On the density of languages representing finite set partitions

Transcription

1 On the density of languages representing finite set partitions Nelma Moreira Rogério Reis Tehnial Report Series: DCC Departamento de Ciênia de Computadores Fauldade de Ciênias & Laboratório de Inteligênia Artifiial e Ciênia de Computadores Universidade do Porto Rua do Campo Alegre, Porto, Portugal Tel: Fax:

2 On the density of languages representing finite set partitions Nelma Moreira Rogério Reis DCC-FC & LIACC, Universidade do Porto R. do Campo Alegre 823, 450 Porto, Portugal August 2004 Abstrat We present a family of regular languages representing partitions of a set N n = {,..., n} in less or equal parts. We determine expliit formulas for the density of those languages and their relationship with well known integer sequenes involving Stirling numbers of seond kind. We also determine their it frequeny. This work was motivated by omputational representations of the onfigurations of some numerial games. The languages L Consider a game where natural numbers are to be plaed, by inreasing order, in a fixed number of olumns, subjet to some speifi onstraints. In these games olumn s order is irrelevant. Game onfigurations an be seen as sequenes of olumns where the suessive integers an be plaed. For instane, the string 23 stands for a onfiguration where, 2, 4 were plaed in the first olumn, 3 was plaed in the seond and 5 was plaed in the third. Beause olumn order is irrelevant, and to have a unique representation for eah onfiguration, it is illegal to plae an integer in the k olumn if the (k )th is still empty, for any k. Blanhard and al.[bhr04] and Reis and al.[rmp04], used this kind of representation for the possible onfigurations of sum-free games. Given olumns, let N = {,..., }. We are interested in studying the set of game onfigurations as strings in (N ), i.e, in the set of finite sequenes of elements of N. We haraterise game onfigurations by the following language L (N ) : L = {a a 2 a k (N ) i N k, a i max{a,..., a i } + }. For = 4, there are only 5 strings in L of length 4, instead of the total possible 256 in (N 4 ) 4 : We are going to show that L are regular languages. Given a finite set Σ, a regular expression (r.e.) α over Σ represents a (regular) language L(α) Σ and is indutively 2

3 defined by: is a r.e and L( ) =, ɛ (empty string) is a r.e and L(ɛ) = {ɛ}; a Σ is a r.e and L(a) = {a}; if α and α 2 are r.e., α + α 2, α α 2 and α are r.e., respetively with L(α + α 2 ) = L(α ) L(α 2 ), L(α α 2 ) = L(α )L(α 2 ) and L(α ) = L(α ) [HMU00]. A regular expression α is unambiguous if for eah w L(α) there is only one path through α that mathes w. Theorem. For all, L is a regular language. Proof. For =, we have L = L( ). We define by indution a family of regular expressions: α =, () α = α + j( + + j), (2) j= where stand for the onatenation operator. It is trivial to see that, α = i= j= where stands for the union operator. For instane, α 4 is i j( + + j), (3) + 2( + 2) + 2( + 2) 3( ) + 2( + 2) 3( ) 4( ). For any, we prove that L = L(α ). L L(α ): If x L(α ) it is obvious that x L L L(α ): By indution on the length of x L : If x = than x L(α ) L(α ). Suppose that for any string x of length n, x L(α ). Let x = n + and x = ya, where a N and y L(α ). Let = max{a i a i y}. If =, obviously x L(α ). If <, then y L(α ), and x L(α +) L(α ). 2 Counting the strings of L The density of a language L over Σ, ρ L (n), is the number of strings of length n that are in L, i.e, In partiular, the density of L is ρ L (n) = L Σ n. ρ L (n) = L N n. We use generating funtions to determine a losed form for ρ L (n). Reall that, a (ordinary) generating funtion for a sequene {a n } is a formal series [GKP94] G(z) = a n z n. i=0 If A(z) and B(z) are generating funtions for the density funtions of the languages represented by unambiguous regular expressions A and B, and A + B, AB and A are also unambiguous r.e., we have that A(z) + B(z), A(z)B(z) and, are the A(z) 3

