Regularity of Iterative Hairpin Completions of Crossing (2, 2)-Words

Transcription

1 International Journal of Foundations of Computer Science Vol. 27, No. 3 (2016) c World Scientific Publishing Company DOI: /S Regularity of Iterative Hairpin Completions of Crossing (2, 2)-Words Kayoko Shikishima-Tsuji Center for Liberal Arts Education and Research Tenri University, 1050 Somanouchi Tenri, Nara , Japan tsuji@sta.tenri-u.ac.jp Received 11 January 2015 Accepted 14 May 2015 Communicated by Szilárd Fazekas Hairpin completion is a formal operation inspired from DNA biochemistry. It is known that the (one step) hairpin completion of a regular language is linear context-free, but not regular in general. Further, it is decidable whether the (one step) hairpin completion of a regular language is regular. However, it is an open question whether the iterated hairpin completion ofaregularlanguage isregular,even ifitisa singleton.ifthe wordisa non-crossing α-word, there are results, but for crossing words there are no results. In this paper, we give necessary and sufficient conditions that the iterated hairpin completion of a given crossing (2, 2)-α-word in ασ ᾱ is regular. Keywords: Iterated hairpin completions; regularity; crossing words. 1. Introduction The mathematical hairpin concept was introduced by Păun et al. in [13]. This operation is inspired by DNA biochemical processes which play an important role in DNA-computing. A DNA-strand is composed of nucleotides which differ from each other in their bases: A (adenine), G (guanine), C (cytosine), and T (thymine). By Watson-Crick base pairing, the complementary bases A and T (resp., G and C) can bond to each other via hydrogen bonds. Two strands can bond to each other if they have opposite orientation and they are base-wise complementary. When a single strand of DNA has a suffix α which is the mirrored complement of a middle subword α of the strand, it bonds to the middle subword part and looks like a hairpin. Then a polymerase chain reaction extends the shorter end to generate a complete double strand. A strand can be seen as a word and a set of strands as a formal language. In this paper we discuss the hairpin manipulations from a language theoretic point of view. The (one step) hairpin completion operation was investigated in [1, 2, 5, 8 12]. Most basic algorithmic questions about (one step) hairpin completion have been answered (see, e.g. Cheptea et al. [1] and Diekert et al. [2]). 375

2 376 K. Shikishima-Tsuji However the situation is different for the iterated version. It is a difficult question whether or not the iterated hairpin completion of a regular language is regular, even if the regular language is a singleton. In the case of bounded hairpin completion, which is a weaker variant of hairpin completion introduced by Ito et al. in [4], the iterated completion of a regular language is regular. When a given word is a non-crossing α-word, there are necessary and sufficient conditions that the iterated hairpin completion of this word is regular (see Kari et al. [5]). In this paper, we give necessary and sufficient conditions that the iterated hairpin completion of a given crossing (2, 2)-α-word in ασ ᾱ is regular. 2. Preliminaries We assume the reader to be familiar with fundamental concepts from Formal Language Theory, such as the classes of the Chomsky hierarchy, finite automaton, regular expressions (e. g., see the textbook by Harrison [3]), as well as fundamental concepts from combinatorics on words (e. g., see the Lothaire textbook [7]). Let Σ be a non-empty finite alphabet with letters as elements. A sequence of letters constitutes a word w Σ and we denote by ε the empty word. Words together with the operation of concatenation form a free monoid, which is usually denoted by Σ for an alphabet Σ. Any subset of Σ is called a language. By Σ + we denote the set of non-empty words over Σ. Repeated concatenation of a word w with itself is denoted by w i for integers i 0 with i representing a power. Furthermore, w is said to be primitive if there exists no non-empty word u such that w = u j for some integer j > 1. Otherwise, we call w a repetition and the smallest such u its root (note that in this case the word u is primitive). The length of a finite word w is the number of not necessarily distinct symbols it consists of and is denoted by w. The ith symbol we write as w[i] and use the notation w[i..j] to refer to the part of a word starting at the ith and ending at the jth position. A word u is a factor of w if there exist integers i,j with 1 i,j w such that u = w[i..j]. We say that u is a prefix of w whenever we can fix i = 1 and denote this by u p w. If j < w, then the prefix is called proper. Suffixes are the corresponding concept, reading at the end of the word to the front. A word w has a positive integer h as a period if for all i,j such that i j(modh) we have w[i] = w[j], whenever both w[i] and w[j] are defined. The number of occurrences of a word u in a word w as a factor is denoted by w u. Let be an antimorphic involution, i.e. : Σ Σ is a function, such that for a = a for all a Σ, and uv = vū for all u,v Σ +. A word w is said to be a pseudopalindrome if w = w. Throughout this paper, let k be a fixed integer. For a word w = γαβα where α Σ k and γ, β Σ, we define right k-hairpin completion of w as γαβᾱ γ. If

