IMFM Institute of Mathematics, Physics and Mechanics Jadranska 19, 1000 Ljubljana, Slovenia. Preprint series Vol. 50 (2012), 1173 ISSN

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1 IMFM Institute of Mathematics, Physics an Mechanics Jaransa 19, 1000 Ljubljana, Slovenia Preprint series Vol. 50 (2012), 1173 ISSN PARITY INDEX OF BINARY WORDS AND POWERS OF PRIME WORDS Alesanar Ilić Sani Klavžar Yoomi Rho Ljubljana, April 12, 2012

2 Parity inex of binary wors an powers of prime wors Alesanar Ilić Faculty of Sciences an Mathematics University of Niš, Serbia Sani Klavžar Faculty of Mathematics an Physics University of Ljubljana, Slovenia an Faculty of Natural Sciences an Mathematics University of Maribor, Slovenia Yoomi Rho Department of Mathematics University of Incheon, Korea Abstract Let f be a binary wor an letf (f) be the set of wors of length which o not contain f as a factor (alias wors that avoi the pattern f). A wor is calle even/o if it contains an even/o number of 1s. The parity inex of f (of imension ) is introuce as the ifference between the number of even wors an the number of o wors inf (f). A wor f is calle prime if every nontrivial suffix of f is ifferent from the prefix of f of the same length. It is prove that if f is a power of a prime wor, then the absolute value of the parity inex of f is at most 1. We conjecture that no other wor has this property an prove the conjecture for wors 0 r 1 s 0 t, r,s,t 1. The conjecture has also been verifie by computer for all wors f of length at most 10 an all 31. Keywors: binary wors, combinatorics on wors, wors avoiing a pattern, parity inex, generalize Fibonacci cubes. AMS Subject Classification (2010): 05A05, 68R15, 68W32. Corresponing author 1

3 1 Introuction Elements of B=0,1 are calle bits an an element of B is a binary wor of length. Since all wors consiere here are binary, we will simply spea about wors. A wor u B will be written in the coorinate form as u=u 1 u 2...u. A wor f is a factor of a wor x if f appears as a sequence off consecutive bits of x. A wor u is calle f-free if it oes not contain f as a factor. For a wor f an positive integer, let F (f)=u B u is f free. The prouct notation will mean concatenation, for example, 1 r is the wor of length r with all bits equal 1. A wor b is a power of a wor c if b=c for some 1. A wor is calle even if it contains an even number of 1s an o otherwise. Suppose that f is a wor an is a positive integer. Then the generalize Fibonacci cube, Q (f), is the graph obtaine from the -imensional cube Q by removing all vertices that contain f as a factor. In other wors, V(Q (f))=f (f), two vertices being ajacent if they iffer in exactly one bit. These graphs were stuie for the first time in [3], but special cases were extensively stuie earlier. The most notable special case is forme by Fibonacci cubes Γ = Q (11), 1, see the survey [4]. The special case of Q (1 s ) was introuce in [2] (uner the same name of generalize cubes) an further investigate in [6, 9]. The efinition of the generalize Fibonacci cubes naturally leas to ifferent problems on wors. The most funamental problem is to etermine the orer of these graphs. This problem was stuie earlier uner the notion of wors avoiing a pattern. Calling f a pattern, then the number of wors avoiing f is just the number of f-free wors. Baccherini, Merlini an Sprugnoli [1] were intereste in the number of f-free wors that contain prescribe numbers of 0s an 1s an establishe that they are closely relate to proper Rioran arrays. This wor was extene in [7]. Another natural problem about generalize Fibonacci cubes is when they embe isometrically into hypercubes. This question naturally leas to the concept of the so calle goo an ba wors. A wor f is sai to be -goo if for any f-free wors u an v of length, v can be obtaine from u by complementing one by one the bits of u on which u an v iffer, such that all intermeiate wors are f-free. Then f is goo if it is -goo for any 1. The main result of [5] asserts that about eight percent of all wors are goo. Our principal motivation for the present paper is a result of [6] asserting that each Q (1 r ) contains a hamiltonian path. This in particular implies that the bipartition of Q (1 r ) is balance. (By the way, it is not ifficult to see that every generalize 2

