Encoding of Partition Set Using Sub-exceeding Function

Transcription

1 International Journal of Contemporary Mathematical Sciences Vol 3, 208, no 2, HIKARI Ltd, wwwm-hiaricom Encoding of Partition Set Using Sub-exceeding Function Luc Rabefihavanana Department of Mathematics and Computer Faculty of Science, University of Antananarivo, Madagascar Copyright c 208 Luc Rabefihavanana This article is distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original wor is properly cited Abstract We introduce in this paper a new method to encode all partitions of any set of n objects using sub-exceeding function So, let n and be two positive integers such that n and let too Ω a set of cardinal n The study of the subset called G n of F n (set of sub-exceeding functions on [n]) leads us to the construction of a bijection between this set and the set of partitions of Ω with -subsets Once the bijection is built, we can thus code all the partitions of the set Ω by sub-exceeding functions and respectively to find an algorithm for the decoding Mathematics Subject Classification: 05A05; 05A8; 05A9; 8C35 Keywords: Sub-exceeding function, Statistical Permutation, Partition set Introduction and Notation Introduction This paper presents the fruit of our results in the article entitled: "Part of a set and sub-exceeding function (Coding and Decoding) [4]" which presents, for a positive integer n, the bijection between the subset H n of F n and the set P(Ω) Here F n represents the set of sub-exceeding functions on [n] and P(Ω) present all parts of the set Ω where Card(Ω) = n

2 64 Luc Rabefihavanana The study of the sub-exceeding functions does not stop to present us surprising results and which we push to enrich our research Let now n and be two positive integers such that n From the bijection between the sets H n and P(Ω), we can encode any subset of Ω with one and only one sub-exceeding function in H n So, from this result, a question arises if it is also possible to encode any partition of Ω which have -subset by one and only one sub-exceeding function in F n In terms of coding, this maes possible to speed up the processing time for any given information sequence To give an answer for this question, we will analyze a set called G n which is an extension of the set H n Our goal for this search will be to encode all partitions of Ω where Card(Ω) = n That is to say, for any partition of Ω into -subsets, there is one and only one sub-exceeding function in G n who represents it, and conversely for a sub-exceeding function in G n, there exists one and only one corresponding subdivision of Ω into subsets Notation Let us recall some notations that allow us to facilitate the reading of this paper [n] : the set {; 2; ; n}, 2 N : the se of positive integer, 3 Ω : a set of n elements such that Ω = {ω ; ω 2 ; ; ω n }, 4 P (Ω) : the set of the parts of Ω which have elements, 5 P (Ω) : the set of partitions of Ω which have any subsets, 6 S n : the set of permutations of [n] 2 Preliminaries 2 The set F n and its properties Definition 2 Let n be a positive integer such that n > 0 and let too f a map from [n] to [n] This function f is called sub-exceeding on [n] if and only if f(i) i for all i in [n] In the whole continuation we will represent f by the word of n letters f()f(2) f(n) Thus we describe f by his images ie f = f()f(2) f(n) Moreover, we denote by F n the set of this sub-exceeding function So F n = {f : [n] [n] f(i) i, i [n]} (2)

3 Encoding of partition set using sub-exceeding function 65 Following this definition, the study of this set F n presents us an important result introduced in the theorem as follow Theorem 2 (Roberto Mantaci and Fanja Raotondrajao [7]) Let φ be the map from F n to S n defined by φ : F n S n f φ(f) = (n f(n))(n f(n ))(2 f(2))( f()) = σ f (22) So, the map φ is bijective Here (i f(i)) is the permutation which transforms i into f(i) and f(i) into i Similarly (i f(i))(j f(j)) is the compose of two permutations (i f(i)) and (j f(j)) So σ f = (n f(n))(2 f(2))( f()) = (n f(n)) (n f(n ) ( f()) (23) Example 22 For n = 3 and f = 22, we have φ(f) = (3 2)(2)() = 32 where the permutation (i, i) is simplified by (i) 22 The set H n and its properties Definition 22 For a positive integers n and such that n, we denote by H n the sub-set of F n such that Hn = {f F n f(i) f(i + ) i n and Im(f) = []} (24) Here, H n is the set of all sub-exceeding function of F n with a quasiincreasing sequence of image formed by all elements of [] Example 23 Tae n = 4 and = 3 We have here f = 23 H4 3 because (f(i)) i [4] is an quasi-increasing sequence and all of the elements of [3] are there Now if we tae f = 33, although the sequence (f(i)) i [4] is quasiincreasing, f = 33 / H4 3 because Im(f) [3] (without 2 among the f(i)) From this definition 22, we denote by H n the set as Example 24 Tae n =, 2, 3 H n = n Hn (25) = H = H 2 = et H 2 2 = 2 H 3 =, H 2 3 = 2, 22 et H 3 3 = 23

