Efficient enumeration of non-equivalent squares in partial words with few holes

Abstract

A word of the form WW for some word $W\in \varSigma ^*$ is called a square. A partial word is a word possibly containing holes (also called don’t cares). The hole is a special symbol $\lozenge \notin \varSigma$ which matches any symbol from $\varSigma \cup \{\lozenge \}$. A p-square is a partial word matching at least one square WW without holes. Two p-squares are called equivalent if they match the same set of squares. A p-square is called here unambiguous if it matches exactly one square WW without holes. Such p-squares are natural counterparts of classical squares. Let $\mathrm {PSQUARES}_k(n)$ and $\mathrm {USQUARES}_k(n)$ be the maximum number of non-equivalent p-squares and non-equivalent unambiguous p-squares in T over all partial words T of length n with at most k holes. We show asymptotically tight bounds:

$$\begin{aligned} \mathrm {PSQUARES}_k(n) = \varTheta (\min (nk^2,\, n^2)),\ \ \mathrm {USQUARES}_k(n) = \varTheta (nk). \end{aligned}$$

We present an algorithm that reports all non-equivalent p-squares in $\mathcal {O}(nk^3)$ time for a partial word of length n with k holes, for an integer alphabet. In particular, it runs in linear time for $k=\mathcal {O}(1)$ and its time complexity near-matches the asymptotic bound for $\mathrm {PSQUARES}_k(n)$. We also show an $\mathcal {O}(n)$-time algorithm that reports all non-equivalent p-squares of a given length. The paper is a full and improved version of Charalampopoulos et al. (in Cao Y, Chen Y (eds) Proceedings of the 23rd international conference on computing and combinatorics, COCOON 2017; Springer, 2017).

1 Introduction

A word is a sequence of letters from a given alphabet $\varSigma$. By $\varSigma ^*$ we denote the set of all words over $\varSigma$. A word of the form $U^2=UU$, for some word U, is called a square. For a word W, a factor is a subword composed of some number of consecutive letters and a square factor is a factor of W which is a square. Enumeration of square factors in words is a well-studied topic, both from a combinatorial and from an algorithmic perspective. Obviously, a word W of length n may contain $\varTheta (n^2)$ square factors (e.g. $W=a^n$), however, it is known that such a word contains only $\mathcal {O}(n)$ distinct square factors (Fraenkel and Simpson 1998; Ilie 2005); currently the best known upper bound is $\frac{11}{6}n$ (Deza et al. 2015).

Moreover, all distinct square factors of a word over an integer alphabet can be listed in $\mathcal {O}(n)$ time using the suffix tree (Gusfield and Stoye 2004; Bannai et al. 2017) or the suffix array and the structure of runs (maximal repetitions) in the word (Crochemore et al. 2014).

A partial word is a sequence of letters from $\varSigma \cup \{\lozenge \}$, where $\lozenge$ denotes a hole, that is, a don’t care symbol. Two symbols $a,b \in \varSigma \cup \{\lozenge \}$ are said to match (denoted as $a \approx b$) if they are equal or one of them is a hole; note that this relation is not transitive. The relation of matching is extended in a natural way to partial words of the same length.

A partial word UV is called a p-square if $U \approx V$. Like in the context of words, a p-square factor of a partial word T is a factor being a p-square; see Blanchet-Sadri et al. (2014b, 2015).

We introduce the notion of equivalence of p-square factors in partial words. Let $\hbox { sq-val}(UV)$ denote the set of squares that match the partial word UV:

$$\begin{aligned} \hbox { sq-val}(UV) = \{WW\,:\,W \in \varSigma ^*,\,WW\approx UV\}. \end{aligned}$$

Example 1.1

Let $\varSigma =\{a,b\}$. Then:

$$\begin{aligned} \hbox { sq-val}(a\lozenge b\ a\lozenge \lozenge )=\hbox { sq-val}(a\lozenge \lozenge \ \lozenge \lozenge b)&=\{(aab)^2, (abb)^2\},\\ \hbox { sq-val}(a\lozenge \lozenge \ \lozenge ab)&=\{(aab)^2\}. \end{aligned}$$

The p-squares UV and $U'V'$ are called equivalent if $\hbox { sq-val}(UV)=\hbox { sq-val}(U'V')$ (denoted as $UV \equiv U'V'$). For example,

$$\begin{aligned} a\lozenge b\ a\lozenge \lozenge \equiv a\lozenge \lozenge \ \lozenge \lozenge b\hbox {, but} \quad a\lozenge b\ a\lozenge \lozenge \not \equiv a\lozenge \lozenge \ \lozenge a b. \end{aligned}$$

Let us assume that $\varSigma$ is non-unary. We say that $X^2=XX$ is the representative (also called general form; see Blanchet-Sadri et al. 2009) of a p-square UV, denoted as $repr (UV)$, if

$$\begin{aligned} XX \approx UV \ \hbox {and}\ \hbox { sq-val}(XX)=\hbox { sq-val}(UV). \end{aligned}$$

(In other words, X is the “most general” partial word that matches both U and V.) It can be noted that the representative of a p-square is unique. Then $UV \equiv U'V'$ if and only if $repr (UV)= repr (U'V')$. A p-square is called unambiguous if its representative does not contain the symbol ${\lozenge }$ and ambiguous otherwise.

Example 1.2

• $repr (a\lozenge b\ a\lozenge \lozenge )=(a\lozenge b)^2$ and the p-square is ambiguous.

• $repr (a\lozenge \lozenge \ \lozenge a b)=(aab)^2$ and the p-square is unambiguous.

The set of non-equivalent p-square factors in a partial word T is denoted by $psquares (T)$. Thus, $psquares (T)$ corresponds to the set of different representatives of p-square factors of T.

Example 1.3

Let $T\,=\,ab \lozenge \lozenge ba \lozenge aaba\lozenge b$.

T contains 4 non-equivalent classes of p-squares of length 4:

• 1. $a\lozenge aa$ with representative $(aa)^2$,

• 2. $ab\lozenge \lozenge \equiv \lozenge ba\lozenge \equiv aba\lozenge$ with representative $(ab)^2$,

• 3. $\lozenge \lozenge ba \equiv ba\lozenge a$ with representative $(ba)^2$, and

• 4. $b \lozenge \lozenge b$ with representative $(bb)^2$.

T contains 4 equivalence classes of p-squares of length 6 with representatives:

$$\begin{aligned} (aab)^2,\ (aba)^2,\ (baa)^2,\ (ba\lozenge )^2; \end{aligned}$$

see also Fig. 1.

Fig. 1

All non-equivalent p-square factors of length 6 with their representatives in an example partial word

Overall, we have $| psquares (T)|=14$. The remaining 6 representatives are:

$$\begin{aligned} \lozenge \lozenge ,\ aa,\ bb,\ (aaba\lozenge )^2,\ (abaab)^2,\ (baaba)^2. \end{aligned}$$

Our work is devoted to enumeration of non-equivalent p-square factors in a partial word with a given number $k>0$ of holes.

Previous results Alongside (Blanchet-Sadri et al. 2009, 2014b, 2015), we define a solid square as a square of a word and a square subword of a partial word T as a solid square that matches a factor of T.

