How Much Different Are Two Words with Different Shortest Periods

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How Much Different Are Two Words with Different Shortest Periods

Mai Alzamel, Maxime Crochemore, Costas Iliopoulos, Tomasz Kociumaka, Ritu Kundu, Jakub Radoszewski, Wojciech Rytter, Tomasz Waleń

To cite this version:

Mai Alzamel, Maxime Crochemore, Costas Iliopoulos, Tomasz Kociumaka, Ritu Kundu, et al.. How Much Different Are Two Words with Different Shortest Periods. 14th IFIP International Conference on Artificial Intelligence Applications and Innovations (AIAI), May 2018, Rhodes, Greece. pp.168-178,

�10.1007/978-3-319-92016-0_16�. �hal-01821339�

How much different are two words with different shortest periods

Mai Alzamel¹, Maxime Crochemore^1,2, Costas S. Iliopoulos¹, Tomasz Kociumaka³, Ritu Kundu¹, Jakub Radoszewski^1,3,

Wojciech Rytter^3,?, and Tomasz Waleń³

1 Department of Informatics, King’s College London, London, UK [mai.alzamel,maxime.crochemore,costas.iliopoulos,ritu.kundu]@kcl.ac.uk

2 Université Paris-Est, France

3 Faculty of Mathematics, Informatics and Mechanics, University of Warsaw, Warsaw, Poland [kociumaka,jrad,rytter,walen]@mimuw.edu.pl

Abstract. Sometimes the difference between two distinct words of the same length cannot be smaller than a certain minimal amount. In par- ticular if two distinct words of the same length are both periodic or quasiperiodic, then their Hamming distance is at least 2. We study here how the minimum Hamming distancedist(x, y)between two wordsx, y of the same length n depends on their periods. Similar problems were considered in [1] in the context of quasiperiodicities. We say that a period pof a wordxisprimitiveifxdoes not have any smaller periodp⁰which dividesp. For integersp, n(p≤n) we definePp(n)as the set of words of lengthnwith primitive periodp. We show several results related to the following functions introduced in this paper forp6=qandn≥max(p, q).

Dp,q(n) = min{dist(x, y) : x∈ Pp(n), y∈ Pq(n)}, Np,q(h) = max{n : Dp,q(n)≤h}.

1 Introduction

Consider a word x of length |x| = n, with its positions numbered 0 through n−1. We say thatxhas a periodpifx_i =x_i+p for all0≤i < n−p. Our work can be seen as a quest to extend Fine and Wilf’s Periodicity Lemma [14], which is a ubiquitous tool of combinatorics on words.

Lemma 1 (Periodicity Lemma [14]). If a word xhas periods pand q and

|x| ≥p+q−GCD(p, q), thenxalso has a period GCD(p, q).

Other known extensions of this lemma include a variant with three [10] and an arbitrary number of specified periods [11,16,17,23], the so-called new periodicity lemma [13,3], a periodicity lemma for repetitions that involve morphisms [19], and extensions into periodicity of partial words [4,5,6,7,8,9,22], into abelian [12]

?Supported by the Polish National Science Center, grant no 2014/13/B/ST6/00770.

andk-abelian [18] periodicity, into bidimenional words [20], and other variations [15,21].

We say that a word xof length n is periodic if it has a period psuch that 2p≤n. For two wordsxandyof lengthn, bydist(x, y)we denote their Hamming distance being the number of positionsi= 0, . . . , n−1 such thatxi 6=yi. The following folklore fact gives a lower bound on how different are two distinct periodic words. Its proof can be found in [1].

Fact 2. If x and y are distinct periodic words of the same length, then dist(x, y)6= 1.

We present several generalizations of this fact.

Results similar to Fact 2 were presented recently in the context of quasiperi- odicity [1]. We say that a word xhas acover uif each position in xis located inside an occurrence of u in x. The word x is called quasiperiodic if it has a cover uother than x. In [1] the following generalization of Fact 2 was shown:

dist(x, y)>1 for any two distinct quasiperiodic wordsx,y of the same length.

This type of fact has potential applications; see [2].

There is a quantitative difference between periods and covers. For example, there are words xandy of length 1024with shortest covers of length 4 and5, respectively, and dist(x, y) = 2:

x= (abaa)²⁵⁶ and y=aaba(abaa)²⁵⁵

with coversabaaandaabaa. However, if xandy are words of length1024with shortest periods4 and5, respectively, then we must havedist(x, y)≥357.

