime})=I\cup I^{\prime}$.

Proof.

Note that $I=\mathsf{Misper}_{k+1}(S,i,i+q-1)$ and $I^{\prime}=\mathsf{Misper}_{k+1}(S^{\prime},i,i+q-1)$. The latter equality follows from the fact that $\mathsf{Misper}_{k+1}(T,s,s+q-1)=\mathsf{Misper}_{k+1}(T,t,t+q-1)$ for any $t\in[s,s+|R|-q]$. We can thus directly apply Lemma 10 to strings $S$ and $S^{\prime}$. ∎

In particular, $|I|+|I^{\prime}|\leq k$. Moreover, $\min(J)<p$ and $p+m-1<\max(J)$ as well as $\min(J^{\prime})<x$ and $x+m-1<\max(J^{\prime})$, since otherwise we would have $|I|\geq k+1$ or $|I^{\prime}|\geq k+1$. In conclusion, this $k$-occurrence will be found in the Periodic-Periodic-Match problem, i.e. $p\in Y$. ∎

The following proposition summarizes the results of this section.

Proposition 2**.**

If $m\leq n\leq 2m$, $k$-CPM can be solved in $\mathcal{O}(n+k^{5})$ time and $\mathcal{O}(n)$ space.

Proof.

We split the pattern into $2k+3$ fragments and choose a sample $\mathcal{S}$ among them in every possible way.

If the sample $\mathcal{S}$ is not periodic, we use the algorithm of Lemma 8 for Sample-Match in $\mathcal{O}(k^{2})$ time (after $\mathcal{O}(n)$-time preprocessing). It returns a representation of $k$-occurrences as a union of $\mathcal{O}(k^{2})$ intervals.

If the sample $\mathcal{S}$ is periodic, we need to find all $\mathcal{S}$-runs in $T$. By Lemma 9, there are $\mathcal{O}(k)$ of them and they can all be computed in $\mathcal{O}(k)$ time (after $\mathcal{O}(n)$-time preprocessing). For every such $\mathcal{S}$-run $R$, we apply the Run-Sample-Matching algorithm. Its correctness follows from Lemma 13. By Lemma 12, it takes $\mathcal{O}(k^{3})$ time and returns $\mathcal{O}(k^{3})$ intervals and $\mathcal{O}(k^{2})$ interval chains, each with difference $\textsf{per}(\mathcal{S})$, of $k$-occurrences of $P$ in $T$ (after $\mathcal{O}(n)$-time preprocessing). Over all $\mathcal{S}$-runs, this takes $\mathcal{O}(k^{4})$ time after the preprocessing and returns $\mathcal{O}(k^{4})$ intervals and $\mathcal{O}(k^{3})$ interval chains.

By Lemma 12, if any interval chains are reported in Run-Sample-Matching, then some rotation of the pattern is $k$-periodic with a $k$-period $\textsf{per}(\mathcal{S})$. Then, at least $k+2$ of the $2k+3$ pattern fragments do not contain misperiods and hence they must have a period $q=\textsf{per}(\mathcal{S})$. This is actually their smallest period, for if one of these fragments $\mathcal{S}^{\prime}$ had a period $q^{\prime}<q$, then $|\mathcal{S}^{\prime}|\geq|\mathcal{S}|-1$ and, by Fine and Wilf’s periodicity lemma [34], $\mathcal{S}^{\prime}$ would have a period $q^{\prime\prime}=\gcd(q,q^{\prime})<q$, which would imply that $Q$ would also have a period $q^{\prime\prime}$ and hence $\mathcal{S}$ as well. Thus, throughout the course of the algorithm, Run-Sample-Matching can only return interval chains of period $\textsf{per}(\mathcal{S})$ by the pigeonhole principle.

In total, Sample-Match takes $\mathcal{O}(k^{4})$ time for a given sample (after preprocessing), $\mathcal{O}(n+k^{5})$ time in total, and returns $\mathcal{O}(k^{5})$ intervals and $\mathcal{O}(k^{4})$ interval chains of $k$-occurrences, each with the same difference $q$. Let us note that an interval is a special case of an interval chain with difference, say, 1. We then apply Lemma 4 to compute the union of all chains of occurrences and the union of all intervals in $\mathcal{O}(n+k^{5})$ total time. In the end we return the union of the two unions.

In order to bound the space required by our algorithm by $\mathcal{O}(n)$, we do not store each interval chain explicitly throughout the execution of the algorithm. Instead, for each interval chain we increment/decrement a constant number of cells in a ($\mathcal{G}_{q}$-shaped for interval chains or $\mathcal{G}_{1}$-shaped for intervals) 2D array of size $\mathcal{O}(n)$ as in the proof of Lemma 4, and compute the union of all such interval chains in the end. ∎

6 An $\mathcal{O}(n+\frac{n}{m}\,{k^{4}})$-time Algorithm

Let us observe that for each non-periodic fragment $\mathcal{S}$ we have to solve $\mathcal{O}(k)$ instances of Pair-Match, while for each periodic fragment $\mathcal{S}$ and each $\mathcal{S}$-run, we obtain two sets $J$ and $J^{\prime}$, each of cardinality $\mathcal{O}(k)$, where each pair of elements in $J\times J^{\prime}$ requires us to solve an instance of Pair-Match. Further recall that each instance of Pair-Match reduces to two calls to our $\mathcal{O}(k)$-time algorithm for Anchor-Match. We thus consider $\mathcal{O}(k^{4})$ Anchor-Match instances in total, yielding a total time complexity of $\mathcal{O}(k^{5})$. As can be seen in the proof of Proposition 2 and the pseudocode, this is the bottleneck of our algorithm, with everything else requiring $\mathcal{O}(n+k^{4})$ time. We will decrease the number of calls to Pair-Match by using a marking trick.

