Squares and Primitivity in Partial Words

Transcription

1 Squares and Primitivity in Partial Words F. Blanchet-Sadri 1 Michelle Bodnar 2 Jordan Nikkel 3 J. D. Quigley 4 Xufan Zhang 5 November 11, 2014 Abstract Recently, Tittmann et al. introduced the subgraph component polynomial and showed that its power for distinguishing graphs is quite different from the power of other graph polynomials that appear in the literature such as the matching polynomial, the Tutte polynomial, the characteristic polynomial, the chromatic polynomial, etc. The subgraph component polynomial enumerates vertex induced subgraphs in a given undirected graph with respect to the number of components. We show the use of the subgraph component polynomial to count the number of primitive partial words of a given length over an alphabet of a fixed size, which leads to a method for enumerating such partial words. We also give a tight bound for the maximum number of holes a primitive partial word can have. Doing so allows us to establish some tight upper and lower bounds on the maximum number of occurrences of primitively-rooted squares, i.e., adjacent occurrences of two compatible substrings with primitive root, in a given partial word with a fixed number of holes over a fixed alphabet size. Keywords: Combinatorics on words; Partial words; Primitive partial words; Subgraph component polynomial; Primitively-rooted squares. This material is based upon work supported by the National Science Foundation under Grant No. DMS Part of this paper was presented at LATA 2013 [6]. 1 Department of Computer Science, University of North Carolina, P.O. Box 26170, Greensboro, NC , USA, blanchet@uncg.edu 2 Department of Computer Science, University of California, San Diego, 9500 Gilman Drive #0112, La Jolla, CA , USA 3 Department of Mathematics, Vanderbilt University, 1326 Stevenson Center, Nashville, TN 37240, USA 4 Department of Mathematics, University of Illinois-Urbana-Champaign, Altgeld Hall, 1409 W. Green Street, Urbana, IL 61801, USA 5 Department of Mathematics, Princeton University, Fine Hall, Washington Road, Princeton, NJ 08544, USA 1

2 1 Introduction Motivated by social and biological networks, Tittmann et al. [21] introduced the subgraph component polynomial Q(G; x, y) of an undirected graph G with n vertices as the bivariate generating function which counts the number of connected components in vertex induced subgraphs. More precisely, Q(G; x, y) = n n i=0 j=0 q ij(g)x i y j, where q ij (G) is the number of vertex induced subgraphs of G with exactly i vertices and j connected components. They related the subgraph component polynomial to other graph polynomials that appear in the literature such as the Tutte polynomial, the universal edge elimination polynomial, etc. (see, for instance, [19] for more information on graph polynomials). They showed several remarkable properties of the subgraph component polynomial, among them is their use to compute the so-called residual connectedness reliability. They also showed that the problem of computing Q(G; x, y) is P -hard, but that it is fixed parameter tractable when restricting to graph classes that have bounded tree-width and to classes of bounded clique-width. Primitivity is a well-studied topic in combinatorics on words (see, for instance, [12]). A word is primitive if it is not a power of another word. It is well-known that the number of primitive words of a given length over an alphabet of a fixed size can be calculated using the Möbius function [17, 20]. Here, we discuss the use of the above subgraph component polynomial to count the number P h,k (n) of primitive partial words with h holes of length n over a k-letter alphabet. Research on primitive partial words was initiated by the first author in [4]. A partial word, also referred to as a string with don t-cares [1, 13], is a sequence which may have some undefined positions, called holes and represented by s, that are compatible with, or match, any letter in the alphabet (a (full) word is a partial word without holes). Partial words allow for incomplete or corrupted data and have been subject of much recent investigation, both combinatorial and algorithmic (see, for instance, [5]). In [5], formulas for h = 1 and h = 2 are given in terms of the formula for h = 0 and some bounds are provided for h > 2, but no exact formulas are given for h > 2. We associate a graph G n,p with any partial word of length n with strong period set P as follows: the vertices represent the positions 0,..., n 1 and the edges are the pairs {i, i + mp}, where 0 i < i + mp n 1, m Z, and p P. It turns out that P h,k (n) can be expressed in terms of the Q(G n,p ; x, y) s. Squares are also well-studied in algorithms and combinatorics on words (see, for instance, [11]). A square in a word is a factor uu for some word u, called the root of the square; uu is primitively-rooted or PR if u is primitive. 2

3 The word contains multiple occurrences of the square (01) 2, whose root 01 is primitive, as well as the square (0101) 2, whose root 0101 is not primitive. The maximum number of occurrences of PR-squares in a word of length n is well-known to be Θ(n log n), a bound which is reached by the Fibonacci word [14], and Crochemore [10] gave in 1981 an optimal algorithm for computing all such occurrences in worst-case time O(n log n) (see also [2, 18]). In the context of partial words, a factor is a consecutive sequence of symbols, whereas a subword is a full word compatible with a factor. Here a factor uv is a square if some completion, i.e., a filling in of the holes with letters from the alphabet, turns uv into a full word that is a square; equivalently, u and v are compatible. Although some works related to the asymptotic behavior of square occurrences and square positions and the counting of distinct squares in partial words [7, 8, 15] appear in recent literature, no work has been dedicated to counting the number of occurrences of PR-squares in a partial word of a given length. This work could be applied to analyze algorithms for detecting all repetitions in partial words. The known algorithms for detecting all PR-square occurrences in full words, such as Crochemore s algorithm [10], are not easily extendible to partial words. One of the most important culprits is that the compatibility relation is not transitive, as opposed to the equality relation being transitive. We have that 0 is compatible with and that is compatible with 1, but 0 is not compatible with 1. So we adopt here an approach, for our bounds, based on the large divisors of the length of the partial word. The contents of our paper are as follows: In Section 2, we review a few basic concepts on partial words. In Section 3, we answer the question How many holes can a primitive partial word of length n over a k-letter alphabet contain?. We show that this number can be expressed in terms of the large divisors of n. In Section 4, we describe a general method for counting primitive partial words. We do so using the subgraph component polynomial. This leads to a method for the enumeration of primitive partial words. We also discuss the computation of N h,k (n) = ( n h) k n h P h,k (n), the number of non-primitive partial words with h holes of length n over a k-letter alphabet, using the subgraph component polynomial. In Section 5, we give tight upper and lower bounds on the number of occurrences of PRsquare factors in partial words. With these bounds in mind, we also give tight upper and lower bounds on the number of occurrences of PR-square subwords. Finally in Section 6, we conclude with some remarks. 3