Previous studies on squares in partial words were mostly focused on combinatorics. They started with the case of $k=1$ (Blanchet-Sadri et al. 2009), in which case distinct square subwords correspond to non-equivalent p-square factors. It was shown that a partial word with one hole contains at most $\frac{7}{2}n$ distinct square subwords (Blanchet-Sadri and Mercaş 2009) (3n for binary partial words; Halava et al. 2010). Also a generalization of the three squares lemma (see Crochemore and Rytter 1995) was proposed for partial words (Blanchet-Sadri and Mercaş 2012). As for a larger number of holes, the existing literature is devoted mainly to counting the number of distinct square subwords of a partial word (Blanchet-Sadri et al. 2009, 2015) or all occurrences of p-square factors (Blanchet-Sadri et al. 2014a, 2015). On the algorithmic side, Manea and Tiseanu (2010) proved that the problem of counting distinct square subwords of a partial word is #P-complete and Diaconu et al. (2009), Manea et al. (2014), and Blanchet-Sadri et al. (2014b) showed quadratic- and nearly-quadratic-time algorithms for finding all occurrences of p-square factors and primitively-rooted p-square factors of a partial word, respectively.

Our combinatorial results Let $\mathrm {PSQUARES}_k(n)$ and $\mathrm {USQUARES}_k(n)$ be the maximum number of non-equivalent p-squares and non-equivalent unambiguous p-squares in T over all partial words T of length n with at most k holes. We show the following bounds:

$$\begin{aligned} \mathrm {PSQUARES}_k(n) = \varTheta (\min (nk^2,\, n^2)),\ \ \mathrm {USQUARES}_k(n) = \varTheta (nk). \end{aligned}$$

This work can be viewed as a generalization of the results on partial words with one hole (Blanchet-Sadri et al. 2009; Blanchet-Sadri and Mercaş 2009; Halava et al. 2010) to k holes.

Our algorithmic results We present an algorithm that reports all elements of the set $psquares (T)$ in a partial word of length n with k holes in $\mathcal {O}(nk^3)$ time. In particular, our algorithm runs in linear time for $k=\mathcal {O}(1)$ and its time complexity near-matches the maximum number of non-equivalent p-square factors. We also show an $\mathcal {O}(n)$-time algorithm that reports all non-equivalent p-squares of a given length. The algorithms assume integer alphabet $\varSigma \subseteq \{1,\ldots ,n^{\mathcal {O}(1)}\}$. We use recently introduced advanced data structures by Kociumaka (2016).

Comparison with the conference version The paper is an extended version of Charalampopoulos et al. (2017). As far as combinatorics of p-squares is concerned, the conference version of the paper derived the bound $\mathrm {PSQUARES}_k(n) = \varTheta (\min (n^2,nk^2))$. Let $\mathrm {ASQUARES}_k(n)$ be the maximum number of non-equivalent ambiguous p-squares in T over all partial words T of length n with at most k holes. The bound was proved by showing that $\mathrm {ASQUARES}_k(n) = \varTheta (\min (n^2,nk^2))$ and that $\mathrm {USQUARES}_k(n) = \mathcal {O}(nk^2)$. As a new contribution here, we present a tight estimation $\mathrm {USQUARES}_k(n) = \varTheta (nk)$. This lets us identify ambiguous p-squares as the ones that attain the bound on $\mathrm {PSQUARES}_k(n)$. On the algorithmic side, Charalampopoulos et al. (2017) presented an algorithm computing the set $psquares (T)$ in $\mathcal {O}(nk^3)$ time. Here the readability of the algorithm has been considerably improved; we also show a linear-time algorithm that reports all non-equivalent p-squares of a specified length.

Structure of the paper After the Preliminaries comes the algorithmic part of the paper, which is followed by the combinatorial part. In Sect. 3 we show an $\mathcal {O}(n)$-time algorithm that reports all non-equivalent p-squares of a specified length and, as an immediate corollary, $\mathcal {O}(nk^2)$-time computation of all non-equivalent ambiguous p-squares. Then in Sect. 4 we give an $\mathcal {O}(nk^3)$-time algorithm for computing all non-equivalent unambiguous p-squares. Asymptotic bounds for ambiguous p-squares and unambiguous p-squares are presented in Sects. 5 and 6, respectively.

2 Preliminaries

For a word $W \in \varSigma ^*$, by $|W|=n$ we denote the length of W, and by $W_i$, for $i=1,\ldots ,n$, the ith letter of W. For $1 \le i \le j \le n$, by [i..j] and (i..j] we denote integer intervals $\{i,\ldots ,j\}$ and $\{i+1,\ldots ,j\}$, respectively. W[i..j] denotes the factor of W equal to $W_i \cdots W_j$; we also use the notation W[I], where I is an integer interval. A factor of the form W[1..j] is called a prefix, a factor of the form W[i..n] is called a suffix.

For a partial word T we use the same notation as for words: $|T|=n$ for its length, $T_{i}$ for the ith letter, T[i..j] for a factor. If T does not contain holes, then it is called solid. The relation $\approx$ of matching on $\varSigma \cup \{\lozenge \}$ is defined as: $a \approx a$, $\lozenge \approx a$, and $a \approx \lozenge$ for all $a \in \varSigma \cup \{\lozenge \}$.

We define an operation $\,\odot \,$ such that: $a \,\odot \, a = a \,\odot \, \lozenge = \lozenge \,\odot \, a = a$ for all $a \in \varSigma \cup \{\lozenge \}$, and otherwise $a \,\odot \, b$ is undefined. Two equal-length partial words S and T are said to match (denoted as $S \approx T$) if $S_i \approx T_i$ for all $i=1,\ldots ,n$. In this case, we denote

$$\begin{aligned} S \,\odot \, T\;=\; S_1 \,\odot \, T_{1},\ldots ,S_n \,\odot \, T_{n}. \end{aligned}$$

Also note that if UV is a p-square, then $repr (UV) = (U \,\odot \, V)^2$.

If $U \approx T[i..i+|U|-1]$ for a partial word U, then we say that U occurs in T at position i.

Two equal-length partial words U and V are called cyclic shifts if there are partial words X, Y such that $U=XY$ and $V=YX$. We denote this as $rot (U,|X|)=W$, where |X| is the shift value.

For a partial word X, by $\#_{\lozenge }(X)$ we denote the number of holes in X. For $1 \le i \le n$ and $0 \le q \le \log n$, we denote $T_{i,q}=T[i..\min (n,i+2^q-1)]$. We say that $T_{i,q}$ is a q-basic factor of the partial word T. In other words, q-basic factors are factors of T of length $2^q$ and suffixes of T of length at most $2^q$. By ${\mathcal B}(T)$ we denote the set of all basic factors of T.

Lemma 2.1

If T is a partial word of length n with k holes, then

$$\begin{aligned} \sum _{W \in {\mathcal B}(T)} \#_{\lozenge }(W) \le 2nk. \end{aligned}$$

Proof

The number of q-basic factors that contain a given position $i \in \{1,\ldots ,n\}$ is at most $2^q$. Thus the total number of basic factors that contain a given hole position i is at most:

$$\begin{aligned} \sum \limits _{q=0}^{\left\lfloor \log n \right\rfloor } 2^q \le 2n. \end{aligned}$$

$\square$

We say that a p-square is an unambiguous p-square (u-square) if its representative is solid and an ambiguous p-square (a-square) otherwise. By $asquares (T)$ and $usquares (T)$ we denote the sets of non-equivalent factors of T being a-squares and u-squares, respectively. Obviously:

Observation 2.2

$psquares (T)= asquares (T) \cup usquares (T)$.

2.1 Periods in solid and partial words

A positive integer q is called a period of a word W if $W_i=W_{i+q} \ \hbox {for all}\ i=1,\ldots ,n-q.$ In this case, W[1..q] is called a string period of W. A word W is called periodic if it has a period q such that $2q \le |W|$.

A quantum period of a partial word T is a positive integer q such that $T_{i} \approx T_{i+q}$ for all $i=1,\ldots ,n-q$. A deterministic period of T is an integer q such that there exists a word W such that $W \approx T$ and W has a period q.