3 Regularity of Iterative Hairpin Completions of Crossing (2, 2)-Words 377 the number k is obvious, we can omit it. By the notation w r w, we mean that w is a right k-hairpin completion of w. The left k-hairpin completion is defined analogously. By the notation w l w, we mean that u is left k-hairpin completion of w. The relation of k-hairpin completion is defined as r or l. We denote the reflexive transitive closure of, r and l by, r and r, respectively. For a language L Σ +, we define the right k-hairpin completion of L and the left k-hairpin completion of L, respectively, as follows: RH k (L) = {w w L, w r w }, LH k (L) = {w w L, w l w }. We alsodefine the k-hairpin completion of L by H k (L) = RH k (L) LH k (L). The iterated k-hairpin completion Hk (L) of L is defined inductively, as follows: let Hk 0(L) = L, Hn k (L) = H k(h n 1 k (L)) for every positive integer n, and Hk (L) = n 0 Hn k (L). 3. Main Results We will restrict out attention to words of the form ασ ᾱ where α = k. This does not restrict generality, because, if any word w ασ β and α = β = k has a one-step hairpin completion w of w, then w is in ασ ᾱ or in βσ β. Regularity of the iterated hairpin completion of the word w depends on regularity of Hk (w ). In the rest of this paper, we investigate regularity of Hk (w) where a given word w is in ασ ᾱ. A word w Σ is said to be an (m, n)-α-word (or simply an (m, n)-word) if [w] α = m and [w]ᾱ = n. We say that an (m,n)-α-word w is non-α-crossing if the rightmost occurrence of α precedes the leftmost occurrence of ᾱ on w. Otherwise, the word w is α-crossing. As an (m,n)-α-word in ασ ᾱ is always non-crossing for m = 1 or n = 1, then if an (m,n)-α-word in ασ ᾱ is crossing, we have m,n 2. We have the following theorem: Theorem 1. Let x,y ασ ᾱ be (1,1)-words and w be a crossing (2,2)-word in xσ Σ y. Then the iterated k-hairpin completion Hk (w) of w is regular, if and only if, w is a pseudopalindrome or x and y are both pseudopalindromes. To prove Theorem 1, we need some propositions and corollaries. First, we consider the case where w xσ y and x, y are (1,1)-wordsin ασ ᾱ such that y x, x. Proposition 2. Let x and y be (1,1)-words in ασ ᾱ such that x y x and v be a word in Σ such that xvy is a crossing (2,2)-word. If x or v is not a pseudopalindrome and y is not a pseudopalindrome, then the iterated k-hairpin completion Hk (xvy) of xvy is not regular. Proof. We prove first that the iterated k-hairpin completion Hk (xvy v x) of xvy v x is not regular. Let R = (xv) + y( v x) + v(xv) ȳ( v x) + and S = {(xv) n1 y( v x) n2 v(xv) n3 ȳ( v x) n4 0 < n 1,n 4, 0 n 3 < n 2 }.

4 378 K. Shikishima-Tsuji Since xvy is a (2,2)-word and y x, x, we have (xv) n1 y( v x) n2 v(xv) n3 ȳ( v x) n4 vy = (xv) n1 y( v x) n2 v(xv) n3 ȳ( v x) n4 vȳ = 1 for all 0 < n 1,n 4, 0 n 3 < n 2. By the assumption, x or v is not a pseudopalindrome, we have v x vx. Using Pumping Lemma, the language S is not regular. We prove that Hk (xvy v x) R = S. Every word z S is of the form z = (xv) n1 y( v x) n2 v(xv) n3 ȳ( v x) n4 where n 1,n 2,n 4 > 0,n 3 0 and n 2 > n 3. There is a sequence: xvy v x r xvy( v x) n2 r xvy( v x) n2 v(xv) n3 ȳ v x (xv) n1 y( v x) n2 v(xv) n3 ȳ( v x) n4 where n 2 > n 3. This yields S H k (xvy v x) R. On the other hand, let L = {(xv) n1 ȳ( v x) n2 v(xv) n3 y( v x) n4 0 < n 1,n 4, 0 n 2 < n 3 }. We have H 1 k ((xv)+ y( v x) + ) {xvy v x} = (xv) + y( v x) + S L, then we have H k(xvy v x) = {xvy v x} H 1 k((xv) + y( v x) + ) H k(s) H k(l). Since (xv) n1 y = (xv) n1 ȳ and (xv) n1 y (xv) n1 ȳ for every positive integer n 1, H k (xvy v x) R = ({xvy v x} H1 k ((xv)+ y( v x) + ) H k (S) H k (L)) R = H 1 k ((xv)+ y( v x) + ) H k (S) R = H 1 k((xv) + y( v x) + ) (S H k(h 1 k(s))) R = H 1 k((xv) + y( v x) + ) S R S. Then Hk (xvy v x) R S. ThereforeHk (xvy v x) R = S is not regular.regularityofthe languager yields that Hk (xvy v x) is not regular. NowweprovethattheiteratedhairpincompletionHk (xvy)ofxvy isnotregular. Since every word z in Hk (ȳ vxvy) has a prefix ȳ, the word z is not contained in R and then we have H k(xvy) R = {xvy} H k(ȳ vxvy) H k(xvy v x) R = H k(xvy v x) R = S. Proposition 3. Let x and y be (1,1)-words in ασ ᾱ such that x y x and v be a word in Σ such that xvy is a crossing (2,2)-word. If x and v are pseudopalindromes and y is not a pseudopalindrome, then the iterated k-hairpin completion Hk (xvy) of xvy is not regular. Proof. Let R = (xv) + y(vx) + vȳv(xv) + yv(xv) + ȳ(vx) + and S = {(xv) n1 y(vx) n2 vȳv(xv) n3 yv(xv) n2 ȳ(vx) n4 0 < n 1,n 2,n 3,n 4 }. Since xvy is a (2,2)-word and y x, x, we have (xv) n1 y(vx) n2 vȳv(xv) n3 yv(xv) n2 ȳ(vx) n4 vy = (xv) n1 y(vx) n2 vȳv(xv) n3 yv(xv) n2 ȳ(vx) n4 vȳ = 2 for all 0 < n 1,n 2,n 3,n 4. Using Pumping Lemma, the language S is not regular.