4 2 Preliminaries Let A be a non-empty finite set, or an alphabet. We consider a partial word w over A as a word over the enlarged alphabet A = A { }, where the additional symbol plays the role of an undefined position or a hole. A letter always refers to an element of the alphabet being considered, while a symbol refers to either a letter or a hole. For 0 i < n, the symbol at position i of w is denoted by w[i] using an indexing starting at 0, and D(w) (respectively, H(w)) is the set of positions that contain letters (respectively, holes). If w[i] A, then i is defined, otherwise i is a hole. If H(w) =, then w is a full word. We denote by w the length of w or the number of symbols in w. Two partial words u and v are compatible, denoted u v, if u = v and for every i D(u) D(v) we have u[i] = v[i], and u is contained in v, denoted u v, if for every i D(u) we have u[i] = v[i]. If u and v are compatible, then u v denotes the partial word such that u u v, v u v, and D(u v) = D(u) D(v). A factor of a partial word w is a consecutive sequence of symbols in w, whereas a subword of w is a full word compatible with a factor in w. We denote by w[i..j) the factor of w that starts at position i and ends at position j 1. A completion of a partial word u is any full word v such that u v. A strengthening of u is a partial word obtained by filling in one hole of u with a letter, and a weakening of u is a partial word obtained by replacing one letter of u with a hole. For a positive integer p, a partial word w of length n has a strong period of p or is strongly p-periodic if for every 0 i < j < n, i mod p = j mod p implies w[i] w[j]. It is weakly p-periodic if for every 0 i < n p, w[i] w[i + p]. For full words, weak periodicity implies strong periodicity, and we may drop the words weakly and strongly when referencing full words. A partial word w is primitive if for every full word v and integer m 2, we have w v m, equivalently, if there is no proper divisor p of w such that w is strongly p-periodic. Clearly, if w is primitive and w v, then v is primitive. Thus if w is primitive, so are all of its completions. We define the root of a square factor uv of a partial word as u v. Now uv is a PR-square factor if its root is a primitive partial word, while uu is a PR-square subword if it is a full square word, with primitive root u, that is compatible to a square factor (not necessarily a PR-square factor). For example, is a PR-square subword occurring at position 0 of the partial word , but the factor is not a PR-square factor. Note that 10 has three occurrences of PR-square factors, 0,, 10, with respective primitive roots 0,, 10. It has four occurrences of PR- 4

5 square subwords, though, namely 1 2, 0 2, and (10) 2, noting that 0 2 occurs twice, once beginning at position 1 and once at position 2. 3 Maximizing the Number of Holes in Primitive Partial Words We give a tight bound for the maximum number of holes a primitive partial word can have. We define the set of large factors, or large divisors, LF (n) of an integer n as the set of divisors of n, distinct from n, whose multiples are either n itself or not divisors of n. Clearly, every distinct prime divisor i of n gives rise to exactly one large divisor of n, namely n/i, and hence the number of distinct prime divisors of n gives LF (n). For example, when n = 30 we have LF (30) = {6, 10, 15}. Proposition 1. Given a primitive partial word of length n which contains the maximum number of holes, set LF (n) = {f 1,..., f m }. For any nonhole positions i and j, we have i j = c 1 f c m f m for some c i Z. Moreover, the fewest number of non-holes which rule out all of the large divisors of n as strong periods is LF (n) + 1. Proof. Let n = p α 1 1 pα 2 2 pαm m be a prime factorization of n, and let f i = n p i be the elements of LF (n), all distinct. We claim that the equality c i f i = c 1 f c i 1 f i 1 + c i+1 f i c m f m never holds for any choice of c j Z with j {1,..., m}\{i} and any choice of c i with p i < c i < p i. To see this, divide both sides of the equality by p α i 1 i. By the construction of the f j s, we know we can always do this. Then every term on the right hand side contains a single p i factor, so the right hand side is divisible by p i. However the left hand side does not contain p i in its prime factorization, so they cannot be equal. Now, to rule out the large divisor f 1, we must use two non-holes i 1, i 2 and they must differ by a multiple of f 1. Note that for each i and j, we have lcm(f i, f j ) = n, so i 1 and i 2 rule out at most one large divisor, i.e., f 1. To rule out the next large divisor f 2, there must exist a non-hole position i 3 that differs from i 1 or i 2, say i 1, by some multiple of f 2. Since i 1 i 2 = c 1 f 1 and i 1 i 3 = c 2 f 2 for some c 1, c 2 Z, we get i 2 i 3 = i 2 i 1 +i 1 i 3 = c 1 f 1 +c 2 f 2. By our claim, the addition of i 3 cannot possibly rule out any large divisor other than f 2. We continue in this way until all large divisors have been ruled out as strong periods. After ruling out the first large divisor with 5