4 Fibonacci cube is connecte.) Clearly, the bipartition sets of Q (f) are forme by even an o wors, respectively. Hence, for a set of wors X, let e(x) an o(x) be the number of even an o wors in X, respectively. Let in aition (X)=e(X) o(x), in particular write (x)= (x) for a wor x. That is, (x)=1 if x is even an (x)= 1 if x is o. Then we efine the parity inex of f of imension as PI (f)= (F (f)). Using this notation, a necessary conition for Q (f) to contain a hamiltonian path is thatpi (f) 1. In the next section we introuce prime wors an prove that if f is a power of a prime wor thenpi (f) 1 hols for any. In Section 3 we consier the parity inex of the wors 0 r 1 s 0 t an prove that for any large enough,pi (0 r 1 s 0 t ) 2. For the special case of 0 r 10 r a more precise result is obtaine, in particular it is note thatpi (010) 3 is the so-calle Paovan sequence. In the final section we pose a conjecture that powers of prime wors are the only wors with the property PI (f) 1 for any an verify the conjecture for all wors of length 10 an for all Powers of prime wors A wor f of length is prime if for any, 1 1, the suffix of f of length is ifferent from the prefix of f of the same length. In particular, wors 0 an 1 are prime, an if 2, then the first bit an the last bit of a prime wor are ifferent. For instance, is a prime wor which easily follows from the fact that the factor 00 appears only at its beginning. On the other han the wor is not prime as it starts an ens with 011. For a wor f of length l lets (f)=b F (f), that is, For i=1,2,..., l+1 let in aition S (f)=b=b 1 b 2...b b contains factor f. S (i) (f)=b=b 1b 2 b b S,b i b i+1 b i+l 1 = f. ThenS (f)=i=1 l+1 S (i) (f). By X we enote the set of all -subsets of the set X. Lemma 2.1 Let f be a wor of length l. Then (S (f))= l+1 =1 ( 1) 1 I N l+1 3 i I S (i) (f).

5 Proof. Let χ A be the characteristic function of a set A: SinceS (f)= l+1 every x S (f), Therefore, i=1 S (i) l+1 =1 χ A (x)= 1; x A, 0; otherwise. (f), the inclusion an exclusion principle implies that for ( 1) 1 (S (f)) = (x) x S (f) = (x) x S (f) = = l+1 =1 l+1 =1 ( 1) 1 ( 1) 1 I N l+1 l+1 =1 I N l+1 I N l+1 χ i I S (i) (f)(x)=1. ( 1) 1 x S (f) I N l+1 χ i I S (i) (f)(x) (x)χ i I S (i) (f)(x) ( i I S (i) (f)). Theorem 2.2 Let f be a power of a prime wor. ThenPI (f) 1 for any 1. Proof. Let 1. Suppose first that f is a prime wor. When <l, we have F (f)=b an if =l, thenf (f) contains all but the wor f. Hence we may assume in the rest that l. Since e(b )=o(b ), we have e(f (f))+e(s (f))= o(f (f))+o(s (f)). Hence PI (f)= (F (f))= (S (f)). It thus suffices to prove that (S (f)) 1. We first note that (S (i) (f))=0. Inee, the first i 1 bits an the last l i+1 bits of the wors froms (i) (f) are arbitrary, hences(i) (f) contains 2 l 1 (f) where even wors an the same number of o wors. Consier now X= i I S (i) I=i 1,i 2,...,i an i 1 < i 2 <<i. Because f is a prime wor, X= as soon as for some inex j, i j+1 i j < l. Moreover, by the same argument as the one use for (S (i) (f)), (X)=0 as soon as for some inex j, i j+1 i j l. Hence (X) can be nonzero only when l= an i j =(j 1)l+1 for each 1 j. Therefore, applying 4