4 66 Luc Rabefihavanana Theorem 25 (Luc RABEFIHAVANANA [4]) Let n and be two integers such that n For =, we see always that H n is the singleton set such that H n = {f = n-fois } 2 For = n, we see also that H n n is the singleton set such that H n n = {f = 23(n )(n)} 3 For all integers such that < < n, we can construct all sub-exceeding functions of H n as follow: (a) Tae all elements of H n and add the integer at the end, (b) Tae also all elements of H n and add the integer at the end So, for the best presentation, we write this construction as H n = { H n } { H n } Here, ( ) means that we add the integer at the and of all elements of ( ) Proposition 2 (Luc RABEFIHAVANANA [4]) Let n and be two integers such that n So, we have the following relations Card Hn = Card Hn + Card Hn ( ) n 2 Card Hn = 3 Card H n = 2 n From this proposition, we have the cardinal table of H n as n

5 Encoding of partition set using sub-exceeding function 67 This table is a modified Pascal triangle More precisely, it is shifted ie instead of N, we have N Moreover, the values of the i th line in this table is the value of (i ) th line of the Pascal triangle Definition 23 Let n and be two integers such that n For a sub-exceeding function f in H n and an integer i such that i [], we define by dpos(i) the last position of i in f Example 26 In H 4 9, tae f = We have here dpos() = 3, dpos(2) = 5 and dpos(3) = 8 Theorem 27 (Luc RABEFIHAVANANA [4]) Let n and be two integers such that n Now let ψ be a map from H n to P (Ω) such that ψ : H n P (Ω) f ψ(f) = { ω dpos() ; ω dpos(2) ; ; ω dpos( ) } (26) where Ω = {ω ; ω 2 ; ; ω n } (27) Then, the mapping ψ is bijective From this theorem, H n represent the set of the parts of Ω to elements Moreover, The map ψ is also bijective from H n to P(Ω) As results, given an arbitrary subset of Ω, we can encode it by using a subexceeding function in H n Reciprocally, if we have a sub-exceeding function in H n, we can decode it and find the subset of Ω that it represents Now, as P(Ω) is stable by the operations, and by passing to the complement, so we present in this section the equivalent of these operations in H n Definition 24 Let f be a sub-exceeding function in H n, we define the indicator of f the set I f such that I f = {dpos(); dpos(2); ; dpos( )} (28) Definition 25 Let f and g be two sub-exceeding functions who belong respectively in Hn and Hn l We denote by (f Rab g) the sub-exceeding function h such that g) Hn m+ avec I h = I f Ig (29) h = (f Rab et m = card I h

6 68 Luc Rabefihavanana Example 28 Let f = H7 4 and g = H7 5 be two subexceeding functions We have here I f = {2; 4; 5} and I g = {; 4; 5; 6} Then, the sub-exceeding function h = ( f Rab g) has his indicator the set = {; 2; 4; 5; 6} From this indicator, we have h H7 6 such that I h h = Theorem 29 (Luc RABEFIHAVANANA [4]) Let f and g be two sub-exceeding functions in H n which respectively represents Ω f and Ω g (two subset of Ω) Then, H n is stable by Rab and (f Rab g) represents Ω f Ω g Definition 26 Let f and g be two sub-exceeding function who belong respectively in Hn and Hn l We denote by (f Rab g) the sub-exceeding function h such that g) Hn m+ avec I h = I f Ig (20) h = (f Rab et m = card I h Example 20 Let f = H7 4 and g = H7 5 be two subexceeding functions We have here I f = {2; 4; 5} and I g = {; 4; 5; 6} Then, the sub-exceeding function h = (f Rab g) has his indicator the set = {4; 5} From this indicator, we have h H7 3 such that I h h = Theorem 2 (Luc RABEFIHAVANANA [4]) Lets f and g be two sub-exceeding functions in H n which respectively represents Ω f and Ω g (two parts of Ω) Then, H n is stable by Rab and (f Rab g) represent Ω f Ω g Definition 27 Let f be the sub-exceeding function defined on Hn We denote by f the complementary sub-exceeding function of f such that f Hn n + with indicator I f = Īf Here, Ī f is the complementary set of I f in [n ] Example 22 Let f = H7 4 we have I f = {2; 4; 5} and Ī f = {; 3; 6} The complement sub-exceeding function of f denoted by f is then f = H 4 7 Theorem 23 (Luc RABEFIHAVANANA [4]) Let f be a sub-exceeding function in H n which represent Ω f So, f represents Ω f the complement of Ω f in Ω