We say that a periodpof a word xisprimitive if no proper divisor ofpis a period ofx, i.e., ifp⁰|pandp⁰ is a period ofx, thenp⁰=p. We define

Pp(n) ={ |x|=n, pis a primitive period ofx}.

The ultimate goal of this work is a characterization of the functionD_p,q defined forp6=q andn≥max(p, q)as:

Dp,q(n) = min{dist(x, y) : x∈ Pp(n), y∈ Pq(n)}.

As Dp,q is non-decreasing for given p, q, it can be described by the following auxiliary function:

Np,q(h) = max{n : Dp,q(n)≤h}.

One can note that Lemma 1 can be equivalently formulated as Np,q(0) <

p+q−GCD(p, q). Similarly, an equivalent formulation of Fact 2 isNp,q(1)<2q.

Fine & Wilf [14] also proved that the boundp+q−GCD(p, q)of Lemma 1 cannot be improved. Consequently, Np,q(0) = p+q−GCD(p, q)−1. On the other hand, we show that Np,q(1) = 2q−1 only for p | q. Hence, the bound Np,q(1)<2qof Fact 2 is not tight in general.

Our results. In Section 2 we consider the case that p | q. In the remaining sections, we only consider the case ofp < qandp-q. In Section 3 we show exact values of the function D_p,q for p+q−GCD(p, q) ≤n ≤ 2q. In Section 4, we show the following bounds valid for abitraryn≥q:

jn−q p

k ≤ Dp,q(n) ≤2l

n−q p

m .

We also prove an alternative boundDp,q(n)≥j

2n p+q

k

valid forn≥p+q.

h N3,4(h) example

1 6 a a b a ab

a a b a aa

2 8 a a b a ab aa

a a b a aa ba

3 10 a a b a ab aaba

a a b a aa baaa

4 11 a a b a ab aabaa

a a b a aa baaab

5 17 a a b a ab aabaa ba a b a a a a b a aa baaab aa a b a a

h N2,3(h) example

0 3 a b a

a b a

1 4 a b ab

a b aa

2 5 a b ab a

a b aa b

3 9 a b ab a ba b a

a b aa b aa b a

4 10 a b ab a ba b ab

a b aa b aa b aa

5 11 a b ab a ba b ab a

a b aa b aa b aa b 6 15 a b ab a ba b ab a ba b a

a b aa b aa b aa b aa b a 7 16 a b ab a ba b ab a ba b ab

a b aa b aa b aa b aa b aa

Fig. 1.Upper table: Values ofN3,4(h)forh= 1, . . . ,5together with pairs of words of lengthN3,4(h)that have the Hamming distanceh. Lower table: Values ofD3,4(n)for n= 6, . . . ,17.

2 Preliminaries

Let us consider a finite alphabet Σ. If xis a word of length |x| = n, then by xi ∈Σ fori = 0, . . . , n−1 we denote its ith letter. We say that a wordv is a

factor of a wordxif there exist wordsuand wsuch thatx=uvw. A factorv is called a prefix ofxifuis an empty word in some such decomposition and a suffix if wis an empty word in some such decomposition. By x[i..j] we denote the factorxi. . . xj.

Ifxi=xi+pfor all0≤i < n−pfor some integerp, thenpis calleda period ofxand the prefix ofxof length pis calleda string period ofx. Ifxhas period p, theny is called aperiodic extension of xwith periodpify also has periodp and hasxas a prefix.

We say that a periodpisprimitiveif no proper divisor of pis a period ofx.

Note that the shortest period (denotedp=per(x)) is always primitive.

We say that a wordxisprimitive if there exists no other worduand integer k > 1 such that x= u^k. Note that pis a primitive period of xif and only if the corresponding string period is a primitive word. Two wordsxandyare each other’s cyclic rotations if there exist wordsuandvsuch thatx=uvandy=vu.

In this case we also say that|u| is the shift betweenxandy.

For a sequence of positive integers (a1, . . . , am), we define a (a1, . . . , am)- decomposition of a wordx as a sequence of consecutive factors ofx of lengths a1, . . . , am, a1, . . . , am, . . .The sequence ends at the lastcompletefactor that can be cut out ofx; see Fig. 2 for an example.

a b a b b a b a b a a b a b a a b a a b

1 2 4 1 2 4 1 2

Fig. 2. The(1,2,4)-decomposition ofababbababaababaabaabisa ba bbab a ba abab a ab.