6 Lemma 2.1, (S (f))= The proof is complete for a prime wor f. 0; l, 1; l, o, f contains o number of 1s, 1; otherwise. Assume now that f=(f ) r, where f is a prime wor an r 2. Letf =l. The proof continues similarly as in the case when f was prime. The only ifference is that now X= as soon as for some inex j, the ifference i j+1 i j is not a multiple of l an so (X) can be nonzero only when is a multiple of l. 3 Non-prime wors In this section we stuy the parity inex of wors consisting of three blocs, that is, of wors 0 r 1 s 0 t, r,s,t 1. Clearly, none of these wors is prime. In our main result (Theorem 3.2) we prove that for no such wor f,pi (f) 1 hols for all. Before that we separately give a more precise result for the special case of 0 r 10 r. The obtaine results in particular imply that Q (0 r 1 s 0 t ) oes not contain a hamiltonian path as soon as is large enough. Theorem 3.1 Let r 1. Then PI (0 r 10 r )= Moreover, for any 4r+ 4,PI (0 r 10 r ) 2. 0; 2r,2r+ 2 3r+ 1, 1; =2r+ 1,3r+ 2 4r+ 3. Proof. Suppose first that 2r. ThenF (0 r 10 r )=B an hence PI (0 r 10 r )=0. SinceF 2r+1 (0 r 10 r )=B 0 r 10 r we have PI 2r+1 (0 r 10 r )=1. Let 2r+ 2. Recall that PI (0 r 10 r ) = (S (f)) = b S (f) (b). By Lemma 2.1, PI (f)= l+1 =1 ( 1) 1 I N l+1 ( i I S (i) (f)). Suppose that for a set X= i I S (i) (f) there exists an inex i such that if w X then also w+ e i X. Then (X)=0. It follows that (X)0 if an only if there exist 0 an r r j 2r for all 1 j, such that X=0 r 10 r 1 10 r 2 10 r 10 r. Moreover, in that case (X)= (0 r 10 r 1 10 r 2 10 r 10 r )=( 1) +1. 5

7 Hence let 0 an r r j 2r, 1 j, an set b=0 r 10 r 1 10 r 2 10 r 10 r. Let v= b r1 +2b t1 +3b be the wor obtaine from b by omitting the first r bits, so that v B r 1 1. Since b has one more bit of 1 than v oes, (v)= (b). Note that v starts with 0 r 1. Then v S (j) r 1 1 (f) if an only if b S(1) j J Now we can compute as follows: PI (0 r 10 r ) = (S (f))= (b) b S (f) = = b S (f) r 1 =r v S r 1 (f) (f) j J S (j+r 1+1) (f). (b)+ (b)++ b S (f) b S (f) r 1 =r+1 r 1 =2r (b) r 1 (v)+ r 2 (v)+ v S r 2 (f) + v S 2r 1 (f) 2r 1 (v) = r 1 (v)+ r 2 (v)+ v S r 1 (f) v S r 2 (f) + 2r 1 (v) v S 2r 1 (f) = r r 1 2r v S r1 1(f) r1 1(v) = r1 1 r r 1 2r v S r1 1(f)v = r1 1(S r1 1(f)) r r 1 2r = PI r1 1(0 r 10 r ). (1) r r 1 2r It follows thatpi (0 r 10 r )= r r1 2r PI r1 1(0 r 10 r ). As the values PI r1 1(0 r 10 r ) have the same sign for all r r 1 2r, from Equation (1) we get PI (0 r 10 r )= PI r1 1(0 r 10 r ). (2) r r 1 2r Set a =PI (0 r 10 r ) for all. We alreay now that a = 0 for all 2r an that a 2r+1 = 1. Let 2r+ 2. If 3r+ 1 an there is a wor b S (f), then there is an inex i such that if w X then also w+ e i X an hence (X)=0. 6