6 two non-holes, LF (n) 1 non-holes are required to rule out the remaining LF (n) 1 large divisors. This necessitates a total of LF (n) + 1 non-hole positions to rule out all large divisors of n. The next theorem is useful because the structure of primitive partial words with maximum number of holes will help in counting PR-square factor occurrences. Theorem 1. The maximum number of holes that a primitive partial word of length n over an alphabet of size k 2 can contain, denoted τ(n), is τ(n) = n LF (n) 1. Moreover, the maximum number of holes a primitive word can contain can be achieved using a binary alphabet. Proof. We begin with a construction over the binary alphabet {0, 1} which shows that this number of holes can always be achieved. Let the word w be defined as w[i] = 0 if i + 1 LF (n), w[i] = 1 if i = n 1, and w[i] = otherwise. There are LF (n) 0 s and one 1, leaving room for exactly n ( LF (n) + 1) holes as desired. We observe that it is unnecessary to check for incompatibilities in smaller strong periods because for any q which divides an element p of LF (n), an incompatibility in a strong period p implies an incompatibility in a strong period q, so we need only check that all strong periods given by our large divisor set do not occur in w. Let p LF (n). Since p divides n we have n = lp for some l. There is a 0 in position p 1 and a 1 in position n 1 = lp 1, so p cannot be a strong period of w because we have two incompatible positions which differ by a multiple of p. We conclude that this construction yields a primitive word, so τ(n) n LF (n) 1. By Proposition 1, we require our word to have at least LF (n) + 1 non-holes, so τ(n) n ( LF (n) + 1). Therefore our construction is maximal. The following corollary easily follows. Corollary 1. If a partial word w has a PR-square factor of root length m 2, then H(w) w LF (m) 1. Proof. Assume that uv is a PR-square factor of w with root length m 2. Then, D(w) D(u v) = m H(u v). Theorem 1 gives D(w) m (m LF (m) 1) = LF (m) + 1. Since D(w) = w H(w), the result follows. The next proposition states how many primitive completions a partial word with h holes over an alphabet of size k can contain. Clearly this 6

7 Figure 1: G n,p where n = 6, P = {2, 3} number is bounded by k h, the number of completions that the word has, and any primitive partial word w meets this bound. But suppose w is not primitive, then how many primitive completions must it have? Proposition 2. Let w be a partial word with h > 0 holes over an alphabet of size k 2. Then there are at least (k 1)k h 1 completions of w which are primitive. Proof. Let w be any partial word with h > 0 holes over a k-letter alphabet A such that k 2. Choose any position i such that i H(w). Let a A, then consider ŵ i,a to be any completion of w with the requirement that ŵ i,a [i] = a, noting that there are k h 1 such completions. For each letter b A, different from a, consider the completion of w that agrees with ŵ i,a except that the letter at position i is b instead of a. By [5, Corollary 6.9], at most one of these completions is non-primitive, and hence for each of the k h 1 completions ŵ i,a, we have at least k 1 unique primitive completions of w. 4 Counting with the Subgraph Component Polynomial We start with the graphical representation G n,p of a partial word of length n with strong period set P where edges indicate compatibility: G n,p = (V, E) is a simple, undirected graph where V = {0,..., n 1}, and {i, j} E if and only if there exists p P such that j = i + mp for m Z. Figure 1 gives an example. Let Q(G; x, y) = n n i=0 j=0 q ij(g)x i y j be the subgraph component polynomial of a graph G with n vertices, where q ij (G) is the number of induced 7

8 subgraphs of G with exactly i vertices that have j connected components. For fixed i, we use the notation Q i (G; x, y) = n j=0 q ij(g)x i y j. For example, the subgraph component polynomial for the graph in Figure 1 is Q(G 6,P ; x, y) = 1+6xy +9x 2 y +6x 2 y 2 +14x 3 y +6x 3 y 2 +15x 4 y +6x 5 y +x 6 y. The coefficient 9 on x 2 y indicates that there are 9 ways to create an induced subgraph of two vertices with a single connected component. This corresponds to the 9 edges in our graph. Lemma 1. The number of partial words of length n with h holes over a k-letter alphabet with strong period set P is Q n h (G n,p ; 1, k). Proof. The expression Q n h (G n,p ; 1, y) gives a polynomial in y whose coefficient on y j is the number of induced subgraphs on n h vertices that have exactly j connected components. Subgraphs with n h vertices represent which compatibilities must be satisfied amongst the symbols in the word. We associate specific letters with these remaining vertices, noting that each connected component must consist of vertices all associated with the same letter. We have k letter choices to associate with each connected component, so we substitute k for y. There is no double counting between terms because hole placements that lead to different numbers of connected components are necessarily distinct. The following theorem gives a formula for the number of non-primitive words of length n with h holes over k letters. Theorem 2. If LF (n) = {f 1,..., f m }, then N h,k (n) = m Q n h (G n,{fi }; 1, k) i=1 i j + ( 1) l+1 Q n h (G n,{fi,f j }; 1, k) + i 1 =i l Q n h (G n,{fi1,...,f il }; 1, k) + + ( 1) m+1 Q n h (G n,{f1,...,f m}; 1, k). Proof. A word of length n is non-primitive if and only if it has a strong period equal to one of n s large divisors. By Lemma 1, the term in the first sum counts all words with h holes and strong period f i. The term in the second sum subtracts those which are double counted because they have strong periods f i and f j. By inclusion-exclusion, the formula counts the total number of words with h holes whose strong period set contains at least one large divisor of n, and thus all non-primitive words. 8

