A new approach to cross-bifix-free sets
aa r X i v : . [ c s . F L ] D ec A new approach to cross-bifix-free sets
S. Bilotta ∗ E. Pergola ∗ R. Pinzani ∗ Abstract
Cross-bifix-free sets are sets of words such that no prefix of any word is a suffix of anyother word. In this paper, we introduce a general constructive method for the sets of cross-bifix-free binary words of fixed length. It enables us to determine a cross-bifix-free wordssubset which has the property to be non-expandable.
In digital communication systems, synchronization is an essential requirement to establishand maintain a connection between a transmitter and a receiver.Analytical approaches to the synchronization acquisition process and methods for the con-struction of sequences with the best aperiodic autocorrelation properties have been the subjectof numerous analyses in the digital transmission.The historical engineering approach started with the introduction of bifix. It denotes asubsequence that is both a prefix and suffix of a longer observed sequence. Rather than to thebifix, much attention has been devoted to a bifix-indicator, an indicator function implying theexistence of the bifix [10]. Such indicators were shown to be without equal in performing variousstatistical analysis, mainly concerning the search process [3, 10]However, an analytical study of simultaneous search for a set of sequences urged the inven-tion of cross-bifix indicators [1, 2] and, accordingly, turned attention to the sets of sequenceswhich avoid cross-bifixes, called cross-bifix-free sets.In [1], the author analyzes some properties of binary words that form a cross-bifix-freeset, in particular, a general constructing method called the kernel method is presented. Thisapproach leads to sets S ( n ) of cross-bifix-free binary words, of fixed length n , having cardinality1 , , , , , , , , , , , ,
233 for n = 3 , , , , , , , , , , , ,
15 respectively.This sequence forms a Fibonacci progression and satisfies the recurrence relation | S ( n ) | = | S ( n − | + | S ( n − | with | S (3) | = 1 and | S (4) | = 1.The problem of determining cross-bifix-free sets is also related to several other scientificapplications, for instance in multiaccess systems, pattern matching and automata theory.The aim of this paper is to introduce a method for the generation of sets of cross-bifix-free binary words of fixed length based upon the study of lattice paths on the Cartesian plane.This approach enables us to obtain cross-bifix-free sets having greater cardinality than the onespresented in [1].The paper is organized as follows. In Section 2 we give some basic definitions and notationrelated to the notions of bifix-free word and cross-bifix-free set. In Section 3 we propose a ∗ Dipartimento di Sistemi e Informatica, Universit`a degli Studi di Firenze, Viale G.B. Morgagni 65, 50134Firenze, Italy. [email protected], [email protected], [email protected] n related tothe parity of n . We are not able to say if such cross-bifix-free sets have maximal cardinality onthe set of bifix-free binary words of fixed length n or not. Let A be a finite, non-empty set called alphabet . The elements of A are called letters . A(finite) sequence of letters in A is called (finite) word . Let A ∗ denote the monoid of all finitewords over A where ε denotes the empty word and A + = A ∗ \ ε . Let ω be a word in A ∗ , then | ω | indicates the length of ω and | ω | a denotes the number of occurrences of a in ω , being a ∈ A .Let ω = uv then u is called prefix of ω and v is called suffix of ω . A bifix of ω is a subsequenceof ω that is both its prefix and suffix.A word ω of A + is said to be bifix-free or unbordered [7, 11] if and only if no strict prefix of ω is also a suffix of ω . Therefore, ω is bifix-free if and only if ω = uwu , being u any necessarilynon-empty word and w any word. Obviously, a necessary condition for ω to be bifix-free is thatthe first and the last letters of ω must be different. Example 2.1