5 Regularity of Iterative Hairpin Completions of Crossing (2, 2)-Words 379 We prove that Hk (xvyvx) R = S. Every word z S is of the form z = (xv) n1 y(vx) n2 vȳv(xv) n3 yv(xv) n2 ȳ(vx) n4 where 0 < n 1,n 2,n 3,n 4. There is a sequence: xvyvx r m1 1 r m2 1 xvy(vx) m1 rxvy(vx) m1 v(xv) m 1ȳvx = xvy(vx) n2 vȳvx xvy(vx) n2 vȳ(vx) m2 rxvy(vx) n2 vȳ(vx) m2 v(xv) m 2 yv(xv) n2 ȳvx=xvy(vx) n2 vȳv(xv) n3 yv(xv) n2 ȳvx (xv) n1 y(vx) n2 vȳv(xv) n3 yv(xv) n2 ȳ(vx) n4 = z, where m 1,m 1,m 2,m 2 are positive integers such that m 1+m 1 = n 2 and m 2 +m 2 = n 3. Therefore, every word z S is contained in H k (xvyvx). On the other hand, let and It is easy to see L 1 = {(xv) n1 y(vx) n2 vȳv(xv) n3 ȳ(vx) n4 0 < n 1,n 2,n 3,n 4 } L 2 = {(xv) n1 yv(xv) n2 y(vx) n3 vȳ(vx) n4 0 < n 1,n 2,n 3,n 4 }. H 1 k((xv) + y(vx) + vȳ(vx) + ) xvy(vx) + vȳvx (xv) + y(vx) + vȳ(vx) + S L 1 L 2 and then we have H k(xvyvx)=h k((xv) + y(vx) + (xv) + ȳ(vx) + vy(vx) + (xv) + y(vx) + vȳ(vx) + ) =H k((xv) + y(vx) + ) H k((xv) + ȳ(vx) + vy(vx) + ) H k((xv) + y(vx) + vȳ(vx) + )) =(xv) + y(vx) + H k((xv) + ȳ(vx) + vy(vx) + ) H k((xv) + y(vx) + vȳ(vx) + )) (xv) + y(vx) + H k((xv) + ȳ(vx) + vy(vx) + ) (xv) + y(vx) + vȳ(vx) + H k(s L 1 L 2 ). Since (xv) n1 y = (xv) n1 ȳ and (xv) n1 y (xv) n1 ȳ for every positive integer n 1, and every word in Hk ((xv)+ ȳ(vx) + vy(vx) + ) has a prefix in (xv) + ȳ, we have Hk ((xv)+ ȳ(vx) + vy(vx) + ) R =. Moreover, since every word z Hk (L 1) has a factor of the form (xv) n1 y(vx) n2 vȳv(xv) n3 ȳ(vx) n4 where n 1,n 2,n 3,n 4 > 0, the word z is not contained in R, that is, Hk (L 1) R =. We also have Hk (L 2) R = in a similar way. It is obvious that Hk (S) R S. Then we have H k(xvyvx) R {(xv) + y(vx) + H k ((xv)+ ȳ(vx) + vy(vx) + ) (xv) + y(vx) + vȳ(vx) + H k (S L 1 L 2 )} R Hk (S) R S. ThereforeHk (xvyvx) R = S is not regular.regularityofthe languager yields that Hk (xvyvx) is not regular. NowweprovethattheiteratedhairpincompletionHk (xvy)ofxvy isnotregular. Since every word z in Hk (ȳvxvy) has a prefix ȳ, the word z is not contained in R and then we have H k(xvy) R = {xvy} H k(ȳvxvy) H k(xvyvx) R = H k(xvyvx) R = S.

6 380 K. Shikishima-Tsuji Proposition 4. Let x and y be (1,1)-words in ασ ᾱ such that x y x and v be a word in Σ such that xvy is a crossing (2,2)-word. If x and y are pseudopalindromes, then the iterated k-hairpin completion Hk (xvy) of xvy is regular. Proof. Let and R = {x(vx) (vy( vx) + (vx) ) vy( vx) + } L = {(y v) + xv((yv) (y v) + xv) (yv) y}. Since these languages are regular and Hk (xvy) = {w} H k (xvy vx) H k (y vxvy), if we show R = Hk (xvy vx) and L = H k (y vxvy), then the proposition is proved. First, we prove R = Hk (xvy vx). Let R n = x(vx) (vy( vx) + (vx) ) n (vy)( vx) + for every nonnegative integer n, and then R = n 0 R n. We show that R n Hk (xvy vx) by induction on n. It is clear that R 0 = x(vx) vy( vx) + Hk (xvy vx) by the following: xvy vx i l x(vx)i vy( vx) j r x(vx)i vy( vx) j+1. Suppose that R n = x(vx) (vy( vx) + (vx) ) n (vy)( vx) + is contained in H k (xvy vx) for some nonnegative integer n. We prove that R n+1 is contained in H k (xvy vx). Every word z in R n+1 can be written as the form z = x(vx) r (vy( vx) m1 (vx) t1 ) (vy( vx) mn+1 (vx) tn+1 )vy( vx) s for some m 1,,m n+1 > 0, t 1,,t n+1 0, and r,s 0. (1) When m 1 > t n+1, let z = x(vy( vx) m1 (vx) t1 ) (vy( vx) mn (vx) tn )vy( vx). We have z = xvy( vx) m1 (vx) t1 vy( vx) mn (vx) tn vy( vx) mn+1 1 r xvy( vx) m1 (vx) t1 vy( vx) mn (vx) tn vy( vx) ( vx) mn+1 1 = xvy v(x v) tn+1 x( vx) m1 tn+1 1 (vx) t1 vy( vx) mn (vx) tn vy( vx) mn+1 r xvy v(x v) tn+1 x( vx) m1 tn+1 1 (vx) t1 vy( vx) mn (vx) tn vy( vx) mn+1 (vx) tn+1 vy vx = xvy( vx) m1 (vx) t1 vy( vx) mn+1 (vx) tn+1 vy( vx) = z. (xv) r xvy( vx) m1 (vx) t1 vy( vx) mn+1 (vx) tn+1 vy( vx) ( vx) s 1 (2) When m 1 t n+1, let z = xvy( vx) m2 (vx) t2 vy( vx) mn+1 (vx) tn+1 vy( vx). We have z = xvy( vx) m2 (vx) t2 vy( vx) mn+1 (vx) tn+1 vy( vx) t1 l (xv) t1 xvy( vx) m2 (vx) t2 vy( vx) mn+1 (vx) tn+1 vy( vx)