9 To investigate methods of computing the desired subgraph component polynomials, we first recall two important facts from [21, Proposition 15,Theorem 12]: If G = K n is the complete graph on n vertices, then Q(K n ; x, y) = y(1+x) n y +1. While if G = G 1 G c is the disjoint union of c graphs, then Q(G; x, y) = c j=1 Q(G j; x, y). Proposition 3. Let P = {p}, where p divides n. Then G n,p is the disjoint union of p graphs isomorphic to K n p and Q(G n,p; x, y) = (y(1+x) n p y+1) p. Proof. Let V i consist of the vertex labeled i and all other vertices which differ from i by a multiple of p. This partitions the set of vertices into exactly p equivalence classes, so each induced subgraph G i is disjoint from every other one. By the definition of G n,p, any two vertices in V i are connected by an edge so G i is a complete graph, but no edge connects a vertex in V i to a vertex in V j for i j. Since there are n/p vertices in V i, we have G i isomorphic to K n/p for 0 i < p. The formula then follows from the abovementioned two facts. As an example, we count the number of non-primitive words of length 6 over a 3-letter alphabet with two holes. Example 1. First note that LF (6) = {2, 3}. By Proposition 3 we have Q(G 6,{2} ; x, y) = (y(1 + x) 3 y + 1) 2 and Q(G 6,{3} ; x, y) = (y(1 + x) 2 y + 1) 3. Multiplying out the polynomials and looking at terms of interest we have Q 4 (G 6,{2} ; x, y) = 15x 4 y 2 and Q 4 (G 6,{3} ; x, y) = 3x 4 y x 4 y 3. Applying Lemma 1, the number of the words that are strongly 2-periodic is Q 4 (G 6,{2} ; 1, 3) = 15(3) 2 = 135 and the number of the ones that are strongly 3-periodic is Q 4 (G 6,{3} ; 1, 3) = 3(3) (3) 3 = 351. Recalling the polynomial associated with G 6,{2,3}, the only term containing x 4 is 15x 4 y. Replacing x with 1 and y with 3, the number of words that are both strongly 2- and strongly 3-periodic is Q 4 (G 6,{2,3} ; 1, 3) = 45. Finally applying Theorem 2, we find that N 2,3 (6) = = 441. The following theorem illustrates how the subgraph component polynomial can arise as a product of smaller polynomials. Theorem 3. If n = c d i=1 p i for some c N and primes p 1,..., p d, then Q(G { } n, n ; x, y) = (Q(G { },..., n n p 1 p d c, n ; x, y)) c.,..., n cp 1 cp d Proof. Begin by creating the induced subgraph G j of G = G n, { n p 1,..., n p d } consisting only of vertices labeled i such that i = j mod c and relabel them 9

10 0 through n c. In G, two vertices are connected by an edge if and only if the distance between them is a multiple of n p 1, or..., or n p d. In G j, two vertices n cp 1, are connected if and only if the distance between them is a multiple of n or..., or cp d. Thus, G j is precisely G { } n c, n. By construction, for,..., n cp 1 cp d i j, we have that G i and G j are disjoint (with respect to vertices and edges before the relabeling). Viewed before the relabeling, each G j is equivalent and G = G 1 G c, so the result follows from the multiplicativity of the subgraph component polynomial. As an example, we compute the number of binary non-primitive words of length 12 with two holes. Example 2. By Theorem 3, Q(G 12,{4,6} ; x, y) = (Q(G 6,{2,3} ; x, y)) 2. To count words with two holes, we must find terms containing x 10. Squaring the polynomial computed for G 6,{2,3}, we find that there is precisely one such term: 66x 10 y 2. Plugging in x = 1 and y = 2, we see that there are 264 words of length 12 with two holes that have strong periods 4 and 6. Applying Proposition 3 we have Q(G 12,{4} ; x, y) = (y(1+x) 3 y +1) 4 and Q(G 12,{6} ; x, y) = (y(1+x) 2 y+1) 6, and computing coefficients using the binomial theorem we have Q 10 (G 12,{4} ; x, y) = 66x 10 y 4 and Q 10 (G 12,{6} ; x, y) = 6x 10 y x 10 y 6. By Lemma 1, there are 66(2) 4 = 1056 words of length 12 with two holes and strong period 4, and 6(2) (2) 6 = 4032 with strong period 6. Finally we apply Theorem 2 to obtain N 2,2 (12) = = 4824 binary non-primitive words of length 12 with two holes. 5 Tight Bounds on the Number of Primitively- Rooted Squares in Partial Words Tables 1 and 2 summarize our results for varying numbers of holes. Note that for a constant number of holes, at most a constant number of PR-square subwords can be introduced into a partial word, and hence the bounds are identical to the full word case. 5.1 PR-square factor occurrences We first discuss the bounds in Table 1. Define psf(w, k) to be the number of occurrences of PR-square factors in a partial word w over an alphabet of fixed size k, and let psf(h, n, k) = max{psf(w, k) : H(w) = h, w = n}. 10