Ifp|q, we can give a simple complete characterization of functionsNp,q and Dp,q.

Fact 3. Ifp|q andp < q, thenDp,q(n) =j

n q

kandN_p,q(h) =q·(h+ 1)−1.

Proof. We first show that Dp,q(n)≥j

n q

k

. Consider a positive integern, words x ∈ Pp(n), y ∈ Pq(n), and the (q)-decompositions of xand y: α1, . . . , αk and β1, . . . , βk. Observe thatα1=. . .=αk andβ1=. . .=βk becauseq is a period of both xand y, but α1 6= β1 because q is a primitive period of y, but not a primitive period ofx. Hence,dist(x, y)≥k.

As for the other inequality onDp,q(n), let us takex= (a^p−1b)^bn/pca^nmodp and letybe the word that is obtained fromxby changing the letters at positions i≡q−1 (modq)from btoc. Thendist(x, y) =j

n q

k .

Finally, the formula forN_p,q(h)follows directly from the other one. ut Henceforth, we will always assume thatp-qandq-p.

3 Exact Values for Small n

Let us start with the following useful lemma.

Lemma 4. Let x be a word of length n and let y by its cyclic rotation by s characters. If x6= y, thendist(x, y) ≥2. Moreover, there are two mismatches between xandy located at leastGCD(n, s)positions apart.

Proof. Note thatyi=x_{(i+s) modn}for0≤i < n. Sincex6=y, we havexa 6=ya= x(a+s) modnfor some positiona. Letkbe the smallest positive integer such that xa = x(a+ks) modn. Due to x(a+s) modn 6= xa and x(a+ns) modn =xa, we have 1 < k≤n. Letb = (a+ (k−1)s) modn. Note thatxb 6=xa =x(b+s) modn = y_b. Hence, a and b are positions of two distinct mismatches between x and y.

Moreover,b≡(a+ (k−1)s) modn≡a (mod GCD(n, s)). Consequently, these two mismatches are indeed located at leastGCD(n, s)positions apart. ut

Lemma 5. Consider positive integersp, q satisfyingp < q andp-q. Let xand y be words of lengthn such thatp+q−GCD(p, q)≤n≤q+pd^q_pe −1, pis a period ofx, andqis a period ofy but not a period of x. Then

dist(x, y)≥j

n−q p

k

+jn−q+GCD(p,q) p

k .

Proof. Letu=x[0..p−1]and letv be the cyclic rotation ofubyq characters.

Note thatuis a string period ofx, sou6=v; otherwise,qwould be a period of x. Consequently, Lemma 4 provides two distinct indicesa, bsuch that ua 6=va, ub6=vb, anda≤b−GCD(p, q)< p−GCD(p, q). Let us define

A=n

kp+a: 0≤k <jn−q+GCD(p,q) p

ko , B=n

kp+b: 0≤k <j

n−q p

ko . Observe that

maxA=jn−q+GCD(p,q) p

k

p−p+a≤n−q+ GCD(p, q)−p+a < n−q and

maxB=j_n−q

p

kp−p+b≤n−q−p+b < n−q.

Moreover,

maxA≤jpdq/pe−1+GCD(p,q) p

kp−p+a=l_q

p

mp−p+a < q+a≤q+ min(A∪B),

and

maxB ≤j_pdq/pe−1

p

k

p−p+b=l_q

p

m

p−p < q≤q+ min(A∪B).

Consequently, for each i ∈A∪B, there are positions xi and xi+q, and all these 2(|A|+|B|) positions are distinct. Moreover, observe that fori ∈ A, we

have x_i =u_a 6= v_a = x_i+q, while for i ∈ B, x_i = u_b 6=v_b = x_i+q. Thus, for i ∈ A∪B, we have x_i 6= x_i+q, but y_i = y_i+q; hence x_i 6= y_i or x_i+q 6=y_i+q. The positions we consider are distinct, so dist(x, y) ≥ |A∪B| = |A|+|B| = j_n−q

p

k

+jn−q+GCD(p,q) p

k

, as claimed. ut

q p

x * u *

a b

* u

a+p

* v *

a+q b+q

* v

a+b+q

Fig. 3. Illustration of the equalities in the bound in Lemma 5 ford^q_pe= 1.