7 Regularity of Iterative Hairpin Completions of Crossing (2, 2)-Words 381 = x(vx) t1 vy( vx) m2 (vx) t2 vy( vx) mn+1 (vx) tn+1 m1+1 (vx) m1 1 vy( vx) l xvy v(x v) m1 1 x(vx) t1 vy( vx) m2 (vx) t2 vy( vx) mn+1 (vx) tn+1 m1+1 (vx) m1 1 vy( vx) = xvy( vx) m1 (vx) t1 vy( vx) m2 (vx) t2 vy( vx) mn+1 (vx) tn+1 vy( vx) (xv) r xvy( vx) m1 (vx) t1 vy( vx) m2 (vx) t2 vy( vx) mn+1 (vx) tn+1 vy( vx) ( vx) s 1 = z. Since z and z are contained in R n Hk (xvy vx), we have z H k (xvy vx). We proved that Hk (xvy vx) R. To prove that Hk (xvy vx) R is easy by induction on n. It is obvious that Hk 0 (xvy vx) = {xvy vx} R. For a nonnegative integer n, suppose that Hk n (xvy vx) R, then we have Hn+1 k (xvy vx) = Hk 1(Hn k (xvy vx)) H1 k (R) R. Therefore, Hk (xvy vx) = R is proved. As Hk (y vxvy) = L is proved in a similar way, we omit it. Second, we consider the case where w = xσ Σ y but not in xσ y and x, y are (1,1)-words in ασ ᾱ such that y x, x. Since x + y α < w < x + y, there exist words x 0,y 0 Σ and a palindrome v Σ + such that x = vx 0 v, y = vy 0 v and w = vx 0 vy 0 v. Proposition 5. Let x and y be (1,1)-words in ασ ᾱ such that x y x and w be a crossing (2,2)-word in xσ Σ y but not in xσ y. If x and y are not pseudopalindromes, then the iterated k-hairpin completion Hk (w) of w is not regular. Proof. Since x and y are in ασ ᾱ such that w < xy and w is a crossing (2,2)- word, there exist factors u,v Σ + of α and x 0,y 0 Σ such that α = vu, ᾱ = ūv, v = v and w = vx 0 vy 0 v. Let R = (vx 0 ) + vy 0 v( x 0 v) + (x 0 v) ȳ 0 v( x 0 v) + and S = {(vx 0 ) n1 vy 0 v( x 0 v) n2 (x 0 v) n3 ȳ 0 v( x 0 v) n4 0 < n 1,n 2,n 4, 0 n 3 < n 2 }. Since w is a (2,2)-word and x 0 y 0,ȳ 0, we have z y = z ȳ = 1 for every z S. Then, the language S is not regular. First, we prove that H k (vx 0vy 0 v x 0 v) R = S. (1) Every word z S is of the form (vx 0 ) n1 vy 0 (v x 0 ) n2 (vx 0 ) n3 vȳ 0 v( x 0 v) n4 where n 1,n 2,n 4 > 0,n 3 0 and n 2 > n 3. There is a sequence: vx 0 vy 0 v x 0 v r vx 0 vy 0 v( x 0 v) n2 r vx 0 vy 0 v( x 0 v) n2 (vx 0 ) n3 vȳ 0 v xv (vx 0 ) n1 vy 0 v( x 0 v) n2 (vx 0 ) n3 vȳ 0 v( x 0 v) n4 = z where n 2 > n 3. This yields S H k (vx 0vy 0 v x 0 v) R.

8 382 K. Shikishima-Tsuji (2) On the other hand, let L = {(vx 0 ) n1 vȳ 0 v( x 0 v) n2 (vx 0 ) n3 vy 0 v( x 0 v) n4 0 < n 1,n 4, 0 n 2 < n 3 }. Since Hk 1((vx 0) + vy 0 v( x 0 v) + ) {(vx 0 )vy 0 v( x 0 v)} = (vx 0 ) + vy 0 v( x 0 v) + S L, we have Hk ((vx 0) + vy 0 v( x 0 v) + )) = (x 0 v) + vy 0 v( x 0 v) + Hk (S) H k (L). It is obvious that (vx 0 ) m y (vx 0 ) n ȳ for every positive integers m and n. Then, we have H k (vx 0vy 0 v x 0 v) R = H k ((vx 0) + vy 0 v( x 0 v) + )) R = {(x 0 v) + vy 0 v( x 0 v) + H k (S) H k (L)} R = Hk (S) R = S R S. Now we prove that H k (vx 0vy 0 v) is not regular. Since every word z in H k (vȳ 0vx 0 vy 0 v) has a prefix ȳ = vȳ 0 v, the word z is not contained in R and then we have H k (vx 0vy 0 v) R = {{vx 0 vy 0 v} H k (vȳ 0vx 0 vy 0 v) H k (vx 0vy 0 v x 0 v)} R = H k (vx 0vy 0 v x 0 v) R = S. Therefore, H k (vx 0vy 0 v) R = S is not regular and then the iterated hairpin completion H k (w) of w = vx 0vy 0 v is not regular. Proposition 6. Let x and y be (1,1)-words in ασ ᾱ such that x y x and w be a crossing (2,2)-word in xσ Σ y but not in xσ y. If x is a pseudopalindrome and y is not a pseudopalindrome, then the iterated k-hairpin completion Hk (w) of w is not regular. Proof. There exist factors u,v Σ + of α and x 0,y 0 Σ such that α = vu, ᾱ = ūv, v = v and w = vx 0 vy 0 v. Let R = (vx 0 ) + vy 0 (vx 0 ) + vȳ 0 v(x 0 v) + y 0 v(x 0 v) + ȳ 0 v(x 0 v) + and S = {(vx 0 ) n1 vy 0 (vx 0 ) n2 vȳ 0 v(x 0 v) n3 y 0 v(x 0 v) n2 ȳ 0 v(x 0 v) n4 0 < n 1,n 2,n 3,n 4 }. Since vx 0 vy 0 v is a (2,2)-word and x 0 y 0,ȳ 0, we have z y0 = z ȳ0 = 2 for every z S. Hence, S is not regular. We prove that H k (x 0vyvx 0 ) R = S. For every word z = (vx 0 ) n1 vy 0 (vx 0 ) n2 vȳ 0 v(x 0 v) n3 y 0 v(x 0 v) n2 ȳ 0 v(x 0 v) n4 S where 0 < n 1,n 2,n 3,n 4, there is a sequence: vx 0 vy 0 vx 0 v r m1 1 vx 0 vy 0 v(x 0 v) m1 r vx 0 vy 0 v(x 0 v) m1 (x 0 v) m 1 vȳ 0 vx 0 v = vx 0 vy 0 (vx 0 ) n2 vȳ 0 vx 0 v m2 1 r vx 0 vy 0 (vx 0 ) n2 vȳ 0 v(x 0 v) m2 r vx 0 vy 0 (vx 0 ) n2 vȳ 0 (vx 0 ) m2 v (x 0 v) m 2 y 0 v(x 0 v) n2 ȳ 0 vx 0 v = vx 0 vy 0 (vx 0 ) n2 vȳ 0 v(x 0 v) n3 y 0 v(x 0 v) n2 ȳ 0 vx 0 v (vx 0 ) n1 vy 0 (vx 0 ) n2 vȳ 0 v(x 0 v) n3 y 0 v(x 0 v) n2 ȳ 0 v(x 0 v) n4 = z,