11 Table 1: Counting the number of PR-square factor occurrences in a partial word with h holes of length n over an alphabet of size k number psf(h, n, k) of holes constant Θ(n log n) n 2 Θ(n 2 ) n constant Θ( n2 log n ) n 1, n Θ(n) Table 2: Counting the number of PR-square subword occurrences in a partial word with h holes of length n over an alphabet of size k number pss(h, n, k) of holes constant Θ(n log n) n 2 Θ(nk n/4 ) n constant Θ(k n/2 ) n Θ(k n/2 ) Note that psf(0, n, k) is Θ(n log n) since this is identical to the full word case. Note also that psf(h, n, k) is O(n 2 ) since the number of PR-square factors in a word does not exceed the number of factors. We first examine some specific values of h for which this bound is not reached, and then give examples of where it is. Theorem 4. Let h and k be constants. Then for all n h, psf(h, n, k) is Θ(n log n). Proof. We first prove that psf(h, n, k) is O(n log n). Let w be any partial word of length n h with h holes, and let the size of the alphabet be k. By the pigeonhole principle, w can be completed in such a way that the number of occurrences of PR-square factors in its completion ŵ is at least 1 psf(w, k), giving that psf(w, k) k h psf(ŵ, k). Since psf(ŵ, k) is k h O(n log n) and k h is a fixed constant, psf(w, k) is O(n log n), and hence psf(h, n, k) is as well. We now prove that psf(h, n, k) is Ω(n log n). Let h be given. We know that psf(0, n, k) is Θ(n log n), so there exist n 0 and c such that for all n n 0, psf(w, k) cn log n for some full word w. Then let n max{h + n 0, 2h}, and choose u to be a full word of length n h such that psf(u, k) c(n h) log(n h). Then since n 2h, psf( h u, k) c(n h) log(n h) c n 2 log n 2. For the n-hole case, every square is of the form ( m ) 2, 1 m n/2, but m is strongly 1-periodic, hence if m 1, the square is not primitive. 11

12 Likewise for the (n 1)-hole case, every square factor has at most one nonhole, and hence is again strongly 1-periodic. For both of these cases then, every position except the last one starts a square of the form, a, or a for some letter a. Thus for all n 1, psf(n, n, k) = psf(n 1, n, k) = n 1. The (n d)-hole case, however, is interesting for 2 d n. Theorem 5. The bound of Θ( n2 log n ) holds for psf(n d, n, k), where d 2 is constant. Proof. To get a lower bound on psf(n d, n, k), let w = 0 d 2 w, where w = n d 2 01 n d 2. We will show that the number of PR-square factors in w is Ω( n2 log n ), and then that the number of additional PR-square factors brought by the d 2 leading 0 s is O(n). Counting PR-square factors in w with root length m, where m is prime, first note that m n d+2 2 since the square must fit inside w. Second, if a factor forms a square with an adjacent factor of the same prime length, then this square factor will be PR. Therefore, there is a limited number of start indices associated with each m that depends on n, which we denote δ n (m). There are two possibilities: either the number of start positions associated with m is limited by having to include the positions in w which have 0 and 1, or is limited by the length n d+2 of w. In the former case, δ n (m) = 2m 1, and in the latter case, δ n (m) = (n d + 2) 2m + 1 = n 2m + 3 d. Thus δ n (m) = min{2m 1, n 2m + 3 d}. So the number of occurrences of PR-square factors in w is at least n d+2 2 m=2 δ n(m), where δ n(m) is δ n (m) if m is prime and 0 otherwise. To find a lower bound, we compare the two possible values of δ n (m) and find that δ n (m) = 2m 1 when m n d+4 4. From this, the sum of all primes less than n d 4 gives a lower bound on this summation. The sum of all primes less than or equal to n is Ω( n2 log n ) [3]. This gives the lower bound for w. Note that the leading d 2 0 s bring at most n(d 2) additional PR-square factors in w = 0 d 2 w. To get an upper bound on psf(n d, n, k), consider the general form of a partial word w = c 0 a 0 c 1 a 1 a d 1 c d. 12

13 We can count the occurrences of PR-square factors in w by their root length m. Using Corollary 1, if a partial word with n d holes of length n has a PR-square factor of root length m, then LF (m) d 1. Moreover, LF (m) is the number of distinct prime divisors of m. Hence either m = 1, m is prime, or m has b distinct prime divisors, 1 b < d, and m is not prime. For the case when m = 1, there are less than n occurrences, one for each position except the last. For the case when m is prime, any such factor must include a i and a i+1 for some i. This can be upper bounded taking d 1 pairs and 2m 1 positions per pair, getting (d 1) (2m 1), m is prime; m n/2 which by [3], is O( n2 log n ). The only remaining case is when m = p l p l b b for 1 b < d, some distinct prime numbers p i s and some positive integers l i s, and l i > 1. If uv is such a PR-square factor then for every large divisor f of m, uv must contain a pair a i, a j, such that a i a j, and the distance between a i and a j is a multiple of f. For a given pair a i, a j corresponding to some large divisor, after permutation if necessary, assume there are x = yp l p l p l b b 1 symbols between a i and a j. Notice that m is determined by p 1,..., p b, l 1,..., l b. There are two cases depending on l 1. If l 1 > 1, then since x+1 has at most log n distinct prime divisors, and each p i can have a power at most log n in the factorization of x+1, the number of choices of p 1,..., p b, l 1,..., l b is at most ( log n ) b (log n) b. If l 1 = 1, then p 1 must divide a large divisor of m, which implies that p 1 divides the distance between a i and a j for some a i a j. There are d(d 1) 2 pairs of a i, a j, and the distance between each pair can have at most log n distinct prime divisors, so p 1 can take at most d(d 1) 2 log distinct values. The number of choices of p 2,..., p b, l 2,... l b is at most ( log n ) b 1 (log n) b 1, and the product gives the upper bound d(d 1) ( log n ) 2 b 1 (log n) b. Summing the two cases (l 1 > 1 and l 1 = 1) gives the upper bound of number of m s related to a specific pair a i, a j and some b, ( ) ( log n (log n) b d(d 1) log n + b 2 b 1 13 ) (log n) b,