Lemma 6. Consider coprime integers p, q satisfying 1 < p < q. Let w be a word of length p+q−2 with periods p and q, but without period 1. Moreover, let n be an integer such that p+q−1 ≤ n ≤ q+d^q_pep−1, and let x and y be periodic extensions ofw of length n preserving periods pandq, respectively.

Thenper(x) =p,per(y) =q, and dist(x, y)≤j

n−q p

k +j

n−q+1 p

k ,

Proof. Claim. If a positionisatisfiesi < qor(i−q) modp < p−2, thenxi=yi. Proof. The claim is clear fori < q+p−2since due to the common prefix ofxand y. Thus, we consider a positioni=q+kp+rwith1≤k <d^q_peand0≤r < p−2.

We have xq+kp+r =xq+r =yq+r =yr =xr =xkp+r=ykp+r =yq+kp+r. This is because positionsr < kp+r < q+rare within the common prefix of xand

y. ut

Consequently,

dist(x, y)≤ {i:q≤i < n∧ (i−q) modp≥p−2}=

{j: 0≤j < n−q∧j modp=p−1}+{j: 0≤j < n−q∧jmodp=p−2}= jn−q

p

k +j

n−q+1 p

k , as claimed. Next, we prove thatp⁰ :=per(x)is equal top. Note thatp⁰ ≤pby definition ofx. For a proof by contradiction, suppose thatp⁰ < p. Note thatw

has periodsp⁰ andq. Moreover,|w|=p+q−2≥p⁰+q−1, soGCD(p⁰, q)is a period ofw. Moreover,n≥p+q−1≥p+GCD(p⁰, q)−1, soGCD(GCD(p⁰, q), p) is a period ofx. However,GCD(GCD(p⁰, q), p) = GCD(p⁰,GCD(q, p)) = 1is not a period ofw, which is a prefix of x.

Similarly, suppose thatq⁰:=per(y)< q. We observe that|w|=p+q−2≥ p+q⁰−1, soGCD(p, q⁰)is a period ofw. Moreover,n≥p+q−1≥GCD(p, q⁰) + q−1, soGCD(GCD(p, q⁰), q) is a period ofy. However,GCD(GCD(p, q⁰), q) = GCD(q⁰,GCD(p, q)) = 1is not a period ofw, which is a prefix ofy. ut

q p−2 2 p−2

p

x u u u

u

p p

6= 6=

y u u u

u q

q

Fig. 4. Illustration of the equalities in the lower bound in Lemma 6 forn=q+ 2p−2.

a b a a b a b a a b a b aa b a a b a a b a b a a b a a ba b a

Fig. 5. A periodic prefix of a Fibonacci word and a power of a Fibonacci word that differ only at two positions.

Theorem 7. If p < q,p-q, andp+q−GCD(p, q)≤n≤q+d^q_pep−1, then

Dp,q(n) = n−q

p

+

n+ GCD(p, q)−q p

. (1)

Proof. Lemma 5 gives a lower bound of Dp,q(n). Our upper bound is based on Lemma 6. Let d = GCD(p, q), p⁰ = ^p_d, q⁰ = ^q_d, and n⁰ = n

d

. Observe that

1 < p⁰ < q⁰ and p⁰+q⁰−1≤n⁰ ≤q⁰+d^q_p⁰0ep⁰−1. Hence, Lemma 6 results in strings x⁰, y⁰ with of lengthn⁰ with shortest periodsp⁰ and q⁰ respectively, and withdist(x⁰, y⁰)≤j

n⁰−q⁰ p⁰

k +j

n⁰−q⁰+1 p⁰

k

=j

n−q p

k

+jn−q+GCD(p,q) p

k .

Letcbe a character occurring neither inx⁰nor iny⁰. Let us definexandyso thatx_id+d−1=x⁰_i andy=_id+d−1=y_i⁰, andxj=yj =cifj modd6=d−1. Note that dist(x, y) = dist(x⁰, y⁰) and |x| =|y| = n. Also, observe that due to the choice of the characterc, all periods ofxandy are larger thandn⁰ or multiples of d. Consequently, per(x) = dper(x⁰) = dp⁰ = p and per(y) = dper(y⁰) =

dq⁰=q. This completes the construction. ut

Corollary 8. The formula (1) of Theorem 7 applies for p+q−GCD(p, q)≤ n≤2q.

Fact 9. The function Dp,q(n) is non-decreasing for n ≥ p+q−GCD(p, q).