We first present a simple application of the marking trick. Suppose that we are in the standard $k$-mismatch problem, where we are to find all $k$-occurrences (not circular ones) of a pattern $P$ of length $m$ in a text $T$ of length $n$, and $n\leq 2m$. Further suppose that $P$ is square-free, or, in other words, that it is nowhere periodic. Let us consider the following algorithm. We split the pattern into $k+1$ fragments of length roughly $m/k$ each. Then, at each $k$-occurrence of $P$ in $T$, at least one of the $k+1$ fragments must match exactly. We then find the $\mathcal{O}(k)$ such exact matches of each fragment in $T$ and each of them nominates a position for a possible $k$-occurrence of $P$. We thus have $\mathcal{O}(k^{2})$ candidate positions in total to verify.

Now consider the following refinement of this algorithm. We split the pattern into $2k$ fragments (instead of $k+1$), each of length roughly $m/(2k)$. Then, at each $k$-occurrence of $P$ in $T$, at least $k$ of the fragments must match exactly; we exploit this fact as follows. Each exact occurrence of a fragment in $T$ gives a mark to the corresponding position for a $k$-occurrence of $P$. There are thus $\mathcal{O}(k^{2})$ marks given in total. However, we only need to verify positions with at least $k$ marks and these are now $\mathcal{O}(k)$ in total. An illustration is provided in Fig. 9.

Let us get back to the $k$-CPM problem. Recall that each $k$-occurrence implies that at least $k+2$ fragments match exactly. We run the algorithm yielding Proposition 2 with a single difference. Instead of processing each instance of Pair-Match separately, we apply the marking trick in order to decrease the exponent of $k$ by one. This is achieved by a reduction in the number of calls to the algorithm that solves Anchor-Match. For each of the $\mathcal{O}(k^{4})$ instances of Pair-Match we mark the two possible anchors for a $k$-occurrence and note that only anchors with at least $k+2$ marks need to be verified; these are $\mathcal{O}(k^{3})$ in total. Finally, for each such anchor we apply our solution to the Anchor-Match problem, which requires $\mathcal{O}(k)$ time, hence obtaining an $\mathcal{O}(n+k^{4})$-time algorithm.

The correctness of this approach follows from the following fact. (Note that Periodic-Periodic-Match is not affected by this modification.) If there is a $k$-occurrence at position $i$ of $T$ that has not been returned by any of the calls to Periodic-Periodic-Match, then, the algorithm would return $i$ through at least $k+2$ calls to Anchor-Match. We arrive at the following result.

Proposition 3**.**

If $m\leq n\leq 2m$, $k$-CPM can be solved in $\mathcal{O}(n+k^{4})$ time and $\mathcal{O}(n)$ space.

Both Propositions 1 and 3 use $\mathcal{O}(n)$ space. Moreover, Proposition 3 assumes that $n\leq 2m$. In order to solve the general version of the $k$-CPM problem, where $n$ is arbitrarily large, efficiently and using $\mathcal{O}(m)$ space, we use the so-called standard trick: we split the text into $\mathcal{O}(n/m)$ fragments, each of length $2m$ (perhaps apart from the last one), starting at positions equal to $0\bmod m$.

We need, however, to ensure that the data structures for answering $\mathsf{lcp}$, $\mathsf{lcs}$, and other internal queries over each such fragment of the text can be constructed in $\mathcal{O}(m)$ time when the input alphabet $\Sigma$ is large. As a preprocessing step we hash the letters of the pattern using perfect hashing. For each key, we assign a unique identifier from $\{1,\ldots,m\}$. This takes $\mathcal{O}(m)$ (expected) time and space [39]. When reading a fragment $F$ of length (at most) $2m$ of the text we look up its letters using the hash table. If a letter is in the hash table we replace it in $F$ by its rank value; otherwise we replace it by rank $m+1$. We can now construct the data structures in $\mathcal{O}(m)$ time and thus our algorithms can be implemented in $\mathcal{O}(m)$ space.

If $\Sigma=\{1,\ldots,n^{\mathcal{O}(1)}\}$, the same bounds can be achieved deterministically. Specifically, we consider two cases. If $m>\sqrt{n}$ we sort the letters of every text fragment and of the pattern in $\mathcal{O}(m)$ time per fragment because $n$ is polynomial in $m$ and $|\Sigma|$ is polynomial in $n$. Then we can merge the two sorted lists and replace the letters in the pattern and the text fragments by their ranks. Otherwise ($m\leq\sqrt{n}$), we construct a deterministic dictionary for the letters of the pattern in $\mathcal{O}(m\log^{2}\log m)$ time [40]. The dictionary uses $\mathcal{O}(m)$ space and answers queries in constant time; we use it instead of perfect hashing in the previous solution.

We combine Propositions 1 and 3 with the above discussion to get our final result.

Theorem 4**.**

Circular Pattern Matching with $k$ Mismatches can be solved in $\mathcal{O}(\min(nk,\;n+\frac{n}{m}\,{\small k^{4}}))$ time and $\mathcal{O}(m)$ space.

Our algorithms output all positions in the text where some rotation of the pattern occurs with $k$ mismatches. It is not difficult to extend the algorithms to output, for each of these positions, a corresponding rotation of the pattern.