14 which is O(n ɛ ) for any ɛ > 0. Since there are d(d 1) 2 pairs of a i, a j and 1 b < d, the number of m s is bounded by d(d 1) 2 d 1 O(n ɛ ) = O(n ɛ ). b=1 For each value of m at most one square factor can occur at each position, giving less than O(n 1+ɛ ) additional occurrences for any ɛ > 0. As mentioned earlier, the number of PR-square factor occurrences is bounded from above by the number of square factor occurrences, which is O(n 2 ), so the question remains as to whether a partial word exists which has this many PR-square factor occurrences. The answer is yes, and the prefix of length n of is such a word. The proof of Theorem 6 analyzes this word s number of PR-square factor occurrences. Theorem 6. psf( n/2, n, k) is Θ(n 2 ). Proof. We have already seen that psf( n/2, n, k) is O(n 2 ), so it remains to prove that psf( n/2, n, k) is Ω(n 2 ). To do this, consider w = and the prefix of w of length n, w n. The first important property w has is that if u is a factor of w which begins at position i, then u occurs in w at position i + 4m for every integer m 1. Even stronger than that, though, is that each of the three factors of length u starting at i + 1, i + 2, and i + 3 are the same as u up to complement and/or reversal, which means that if any of them begins a PR-square factor of length j, they all do, and hence if any position begins a PR-square factor of length j, then every position of w begins a PR-square factor of length j. Let uv be a prefix of w with u = v = 2i + 1 for some i 0. Clearly u has a letter at u[2j] for all integers j such that 2i 2j 0. Since u has odd length, v has a letter at v[2j + 1] for all integers j such that 2i 2j This implies that j D(u) if and only if j H(v). Thus u v, and either u v = (0011) 2i+1 4 or u v = (0110) 2i+1 4. However, 2i+1 4 is not an integer, and since 0011 and 0110 are primitive, u v is primitive. Thus a PR-square factor of every odd root length occurs at position 0, and hence occurs at every position in w. Therefore, for each odd i from 1 to n 2 a PR-square factor of root length i occurs in w n at every position from 0 to n 2i. To simplify this to a sum, there are n+2 4 odd integers less than or equal to n 2, and if j starts at 1 14

15 and goes to n+2 4, then the jth odd integer is 2j 1. Thus psf(w n, k) n+2 4 j=1 which implies that psf(w n, k) is Ω(n 2 ). (n 2(2j 1) + 1) = (n 2 n PR-square subword occurrences + 1) n + 2, 4 We now discuss the bounds in Table 2. Denote pss(w, k) to be the number of PR-square subword occurrences in a partial word w over an alphabet of fixed size k, and pss(h, n, k) to be max{pss(w, k) : H(w) = h, w = n}. An obvious upper bound on pss(h, n, k) comes from counting all occurrences of square subwords in the partial word n. In this case, every square subword of root length i occurs at positions 0 through n 2i, giving a total of n 2 (n 2i + 1)k i, i=1 which is Θ(k n 2 ). Except for the case when h = n, though, pss(h + 1, n, k) pss(h, n, k) because any full word compatible with a factor of a partial word w would also be compatible with a weakening of w. Thus pss(h + 1, n, k) is a good upper bound for pss(h, n, k), so for all n 1 and k 2, the sequence {pss(h, n, k)} 0 h n is monotone increasing. For h = 0, pss(h, n, k) = psf(h, n, k), so pss(0, n, k) is Θ(n log n). In fact, for any fixed h and k, the same argument as for the O bound in Theorem 4 holds. Moreover, for any occurrence of a PR-square factor in w, there is an occurrence of a PR-square subword, hence psf(h, n, k) pss(h, n, k), which implies that the Ω bound in Theorem 4 holds for pss(h, n, k) as well. For h = n, n 2 pss(n, n, k) = (n 2i + 1)P (i), i=1 where P (i) is the number of primitive full words of length i over the k-ary alphabet. However, by Proposition 2, there are at least k i 1 (k 1) k-ary primitive full words of length i. Thus n 2 pss(n, n, k) (n 2i + 1)k i 1 (k 1), i=1 15

16 which is Θ(k n 2 ). This proves that the order of pss(n, n, k) is Θ(k n 2 ). For a constant d, pss(n, n, k) and pss(n d, n, k) differ by at most nkd of PR-square subword occurrences, and hence the bound on pss(n d, n, k) is identical to the bound on pss(n, n, k). Theorem 7. Let d and k be fixed. Θ(k n 2 ). Then for n d, pss(n d, n, k) is Proof. Since, as mentioned above, for all n 1, the sequence {pss(h, n, k)} 0 h n is monotone increasing, pss(n d, n, k) pss(n, n, k), hence pss(n d, n, k) is O(k n 2 ). Letting w = n, by the pigeonhole principle, the first d positions of w can be strengthened to form w in such a way that pss(w, k) k d pss(w, k). Since d and k are fixed, 1 k d is constant, and thus pss(w, k) being Ω(k n 2 ) implies pss(w, k) is Ω(k n 2 ). The question that remains is how pss(h, n, k) behaves when h is between these extremes, i.e., h is more than a constant away from both 0 and n. We start with a lower bound for pss( n 2, n, k). Theorem 8. pss( n 2, n, k) is Ω(nkn/4 ). Proof. Without loss of generality, assume 12 divides n. Let w have the form u n 4 u n 4 u, where u = 1 n 6. Consider a factor w[i..i + 5n 6 ) of w with 0 i < n 6. Its two halves are v 1 = w[i..i + 5n 12 ) and v 2 = w[i + 5n 12..i + 5n 6 ). Each half contains holes, and if we align them, n 4 v 1 = v 2 = , we can see v 1 = v 2 with ( n 6 i) 1 s at the beginning. The number of PRsquare subwords compatible with v 1 v 2 is given by the number of primitive completions of v 1, which is at least k n 4 1 (k 1) by Proposition 2. Since i can range from 0 to n 6 1, pss(w, k) n 6 k n 4 1 (k 1) = Ω(nk n 4 ). 16