The partial word T is called quantum (deterministically) periodic if it has a quantum (deterministic) period q such that $2q \le n$.

For a partial word U and integer $\delta >0$, we denote

$$\begin{aligned} \mathsf {Mis}_\delta (U)&= \{i \in [\delta +1..|U|]\,:\,U_{i-\delta } \not \approx U_i\},\\ \mathsf {Holes}(U)&= \{i \in [1..|U|]\,:\,U_i = \lozenge \}. \end{aligned}$$

We say that p is a d-approximate quantum period of a partial word T if $|\mathsf {Mis}_d(T)| \le d$. Note that a 0-approximate quantum period is exactly a quantum period.

Lemma 2.3

Assume that $U \approx V$.

• (a) If $i \in \mathsf {Mis}_\delta (U)$, then $i \in \mathsf {Mis}_\delta (V)$ or $i \in \mathsf {Holes}(V)$ or $i-\delta \in \mathsf {Holes}(V)$.

• (b) $|\mathsf {Mis}_\delta (U)| \le |\mathsf {Mis}_\delta (V)| + 2|\mathsf {Holes}(V)|$.

• (c) If $\delta \ge \frac{1}{2}|U|$, then $|\mathsf {Mis}_\delta (U)| \le |\mathsf {Mis}_\delta (V)| + |\mathsf {Holes}(V)|$.

Proof

(a): We have $V_{i-\delta } \approx U_{i-\delta } \not \approx U_i \approx V_i$. This means that $U_{i-\delta },U_i \in \varSigma$. Hence, if $i \not \in \mathsf {Holes}(V)$ and $i-\delta \not \in \mathsf {Holes}(V)$, then $V_{i-\delta } = U_{i-\delta }$ and $V_i = U_i$, so $i \in \mathsf {Mis}_\delta (V)$.

Point (b) follows from point (a). Also point (c) follows from point (a). Indeed, if $i \in \mathsf {Mis}_\delta (U)$ in this case, then for each of the positions i, $i-\delta$ in V, if it contains a hole, then it is counted only for the index i. $\square$

3 Computing all p-squares of specified length and non-equivalent ambiguous p-squares

In this section we develop an $\mathcal {O}(n)$-time algorithm that enumerates all non-equivalent p-squares of a half length d in a partial word T of length n. As a corollary, we obtain a simple computation of all non-equivalent ambiguous p-squares in optimal time.

For a partial word T, we denote by $T'$ a partial word of length $n-d$ such that $T'[i]=T[i] \odot T[i+d]$ for each $i=1,\ldots ,n-d$. If $T[i] \odot T[i+d]$ is undefined (since $T[i] \not \approx T[i+d]$), we set the value to a symbol $\# \not \in \varSigma$.

Observation 3.1

• (a) $T[i..i+2d-1]$ is a p-square if and only if $T'[i..i+d-1]$ does not contain the symbol $\#$.

• (b) If $T[i..i+2d-1]$ is a p-square, then $repr (T[i..i+2d-1]) = (T'[i..i+d-1])^2$.

Proof

• (a) If $T'[i..i+d-1]$ contains the symbol $\#$, this means that $T[j] \not \approx T[j+d]$ for some $j \in [i..i+d-1]$. Hence, $T[i..i+2d-1]$ is not a p-square. Otherwise, $T[j] \approx T[j+d]$ for all $j \in [i..i+d-1]$. Hence, $T[i..i+2d-1]$ indeed is a p-square.

• (b) If $T[i..i+2d-1]$ is a p-square, then

$$\begin{aligned} (T'[i..i+d-1])^2 = (T[i..i+d-1] \odot T[i+d..i+2d-1])^2 = repr (T[i..i+2d-1]). \end{aligned}$$

$\square$

Example 3.2

Let us consider the partial word $T\,=\,ab \lozenge \lozenge ba \lozenge aaba\lozenge b$ from Example 1.3. For $d=2$ we construct the following partial word $T'$:

$$\begin{aligned} abbabaa\#ab\# \end{aligned}$$

from which we conclude that T contains p-squares of half length 2 with representatives:

$$\begin{aligned} (ab)^2,\, (bb)^2,\, (ba)^2,\, (ab)^2,\, (ba)^2,\, (aa)^2,\, (ab)^2. \end{aligned}$$

For $d=3$ we construct the partial word $T'$:

$$\begin{aligned} aba\lozenge \#abaab \end{aligned}$$

which means that T contains p-squares of half length 3 with representatives:

$$\begin{aligned} (aba)^2,\, (ba\lozenge )^2,\, (aba)^2,\, (baa)^2,\, (aab)^2. \end{aligned}$$

Theorem 3.3

All non-equivalent p-squares of half length d in a partial word of length n can be reported (as factors of the partial word) in $\mathcal {O}(n)$ time.

Proof

Let T be a partial word of length n. In $\mathcal {O}(n)$ time we compute $T'$. Let $S_1,\ldots ,S_q$ be a partition of $T'$ into maximal factors that do not contain the symbol $\#$. By Observation 3.1, our task is equivalent to reporting all distinct factors of length d of the partial words $S_j$. This can be performed by listing all nodes (implicit and explicit) at depth d in the generalized suffix tree$\mathcal {T}$ of $S_1,\ldots ,S_q$, that is, in the suffix tree of $S_1 \#_1 \ldots S_q \#_q$, where $\#_1,\ldots ,\#_q \not \in \varSigma$ are distinct symbols. For details, see Gusfield (1997). As the suffix tree of a word of length n can be constructed in $\mathcal {O}(n)$ time (Farach 1997), the whole algorithm works in $\mathcal {O}(n)$ time. $\square$

As a corollary we obtain efficient computation of non-equivalent a-squares.

Theorem 3.4

For a partial word T of length n with k holes, all elements of the set $asquares (T)$ can be reported in $\mathcal {O}(nk^2)$ time.

Proof

There are at most $k^2$ possible lengths of ambiguous p-squares. For each length we use the algorithm of Theorem 3.3 to report all non-equivalent p-squares. This takes $\mathcal {O}(nk^2)$ time. In the end, for each length we need to filter out unambiguous p-squares. For a specified half length d, it suffices to check, for each p-square $T[i..i+2d-1]$ found, if $T'[i..i+d-1]$ contains a hole. This condition can be checked in $\mathcal {O}(1)$ time if the prefix sums of the sequence $a_i=[T'[i]={\lozenge }]$ are stored. $\square$

4 Computing all non-equivalent unambiguous p-squares

We start the description of the algorithm by an abstract lemma that lets us efficiently generate all distinct squares induced by a special family of (solid) words.

4.1 Computing squares induced by a family of words

For a word S, we define its primitive rootU as the shortest word such that $U^k = S$ for some integer $k \ge 1$. The Lyndon root$\lambda$ of a word U is the minimal cyclic shift of the shortest string period of U. The notion of a Lyndon root was introduced in the context of runs by Crochemore et al. (2014).

Example 4.1

The Lyndon root of $U=abaababaababa$ is aabab. The word U is periodic and its shortest period is 5.

For a word W and its period q, by $squares (W,q)$ we denote the set of square factors of W of length 2q. We say that $squares (W,q)$ is the set of squares induced by the word W with the period q. Each square factor in $squares (W,q)$ can be represented in $\mathcal {O}(1)$ space by specifying its occurrence in W.