In the monoid { , } ∗ , the word of length n = 9 is bifix-free, while theword contains two bifixes, and . Let BF q ( n ) denote the set of all bifix-free words of length n over an alphabet of fixed size q . The following formula for the cardinality of BF q ( n ), denoted by | BF q ( n ) | , is well-known [11]. | BF q (1) | = q | BF q (2 n + 1) | = q | BF q (2 n ) || BF q (2 n ) | = q | BF q (2 n − | − | BF q ( n ) | (2.1)The number sequences related to this recurrence can be found in Sloane’s database of integersequences [12]: sequences A003000 ( q = 2), A019308 ( q = 3) and A019309 ( q = 4).Table 2.1 lists the set BF ( n ), 2 ≤ n ≤
6, the last row reports the cardinality of each set.n=2 n=3 n=4 n=5 n=610 01 100 001 1000 0001 10000 00001 100000 000001110 011 1100 0011 10100 00101 101000 0001011110 0111 11000 00011 101100 00110111100 00111 110000 00001111010 01011 110100 00101111110 01111 111000 000111111100 001111110010 010011111010 010111111110 0111112 4 6 12 20Table 2.1: The set BF ( n ), 2 ≤ n ≤ q > n > ω, ω ′ ∈ BF q ( n ) are said to be cross-bifix-free if and only if no strict prefix of ω is also a suffix of ω ′ and vice-versa. Example 2.2
The binary words and in BF (9) are cross-bifix-free, whilethe binary words and in BF (9) have the cross-bifix . A subset of BF q ( n ) is said to be cross-bifix-free set if and only if for each ω, ω ′ , with ω = ω ′ ,in this set, ω and ω ′ are cross-bifix-free. This set is said to be non-expandable on BF q ( n ) if andonly if the set obtained by adding any other word is not a cross-bifix-free set. A non-expandablecross-bifix-free set on BF q ( n ) having maximal cardinality is called maximal cross-bifix-free set on BF q ( n ).Each word ω ∈ BF ( n ) can be naturally represented as a lattice path on the Cartesianplane, by associating a rise step , defined by (1 ,
1) and denoted by x , to each 1’s in BF ( n ), anda fall step , defined by (1 , −
1) and denoted by x , to each 0’s in BF ( n ), running from (0 ,
0) to( n, h ), − n < h < n .From now on, we will refer interchangeably to words or their graphical representations onthe Cartesian plane, that is paths.The definition of bifix-free and cross-bifix-free can be easily extended to paths. Figure 2.1shows the two paths corresponding to the cross-bifix-free words of Example 2.2. Figure 2.1:
Two paths in BF (9) which are cross-bifix-free A lattice path on the Cartesian plane using the steps (1 ,
1) and (1 , −
1) and running from(0 ,
0) to (2 m, m ≥
0, is said to be
Grand-Dyck or Binomial path (see [5] for furtherdetails). A
Dyck path is a sequence of rise step and fall steps running from (0 ,
0) to (2 m,
0) andremaining weakly above the x -axis (see Figure 2.2). The number of 2 m -length Dyck paths isthe m th Catalan number C m = 1 / ( m + 1) (cid:0) mm (cid:1) , see [13] for further details. m=3m=1m=2 Figure 2.2:
The 2 m -length Dyck paths, 1 ≤ m ≤ In this paper, we are interested in investigating a possible non-expandable cross-bifix-freeset, that is the set
CBF S ( n ) of cross-bifix-free words of fixed length n > { , } ∗ . In order to do so, we focus on the set ˆ BF ( n ) of bifix-free binary words of fixed length n having 1 as the first letter and 0 as last letter or equivalently the set of bifix-free lattice paths on3he Cartesian plane using the steps (1 ,
1) and (1 , − ,
0) to ( n, h ), − n < h < n ,beginning with a rise step and ending with a fall step. Of course ˇ BF ( n ) = BF ( n ) \ ˆ BF ( n ) isobtained by switching rise and fall steps.Let ˆ BF h ( n ) denote the set of the paths in ˆ BF ( n ) having h as the ordinate of their endpoint, − n < h < n . C BF S ( n ) In order to prove that
CBF S ( n ) is a non-expandable cross-bifix-free set on BF ( n ) wehave to distinguish the following two cases depending on the parity of n . CBF S (2 m + 1) Let
CBF S (2 m + 1) = { xα : α ∈ D m } that is the set of paths beginning with a rise steplinked to a 2 m -length Dyck path (see Figure 3.3). Note that CBF S (2 m + 1) is a subset ofˆ BF (2 m + 1), m ≥ CBFS (2m+1) = α , α D Figure 3.3:
A graphical representation of
CBF S (2 m + 1), with m ≥ Of course | CBF S (2 m + 1) | = C m , being C m the m th Catalan number, m ≥ CBF S (7), with | CBF S (7) | = C = 5. CBFS (7) ,,,, = Figure 3.4:
A graphical representation of
CBF S (7) Proposition 3.1
The set