4 generating funtions for the density funtions of the orresponding languages (see [SF96], page 378). As α are unambiguous regular expressions, from (3), we obtain the following generating funtion for {ρ L (n)}: Notie that T (z) = i i= j= z ( jz) = z i i= z i i j= ( jz). S i (z) = i j= ( jz) are the generating funtions for the Stirling numbers of seond kind S(n, i) = i ( ) i ( ) j (i j) n, i! j j=0 i.e, the number of ways of partitioning a set of n elements into i nonempty sets [GKP94]. So, a losed form for the density of L, ρ L (n), is given by ρ L (n) = S(n, i). (4) i= In Table 2 we present ρ L (n), for =..8 and n =..3. For some sequenes, we also indiate the orrespondent number in Sloane s On Line Enylopedia of Integer Sequenes [Slo03]. The losed forms were alulated using the Maple omputer algebra system [He03]. Moreover we an express (4) as a generi linear ombination of n powers of i, for i N. Let S j (n, i) denote the jth term in the summation of a Stirling number S(n, i), i.e, S j (n, i) = ( ) i i! ( )j (i j) n. j Lemma. For all n, i 0, Proof. S (n, i + ) = S 0 (n, i) = S (n, i + ). (5) (i + )! ( ) = ( ) i! in = S 0 (n, i). ( i + ) i n Applying (5) in the summation (4) of ρ L (n), eah term S(n, i) simplifies the subterm S (n, i) with the subterm S 0 (n, i ) of S(n, i), for i 2. So, we have ρ L (n) = S 0 (n, ), ρ L (n) = S 0 (n, 2), ρ L (n) = S 0 (n, ) + = n i! + i=3 i S j (n, i), for > 2 i=3 S j (n, i), for > 2, 4

5 ρ L (n) OISE ,,,,,,,,,,,,,, n, 2, 4, 8, 6, 32, 64, 28, 256, 52, 024, 2048, 4096, 892, n + 2, 2, 5, 4, 4, 22, 365, 094, 328, 9842, 29525, 88574, 26572,... A n + 4 2n + 3, 2, 5, 5, 5, 87, 75, 2795, 05, 43947, 75275, , ,... A n + 2 3n + 6 2n + 3 8, 2, 5, 5, 52, 202, 855, 3845, 8002, 86472, , , ,... A n n + 8 3n n + 30, 2, 5, 5, 52, 203, 876, 4, 20648, 09299, 60492, , ,... A n n n + 6 3n n , 2, 5, 5, 52, 203, 877, 439, 20, 579, , , n n n n n n , 2, 5, 5, 52, 203, 877, 440, 246, 5929, , , ,... Table : Density funtions of L, for =..8. 5

6 and, if the sums are rearranged suh as that i = k + j, we have where ρ L (n) = n 2! + k S j (n, k + j) = Replaing (8) into (6) we get ρ L (n) = n 2! + k= S j (n, k + j), (6) k n ( ) k + j (k + j)! ( )j j (7) = kn k!j! ( )j. (8) k= k n k! k ( ) j. (9) j! In equation (9), the oeffiients of k n, k, an be alulated using the following reurrene relation: And, we obtain Theorem 2. For all, γ = ; γ = γ + ( ), for > ; ( )! γk = γ k, for > and 2 k. k ρ L (n) = γkk n. (0) Finally, the frequeny of strings in L an be ompared with that of (N ). Notie that ρ (N) (n) = n. So, as n ( k ) n = 0, we have Theorem 3. For all, n k= ρ L (n) ρ (N) (n) =!. 3 A bijetion between strings of L and partitions of finite sets The onnetion between the density of L and Stirling numbers of seond kind is not aidental. Eah string of L with length n orresponds to a partition of N n with no more than parts. This orrespondene an be defined as follows. Let a a 2 a n be a string of L. This string orresponds to the partition {A j } j N of N n with = max{a,..., a n }, suh that for eah i N n, i A j. For example, the string 23 orresponds to the partition {{, 2}, {3}, {4}} of N 4 into 3 parts. This defines a bijetion. That eah string orresponds to a unique partition is obvious. Given a partition {A j } j N of N n with, we an onstrut the string b b n, suh that for i N n, b i = j if i A j. For the partition {{, 2}, {3}, {4}} we obtain 23. 6