Lemma 4.2

Assume we have a family of possibly unknown words $W_1,W_2,\ldots ,W_N$ with periods $q_1,\ldots ,q_N$, a positive integer k and positive integers $n_i, first _i,\ell _i$ for $i=1,\ldots ,N$, such that:

• (1) $n_i\le n$ is the length of $W_i$ and $2q_i \le n_i$;

• (2) all the words $W_i$ for which $2q_i = n_i$ (so-called short words) are distinct;

• (3) for a given $q_i$, the number of words $W_i$ for which $2q_i < n_i$ (so-called long words) is at most k;

• (4) $first _i$ is the starting position of the first occurrence of the Lyndon root $\lambda _i$ of $W_i$ in $W_i$ and $\ell _i$ is its length;

• (5) any two Lyndon roots $\lambda _i$, $\lambda _j$ can be compared in O(k) time.

Then we can compute the cardinality of the set $SQ\,=\, \bigcup _i\, squares (W_i,q_i)$ and its representation (as sets of intervals in $W_i$’s) in $O(Nk^2 + nk^3 +|SQ|)$ time.

Proof

Let us start with the following observation; see also Fig. 2. The same type of observation was used by Crochemore et al. (2014).

Observation 4.3

For every i, the set $squares (W_i,q_i)$ equals

$$\begin{aligned} \{ rot (\lambda _i^{2q_i/\ell _i},a)\,:\,a = (1- first _i) \bmod \ell _i, \ldots , (1+n_i-2q_i- first _i) \bmod \ell _i\}. \end{aligned}$$

The above set of integers is denoted by $I_i$. Note that it forms one cyclic subinterval of $[0..\ell _i-1]$ (composed of up to two standard intervals) and that it can be computed in $\mathcal {O}(1)$ time. Each of the elements $a \in I_i$ represents a unique square that is induced by $W_i$ and $q_i$.

Fig. 2

In this case $n_i=32$, $q_i=14$, $\ell _i=7$, $first _i=3$. Hence, $W_i$ induces 5 squares being cyclic shifts of $\lambda _i^4$, that is, $I_i = [0, 2] \cup [5, 6]$

We make two transformations of the set of intervals $I_i$ so that, in the end, each square from the set SQ is induced by exactly one word $W_i$ with period $q_i$. If any of the intervals is made empty, this corresponds to removing the word as unnecessary. The first transformation deals with the long words $W_i$; by definition, at most k of them share the same period $q_i$.

First transformation For every pair $W_i,q_i$ and $W_j,q_j$ of long words such that $i \ne j$ and $q_i=q_j$, we check if $\lambda _i=\lambda _j$. If $I_i \subseteq I_j$, we dispose of $W_i$. Likewise, if $I_j \subseteq I_i$, we remove $W_j$. If none of the two cases holds and still $I_i \cap I_j \ne \emptyset$, we trim $I_j$ to make it disjoint with $I_i$.

Complexity All long words can be sorted by their periods in $\mathcal {O}(N+n)$ time by bucket sort. There are n / 2 buckets and each bucket contains at most k words. For each of the $k(k-1)/2$ pairs of long words in a bucket, we check equality of their Lyndon roots, which takes $\mathcal {O}(k)$ time per pair and $\mathcal {O}(nk^3)$ time overall. The time complexity of trimming of cyclic intervals is dominated by this step.

Second transformation For every short word $W_i$ with period $q_i$ and long word $W_j$ with period $q_j=q_i$, we check if $\lambda _i=\lambda _j$. If so and $I_i \subseteq I_j$, we remove $W_i$. Note that $I_i$ is a singleton.

Complexity All words can be sorted by their periods in $\mathcal {O}(N+n)$ time by bucket sort. For each short word $W_i$, we need to inspect at most k long words and check if their Lyndon roots are equal. This takes $\mathcal {O}(k^2)$ time per short word, $\mathcal {O}(Nk^2)$ time overall. Checking inclusion of elements in cyclic intervals is dominated by this step.

The two transformations take $\mathcal {O}(Nk^2+nk^3)$ time in total. Afterwards each square is induced by exactly one interval $I_i$ for a word $W_i$ and period $q_i$, so we can list all the distinct squares in $\mathcal {O}(|SQ|)$ time. $\square$

For a partial word T, by $ssquares (T)$ we denote the set of distinct solid factors of T being squares. The following fact was already mentioned in Sect. 1.

Fact 4.4

(Bannai et al. 2017; Crochemore et al. 2014; Gusfield and Stoye 2004) All distinct squares in a word of length n can be computed in $\mathcal {O}(n)$ time.

By substituting all holes in a partial word with distinct symbols $\#_1,\ldots ,\#_k$, we obtain the following corollary.

Corollary 4.5

For partial word T of length n, the set $ssquares (T)$ can be computed in $\mathcal {O}(n)$ time.

The algorithm of Crochemore et al. (2014) actually computes the set $ssquares (T)$ together with all the data in assumption of Lemma 4.2. These are the short words in the construction.

In the following section we construct a family ${\mathcal F}$ of words (called sealed fragments) that represent the u-squares that contain a hole and compute for them the data required in Lemma 4.2. These are the long words in the construction. Afterwards we list all distinct representatives of u-squares using Lemma 4.2. Then non-equivalent u-squares are extracted from their representatives.

4.2 Computing a special family of sealed fragments

If T is a partial word, then U is a sealed fragment of T if U is a factor of T with holes substituted by solid symbols. By $unseal (U)$ we denote the original factor of the partial word.

A sealed fragment is always solid. Obviously, a sealed fragment can be represented in space proportional to the number of holes that were substituted. For example, if $T[i..i+2q-1]$ is a u-square, then $repr (T[i..i+2q-1])$ is a sealed fragment.

If W is a (solid) word, then by a d-fragment we mean a concatenation of d factors $W[i_1..j_1] \dots W[i_d..j_d]$. A d-fragment can be represented in $\mathcal {O}(d)$ space. Kociumaka (2016) showed that several types of operations on d-fragments can be performed in $\mathcal {O}(d)$ or $\mathcal {O}(d^2)$ time after $\mathcal {O}(n)$-time preprocessing. We notice here that a sealed fragment of a partial word T with k holes corresponds to a d-fragment with $d=\mathcal {O}(k)$ in a word that corresponds to T where ${\lozenge }$ is treated as an alphabet symbol. Thus the following simple fact is a consequence of Observation 18 from Kociumaka (2016) that was stated in terms of d-fragments.

Fact 4.6

(Kociumaka 2016) For a partial word of length n with k holes, after $\mathcal {O}(n)$-time preprocessing, the length of the longest common prefix (or suffix) of any two sealed fragments can be computed in $\mathcal {O}(k)$ time. In particular, equality of sealed fragments can be checked within the same time complexity.

Definition 4.7

A family of pairs $(W_i,q_i)$, where each $W_i$ is a sealed fragment of a partial word T of length n with k holes and $q_i$ is a positive integer, is called an S-family if it satisfies the following properties:

• (a) For every i, $q_i$ is a period of $W_i$ and $|W_i| \ge 2q_i$.

• (b) For every i, there are no two holes in $unseal (W_i)$ at distance $q_i$.

• (c) For every $q=1,\ldots ,n$, there are $\mathcal {O}(k)$ sealed fragments with $q_i=q$.

• (d) If X is a non-solid u-square in T, then X is a factor of $unseal (W_i)$ for some $W_i$ with $q_i =\frac{1}{2}|X|$.

The size of an S-family follows from point (c).

Observation 4.8

An S-family contains $\mathcal {O}(nk)$ elements and thus can be represented in $\mathcal {O}(nk^2)$ space.

In the following lemma we provide an algorithm for constructing an S-family. Our approach resembles computing anchored squares in the Main-Lorentz algorithm (Main and Lorentz 1984).

Lemma 4.9

For a partial word T of length n with k holes, an S-family can be computed in $\mathcal {O}(nk^2)$ time.