9 Regularity of Iterative Hairpin Completions of Crossing (2, 2)-Words 383 where m 1,m 1,m 2,m 2 are positive integers such that m 1 +m 1 = n 2 and m 2 +m 2 = n 3. Therefore, every word z S is contained in H k (x 0vyvx 0 ). On the other hand, let and It is easy to see L 1 = {(x 0 v) n1 y(vx 0 ) n2 vȳv(x 0 v) n3 ȳ(vx 0 ) n4 0 < n 1,n 2,n 3,n 4 } L 2 = {(x 0 v) n1 yv(x 0 v) n2 y(vx 0 ) n3 vȳ(vx 0 ) n4 0 < n 1,n 2,n 3,n 4 }. H 1 k ((vx 0) + vy 0 (vx 0 ) + vȳ 0 v(x 0 v) + ) (vx 0 ) + vy 0 (vx 0 ) + vȳ 0 v(x 0 v) + S L 1 L 2. Since (x 0 v) n1 y = (x 0 v) n1 ȳ and (x 0 v) n1 y (x 0 v) n1 ȳ for every positive integer n 1 and every word in Hk ((x 0v) + ȳ(vx 0 ) + vy(vx 0 ) + ) has a prefix in (x 0 v) + ȳ, we have Hk ((x 0v) + ȳ(vx 0 ) + vy(vx 0 ) + ) R =. Moreover,since everywordz Hk (L 1) has a factor of the form (x 0 v) n1 y(vx 0 ) n2 vȳv(x 0 v) n3 ȳ(vx 0 ) n4 where n 1,n 2,n 3,n 4 > 0, the word z is not contained in R, that is, Hk (L 1) R =. We also have Hk (L 2) R = in a similar way. It is obvious that Hk (S) R S. Then we have H k (vx 0vy 0 v) R = H k ({vx 0vy 0 v,vx 0 vy 0 vx 0 v,vȳ 0 vx 0 vy 0 v}) R = H k (vx 0vy 0 vx 0 v) R = H k((vx 0 ) + vy 0 v(x 0 v) + (vx 0 ) + vȳ 0 (vx 0 ) + vy 0 v(x 0 v) + (vx 0 ) + vy 0 v(x 0 v) + ȳ 0 v(x 0 v) + ) R = (vx 0 ) + vy 0 v(x 0 v) + H k ((vx 0) + vȳ 0 (vx 0 ) + vy 0 v(x 0 v) + (vx 0 ) + vy 0 v(x 0 v) + ȳ 0 v(x 0 v) + ) R = H k ((x 0v) + y(vx 0 ) + vȳ(vx 0 ) + ) R (x 0 v) + y(vx 0 ) + vȳ(vx 0 ) + H k (S) H k (L 1) H k (L 2) R = H k(s) R S. Therefore H k (x 0vy 0 vx 0 ) R = S is not regular, that is, H k (x 0vy 0 vx 0 ) is not regular. Proposition 7. Let x and y be (1,1)-words in ασ ᾱ such that x y x and w be a crossing (2,2)-word in xσ Σ y but not in xσ y. If x and y are pseudopalindromes, then the iterated k-hairpin completion Hk (w) of w is regular. Proof. There exist factors u,v Σ + of α and x 0,y 0 Σ such that α = vu, ᾱ = ūv, v = v and w = vx 0 vy 0 v. Since x and y are pseudopalindromes, then x 0 and y 0 are also pseudopalindromes. Now we consider regular languages R = (vx 0 ) + (vy 0 (vx 0 ) + ) vy 0 (vx 0 ) + v and L = v(y 0 v) + x 0 v((y 0 v) + x 0 v) (y 0 v) +. Since Hk (w) = {w} H k (vx 0vy 0 vx 0 v) Hk (vy 0vx 0 vy 0 v), if we prove R = Hk (vx 0vy 0 vx 0 v) and L = Hk (vy 0vx 0 vy 0 v), then Hk (w) is regular.