17 Define the number of pairs of holes in a square factor uv to be {i [0.. u ) : u[i] = v[i] = }. For example, w = has length 20, has 10 holes, and w[0..14) is a square factor with 5 pairs of holes. We can find an upper bound of cnk n/4, where c is a constant, on the number of square subword occurrences in any partial word with n/2 holes of length n over a k-letter alphabet, where n is a sufficiently large multiple of 4, having a square factor uu with n/4 pairs of holes (all holes are in uu). The proof is a particular case of the following proof of an upper bound for pss( n 2, n, k). Theorem 9. pss( n 2, n, k) = O(nkn/4 ). Proof. For sufficiently large n, suppose w is a partial word with the maximum number of occurrences of square subwords in {u : u = n, H(u) = n 2 }. Among the partial words with the same length and the same hole positions, the partial word with 1 s in all non-hole positions has the most occurrences of square subwords, so we can assume w[i] = 1 for all i D(w). Let u 1 u 2 be a square factor of w with the maximum number of pairs of holes. We write the number of pairs of holes in u 1 u 2 as ( 1 4 ɛ)n, 0 ɛ < 1 4. We strengthen w by replacing all the s in w which are not in any pair of holes in u 1 u 2 with 1 and call this new partial word w (u 1 u 2 becomes u u, say). Thus the partial word w has length n, has 2( 1 4 ɛ)n holes, and has a square factor u u with ( 1 4 ɛ)n pairs of holes. We show that w has O(nk ( 1 4 ɛ 2 )n ) occurrences of square subwords. Let m = u. We refer to the first u as u 1, and the second u as u 2. Any square factor with root length m is compatible with at most k ( 1 4 ɛ)n occurrences of square subwords. Now consider a square factor v 1 v 2 with a different root length m. There are two cases to consider. First, consider the case where m > m. We align v 1 and v 2. At one position i [0..m ), if v 1 [i] or v 2 [i] is a letter, then a subword v compatible with v 1 v 2 must have this letter at its positions i and i+m. Otherwise, there are k possibilities to choose v [i] = v [i + m ]. Therefore, if the number of pairs of holes in v 1 v 2 is t, then the number of occurrences of square subwords compatible with v 1 v 2 is kt. Evidently, t is at most the number of pairs of s in w which are m -apart. Two s are m -apart if and only if the first is u 1 [i] and the second is u 2 [j] such that j i = m m. Hence t is also the number of such pairs i, j in u which are (m m)-apart. We associate each such pair in u with its first, then clearly each can be associated with at most one pair. 17

18 We define an equivalence relation on [0..m) by i j if and only if m m divides j i. This relation partitions [0..m) into m m equivalence classes, and each pair of holes that are (m m)-apart must belong to the same class. We say that a class contains a if it contains i such that u [i] =. Then in such a class, there is a largest i such that u [i] =, and this is not associated with any pair. A lower bound on the number of classes that ( 1 4 ɛ)n contain a is m/(m m), where m/(m m) is the maximum size of a class. This gives a lower bound on the number of s that are not associated with any pair. Hence t ( 1 4 ɛ)n ( 1 4 ɛ)n m/(m m) ( 1 4 ɛ)n ( 1 4 ɛ)n n 2(m m) +1 ( 1 4 ɛ)n ( 1 4 ɛ)(m m) ( 1 4 ɛ 2 )n 1 4 (m m). So the number of occurrences of square subwords in w with root length greater than m is at most nk ( 1 4 ɛ 2 )n 1 4 (m m) cnk ( 1 4 ɛ 2 )n (1) m<m n/2 for some constant c that does not depend on ɛ, and where the leading n comes from the number of possible starting positions for v 1 v 2. Next, consider the case where m < m. By symmetry, we consider only the cases when v 1 does not overlap with u 2 and v 1 is to the left of u 2. We define the distance between the rightmost position of v 1 and the rightmost position of u 1 as y, and define the number of s in u 1 to the right of v 1 as x. There are two cases to consider. First, suppose that x ( 1 4 ɛ) n 2. Then v 1 has at most ( 1 4 ɛ) n 2 s, so the number of square subwords compatible with v 1 v 2 is at most k( 1 4 ɛ) n 2, and the total number of such occurrences of square subwords summing over all m and starting positions of subwords is at most n 2 k ( 1 4 ɛ) n 2, which is less than cnk ( 1 4 ɛ 2 )n for sufficiently large n and constant c. Now, suppose that x < ( 1 4 ɛ) n 2. If v 2 does not overlap with u 2, then v 2 has at most ( 1 4 ɛ) n 2 s, so as above, the total number of such occurrences of square subwords is at most n 2 k ( 1 4 ɛ) n 2 < cnk ( 1 4 ɛ 2 )n for sufficiently large n and constant c. So we assume v 2 overlaps with u 2. We now argue similarly to the first case. Recall that t is the number of i [0..m ) such that both v 1 [i] and v 2 [i] are s. We associate each pair of holes with the one and only one of the two which is in u 1 [m y..m) or u 2 [0..m y). Since we want to 18