8 Assume 3r+ 2. When 3r+ 2 4r+ 2, PI (0 r 10 r )= (0 r 10 2r 2 10 r )=1 or 1 an hence a = 1. When =4r+3, a = a 2r+2 ++a 3r+2 = 1. Let =4r+3+u for some u 1. Then by Equation (2), a 4r+3+u = a 2r+2+u ++a 3r+2+u. If u r, then 2r+2+u 3r+2<3r+2+u 4r+2 an therefore a 2. Assume u r+1. Let u = u r. Then =5r+ 3+u for u 1. By Equation (2), a 5r+3+u = a 3r+2+u ++a 4r+2+u. If u r, then 3r+ 2 3r+ 2+u < 4r+ 3 4r+ 2+u an therefore a 2. Assume u r+ 1. Then let u = u r. Then =6r+ 3+u for u 1. By Equation (2), a 6r+3+u = a 4r+2+u ++a 5r+2+u. As 3r+ 2 4r+ 2+u < 5r+ 2+u, a 2. Thus when 4r+ 4, a 2. case, The special case of Theorem 3.1 when r= 1 eserves a special attention. In that PI (010)=PI 2 (010)+PI 3 (010) with initial conitionspi 3 (010)=1,PI 4 (010)=0,PI 5 (010)=1which is the Paovan sequence, see sequence A from [8]. Theorem 3.2 Let r,s,t 1. Let z be the integer such that(z 1)t+2 r+s zt+1. Then PI (0 r 1 s 0 t ) = 0; <r+ s+t, y(r+ s+t)<<(y+ 1)(r+ s)+t for 1 y z, 1; =r+ s+t, (y+ 1)(r+ s)+t (y+ 1)(r+ s+t) for 1 y z, =(z+ 1)(r+ s+t)+1. Moreover, for any (z+ 1)(r+ s+t)+2,pi (0 r 1 s 0 t ) 2. Proof. Since PI (0 r 1 s 0 t )=PI (0 t 1 s 0 r ), it suffices to prove the result for wors 0 r 1 s 0 t with r t. By the same argument as in the proof of Theorem 3.1, (X)0 if an only if there exist 0 an r r j r+ t for all 1 j such that where (X)=( 1) (+1)s. Also X=0 r 1 s 0 r 1 1 s 0 r 2 1 s 0 r 1 s 0 t PI (0 r 1 s 0 t )= PI r1 s(0 r 1 s 0 t ). (3) r r 1 r+t Set a =PI (0 r 1 s 0 t ) for all. We alreay now that a = 0 for all <r+ s+t an that a r+s+t = 1. Let r+ s+t+1. In the first part of the proof, we prove the theorem for (z+ 1)r+(z+ 1)s+(z+ 1)t by inuction on y for 1 y z. Then we prove the theorem for (z+ 1)r+(z+ 1)s+(z+ 1)t+1. The iea of the proof 7