10 384 K. Shikishima-Tsuji Let R n = (vx 0 ) + (vy 0 (vx 0 ) + ) n vy 0 (vx 0 ) + v for every nonnegative integer n, then it is obvious that R = n 0 R n. We show that R n H k (vx 0vy 0 vx 0 v) by induction on n. Suppose that R n is containedin H k (vx 0vy 0 vx 0 v) forsomenonnegativeintegern.everywordz in R n+1 is written of the form z = (vx 0 ) r (vy 0 (vx 0 ) m1 ) (vy 0 (vx 0 ) mn+1 )vy 0 (vx 0 ) s v where m 1,,m n+1 > 0 and r,s > 0. Let z = vx 0 (vy 0 (vx 0 ) m1 ) (vy 0 (vx 0 ) mn )vy 0 (vx 0 )v. We have z = vx 0 (vy 0 (vx 0 ) m1 ) (vy 0 (vx 0 ) mn )vy 0 vx 0 v mn+1 1 l vx 0 (vy 0 (vx 0 ) m1 ) (vy 0 (vx 0 ) mn )vy 0 vx 0 v(x 0 v) mn+1 1 = vx 0 vy 0 (vx 0 )(vx 0 ) m1 vy 0 (vx 0 ) mn vy 0 (vx 0 ) mn+1 v r vx 0 vy 0 (vx 0 ) m1 vy 0 (vx 0 ) mn vy 0 (vx 0 ) mn+1 v y 0 vx 0 v (vx 0 ) r vy 0 (vx 0 ) m1 vy 0 (vx 0 ) mn vy 0 (vx 0 ) mn+1 vy 0 v(x 0 v) s = z. Since z is in R n Hk (vx 0vy 0 vx 0 v), we have z Hk (z ) Hk (R) Hk (H k (vx 0vy 0 vx 0 v)) = Hk (vx 0vy 0 vx 0 v). On the other hand, for a nonnegative integer n, suppose that Hk n(vx 0vy 0 vx 0 v) R, then we have H n+1 k (vx 0 vy 0 vx 0 v) = Hk 1(Hn k (vx 0vy 0 vx 0 v)) Hk 1 (R) R. Since Hk (vy 0vx 0 vy 0 v) = L is proved in a similar way, we omit it. From the above propositions, we have the following. Corollary 8. Let x,y ασ ᾱ be (1,1)-words such that x y x and w be a crossing (2,2)-word in xσ Σ y. The iterated k-hairpin completion Hk (w) of w is regular, if and only if x and y are pseudopalindromes. Now we consider the case where x or y are not in ασ ᾱ. For a word w xσ Σ y, if x,y / ασ ᾱ, then there are no k-hairpin completion of w and then the iterated k-hairpin completion Hk (w) = {w} is regular. We may assume that one of x and y is in ασ ᾱ. The following lemma is usefull. Lemma 9. If x is a (1,1)-word in ασ Σ ᾱ but not in ασ ᾱ, then x is a pseudopalindrome. Proof. Since x is (1,1)-word in ασ Σ ᾱ but not in ασ ᾱ, we have w < xx and there exist u,v Σ + of α such that α = uv, x = uᾱ. Then, we have x = uvū = αū = uᾱ = x. Proposition 10. Let x be (1,1)-word in ασ Σ ᾱ but not in ασ ᾱ, y be (1,1)- word in ασ ᾱ and v be a word in Σ such that xvy is a crossing (2,2)-word. For w = xvy or yvx, the iterated k-hairpin completion Hk (w) is regular, if and only if y is a pseudopalindrome.

11 Regularity of Iterative Hairpin Completions of Crossing (2, 2)-Words 385 Proof. By Lemma 9, x is a pseudopalindrome. Since the proof for yvx is similar to the proof for xvy, we only show in the case w = xvy. First, we prove that if y is not a pseudopalindrome, then the iterated k-hairpin completion Hk (xvy) is not regular. Suppose v is not a pseudopalindrome. Since Hk 1 (xvy) = {xvy vx}, we have H k (xvy)={xvy} H k ((xv) xvy vx( vx) ) and ={xvy} (xv) xvy vx( vx) H k({(xv) n1 xvy vx vx( vx) n2 (vx) n3 ȳ vx( vx) n4 0 n 1,n 4,0 n 3 n 2 }) H k({(xv) n1 ȳ v(x v) n2 (xv) n3 xvxvy vx( vx) n4 0 n 1,n 4,0 n 2 n 3 }) H k((xv) xvy vx vx( vx) (vx) ȳ( vx) + ) H k ((xv)+ ȳ(x v) (xv) xvxvy vx( vx) ) =. In a similar way as the proof for Proposition 2, it is proved that Hk (xvy) of xvy is not regular. Supposev isapseudopalindrome.inasimilarwayastheproofforproposition3, it is proved that Hk (xvy) of xvy is not regular. Now, we prove that if y is a pseudopalindrome, then the iterated k-hairpin completion Hk (xvy) is regular. It is easy to see that H k(xvy) = {xvy} H k((xv) xvy vx( vx) ) = {xvy} ((xv) xvy v(x v) ) (xv) xvy vx( vx) (v(xv) y vx( vx) ). Therefore, Hk (xvy) is regular, even if v is not a pseudopalindrome. Proposition 11. Let x be (1,1)-word in ασ Σ ᾱ but not in ασ ᾱ, y be (1,1)-word in ασ ᾱ and w be a crossing (2,2)-word in xσ Σ y or in yσ Σ x. The iterated k-hairpin completion Hk (w) is regular, if and only if y is a pseudopalindrome. Proof. By Lemma 9, x is a pseudopalindrome. First we consider the case: w is in xσ Σ y but not in xσ y. There is a word x 0 Σ such that w = x 0 y and an integer i such that x i 0 x ασ ᾱ and x i 1 0 x / ασ ᾱ. We have H k (w) = H k (x 0y) = x + 0 yx 0 H k (x+ 0 yx 0x i 0 x 0ȳx + 0 ) H k (x 0ȳx 0 x i 0x 0yx 0). If y is not a pseudopalindrome, it is proved that Hk (w) is not regular, in a similar way as the proof for Proposition 6. If y is a pseudopalindrome, it is proved that Hk (w) = (x+ 0 yx 0 xi 0 )+ is regular, in a similar way as the proof for Proposition 7. In a similar way as the above, we can prove in the case: w yσ Σ x.