Moreover:

Dp,q(n) =h ⇐⇒ Np,q(h−1)< n≤Np,q(h) Np,q(h) =n ⇐⇒ Dp,q(n) =h <Dp,q(n+ 1)

4 Bounds for D

(n) for Arbitrary n

Lemma 10. Let p, q be integers such that p < q andp-q. Moreover, letxand y be words of lengthn ≥q such that pis a period of x, and q is a period of y but not ofx. Thendist(x, y)≥j_n−q

p

k.

Proof. Since q is not a period of x, we have x_i 6= x_i+q for some position i, 0≤i < n−q. Consider a setJ ={j : 0≤j < n−q∧j≡i (modp)}. Sincepis a period of x, we have xj 6=xj+q for eachj∈J; on the other hand,yj =yj+q, so xj 6=yj or xj+q 6=yj+q. Moreover,p -q implies that the positions j, j+q acrossj ∈J are pairwise distinct. Consequently,dist(x, y)≥ |J| ≥j

n−q p

k . ut

Theorem 11. Ifp < q,p-q, andn≥p+q, thenj_n−q

p

k≤ Dp,q(n)≤2l_n−q

p

m . Proof. The lower bound follows directly from Lemma 10. The upper bound is obtained using words (x, y) ∈ Sp,q,n with string periodsa^p−1b and (a^p−1b)^ka^r whereq=kp+r. Indeed,xandyagree on the firstqpositions. After that, inside each pair of corresponding fragments of length at most pthey have at most 2

mismatches. ut

Remark 12. In general, it is not true thatD_p,q(n)≥ d^n−q_p e. For example, words x=cacbcacbcacbcacbcacandy=cacbcacacbcacacbcacof length 19, with shortest periods p= 4and q= 6, respectively, satisfydist(x, y) = 3<d¹⁹⁻⁶₄ e.

Theorem 13. Ifp < q andp-q andn≥p+q, thenD_p,q(n)≥j

2n p+q

k .

Proof. We use the following claim:

Claim. Ifp < q andp-q, then (a) Np,q(1) =q+p−1,

(b) N_p,q(2) =q+ 2p−GCD(p, q)−1.

Proof. Observe that q+p≤q+ 2p−GCD(p, q)≤q+d^q_pep−1, so the values below are within the scope of Theorem 7. We have:

Dp,q(q+p−1) =j_p−1

p

k+jp+GCD(p,q)−1 p

k= 0 + 1 = 1

D_p,q(q+p) =j

p p

k

+j_p+GCD(p,q)

p

k

= 1 + 1 = 2.

This concludes the proof of part (a).

D_p,q(q+ 2p−GCD(p, q)−1) =j

2p−GCD(p,q)−1 p

k +j

2p−1 p

k

= 1 + 1 = 2 Dp,q(q+ 2p−GCD(p, q)) =j2p−GCD(p,q)

p

k +j

2p p

k

= 1 + 2 = 3.

This concludes the proof of part (b). ut

Consider words x ∈ Pp(n), y ∈ Pq(n) and their (p+q)-decompositions:

α1, . . . , αk and β0, . . . , β_k−1. If αi = βi for some 1 ≤i ≤ k, then, by the Pe- riodicity Lemma, bothαi,βi have periodGCD(p, q); consequently, both xand y have periodGCD(p, q), a contradiction. Hence, part (a) of the claim implies dist(αi, βi)≥2for each i= 1, . . . , k.

Let αk+1 and βk+1 be the suffixes of x and y starting immediately after the last factors of the corresponding decompositions. If |αk+1| < ^p+q₂ , then we already have that

dist(x, y)≥2j

n p+q

k

=j

2n p+q

k .

Otherwise, by part (b) of the claim applied for the wordsα_kα_k+1 andβ_kβ_k+1, we have

dist(x, y)≥2j

n p+q

k−1

+ 3 = 2j

n p+q

k

+ 1 =j

2n p+q

k .

In both cases we obtain the desired inequality. ut

5 Conclusions

The paper studies the following general type of question:

How much dissimilar in a whole should be two objects which are different in some specific aspect?

The answer to this type of question heavily depends on the studied type of the objects. Thus sometimes the answer is completely trivial; for example, two different strings of the same length may differ at only a single position. In this work we show that if we consider different strings of the same length that are additionally periodic, then the implied number of positions where the two strings must differ can be large. The exact number depends on the length of the strings and on their periods.

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