Proof

Each non-solid u-square X contains a hole in the first half or in the second half. Below, we construct an S-family for u-squares containing a hole in the second half. A symmetric procedure deals with the u-squares containing a hole in the first half.

For a hole h and integer q, we define the family $\mathcal {S}(q,h)$ of u-squares of length 2q, which contain h as the leftmost hole in the second half. For each non-empty set $\mathcal {S}(q,h)$, we shall construct a sealed fragment W with period q so that each u-square $X\in \mathcal {S}(q,h)$ is a factor of $unseal (W)$.

First, let us seal the text consistently with the representatives of u-squares in $\mathcal {S}(q,h)$. A hole at position $i<h$ may only be contained in the first half, while a hole at position $i\ge h$ may only be contained in the second half of such a u-square. Thus, we seal the hole $T[i]=\lozenge$ with $T[i+q]$ if $i<h$, and with $T[i-q]$ if $i\ge h$. Any remaining hole is sealed with a unique marker (distinct for every hole). This produces a sealed fragment $\overline{T}$ that covers the whole partial word T; see Fig. 3. Let z be the distance between h and the position of the preceding hole ($z=+\infty$ if there is none). We define W as a maximal fragment of $\overline{T}$ which contains $\overline{T}[h-q..h]$, is contained in $\overline{T}[h-q-\min (q-1,z-1)..h+q-1]$, and has period q. If $|W|<2q$, there is no u-square of the desired type and we can discard W.

Fig. 3

A partial word T with 5 holes and the corresponding sealed text $\overline{T}$ with holes sealed by 5 (solid) symbols implied by the value of q. The rightmost hole is filled by a special unique marker denoted by $

Fig. 4

The fragment W with period q anchored at h is computed using an operation of modified longest common extension and its reversed version. We have $|S| \le \min (q-1,z-1)$, where z is the distance between h and the position of the preceding hole, and $|P| \le q-1$

The fragment W is unique and it can be retrieved in $\mathcal {O}(k)$ time using Fact 4.6. Indeed, it suffices to compute the longest common prefix P of $\overline{T}[h-q..n]$ and $\overline{T}[h..n]$, the longest common suffix S of $\overline{T}[1..h-q]$ and $\overline{T}[1..h]$, and take the possibly trimmed fragment $S\,\overline{T}[h-q+1..h-1]\,P$; see Fig. 4. We may need to trim S so that its length exceeds neither $q-1$ (so that the hole at position h is contained in the right half of the square) nor $z-1$ (so that h is the leftmost hole in the right half). Similarly, we may need to trim P to the length $q-1$. In total, the construction takes $\mathcal {O}(nk^2)$ time.

Let us verify that this construction indeed satisfies the condition of Definition 4.7. For each hole we construct just one sealed fragment, so the condition (c) is satisfied. Clearly, W has period q and $|W| \ge 2q$, which yields point (a). Moreover, if $X=T[i..j]\in \mathcal {S}(q,h)$, then $repr (X)=\overline{T}[i..j]$, so (by maximality) $repr (X)$ is contained in W, and X is contained in $unseal (W)$. This gives point (d). Finally, we shall prove that $unseal (W)$ does not contain two holes at distance q (condition (b)). Suppose that the holes are at positions i and $i+q$. Observe that one of the holes is sealed with a unique marker, which contradicts $\overline{T}[i]=\overline{T}[i+q]$. This completes the proof. $\square$

Example 4.10

Consider the partial word $T\,=\,ab \lozenge \lozenge ba \lozenge aaba\lozenge b$ from Example 1.3 and $q=2$. For the first hole we obtain the following word $\overline{T}$:

$$\begin{aligned} ab |\underline{a}\underline{b} ba \underline{b} aaba\underline{b}b \end{aligned}$$

with the original positions of holes underlined. The computed sealed fragment is $W=abab$. For the second hole we obtain the word $\overline{T}$:

$$\begin{aligned} ab \underline{b}|\underline{b} ba \underline{b} aaba\underline{b}b \end{aligned}$$

and the sealed fragment bbbb. For the third hole $\overline{T}$ equals:

$$\begin{aligned} ab \underline{b}\underline{a} ba |\underline{b} aaba\underline{b}b \end{aligned}$$

and the sealed fragment is ababa so $\mathcal {S}(2,8)=\{\lozenge ba\lozenge , ba\lozenge a\}$. Finally, for the fourth hole $\overline{T}$ equals:

$$\begin{aligned} ab \underline{b}\underline{a} ba \underline{a} aaba|\underline{b}b \end{aligned}$$

and the sealed fragment is abab.

Henceforth we denote by ${\mathcal F}$ the S-family constructed in Lemma 4.9. In order to transform it into an instance of Lemma 4.2, we need to compute the Lyndon roots of the sealed fragments $W_i$ (that is, the values $first _i$ and $\ell _i$).

4.3 Lyndon roots of sealed fragments

We will show how to compute Lyndon roots $\lambda _i$ of sealed fragments $(W_i,q_i) \in {\mathcal F}$. Obviously, a Lyndon root of a sealed fragment can be represented in the same space complexity as the sealed fragment itself.

Let us start with the following fact that encapsulates Theorems 20 and 23 from Kociumaka (2016).

Fact 4.11

(Kociumaka 2016) For a word of length n, after $\mathcal {O}(n)$-time preprocessing,

• (a) the length of the lexicographically minimal suffix of a d-fragment can be computed in $\mathcal {O}(d^2)$ time;

• (b) the shift value of the minimal cyclic shift of a d-fragment can be computed in $\mathcal {O}(d^2)$ time.

As a consequence of Fact 4.11(a) we obtain:

Observation 4.12

For a word of length n, after $\mathcal {O}(n)$-time preprocessing, the length of the lexicographically maximal suffix of a d-fragment can be computed in $\mathcal {O}(d^2)$ time.

Proof

To compute the maximal suffix instead of the minimal suffix, we reverse the lexicographic order on the alphabet and append the d-fragment in question with a letter that is greater than all the letters from $\varSigma$. $\square$

Fact 4.11(a) and Observation 4.12 provide us with the following toolbox for sealed fragments.

Lemma 4.13

(Kociumaka 2016) For a partial word of length n with k holes, after $\mathcal {O}(n)$-time preprocessing,

• (a) the length of the lexicographically maximal suffix of a sealed fragment can be computed in $\mathcal {O}(k^2)$ time.

• (b) the shift value of the minimal cyclic shift of a sealed fragment can be computed in $\mathcal {O}(k^2)$ time.

Lemma 4.14

If W is a periodic sealed fragment and q is its period (not necessarily shortest) such that $2q \le |W|$, then the length of the Lyndon root of W and its first occurrence in W can be computed in $\mathcal {O}(k^2)$ time after $\mathcal {O}(n)$-time preprocessing.

Proof

Let $s \in [0..q-1]$ be the shift value of the minimal cyclic shift of W[1..2q] and $i=s+1$. It can be computed in $\mathcal {O}(k^2)$ time using Lemma 4.13(a). We know that the Lyndon root $\lambda$ of W starts at the position s and that its length $\ell$ divides q.