12 386 K. Shikishima-Tsuji Now, we prove the case where w = xσ Σ y and x, y are (1,1)-words in ασ ᾱ such that y = x or y = x. Proposition 12. Let x be (1,1)-words in ασ ᾱ and v be a word in Σ such that xvx is acrossing (2,2)-word. The iteratedk-hairpin completions Hk (xvx) is regular, if and only if x and v are pseudopalindromes,. Proof. If x and v are pseudopalindromes, then it is obvious that Hk (xvx) = (xv) + x is regular. Now, we show that if x or v are not pseudopalindromes then Hk (xvx) is not regular. (1) Suppose that v = v and x x. Let R be a regular language (xv)( xv) + (xv) + x. Since R xσ x, H k ( xvxvx) xσ x and xσ x xσ x =, we have H k (xvx) R = {xvx} H k (xvxv x) H k ( xvxvx) R = H k (xvxv x) R = (xv) xvxv x(v x) H k(h 1 k((xv) xvxv x(v x) )) R. For two nonnegative integers m and n, we have H 1 k ((xv)m xvxv x(v x) n ) = {(xv) m+i xvxv x(v x) n+j 0 i,j and 0 i+j} {(xv) m xvxv x(v x) n (vx) n (v x) m+2 0 n n} {(xv) n xv( xv) m+1 (xv) m xvxv x(v x) n 0 m m}. Since every word in R has a prefix xv x, we have H k (xvx) R=H k (xvxv x) R ={(xv) n xv( xv) m +1 (xv) m xvxv x(v x) n 0 n and 0 m m} R ={xv( xv) m +1 (xv) m xvxv xv x 0 m m}. It is clear that Hk (xvx) is not regular. (2) Suppose that v v and x = x. Let R be a regular language (xvx)( vx) + (vx) + vx. Since R xvσ vx, Hk (x vxvx) x vσ vx and xvσ vx x vσ vx =, we have Hk (xvx) R = {xvx} H k (x vxvx) H k (xvx vx) R = H k (xvx vx) R = (xv) xvx vx( vx) Hk(H k((xv) 1 xvx vx( vx) )) R. For two nonnegative integers m and n, we have Hk 1((xv)m xvx vx( vx) n ) = {(xv) m+i xvx vx( vx) n+j 0 i,j and i+j 0} {(xv) m xvx vx( vx) n (vx) n ( vx) m+2 0 n n} {(xv) n xv(x v) m+1 (xv) m xvx vx( vx) n 0 m m}.

13 Regularity of Iterative Hairpin Completions of Crossing (2, 2)-Words 387 Since every word in R has a suffix x vx, we have Hk (xvx) R=H k (xvx vx) R ={(xv) n xv(x v) m +1 (xv) m xvx vx( vx) n 0 n and 0 m m} R ={xv(x v) m +1 (xv) m xvx vx vx 0 m m}. It is clear that Hk (xvx) is not regular. (3) Suppose that v v and x x. Let R be a regular language xvx v( x v) + (xv) + (xv x v x). Since R Σ v x and Hk ( x vxvx) Σ vx, we have H k(xvx) R = {xvx} H k(xvx v x) H k( x vxvx) R = H k(xvx v x) R. Let F = {(xv) (xvx v x)( v x) n (vx) n v x v x( v x) 0 n n} and L = {(xv) xv x v( x v) m (xv) m (xvx v x)( v x) 0 m m}. For two nonnegative integers m and n, we have H 1 k((xv) m (xvx v x)( v x) n ) (xv) (xv) m (xvx v x)( v x) n ( v x) F L. Since Hk (L) (xv) xv x vσ and R xvx vσ, we have Hk (L) R =. The following equation is easy to see: H k(xvx) R=H k(xvx v x) R=H k(h 1 k((xv) xvx v x( v x) )) R H k ((xv) (xv) m (xvx v x)( v x) n ( v x) F L) R =H k (F) R=F R={xvx v( x v)n (xv) n (xvx v x) 0 n n}. On the other hand, it is obvious that {xvx v( x v) m (xv) m (xvx v x) 0 m m} Hk (xvx) R. Therefore, in this case, H k (xvx) R is not regular and then Hk (xvx) is not regular. Proposition 13. Let x be (1,1)-words in ασ ᾱ such that x x and w be a crossing (2,2)-word such that w = xv x where v Σ. The iterated k-hairpin completion Hk (xv x) is regular, if and only if v is a pseudopalindrome. Proof. (1) Suppose that v is a pseudopalindrome. Let w = xv x and R = ((xv) w(v x) ) +. We prove that Hk (w) = R. Since w is a pseudopalindrome, we have Hk 1(R) R and then H k (w) R. In a similar way as the proof for Proposition 4, it is easy to prove that R Hk (w). Then, we omit it. (2) Suppose that v is not a pseudopalindrome. Let R be a regular language xvx v( x v) + (xv) + (xv x v x). Since R Σ v x and Hk (x vxv x) Σ v x, we have H k(xv x) R = {xv x} H k(xv x v x) H k( x vxv x) R = H k(xv x v x) R.