CBF S (2 m + 1) is a cross-bifix-free set on BF (2 m + 1) , m ≥ .Proof. The proof consists of two distinguished steps. The first one proves that each ω ∈ CBF S (2 m + 1) is bifix-free and the second one proves that CBF S (2 m + 1) is a cross-bifix-freeset. Each ω ∈ CBF S (2 m + 1) can be written as ω = vwu , being v, u any necessarily non-emptyword while w can also be an empty word. For each prefix v of ω we have | v | > | v | and for eachsuffix u of ω we have | u | ≤ | u | . Therefore v = u , ∀ v, u ∈ ω so ω is bifix-free.The proof that, for each ω, ω ′ ∈ CBF S (2 m + 1) then ω and ω ′ are cross-bifix-free, followsthe logical steps described above. (cid:4) Proposition 3.2
The set
CBF S (2 m +1) is a non-expandable cross-bifix-free set on BF (2 m +1) , m ≥ . roof. It is sufficient to prove that the set
CBF S (2 m + 1) is a non-expandable cross-bifix-freeset on ˆ BF (2 m + 1), as each ω ∈ CBF S (2 m + 1) and ϕ ∈ ˇ BF (2 m + 1) match on the lastletter of ω and the first one of ϕ at least.Let m ≥ CBF S (2 m + 1) is a non-expandable cross-bifix-freeset on ˆ BF h (2 m + 1) by distinguishing h > h < • h > γ in ˆ BF h (2 m + 1) \ CBF S (2 m + 1) can be written as γ = φxα xα x . . . xα r (seeFigure 3.5, where n = 2 m + 1), being φ a Grand-Dyck path beginning with a rise step, x a rise step, α l Dyck paths, 1 ≤ l ≤ r −
1, and α r a necessarily non-empty Dyck path.Therefore, we can find paths in CBF S (2 m + 1) having a prefix which matches with asuffix of γ . It is sufficient to consider the path ω = xα r α s , being α s a Dyck path ofappropriate length. φ r ααα Figure 3.5:
A graphical representation of a path γ in ˆ BF h ( n ) , h > • h < γ in ˆ BF h (2 m + 1) can be written as γ = α r xα r − x . . . xα xφ (see Figure 3.6, where n = 2 m + 1), being α r a necessarily non-empty Dyck path, x a fall step, α l Dyck paths,1 ≤ l ≤ r −
1, and φ a Grand-Dyck path. Therefore, we can find paths in CBF S (2 m + 1)having a suffix which matches with a prefix of γ . It is sufficient to consider the path ω = xα s α r , being α s a Dyck path of appropriate length. φααα r r− Figure 3.6:
A graphical representation of a path γ in ˆ BF h ( n ) , h < (cid:4) CBF S (2 m + 2) In this case we have to distinguish two further subcases depending on the parity of m > m is an even number then CBF S (2 m +2) = { αxβx : α ∈ D i , β ∈ D m − i ) , ≤ i ≤ m } ,that is the set of paths consisting of the following consecutive sub-paths: a 2 i -length Dyck path,5 rise step, a 2( m − i )-length Dyck path, a fall step, where 0 ≤ i ≤ m (see Figure 3.7). Notethat CBF S (2 m + 2) is a subset of ˆ BF (2 m + 2), for any even number m > ≤ , β D β CBFS (2m+2) = α , α D ≤ Figure 3.7:
A graphical representation of
CBF S (2 m + 2), for any even number m > Of course | CBF S (2 m + 2) | = P m/ i =0 C i C m − i , C m is the m th Catalan Number, for any evennumber m >
1. Figure 3.8 shows the set
CBF S (10), with | CBF S (10) | = C + C C + C C =23. ,, ,, =CBFS (10) , , , , ,, , , , ,, , , ,,, , , Figure 3.8:
A graphical representation of
CBF S (10) Proposition 3.3
The set
CBF S (2 m + 2) is a cross-bifix-free set on BF (2 m + 2) , for anyeven number m > .Proof. The proof consists of two distinguished steps. The first one proves that each ω ∈ CBF S (2 m + 2) is bifix-free and the second one proves that CBF S (2 m + 2) is a cross-bifix-freeset. Each ω ∈ CBF S (2 m + 2) can be written as ω = vwu , being v, u any necessarily non-emptyword while w can also be an empty word. Let m > i = 0 and in the second one 0 < i ≤ m .If i = 0 then ω ∈ { xβx : β ∈ D m } , and for each prefix v of ω we have | v | > | v | and foreach suffix u of ω we have | u | < | u | . Therefore v = u , ∀ v, u ∈ ω and ω is bifix-free.6therwise, ω ∈ { αxβx : α ∈ D i , β ∈ D m − i ) , < i ≤ m } , then for each prefix v of ω we have | v | ≥ | v | and for each suffix u of ω we have | u | ≤ | u | . If | v | > | v | then v = u , ∀ v, u ∈ ω and therefore ω is bifix-free. Let i be fixed, if | v | = | v | then the path v is a 2 k -lengthDyck path, 1 ≤ k ≤ i . In this case, both u = µx , where µ is any suffix of β , and u = µ ′ xβx ,where µ ′ is any suffix of α \ v . If u = µx then | u | < | u | , therefore v = u , ∀ v, u ∈ ω and therefore ω is bifix-free. If u = µ ′ xβx then v does not match with xβx , therefore v = u , ∀ v, u ∈ ω andtherefore ω is bifix-free.The proof that, for each ω, ω ′ ∈ CBF S (2 m + 2) then ω and ω ′ are cross-bifix-free, followsthe logical steps described above. (cid:4) Proposition 3.4
The set