7 Encoding of partition set using sub-exceeding function 69 3 Main result: bijection between the set G n and P (Ω) 3 The set G n and its properties Definition 3 Let n be a not zero positive integer A sub-exceeding function f on [n] satisfies the condition denoted (E) if for any integer m in [n] such that m > and m {f(i)} i [n], by posing m = f(j), there is another integer j where j < j such that f(j ) = m In another way, a sub-exceeding function f on [n] satisfies the condition (E) if and only if for any integer m (m > ) in the sequence of image (f(i)) i [n], all integers from to m must be found before m in this sequence Example 3 In F 7, let s tae two sub-exceeding functions f and g such that f = 2324 and g = 2325 Here the function f satisfies the condition (E) but g is not Taing g(7) which is equal to 5 but all integers to 4 are not found before this value (the integer 4 is not found in the sequence 232 ) Definition 32 Let n and be two positive integers such as n We define by Gn the subset of F n : { } Gn = f F n such that {f(i)} i [n] = [] and f satisfies (E), From this definition, G n groups all sub-exceeding functions of F n having as image all the elements of [] and which checs the condition (E) Example 32 Tae f = 23 G 3 4 a sub-exceeding function We have f G 3 4 because {f(i)} i [4] = [3] and that f chec the condition (E) Now, if we tae g = 32, even if {f(i)} i [4] = [3], we have g = 32 / G 3 4 because the condition (E) is not checed (no 2 before 3) Corollary 33 Let n and be two positive integers such that n We have the relation H n G n (3) Proof Since Hn groups all sub-exceeding function of F n which have a quasiincreasing image formed by all elements of [], that is to say { } Hn = f F n f(i) f(i + ) for all i [n] such that {f(i)} i [n] = [] From this definition of H n and from the condition f(i) f(i + ) for all i [n], all elements of H n satisfy the condition (E) So, we find that H n G n

8 70 Luc Rabefihavanana 32 Iterative construction of G n Example 34 To initialize the construction of the set G n, let s start with n = and n = 2 So, we have G = { } G 2 = { } and G 2 2 = { 2 } Theorem 35 Let n and be two integers such that n For =, we always see that G n is the singleton set such as G n = {f f = n-fois } 2 For = n, we see also that G n n is the singleton set such as G n n = {f f = 23(n )(n)} 3 For all integer such that < < n, we can construct all elements of G n as follow: Proof (a) Tae all elements of G n and add the integer at the end, (b) Tae also all elements of G n and add all integer in [] one by one at the end So, for the best presentation, we write this construction as Gn = { Gn } Gn (32) 2 Here ( ) (resp ( ) ) means that we add the integer 2 (resp add one by one all elements of []) at the end of all elements of ( ) For =, by definition of Gn, we have { } Gn = f F n {f(i)} i [n] = [] and that f satisfie (E), ie f(i) = for all i [n] Then, G n = {f = n-fois }

9 Encoding of partition set using sub-exceeding function 7 2 For = n, still by the definition of Gn, we have { } Gn n = f F n {f(i)} i [n] = [n] and that f satisfie (E) However, by definition of a sub-exceeding function on [n], it is only the function f = 23n which is the sub-exceeding function that has an image sequence {f(i)} i [n] that is equal to the set [n] G n n = {f = 23(n )(n)} 3 For all integer such that < < n, prove by recurrence on n that Gn = { Gn } Gn (33) 2 At order 2, our statement 33 is true (see example 34) but we still try to see for the order 3 Let s prove that G 2 3 First, we find that G3 = for = 2, 2, 22 for = 2 23 for = 3 = {G 2 2} { G 2 2 As G 2 = and that G 2 2 = 2, directly we find that the set G 2 3 are formed by: (a) the element of G 2 by adding integer 2 at the end (b) the element of G 2 2 by adding one by one the integers and 2 at the end Suppose that the statement 33 is true up to the order n that is to say for any integer such that < < n, we have Gn = { Gn 2 } Gn 2 2 Let s show now that our proposition is still true to the order n So, let be an integer such that < < n, ( 2 )}