9 is as follows. From the first part of the proof, we notice that for each y 1, a = 0 for r+ s (y 1)t 1 consecutive numbers of an then a 1 for the next yt+1 consecutive numbers of. As y increases, r+ s (y 1)t 1 ecreases to zero an yt+1 increases. While, by Equation (3), a = a r s t ++a r s, which is a sum of t+1 consecutive numbers, where t+1 is a constant for given 0 r 1 s 0 t. Therefore for large enough, a 2. By a similar argument as in the proof of Theorem 3.1, a = 0 if <2r+2s+t an a = 1 if 2r+ 2s+t 2r+ 2s+2t. Thus the statement is true for y= 1. Let y 2. Suppose the statement is true for all 1 y 0 < y. Let =yr+ys+yt+u for some u 1. Then by Equation (3), a yr+ys+yt+u = a (y 1)r+(y 1)s+(y 1)t+u ++a (y 1)r+(y 1)s+yt+u. When <(y+ 1)r+(y+ 1)s+t, i.e., u<r+s (y 1)t,(y 1)r+(y 1)s+(y 1)t< (y 1)r+(y 1)s+(y 1)t+u<(y 1)r+(y 1)s+yt+u<yr+ys+t an hence by the inuction assumption, a = 0. When =(y+1)r+(y+1)s+t, i.e., u=r+s (y 1)t, (y 1)r+(y 1)s+yt+u=yr+ys+t an hence by the inuction assumption, a a yr+ys+t 1. Assume (y+ 1)r+(y+ 1)s+t+1, i.e., u r+ s (y 1)t+1. Let u = u r s+(y 1)t. Then =(y+ 1)r+(y+ 1)s+t+u where u 1. By Equation (3), a = a yr+ys+u ++a yr+ys+t+u. If (y+ 1)r+(y+ 1)s+(y+ 1)t, i.e., u yt, then yr+ ys+u yr+ ys+yt. Consiering that yr+ ys+t<yr+ ys+t+u, a a yr+ys+t + a yr+ys+t+1 or a a yr+ys+u + a yr+ys+u +1 epening on whether yr+ ys+u < yr+ys+t or not. Therefore by the inuction assumption, a 1. Thus the theorem is prove for all (z+ 1)r+(z+ 1)s+(z+ 1)t. Assume (z+ 1)r+(z+ 1)s+(z+ 1)t+1 (z+ 2)r+(z+ 2)s+t. Let = (z+2)r+(z+2)s+t+u where u 0. Then by Equation (3), a = a (z+2)r+(z+2)s+t+u = a (z+1)r+(z+1)s+u ++a (z+1)r+(z+1)s+t+u. Note that(z+1)r+(z+1)s+t+u (z+1)r+ (z+1)s+t. Assume =(z+2)r+(z+2)s+t, i.e., u = 0. Then(z+1)r+(z+1)s+t+u = (z+ 1)r+(z+ 1)s+t. If r+szt, then(z+ 1)r+(z+ 1)szr+zs+zt an hence a = a (z+1)r+(z+1)s+t 1. If r+ s zt, then(z+ 1)r+(z+ 1)s zr+ zs+zt an hence a a zr+zs+zt + a (z+1)r+(z+1)s+t 2. Let (z+ 2)r+(z+ 2)s+t, i.e., u 0. Then(z+ 1)r+(z+ 1)s+t+u (z+1)r+(z+1)s+t. First assume <(z+2)r+(z+2)s+(z+2)t, i.e., u <(z+1)t. Then (z+1)r+(z+1)s+u <(z+1)r+(z+1)s+(z+1)t. Therefore a a (z+1)r+(z+1)s+t + a (z+1)r+(z+1)s+t+1 or a a (z+1)r+(z+1)s+u +a (z+1)r+(z+1)s+u +1 epening on whether (z+1)r+(z+1)s+u <(z+1)r+(z+1)s+t or not. Thus a 2 in any case. Secon assume =(z+2)r+(z+2)s+(z+2)t, i.e., u =(z+1)t. Then(z+1)r+(z+1)s+u = (z+ 1)r+(z+ 1)s+(z+ 1)t an hence a a (z+1)r+(z+1)s+u + a (z+1)r+(z+1)s+u +1 2 consiering that(z+1)r+(z+1)s+(z+1)t<(z+1)r+(z+1)s+u +1<(z+2)r+(z+ 2)s+(z+2)t. Finally assume (z+2)r+(z+2)s+(z+2)t, i.e., u (z+1)t. Suppose 8