CBF S (2 m +2) is a non-expandable cross-bifix-free set on BF (2 m +2) , for any even number m > .Proof. It is sufficient to prove that the set
CBF S (2 m + 2) is a non-expandable cross-bifix-freeset on ˆ BF (2 m + 2), as each ω ∈ CBF S (2 m + 2) and ϕ ∈ ˇ BF (2 m + 2) match on the lastletter of ω and the first one of ϕ at least.Let m > CBF S (2 m + 2) is a non-expandable cross-bifix-free set on ˆ BF h (2 m + 2), h >
0, in the second one we prove that
CBF S (2 m + 2) is a non-expandable cross-bifix-free seton ˆ BF h (2 m + 2), h <
0, and in the last one we prove that
CBF S (2 m + 2) is a non-expandablecross-bifix-free set on ˆ BF (2 m + 2). • h > γ in ˆ BF h (2 m + 2) can be written as γ = φxα xα x . . . xα r (see Figure 3.5, where n = 2 m + 2), being φ a Grand-Dyck path beginning with a rise step, x a rise step, α l Dyckpaths, 1 ≤ l ≤ r −
1, and α r a necessarily non-empty Dyck path. Therefore, we can findpaths in CBF S (2 m + 2) having a prefix which matches with a suffix of γ . It is sufficientto consider the path ω = xα r α s x , being α s a Dyck path of appropriate length. • h < γ in ˆ BF h (2 m + 2) can be written as γ = α r xα r − x . . . xα xφ (see Figure 3.6, where n = 2 m + 2), being α r a necessarily non-empty Dyck path, x a fall step, α l Dyck paths,1 ≤ l ≤ r −
1, and φ a Grand-Dyck path. Therefore, we can find paths in CBF S (2 m + 2)having a suffix which matches with a prefix of γ . It is sufficient to consider the path ω = xα s α r x , being α s a Dyck path of appropriate length. • h = 0 : a path γ in ˆ BF (2 m + 2) \ CBF S (2 m + 2) either never falls below the x -axis or crosses the x -axis. In the first case, it can be written as γ = α xβ x , where α is a necessarily non-empty 2 k -length Dyck path and β is a 2( m − k )-length Dyck path, with m + 1 ≤ k ≤ m ,see Figure 3.9 a). Therefore, we can find paths in CBF S (2 m + 2) having a prefix whichmatches with a suffix of γ . It is sufficient to consider the path ω = xβ xxβx , since xβ x ∈ D i being i = m − k + 1.If a path γ in ˆ BF (2 m + 2) \ CBF S (2 m + 2) crosses the x -axis then it can be writtenas γ = α φ where α is a necessarily non-empty 2 k -length Dyck path, 1 ≤ k ≤ m , and φ is a necessarily non-empty Grand-Dyck beginning with a fall step, see Figure 3.9 b).Therefore, we can find paths in CBF S (2 m + 2) having a suffix which matches with aprefix of γ . It is sufficient to consider the path ω = xα s α x , being α s a Dyck path ofappropriate length. 7 α b)a) φ αβ Figure 3.9:
The two possible configurations for a path γ in ˆ BF (2 m + 2) \ CBF S (2 m + 2), for any evennumber m > (cid:4) If m is an odd number then CBF S (2 m + 2) = { αxβx : α ∈ D i , β ∈ D m − i ) , ≤ i ≤ m +12 }\{ xα ′ xxβ ′ x : α ′ , β ′ ∈ D m − } , that is the set of paths consisting of the following consecutivesub-paths: a 2 i -length Dyck path, a rise step, a 2( m − i )-length Dych path, a fall step, where0 ≤ i ≤ m +12 , and excluding those consisting of the following consecutive sub-paths: a rise step,a ( m − m − α ′ = β ′ then the excluded paths are not bifix-free, otherwise if α ′ = β ′ thenthe excluded paths match with the paths { αxβx : α ∈ D m +1 , β ∈ D m − } in CBF S (2 m + 2).Note that CBF S (2 m + 2) is a subset of ˆ BF (2 m + 2), for any odd number m ≥ ≤ =(2m+2)CBFS m−1 ’’ D β , α , , ’’ β α m+12i0 , β D β α α D ≤ Figure 3.10:
A graphical representation of
CBF S (2 m + 2), for any odd number m ≥ Of course | CBF S (2 m + 2) | = ( P m +12 i =0 C i C m − i ) − ( C m − ) , C m is the m th Catalan Number,for any odd number m ≥
1. Figure 3.11 shows the set
CBF S (8), with | CBF S (8) | = ( C + C C + C C ) − ( C ) = 8. Proposition 3.5
The set
CBF S (2 m + 2) is a cross-bifix-free set on BF (2 m + 2) , for any oddnumber m ≥ . Proposition 3.6
The set
CBF S (2 m +2) is a non-expandable cross-bifix-free set on BF (2 m +2) , for any odd number m ≥ . The proof of Proposition 3.5 follows the logical steps as far Proposition 3.3 and the proofof Proposition 3.6 follows the logical steps as far Proposition 3.4.8 =CBFS (8) ,,, ,,,,, Figure 3.11:
A graphical representation of the set
CBF S (8) Therefore, the presented constructing method gives sets
CBF S ( n ) of cross-bifix-free bi-nary words, of fixed length n , having cardinality 1 , , , , , , , , , , , ,
429 for n = 3 , , , , , , , , , , , ,
15 respectively.