10 72 Luc Rabefihavanana (a) First, let f be a sub-exceeding function in Gn, that is to say that f satisfies the condition (E) and that {f(i)} i [n ] = [ ] By adding the integer to the end of f, we have a new subexceeding function f on [n] such that f = f()f(2)f(n ) with = f(n) Directly we find that {f(i)} i [n] = [] and that the condition (E) is still checed (b) Next, let f be a sub-exceeding function in Gn that is, f satisfies the condition (E) and that {f(i)} i [n ] = [] By adding an element of the integers, 2, 3,, at the end of f, we have a new sub-exceeding function on [n] denoted f such that f = f()f(2)f(n )j with j = f(n) where j [] Directly we find that {f(i)} i [n] still remains equal to [] and that the condition (E) remains checed Then in both cases (a) and (b), the new function under f is an element of Gn So { G n } Gn Gn 2 (34) Conversely, let f be a subexcessing function in G n, we have {f(i)} i [n] = [] and that f also checs the condition (E) By removing f(n) from {f(i)} i [n], we have a new sub-exceeding function f on [n ] such that f = f()f(2)f(n ) Two cases occur, either we have the equality {f(i)} i [n ] = [ ] or we have {f(i)} i [n ] = [] So,in both cases, the set {f(i)} i [n ] checs again (E) So f G n G n consequently, { G n } Gn 2 G n (35)

11 Encoding of partition set using sub-exceeding function 73 In sum, by the relation (34) and (35) { G n } Gn 2 = G n Example 36 Here is a table which represents the set G n for n =, 2, 3, 4 n Corollary 37 Let n and be two integers such that n For = and = n, we have Card G n = 2 For < < n, we have Card G n = Card G n + Card G n (36) Proof From the construction of the set G n above (in the theorem (35)), we have these relations Corollary 38 The cardinal table of G n is as follows based on n and n

12 74 Luc Rabefihavanana Proof As G =, G2 = et que G2 2 = 2, by combining with the equation (36) which says that Card Gn = Card Gn + Card Gn, we have this table directly Looing at this table and the equation (36), we find that our cardinal table for G n is none other than the second-type Stirling table So by choosing two not-zero positive integers n and where n, the integer obtained on this table represents the possible number of cases for the partition of any set of n elements into subsets 33 Bijection between G n and P (Ω) Notation 39 Let f be a sub-exceeding function in G n For all i [], we denote by A i the set of antecedents of i per f That is to say A i = f (i) Example 30 Let f an element of G 4 9 such that f = So, A = {, 2, 3, 8}, A 2 = {4, 5, 7}, A 3 = {6} and A 4 = {9} Proposition 3 Let f be a sub-exceeding function in G n represented by these sets of antecedents {A, A 2,, A } Let now i an integer in [], we find that the first integer i of f is placed at the j ème place where j = MinA i Proof Let f be a sub-exceeding function in G n represented by these sets of antecedents {A, A 2,, A } Let now i an integer in [], as A i = f (i), so A i groups all the positions of i in f Consequently the first integer i in f is at the j ème position where j = MinA i Example 3 Let f = an element of G 4 9 with A = {, 2, 3, 8}, A 2 = {4, 5, 7}, A 3 = {6} and A 4 = {9} Here we easily find that the first integer 2 of f is placed at 4 th position which is none other than MinA 2 and this is the even for other integers in [4] Theorem 32 Let ϕ be the map from G n to P (Ω) defined by ϕ : G n P (Ω) f ϕ(f) = { {ω i } i A, {ω i } i A2, {ω i } i A }, (37) So, the map ϕ is bijective Proof As we have seen above that the cardinal of G n and P (Ω) are the same and that it is the value of second-type Stirling table at the n th line and th colonm We have to show that the map ϕ is injective