Fig. 5

We have $|W|=18$ and $q=8$. The minimal cyclic shift of W[1..16] is $(abcb)^4$ and starts, e.g., at position $i=7$, so the shift value is $s=6$. Then, the maximal suffix over reversed alphabet of W[8..18] starts at position $i'=11$. We have $\ell =i'-i=4$ and $s\bmod \ell =2$. The Lyndon root of W is abcb

We then use Lemma 4.13(b) to find the starting position $i'$ of the maximal suffix of $W[i+1..|W|]$ with the reversed lexicographic order of the alphabet. If $W[i'..|W|]$ is a prefix of W[i..|W|], then $\ell =i'-i$, and otherwise $\ell =q$. We check this condition in $\mathcal {O}(k)$ time using Fact 4.6. Finally, we return $s \bmod \ell$ and $\ell$; see Fig. 5. $\square$

By point (a) of the definition of an S-family we immediately obtain:

Corollary 4.15

The Lyndon roots of all sealed fragments $(W_i,q_i) \in {\mathcal F}$ can be computed in $\mathcal {O}(nk^3)$ time after $\mathcal {O}(n)$-time preprocessing.

With this missing puzzle we are ready to conclude the algorithm for reporting all unambiguous p-square factors of a partial word.

Theorem 4.16

For a partial word T of length n with k holes, all elements of the set $usquares (T)$ can be reported in $\mathcal {O}(nk^3)$ time.

Proof

We construct a family of sealed fragments that consists of the solid p-squares $ssquares (T)$ and an S-family ${\mathcal F}$. By Corollary 4.5 and Lemma 4.9, this family can be constructed in $\mathcal {O}(nk^2)$ time. We compute Lyndon roots of all the sealed fragments in $\mathcal {O}(nk^3)$ time using Corollary 4.15. For each solid p-square we may compute its Lyndon root in $\mathcal {O}(k^2)$ time using Lemma 4.14; we can also use the Lyndon roots as computed in Crochemore et al. (2014).

The constructed family satisfies the assumption of Lemma 4.2 with $N=\mathcal {O}(nk)$. (Actually, if for any sealed factor $(W_i,q_i)$ of the S-family ${\mathcal F}$ we have $|W_i|=2q_i$, we need to check if it equals any of the solid squares of the same length and, if so, remove it, so that no two short words repeat.) This lemma lets us report all the distinct representatives of u-squares in $\mathcal {O}(nk^3+|SQ|)$ time. The total number of u-squares that will be generated is $\mathcal {O}(nk)$ due to Theorem 6.6. This gives the final complexity of the algorithm. $\square$

5 Combinatorial bounds for ambiguous p-squares

Let T be a partial word of length n with k holes. The upper bound in the case of a-squares is straightforward.

Theorem 5.1

If T is a partial word of length n with k holes, then $asquares (T)=\mathcal {O}(nk^2)$.

Proof

The number of possible lengths of a-squares is at most $\left( {\begin{array}{c}k\\ 2\end{array}}\right)$, since we have $\left( {\begin{array}{c}k\\ 2\end{array}}\right)$ possible distances between the k holes. Consequently, the number of p-squares with such lengths is at most $nk^2$. $\square$

Let us proceed to the lower bound proof. We say that a set A of positive integers is an (m, t)-cover if the following conditions hold:

• (1) For each $d\ge m$, A contains at most one pair of elements with difference d;

• (2) $|\,\{\,|j-i|\ge m\;:\; i,j \in A\,\}\,|\,\ge \, t$.

Example 5.2

• $\{1,2,3,6,9,12\}$ is a (3, 9)-cover.

• $\{1,2,3,11,14,17\}$ is a (8, 9)-cover.

For a set $A\subseteq [1..n]$ we denote by $W_{A,n}$ the partial word of length n over the alphabet $\varSigma$ such that $W_{A,n}[i]=\lozenge \Leftrightarrow i\in A$, and $W_{A,n}[i]=a$ otherwise.

Lemma 5.3

Assume that $A\subseteq [1..n]$ is an (m, t)-cover such that $m=\varTheta (n)$, $|A|=k$, and $t=\varOmega (k^2)$. Let $\varSigma =\{a,b\}$ be the alphabet. Then

$$\begin{aligned} asquares (a^{n-2}\cdot W_{A,n}\cdot a^{n-2})\;=\;\varOmega (n\cdot k^2). \end{aligned}$$

Proof

Each even-length factor of $a^{n-2}\cdot W_{A,n}\cdot a^{n-2}$ is a p-square. Let $\mathcal {Z}$ be the set of these factors X which contain two positions $i<j$ containing holes with $j-i\ge m$ and $|X|=2(j-i)$. As A is an (m, t)-cover, i and j are determined uniquely by $d=j-i$. Then all elements of $\mathcal {Z}$ are pairwise non-equivalent a-squares. The size of $\mathcal {Z}$ is $\varOmega (mt)$ which is $\varOmega (n\cdot k^2)$. $\square$

Theorem 5.4

For every positive integer n and $k\le \sqrt{2n}$, there is a partial word of length n with k holes that contains $\varOmega (nk^2)$ non-equivalent a-square factors.

Proof

Due to Lemma 5.3, it is enough to construct a suitable set A. By monotonicity, we may assume that k and n are even. We take:

$$\begin{aligned} A\;=\; \left[ 1..\tfrac{k}{2}\right] \,\cup \, \left\{ j\cdot \tfrac{k}{2}+\tfrac{n}{2}\;:\; 1 \le j\le \tfrac{k}{2}\right\} . \end{aligned}$$

We claim that A is an $(\tfrac{n}{2},t)$-cover for $t=\varOmega (k^2)$. Indeed, take any $i,j \in [1..\tfrac{k}{2}]$. Then $j\cdot \tfrac{k}{2}+\tfrac{n}{2}-i\ge \tfrac{n}{2}$ and all such values are distinct; hence, $t=\tfrac{k^2}{4}$. The thesis follows from the claim. $\square$

Example 5.5

Let us consider the (8, 9)-cover from Example 5.2, which is a subset of [1..n] for $n=17$, and the partial word $a^{n-2}\cdot W_{A,n}\cdot a^{n-2}$:

$$\begin{aligned} a^{15} {\lozenge }{\lozenge }{\lozenge }aaaaaaa{\lozenge }aa{\lozenge }aa{\lozenge }a^{15}. \end{aligned}$$

This partial word contains all a-squares with representatives being cyclic shifts of $({\lozenge }a^i)$ for $i=7,\ldots ,15$.

6 Combinatorial bounds for unambiguous p-squares

The following theorem shows a lower bound construction. Afterwards we design an upper bound that asymptotically matches this lower bound.

Theorem 6.1

For every positive integers n and k, $k \le \frac{1}{3} n$, there is a partial word of length n with k holes that contains $\varOmega (nk)$ non-equivalent u-square factors.

Proof

Let us consider the following partial word over the alphabet $\{a,b\}$:

$$\begin{aligned} W_m\,=\,a^{m-1} b a^{m-k} \lozenge ^k a^{m-1}. \end{aligned}$$

Then for every $i \in [1..k]$, $W_m$ has $m-k+i$ u-square factors of half length $m-k+i$ containing the letter b; see also Fig. 6. Altogether the number of such u-squares is:

$$\begin{aligned} \sum _{i=1}^k\; m-k+i\;=\; \varOmega (nk), \end{aligned}$$

where $n=3m-1=|W_m|$. If n gives a different remainder modulo 3, we can pad $W_m$ with the letter a. $\square$

Fig. 6

The u-square factors of half length 4 with the letter b in $W_m\,=\,a^{m-1} b a^{m-k} \lozenge ^k a^{m-1}$ for $m=4$, $k=2$

If X is a partial word, then by $LONG (X)$ we denote the set of all p-squares of length at least $\frac{1}{2}|X|$ which occur in X as a prefix.

If A is a set of numbers, $|A| \ge 2$, then we denote

$$\begin{aligned} \mathrm {mingap}(A)=\min \{|b-a|\,:\,a,b\in A,\,a \ne b\}. \end{aligned}$$

If $\mathcal {Z}$ is a set of partial words, then $\mathrm {mingap}(\mathcal {Z})$ denotes $\mathrm {mingap}(\{|S|\,:\,S \in \mathcal {Z}\})$.