In this paper, we introduce a general constructing method for the sets of cross-bifix-freebinary words of fixed length n based upon the study of lattice paths on the Cartesian plane.This approach enables us to obtain the cross-bifix-free set CBF S ( n ) having greater cardinalitythan the ones presented in [1] based upon the kernel method.Moreover, we prove that CBF S ( n ) is a non-expandable cross-bifix-free set on BF ( n ), i.e. CBF S ( n ) ∪ γ is not a cross-bifix-free set on BF ( n ), for any γ ∈ BF ( n ) \ CBF S ( n ). Thenon-expandable property is obviously a necessary condition to obtain a maximal cross-bifix-freeset on BF ( n ), anyway we are not able to find and prove a sufficient condition.Further studies to prove that could investigate both the nontrivial subsets of BF ( n ) inwhich CBF S ( n ) is a maximal cross-bifix-free set, and the study of other non-expandable cross-bifix-free sets on BF ( n ).Another approach to reach the goal could be to find a different characterization of bifix-freewords which could be obtained through bijective methods between particular bifix-free subsetsand other well-known discrete structures.Successive studies should take into consideration the general study of cross-bifix-free sets on BF q ( n ), where q is grater than 2. References [1] D. Bajic. On Construction of Cross-Bifix-Free Kernel Sets. 2nd MCM COST 2100,TD(07)237, February 2007, Lisbon, Portugal.[2] D. Bajic, D. Drajic. Duration of search for a fixed pattern in random data: Distributionfunction and variance.
Electronics letters , Vol. 31, No. 8, 631-632, 1995.[3] D. Bajic, J. Stojanovic. Distributed Sequences and Search Process.
IEEE InternationalConference on Communications ICC2004 , Paris, 514-518, June 2004.94] R. H. Barker. Group synchronizing of binary digital systems.
Communication theory , W.jackson, Ed. London, U.K.: Butterworth, 273-287, 1953.[5] L. Comtet. Advanced Combinatorics: The Art of Finite and Infinite Expansions, D. ReidelPublishing Company, 1974.[6] E. N. Gilbert. Synchronization of Binary Messages.
IRE Trans. Inform. Theory , vol. IT-6,470-477, 1960.[7] T. Harju, D. Nowotka. Counting bordered and primitive words with a fixed weight.
Theo-retical Computer Science : 340 (2005) 273-279.[8] M. Lothaire. Combinatorics on Words. Encyclopedia of Mathematics and its Applications,Vol. 17, Addison-Wesley Publishing Co., Reading, MA, 1983.[9] J. L. Massey. Optimun frame synchronization.
IEEE Transactions on Commununications ,vol. COM-20, 115-119, February 1972.[10] P. T. Nielsen. On the Expected Duration of a Search for a Fixed Pattern in Random Data.
IEEE Trans. Inform. Theory , vol. IT-29, 702-704, September 1973.[11] P. T. Nielsen. A Note on Bifix-Free Sequences.
IEEE Trans. Inform. Theory , vol. IT-29,704-706, September 1973.[12] N. J. A. Sloane. On-line encyclopedia of integer sequences, http://oeis.org/.[13] R. P. Stanley. Enumerative Combinatorics, volume 2. Cambridge University Press, Cam-bridge, 1999.[14] A. J. de Lind van Wijngaarden, T. J. Willink. Frame Synchronization Using DistributedSequences.