13 Encoding of partition set using sub-exceeding function 75 Let f and g be two sub-exceeding functions in Gn such that ϕ(f) = ϕ(g) That is to say { } { } {ω i } i A (f), {ω i} i A2 (f),, {ω i} i A (f) = {ω i } i A (g), {ω i} i A2 (g),, {ω i} i A (g) (38) By the definition of a sub-exceeding function on [n], we have always f() = (resp g() = ) As ( ) j [] {ω i } i Aj (f) = and ( ) j [] {ω i } i Aj (g) =, we have necessary {ω i } i A (f) = {ω i} i A (g) (39) Let now p be the first integer of [] such that {ω i } {ω i Ap(f) i} i Ap(g) ie for all integer j such that j < p, we have {ω i } i Aj (f) = {ω i} i Aj (g) but {ω i} i Ap(f) {ω i} i Ap(g) (30) From the equality in (38), there exists an integer p such that p > p and {ω i } i Ap(f) = {ω i} i Ap (g) (3) Note by m the integer such that m = MinA p So the first integer p in the sequence of image of f is placed et the m th position By definition of a subexceeding function in G n, all integers in [p ] are found before p Consequently, we can write f = f() f(2) f(m ) p f(m + )f(n) where {f(i)} i [m ] = [p ] (32) As {ω i } = {ω i Ap(f) i} i Ap (g), we have A p(f) = A p (g) Consequently, we have also the equality m = MinA p So, the first integer p of g is placed also at the m th position As {ω i } i Aj (f) = {ω i} i Aj (g) for all integer j such that j < p, so f(i) = g(i) for all integer i such that i < m Then, from the relation (32) which say that {f(i)} i [m ] = [p ], we can write {g(i)} i [m ] = [p ] Consequently, g = g() g(2) g(m ) p g(m+)g(n) où {g(i)} i [m ] = [p ] (33) Or the integer p is strictly more than p, so it is not possible if g is an element of Gn because the integer p is not found before p in the sequence of image {g(i)} i [n] Necessary, if f and g are two sub-exceeding functions in Gn such that ϕ(f) = ϕ(g), we have always the equality {ω i } i Aj (f) = {ω i} i Aj (g) for all integer j such that j [] Finally, we have f = g and the map ϕ is injective In sum ϕ is bijective

14 76 Luc Rabefihavanana 34 Algorithm for the construction of the sub-exceeding function f corresponding of any partition in P (Ω) Let n and be two integers such that n Let also Ω an any set of cardinality n such that Ω = {ω, ω 2,, ω n } Tae now S = {Ω, Ω 2,, Ω } a partition of Ω into subsets To construct the corresponding sub-exceeding function f of S in G n, we have necessary three steps: Search the set J Ωi which is the set of all position of the elements of Ω i in Ω 2 We must order S as we have S = {Ω, Ω 2,, Ω } where Min J Ω < Min J Ω 2 < < Min J Ω (34) 3 Finally, we have J Ω = f () J Ω 2 = f (2) J Ω = f () (35) Example 33 Let Ω be the set such that Ω = {a, b, c, d, e, f, g, h} and S = {{h} ; {b, c} ; {a, e, g} ; {d, f}} Here we have Ω = {h} Ω 2 = {b, c} Ω 3 = {a, e, g} Ω 4 = {d, f} Now, let s present the three step to construct the corresponding sub-exceeding f which is an element of G 4 8 First, we have and that J Ω = {8} J Ω2 = {2, 3} J Ω3 = {, 5, 7} J Ω4 = {4, 6} Min J Ω = 8 Min J Ω2 = 2 Min J Ω3 = Min J Ω4 = 4

15 Encoding of partition set using sub-exceeding function 77 2 Secondly, after calculating MinJ Ai above for all i, we can write S = {Ω, Ω 2, Ω 3, Ω 4} such that 3 Finally, we found that Ω = Ω 3 = {, 5, 7} Ω 2 = Ω 2 = {2, 3} Ω 3 = Ω 4 = {4, 6} Ω 4 = Ω = {8} J Ω = f () = {, 5, 7} J Ω 2 = f (2) = {2, 3} J Ω 3 = f (3) = {4, 6} J Ω 4 = f (4) = {8} Consequently f = Acnowledgements The authors would lie to than the anonymous referees for their careful reading of our manuscript and helpful suggestions Ethics The authors declare that there is no conflict interests regarding the publication of this manuscript References [] J Désarménien, Une autre interprétation du nombre de dérangements, Actes 8e Sém Lothar Comb, 9 (984), 6 [2] DH Lahmer, Teaching combinatorial trics to a computer, Proc Sympos Appl Math Combinatorial Analisis, Vol 0, Amer Math Soc, Providence, 960, [3] D Knuth, The Art of Computer Programming, Sorting and Searching, 2nd edn, Vol 3, Addison-Wesley, Reading, 98 [4] Luc Rabefihavanana, Partition set and sub-exceeding fonction: Coding and Decoding, Bulletin of Mathematics and Statistics Research, 5 (207), no 3, [5] M Hall, Proc Sympos Appl Math, Combinatorial Analysis, Vol 6, Math Soc, Providence, RI, and Following 956, 203 [6] F Raotondrajao, Thèse de Doctorat, Université d Antananarivo, Madagascar, November, 999

16 78 Luc Rabefihavanana [7] R Mantaci and F Raotondrajao, A permutation that nows what Eulerian means, Discrete Mathematics and Theortical Computer Science, 4 (200), no, 0-08 [8] R Mantaci, Sur la distribution des anti-excédances dans le groupe symétrique et dans ses sous-groupes, Theoret Comp Science, 7 (993), Received: January 2, 208; Published: March 20, 208