Lemma 6.2

(Three p-Squares Lemma) Let X be a partial word with k holes. Assume that the set $LONG (X)$ contains at least three elements. Then $\delta =\mathrm {mingap}( LONG (X))/2$ is a 12k-approximate quantum period of the longest p-square in $LONG (X)$.

Proof

Let $B,C \in LONG (X)$ be p-squares such that $|B|-|C|=2\delta$. Also let A and D be the longest and the shortest element of $LONG (X)$, respectively. Let $|A|=2a$, $|B|=2b$, $|C|=2c$, $|D|=2d$. We aim to show that $\mathsf {Mis}_{\delta }(A) \le 12k$. We consider two cases, depending on whether $B \ne A$ or $B = A$.

Case $B \ne A$: Let us consider the following intervals (see Fig. 7):

$$\begin{aligned} I_1 = (0..c]\quad I_2 = (c-\delta ..b] \quad I_3 = (c.. b+c] \quad I_4 = (2c.. 2b] \quad I_5 = (b+c.. 2a]. \end{aligned}$$

Fig. 7

$B \ne A$

Let $m_q = |\{i \in I_q\,:\,i-\delta \in I_q,\ X_{i-\delta } \not \approx X_i\}|$. We show the following inequalities:

• (I) $m_1 \le k$:

  Assume that $i \in \mathsf {Mis}_\delta (A) \cap I_1$. Note that $X_i\approx X_{i+c}\approx X_{i+c-b} = X_{i-\delta }$ due to p-squares B and C, respectively. Hence, $X_i\not \approx X_{i-\delta }$ may hold only if $X_{i+c}=\lozenge$.

• (II) $m_3 \le k$:

  Assume that $i \in \mathsf {Mis}_\delta (A) \cap I_3$. Note that $b < i \le b+c$. Hence, $X_i\approx X_{i-b}\approx X_{i-b+c} = X_{i-\delta }$ due to p-squares B and C, respectively. Consequently, $X_i\not \approx X_{i-\delta }$ may hold only if $X_{i-b}=\lozenge$.

• (III) $m_4 \le m_1+k$:

  Assume that $i \in \mathsf {Mis}_\delta (A) \cap I_4$. Note that $a< i-\delta < i \le 2a$. Let $J=(2c-a..2b-a]$. Note that $X[I_4] \approx X[J]$ due to p-square A and that $J \subseteq I_1$. We apply Lemma 2.3(c) to $X[I_4]$ and X[J] to conclude.

• (IV) $m_2 \le m_4+k$:

  Assume that $i \in \mathsf {Mis}_\delta (A) \cap I_2$. Note that $c-\delta< i-\delta < i \le b$. Note that $X[I_2] \approx X[I_4]$ due to p-square B. We apply Lemma 2.3(c) to $X[I_2]$ and $X[I_4]$ to conclude.

• (V) $m_4+m_5 \le m_1+m_2+m_3+2k$:

  Assume that $i \in \mathsf {Mis}_\delta (A) \cap (I_4\cup I_5)$. Note that $a< i-\delta < i \le 2a$. Let $J=(2c-a..a]$. Note that $X[I_4 \cup I_5] \approx X[J]$ due to p-square A and that $J \subseteq I_1\cup I_2 \cup I_3$. We apply Lemma 2.3(b) to $X[I_4 \cup I_5]$ and X[J] to conclude.

We conclude that $|\mathsf {Mis}_{\delta }(A)| = m_1+m_2+m_3+(m_4+m_5) \le k+3k+k+7k = 12k$.

Case $B = A$: Let us consider the following intervals (see Fig. 8):

$$\begin{aligned} I'_1 = (0..c]\quad I'_2 = (c-\delta ..b] \quad I'_3 = (c.. b+c] \quad I'_4 = (2c.. 2b]. \end{aligned}$$

Fig. 8

$B = A$

Let $m'_q = |\{i \in I'_q\,:\,i-\delta \in I'_q,\ X_{i-\delta } \not \approx X_i\}|$. We show the following inequalities:

• (I) $m'_1 \le k$:

  Assume that $i \in \mathsf {Mis}_\delta (A) \cap I'_1$. Note that $X_i\approx X_{i+c}\approx X_{i+c-b}=X_{i-\delta }$ due to p-squares B and C, respectively. Hence, $X_i\not \approx X_{i-\delta }$ may hold only if $X_{i+c}=\lozenge$.

• (II) $m'_3 \le k$:

  Assume that $i \in \mathsf {Mis}_\delta (A) \cap I'_3$. Note that $b< i < b+c$. Hence, $X_i\approx X_{i-b}\approx X_{i-b+c}=X_{i-\delta }$ due to p-squares B and C, respectively. Consequently, $X_i\not \approx X_{i-\delta }$ may hold only if $X_{i-b}=\lozenge$.

• (III) $m'_2 \le m'_1+k$:

  Assume that $i \in \mathsf {Mis}_\delta (A) \cap I'_2$. Note that $d< c-\delta< i-\delta < i \le b \le 2d$. Let $J=(c-\delta -d..b-d]$. Note that $X[I'_2] \approx X[J]$ due to p-square D and that $J \subseteq I'_1$. We apply Lemma 2.3(c) to $X[I'_2]$ and X[J] to conclude.

• (IV) $m'_4 \le m'_2 +k$:

  Assume that $i \in \mathsf {Mis}_\delta (A) \cap I'_4$. Note that $X[I'_4] \approx X[I'_2]$ due to p-square B. We apply Lemma 2.3(c) to $X[I'_4]$ and $X[I'_2]$ to conclude.

We conclude that $|\mathsf {Mis}_{\delta }(A)| = m'_1+m'_2+m'_3+m'_4 \le k+2k+k+3k = 7k$. $\square$

Recall that a deterministic period of a partial word X is an integer q such that there exists a (solid) word W such that $W \approx X$ and W has a period q. In the following lemma we show that if the set $LONG (X)$ is large enough, then the majority of its elements have strong periodic properties.

Lemma 6.3

Let X be a partial word with k holes. Assume that the set $LONG (X)$ contains at least $16k+3$ elements. Then $\delta =\mathrm {mingap}( LONG (X))/2$ is a deterministic period of all p-squares from $LONG (X)$ excluding possibly the $2k+1$ longest ones.

Proof

Let $LONG '(X)$ be the set $LONG (X)$ without the $2k+1$ longest elements, A be the longest p-square in $LONG (X)$, and B be the longest p-square in $LONG '(X)$. We start by a proof of a weaker property. In the proof we will use the fact that $|\mathsf {Mis}_\delta (A)| \le 12k$ (Lemma 6.2).

Claim

$\delta$ is a quantum period of B.

Proof

Assume to the contrary that B does not have quantum period $\delta$, i.e., that $\mathsf {Mis}_\delta (B) \ne \emptyset$. Let i be the minimum index in $\mathsf {Mis}_\delta (B)$.

• 1. Let us count the p-squares from $LONG (X)$ that contain the position i in the first half. Let $C \in LONG (X)$, $|C|=2c$, be such a p-square. Then $X_{i+c} \approx X_i \not \approx X_{i-\delta } \approx X_{i+c-\delta }$. Hence, either at least one of the positions $X_{i+c}$ and $X_{i+c-\delta }$ contains a hole (2k possibilities), or $X_{i+c} \not \approx X_{i+c-\delta }$ which means that $i \in \mathsf {Mis}_\delta (A)$ (12k possibilities due to Lemma 6.2). Therefore, there can be at most 14k such p-squares.