7 4 Counting the strings of L of length equal or less than a ertain value Although strings of L of arbitrary length represent game onfigurations, for omputational reasons we onsider all game onfigurations with the same length padding with zeros the positions of integers not yet in one of the olumns. In this way, we obtain the languages L 0 = L {0 }. So, to determine the number of strings of length equal or less than n that are in L, is tantamount to determine the density of L {0 }, i.e ρ L 0 (n) = L 0 ({0} N ) n. As seen in Setion 2, and beause L {0 } = L(α 0 ), the generating funtion T (z) of ρ L 0 (n), an be obtained using the generating funtion for ρ L (n), T (z), and one for ρ {0} (n). For the latter we have just So, the generating funtion for {ρ L 0 (n)} is z. T (z) = T (z) z = z i ( z) i j= ( jz) i= and a losed form for ρ L 0 (n) is (as exepted) ρ L 0 (n) = n m= i= where m starts in beause S(m, i) = 0, for i > m. Using expression (9) in () we have S(m, i), () ρ L 0 (n) = n, ρ L2{0} (n) = 2n, ρ L 0 (n) = n m 2 +! m= k= = n+ ( )! + n = k ( ) j, for > 2 j! ( ) j 2 k n+ k k + ( ) j j! (k )k! j! ( ) j 2 k n k + ( ) j. j! (k )(k )! j! k m k! n ( )( )! + n If we use the equation (0) we obtain Theorem 4. For all, ρ L 0 (n) = nγ + The data strutures are arrays of fixed length. γk (kn+ k). k 7

8 Proof. ρ L 0 (n) = = n m= k= γ k km n γk k m k= m= = nγ + γk (kn+ k). k In the Table 2 we present the values of ρ L 0 (n), for =..6 and n =..3. ρ L 0 (n) OISE n, 2, 3, 4, 5, 6, 7, 8, 9, 0,, 2, 3, n , 3, 7, 5, 3, 63, 27, 255, 5, 023, 2047, 4095, 89, A n + 2 n 4, 3, 8, 22, 63, 85, 550, 644, 4925, 4767, 44292, 32866, ,... A n + 2 2n + 3 n 5 9, 3, 8, 23, 74, 26, 976, 377, 4822, 58769, , 9349, , n + 8 3n + 3 2n n 5 32, 3, 8, 23, 75, 277, 32, 4977, 22979, 0945, 53456, 26093, , n n + 2 3n n + 30 n , 3, 8, 23, 75, 278, 54, 5265, 2593, 3522, , 43983, , n n n n n n 3 64, 3, 8, 23, 75, 278, 54, 5265, 2593, 3522, , 43983, , n n n n n n n , 3, 8, 23, 75, 278, 55, 5295, 2644, 42370, 89729, , ,... Table 2: Density funtions of L 0, for =..8. Finally, we determine the frequeny of strings in L 0 in (N ) {0}, for >. Notie that ρ (N) {0} (n) = n+. as it is a sum of the first n terms of a geometri progression of ratio. 8

9 We have, Theorem 5. For all, Proof. n ρ L 0 (n) ρ (N) {0} (n) = n n ρ L 0 (n) ρ (N) {0} (n) =!. (n+ )( ) n( ) ( ) j ( )!( n+ + ) n+ j! 2 + (k n+ k k)( ) ( ) j n (k )k!( n+ ) j! = ( )! n n+ n n+ ( ) j ( ) n( ) + j! n n+ =!. 2 + ( ) k ( ) j (k )(k )! j! n ( ( k )n n+ ) n+ 5 Conlusion In this note we presented a family of regular languages representing finite set partitions and studied their densities. Although it is well known that the number of partitions of a set of n elements into (no more than) nonempty sets is given by (the sum of) Stirling numbers of seond kind, S(n, ), we determined expliit formulas for the number of partitions of a set of n elements with no more than. We also determined the it density of those languages. Aknowledgement We thank Miguel Filgueiras for helpful omments. Referenes [BHR04] Peter Blanhard, Frank Harary, and Rogério Reis. Partitions into sum-free sets. To appear, [GKP94] Roland Graham, Donald Knuth, and Oren Patashnik. Conrete Mathematis. AW, 994. [He03] André Hek. Introdution to Maple. SV, 3rd edition, [HMU00] John E. Hoproft, Rajeev Motwani, and Jeffrey D. Ullman. Introdution to Automata Theory, Languages and Computation. Addison Wesley, 2nd edition,

10 [RMP04] Rogério Reis, Nelma Moreira, and João Pedro Pedroso. Eduated brute-fore to get h(4). Tehnial Report DCC , DCC-FC& LIACC, Universidade do Porto, July [SF96] Robert Sedgewik and Philippe Flajolet. Analysis of Algorithms. AW, 996. [Slo03] N.J.A. Sloane. The On-line Enylopedia of Integer Sequenes, njas/sequenes Mathematis Subjet Classifiation: Primary 05A5; Seondary 05A8, B73, 68Q45. Keywords: Partions of sets, Stirling numbers, regular languages, enumeration (Conerned with sequenes A00705 A00758 A A A A ) 0