10 there is u (z+ 1)t such that a 1. Let u 0 be the smallest such an integer an 0 =(z+ 2)r+(z+ 2)s+t+u 0. Then a 0 = a (z+1)r+(z+1)s+u ++a 0 (z+1)r+(z+1)s+t+u. 0 Since(z+1)r+(z+1)s+u 0(z+1)r+(z+1)s+(z+1)t an(z+1)r+(z+1)s+t+u 0 < 0, a 0 2, which is a contraiction. Thus the statement is true for all. 4 Computer evience an conjecture Using computer we obtaine the parity inex for all wors f of length at most 10 an all 31. Since Q (f) is isomorphic to Q (f), where f is the binary complement of f, we have restricte the computation to wors f that contain not more 1s than 0s. From the same reason reverse wors nee not to be consiere. In Table 1 all wors f of length at most 8 an withpi (f) 1 for 31 are collecte. length f , 0011, , 00011, , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , Table 1: List of wors f withf 8 anpi (f) 1 for 31 It can be chece that every wor from the table is a power of a prime wor. Moreover, the same was verifie also for the obtaine wors of length 9 an 10 (not given in the table). Base on this experiment an Theorems 2.2 an 3.2 we pose: Conjecture 4.1 Let f be a wor such thatpi (f) 1 hols for any. Then f is a power of a prime wor. 9

11 A possible approach to the conjecture woul be to prove that if f is not a power of a prime wor, then the sequencepi (f) satisfies a certain recurrence relation from which we can euce the behavior of the sequence. For instance, one can establish the recurrent formula PI (01110)=PI 4 (01110)+PI 5 (01110), with initial conitionspi 5 (01110)=1,PI 6 (01110)=PI 7 (01110)=PI 8 (01110)= 0 anpi 9 (01110)=1. Similarly, either by applying Equation (3) or by a teious case analysis yiels, one can get: PI ( ) = PI 6 ( )+PI 7 ( ) +PI 8 ( )+PI 9 ( ), with initial conitionspi 9 ( )= 1,PI 10 ( )=PI 11 ( )= PI 12 ( )=PI 13 ( )=PI 14 ( )= 0,PI 15 ( )= PI 16 ( )=PI 17 ( )= 1. In Fig. 1 the values ofpi (01110) anpi ( ) for 5 55 are plotte. Note that the sequencepi (f) oes not nee to be monotone, but it seems that starting from some large enough imension the sequence is strictly increasing. Figure 1: Values ofpi (f) for f= an f=

12 Acnowlegements This wor was supporte by the research Grants an of Serbian Ministry of Eucation an Science, by the research grant P of Ministry of Higher Eucation, Science an Technology Slovenia, an by Basic Science Research Program through the National Research Founation of Korea fune by the Ministry of Eucation, Science an Technology grant References [1] D. Baccherini, D. Merlini, R. Sprugnoli, Binary wors excluing a pattern an proper Rioran arrays, Discrete Math. 307 (2007) [2] W.-J. Hsu, Fibonacci cubes a new interconnection technology, IEEE Trans. Parallel Distrib. Syst. 4 (1993) [3] A. Ilić, S. Klavžar, Y. Rho, Generalize Fibonacci cubes, Discrete Math. 312 (2012) [4] S. Klavžar, Structure of Fibonacci cubes: a survey, to appear in J. Comb. Optim., DOI: /s z. [5] S. Klavžar, S. Shpectorov, Asymptotic number of isometric generalize Fibonacci cubes, European J. Combin. 33 (2012) [6] J. Liu, W.-J. Hsu, M. J. Chung, Generalize Fibonacci cubes are mostly Hamiltonian, J. Graph Theory 18 (1994) [7] D. Merlini, R. Sprugnoli, Algebraic aspects of some Rioran arrays relate to binary wors avoiing a pattern, Theoret. Comput. Sci. 412 (2011) [8] N. J. A. Sloane, The On-Line Encyclopeia of Integer Sequences, Publishe electronically at [9] N. Zagaglia Salvi, On the existence of cycles of every even length on generalize Fibonacci cubes, Matematiche (Catania) 51 (1996)