• 2. Let us count the p-squares from $LONG (X)$ that contain $i-\delta$ in the first half and i in the second half. There can be at most one such p-square. Otherwise there would be two p-squares in $LONG (X)$ whose halves’ lengths differ by less than $\delta$, contradicting the definition of $\delta$.

• 3. Let us count the p-squares from $LONG (X)$ that contain both positions $i-\delta$ and i in the second half. Let $C \in LONG (X)$, $|C|=2c$, be such a p-square. Then $X_{i-c} \approx X_i \not \approx X_{i-\delta } \approx X_{i-c-\delta }$. Hence, at least one of the positions $X_{i-c}$ and $X_{i-c-\delta }$ contains a hole (they cannot form a mismatch, as i was selected as the minimal index). This gives 2k possibilities for such a p-square.

• 4. We will show that there are no p-squares from $LONG (X)$ that do not contain the position i. If such a p-square existed, then we would have $|X|/2< i-\delta < i \le |B|$, so $i-\delta$ and i would be contained in right halves of all p-squares that are at least as long as B. There are $2k+1$ of them, which contradicts point 3.

Each p-square in $LONG (X)$ accounts to one of the categories 1-4. We have shown that there can be at most $16k+1$ p-squares in $LONG (X)$ which contradicts the assumptions of the lemma. This completes the proof of the claim.$\square$

Now we strengthen the previous claim and prove that $\delta$ is a deterministic period of B. This will conclude the proof since all the p-squares in $LONG '(X)$ are prefixes of B.

Assume that this is not true and let d be minimal such that $B_{i-d\delta } \not \approx B_i$ and let i be the minimal such index i. Hence, $B_{i-\delta }=\ldots =B_{i-(d-1)\delta }={\lozenge }$. Therefore, $d \le k+1$, and by the claim, $d \ge 2$. Moreover, $k>0$.

• 1. Let us count the p-squares $C \in LONG (X)$, $|C|=2c$, that contain i in the first half. Let $j=i+c$. If $j>|B|$, then $C \in LONG (X) \setminus LONG '(X)$ and there are $2k+1$ such p-squares. Otherwise, there can be at most 3k p-squares $C \in LONG '(X)$ for which any of the positions $j-d\delta$, $j-\delta$, j contains a hole. Assume otherwise. Then $B_{j-d\delta } = B_{i-d\delta } \not \approx B_i = B_j$ and $B_{j-\delta } \ne {\lozenge }$. Hence, $B_{j-\delta } \not \approx B_{j-d\delta }$ or $B_{j-\delta } \not \approx B_j$, either of which contradicts the way d was selected. In total, there can be $5k+1$ of the considered p-squares.

• 2. Let us count the p-squares from $LONG (X)$ that contain $i-d\delta$ in the first half and i in the second half. There can be at most d of them, as otherwise there would be two p-squares in $LONG (X)$ whose halves’ lengths differ by less than $\delta$, a contradiction. Hence, the number of such p-squares is at most $k+1$.

• 3. Let us count the p-squares $C \in LONG (X)$, $|C|=2c$, that contain both positions $i-d\delta$ and i in the second half. Let $j=i-c$. There can be at most 2k such p-squares C for which any of the positions $j-d\delta$, j contains a hole. Assume otherwise. Then $B_{j-d\delta } \not \approx B_j$ which contradicts the definition of i.

• 4. Let us count the p-squares that contain the position $i-d\delta$ in the second half and do not contain the position i. Using the same argument as in 2, we see that there are at most $k+1$ of them.

• 5. Finally, we will show that there are no p-squares in $LONG (X)$ that do not contain the position $i-d\delta$. If such a p-square existed, then both positions $i-d\delta$ and i would be contained in right halves of all p-squares from $LONG (X) \setminus LONG '(X)$. There are $2k+1$ of them, which contradicts point 3.

Each p-square in $LONG (X)$ accounts to one of the categories 1-5. We have shown that there can be at most $9k+3$ p-squares in $LONG (X)$ which contradicts the assumptions of the lemma, as $k>0$. This completes the proof of the lemma.$\square$

By $\mathcal {U}\text{- } Pref (X)$ we denote the set of unambiguous p-squares in $LONG (X)$ that occur in X only as a prefix.

Lemma 6.4

Let X be a partial word with k holes. Then $|\mathcal {U}\text{- } Pref (X)| < 16k+3$.

Proof

Assume to the contrary that $|\mathcal {U}\text{- } Pref (X)| \ge 16k+3$. Let us recall that $\mathcal {U}\text{- } Pref (X) \subseteq LONG (X)$ so the assumptions of Lemma 6.3 are satisfied.

Let $\mathcal {U}\text{- } Pref '(X)$ be the set $\mathcal {U}\text{- } Pref (X)$ without the $2k+1$ longest elements. By Lemma 6.3, each p-square in $\mathcal {U}\text{- } Pref '(X)$ has a deterministic period $\delta = \mathrm {mingap}( LONG (X))/2$.

Let us assume that $B=X[1..2a] \in \mathcal {U}\text{- } Pref '(X)$ and let $W^2$ be its (solid) representative. Then $C=X[1+\delta ..2a+\delta ]$ is a p-square, as it matches $W^2$ due to the deterministic period $\delta$. If $X[2a+1..2a+\delta ]$ did not contain a hole, then C would be another occurrence of a u-square with representative $W^2$. This would contradict the assumption that $B \in \mathcal {U}\text{- } Pref (X)$.

Note that the fragments of the form $X[2a+1..2a+\delta ]$ for $X[1..2a] \in \mathcal {U}\text{- } Pref '(X)$ are pairwise disjoint due to the definition of $\delta$. What follows is that $|\mathcal {U}\text{- } Pref '(X)| \le k$ and $|\mathcal {U}\text{- } Pref (X)| \le 3k+1$, a contradiction. $\square$

We say that a solid square $W^2$ has a solid occurrence in T if T contains a factor equal to$W^2$. By the following fact, there are at most 2n non-equivalent p-square factors of T with solid occurrences.

Fact 6.5

(Fraenkel and Simpson 1998; Ilie 2005; Deza et al. 2015) Every position of a (solid) word contains at most two rightmost occurrences of squares.

In the proof of the upper bound on the number of u-squares we separately count u-squares that have a solid occurrence and those that do not. In the latter case, we use Lemma 6.4, which lets us bound $|\mathcal {U}\text{- } Pref (X)|$ by 19k in case that $k>0$.

Theorem 6.6

If T is a partial word of length n with k holes, then $usquares (T)=\mathcal {O}(nk)$.

Proof

Let us recall that by ${\mathcal B}(T)$ we denote the set of all basic factors of T. If $T[i..i+\ell -1]$ is a rightmost occurrence of a u-square V, then $V \in \mathcal {U}\text{- } Pref (W)$ for some $W \in {\mathcal B}(T)$. (In particular, $W=T_{i,q}$ for $q=\left\lceil \log \ell \right\rceil$.) This lets us bound $usquares (T)$ as follows:

$$\begin{aligned} usquares (T)&\le \quad \sum _{W \in {\mathcal B}(T)} |\mathcal {U}\text{- } Pref (W)| \\ \{\hbox {Fact }15\}\quad&\le \quad 2n + \sum _{W \in {\mathcal B}(T):\#_{\lozenge }(W)>0} |\mathcal {U}\text{- } Pref (W)| \\ \{\hbox {Lemma }~10\}\quad&<\quad 2n + 19 \sum _{W \in {\mathcal B}(T)}\#_{\lozenge }(W) \quad \\ \{\hbox {Lemma }~1\}\quad&\le \quad 2n + 38nk. \end{aligned}$$

This concludes the proof. $\square$