19 find an upper bound on t, and to do so we want to find a lower bound on the number of s which are not associated with any pair, it will be enough to consider only pairs of holes whose right holes are contained in u 2 [0..m y) and omit from consideration possible pairs of holes whose right holes are contained in u 1 [m y..m). Notice that if 0 j + y < m, u 2 [j] and v 2 [j + y] refer to the same position in w. A at u 2 [j] with j [0..m y) is paired if and only if j + y < m and v 1 [j + y] =. Notice that under such conditions, j + m m < m y, v 1 [j + y] and u 1 [j + m m ] refer to the same position in w, so u [j] = u [j + m m ] =. So for any j [0..m y), if either (1) j + m m m y, or (2) j + m m < m y and u [j] or u [j + m m ] is not a, then there is no pair of holes associated to u 2 [j]. We partition [0..m y) into classes using the equivalence relation i j if and only if m m divides j i. This partitions [0..m y) into m m equivalence classes. We say that a class contains a if it contains i such that u 2 [i] =. In such a class, there is a largest i such that u 2 [i] =. Then either (1) i + m m m y, or (2) i + m m < m y and u 2 [i + m m ] is not a, so u 2 [i] is not associated with any pair. A lower bound on the number of ( classes which contain a is 1 4 ɛ)n x (m y)/(m m ), where ( 1 4 ɛ)n x is the number of s in u 2 [0..m y). This gives a lower bound on the number of s which are not associated with any pair. Therefore, t ( 1 4 ɛ)n ( 1 4 ɛ)n x (m y)/(m m ) ( 1 4 ɛ)n ( 1 4 ɛ)n ( 1 4 ɛ) n 2 n/2 (m m ) +1 ( 1 4 ɛ)n 1 2 ( 1 4 ɛ)(m m ) ( 1 4 3ɛ 4 )n 1 8 (m m ). So the number of occurrences of square subwords in w with root length less than m is at most nk ( 1 4 3ɛ 4 )n 1 8 (m m ) c nk ( 1 4 3ɛ 4 )n (2) 1 m <m for some constant c that does not depend on ɛ, where the leading n is the number of possible starting positions for v 1 v 2. 19

20 We have shown that any square factor with root length m = m is compatible with at most k ( 1 4 ɛ)n occurrences of square subwords, by (1) that any square factor with root length m > m is compatible with at most cnk ( 1 4 ɛ 2 )n occurrences of square subwords for some constant c, and by (2) that any square factor with root length m < m is compatible with at most c nk ( 1 4 3ɛ 4 )n occurrences of square subwords for some constant c. Thus w has at most c 1 nk ( 1 4 ɛ 2 )n occurrences of square subwords for some constant c 1. In fact, we observe that the number of occurrences of square subwords in w is bounded from above by G(i, m ), 0 i<n 1 m n 2 where G(i, m ) is an upper bound for the number of square subwords compatible with a square factor w [i..i + 2m ) (G(i, m ) = 0 if i + 2m > n), and apart from G(i, m ) equal to k ( 1 4 ɛ)n, 1 m n 2 G(i, m ) is a geometric sum. Compared to w, w has 2ɛn extra s. By weakening w back to w, we replace G(i, m ) by min{k ( 1 4 ɛ)n, G(i, m )k 2ɛn }, a minimum between two estimates: (1) the maximum number of pairs of holes in a square factor of w is ( 1 4 ɛ)n, (2) weakening w can increase the number of pairs of holes in a square factor by at most 2ɛn. Therefore the number of occurrences of square subwords in w is bounded from above by c 1 nk ( 1 4 ɛ 2 )n + c 2 ɛn 2 k ( 1 4 ɛ)n for some constants c 1 and c 2. For the case when k > 1, since ɛn 1, kɛn it follows that the number of occurrences of square subwords in w is bounded from above by c 3 nk n/4 = O(nk n/4 ) for some constant c 3. Since pss( n 2, n, k) is bounded by the number of occurrences of square subwords, we conclude that pss( n 2, n, k) = O(nkn/4 ) for k > 1. For the case when k = 1, the only PR-square subwords compatible with factors of w are of the form 11, so pss( n 2, n, 1) = O(nkn/4 ). 20

21 We end this section with the following remarks. Remark 1. We can easily generalize Theorems 8 and 9 to estimate the number pss(h, n, k) for some other h. Remark 2. From analysis, it should be noted that many of the bounds for pss(h, n, k) hold for square subword occurrences, because by [5, Proposition 6.15], the majority of the completions of a partial word are primitive. 6 Conclusion The compatibility graphs introduced in Section 4 also appear in the literature under the name circulant graphs (see, for instance, [9]). A future topic for research would be to investigate special properties of subgraph counting polynomials of circulant graphs to make the computation of N h,k (n) more efficient. We also considered occurrences of primitively-rooted squares in a given partial word over an alphabet of a fixed size k of a fixed length n and a fixed number of holes h. We established tight upper and lower bounds for their maximal number for a range of parameters. Referring to Section 5, one interesting thing to note is that as h increases, psf(h, n, k) goes from Θ(n log n) to Θ(n 2 ) to Θ( n2 log n ) to Θ(n), and we conjecture exactly one pivot point. Conjecture 1. Let n be given. As h goes from 1 to n, psf(h, n, k) goes from monotonicly increasing to monotonicly decreasing with exactly one pivot point. Another suggestion for future research is to develop efficient algorithms for generating the PR-square factor and subword occurrences in partial words. Another one is to develop efficient algorithms for detecting all repetitions. Our tight bounds from Sections 3 and 5 could be applied to the analysis of these algorithms. Acknowledgements Thanks to the referees of a preliminary version of this paper for their very valuable comments and suggestions. Thanks to Nathan Fox from Rutgers University, Joe Hidakatsu from the University of Michigan at Ann Arbor, and Sinziana Munteanu from Carnegie Mellon University for their help in refining the ideas in this paper. Also, thanks to Justin Lazarow from Microsoft for his consistent encouragement and willingness to listen. 21

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