14 388 K. Shikishima-Tsuji Let F = {(xv) (xv x v x)( v x) n (vx) n vx v x( v x) 0 n n} and L = {(xv) xvx v( x v) m (xv) m (xv x v x)( v x) 0 m m}. For two nonnegative integers m and n, we have H 1 k((xv) m (xv x v x)( v x) n ) = {(xv) m+i (xv x v x)( v x) n+j 0 i,j and 0 i+j} F S. Since Hk (F) (xv) xv x vσ and R xvx vσ, we have Hk (F) R =. The following equation is easy to see: H k(xv x) R=H k(xv x v x) R=H k(h 1 k((xv) xv x v x( v x) )) R H k((xv) (xv) m (xv x v x)( v x) n ( v x) F L) R =H k (L) R=L R={xvx v( x v)m (xv) m (xv x v x) 0 m m}. On the other hand, it is obvious that {xvx v( x v) m (xv) m (xv x v x) 0 m m} Hk (xv x) R. Therefore, in this case, H k (xv x) R is not regular and then Hk (xv x) is not regular. Proposition 14. Let x,y be (1,1)-word in ασ ᾱ such that y = x or y = x and w be a crossing (2,2)-word in xσ Σ y but not in xσ y. Then the iterated k-hairpin completion Hk (w) is regular. Proof. Since x is in ασ ᾱ, w is a crossing (2,2)-word and w is not in xσ y, there exist u,v Σ + of α and x 0 Σ such that α = vu, ᾱ = ūv, x = vx 0 v and v is a pseudopalindrome. Then, we have w = vx 0 vx 0 v or vx 0 v x 0 v. If x is a pseudopalindrome, then x 0 is also a pseudopalindrome. It is clear that H k (w) = H k (vx 0vx 0 v) = (vx 0 ) + vx 0 v is regular. If x is not a pseudopalindrome, then x 0 is not a pseudopalindrome. First, we prove that H k (w) = H k (vx 0vx 0 v) is regular. Let and R = ((vx 0 ) + (v x 0 ) + ) (vx 0 ) vx 0 vx 0 v x 0 (v x 0 ) ((vx 0 ) + (v x 0 ) + ) v L = v(( x 0 v) + (x 0 v) + ) (x 0 v) ( x 0 v) x 0 vx 0 vx 0 v((x 0 v) + ( x 0 v) + ). Since Hk (R) = R, we have H k (vx 0vx 0 v x 0 v) Hk (R) = R. Inclusion R Hk (vx 0vx 0 v x 0 v) is proved in a similar way as the proof for Proposition 7. Then we have R Hk (vx 0vx 0 v x 0 v). In a similar way, we have L Hk (v x 0vx 0 vx 0 v). Therefore, is regular. H k(w) = H k(vx 0 vx 0 v) = {vx 0 vx 0 v} H k(h 1 k(vx 0 vx 0 v) = {vx 0 vx 0 v} H k (vx 0vx 0 v x 0 v) H k (v x 0vx 0 vx 0 v) = {vx 0 vx 0 v} R L

15 Regularity of Iterative Hairpin Completions of Crossing (2, 2)-Words 389 Now, we prove that H k (w) = H k (vx 0v x 0 v) is regular. It is easy to see H k (vx 0v x 0 v) = {vx 0 v x 0 v} H k (H1 k (vx 0v x 0 v)) Therefore, Hk (w) is regular. = {vx 0 v x 0 v} H k(vx 0 v x 0 v x 0 v,vx 0 vx 0 v x 0 v) = ((vx 0 ) + (v x 0 ) + ) (vx 0 ) vx 0 v x 0 v( x 0 v) ((x 0 v) + ( x 0 ) + v). Propositions 12, 13, 14 yield the following. Corollary 15. Let x,y be (1,1)-word in ασ ᾱ such that y = x or y = x. For a crossing (2,2)-word w in xσ Σ y, the iterated k-hairpin completions Hk (w) of w is regular, if and only if x or w are pseudopalindromes. Let x,y be (1,1)-word in ασ Σ ᾱ such that y = x or y = x and w be a crossing (2,2)-word in xσ Σ y. If x is not in ασ ᾱ, then y is also not in ασ ᾱ, and then we have Hk (w) = {w}. Therefore, we have Theorem 1. References [1] D. Cheptea, C. Martín-Vide and V. Mitrana, A new operation on words suggested by DNA biochemistry: Hairpin completion, Transgressive Computing (2006), pp [2] V. Diekert, S. Kopecki and V. Mitrana, On the Hairpin Completion of Regular Languages, ICTAC, Ed. by M. Leucker and C. Morgan (Springer, 2009), pp [3] M. A. Harrison, Introduction to Formal Language Theory (Addison-Wesley, Reading, Massachusetts, 1978). [4] M. Ito, P. Leupold, F. Manea and V. Mitrana, Bounded hairpin completion, Information and Computation 209 (2011) [5] L. Kari, S. Kopecki and S. Seki, Iterated hairpin completions of non-crossing words, SOFSEM, Ed. by M. Bieliková, G. Friedrich, G. Gottlob, S. Katzenbeisser and G. Turán (Springer, 2012), pp [6] S. Kopecki, On iterated hairpin completion, Theoretical Computer Science 412 (2011) [7] M. Lothaire, Combinatorics on Words (Cambridge University Press, 1997). [8] F. Manea, C. Martín-Vide and V. Mitrana, On some algorithmic problems regarding the hairpin completion, Discrete Applied Mathematics 157 (2009) [9] F. Manea, C. Martín-Vide and V. Mitrana, Hairpin lengthening, CiE, Ed. by F. Ferreira, B. Löwe, E. Mayordomo and L. M. Gomes (Springer, 2010), pp [10] F. Manea and V. Mitrana, Hairpin Completion Versus Hairpin Reduction, CiE, Ed. by S. B. Cooper, B. Löwe and A. Sorbi (Springer, 2007), pp [11] F. Manea, V. Mitrana and T. Yokomori, Two complementary operations inspired by the DNA hairpin formation: Completion and reduction, Theoretical Computer Science 410 (2009) [12] F. Manea, V. Mitrana and T. Yokomori, Some remarks on the hairpin completion, International Journal on Foundations of Computer Science 21 (2010) [13] G. Păun, G. Rozenberg and T. Yokomori, Hairpin languages, International Journal of Foundations of Computer Science 12 (2001)