On the work performed by a transformation semigroup

Transcription

1 AUSTRALASIAN JOURNAL OF COMBINATORICS Volume 49 (2011, Pages On the wor performed by a transformation semigroup James East School of Mathematics and Statistics University of Sydney Sydney, NSW 2006 Australia jameseast@sydneyeduau Peter J McNamara Department of Mathematics Stanford University Stanford, CA USA petermc@mathstanfordedu Abstract A (partial transformation α on the finite set {1,,n} moves an element i of its domain a distance of i iα units The wor w(α performed by α is the sum of all of these distances We derive formulae for the total wor w(s = α S w(α performed by various semigroups S of (partial transformations One of our main results is the proof of a conjecture of Tim Lavers which states that the total wor performed by the semigroup of all order-preserving functions on an n-element chain is equal to (n 12 2n 1 Introduction Fix a positive integer n and write n = {1,,n} The partial transformation semigroup PT n is the semigroup of all partial transformations on n; ie all functions between subsets of n (Note that the use of the word partial does not imply that the domain is necessarily a proper subset of n Inthisway,PT n also includes all full transformations of n; ie all functions n n A partial transformation α moves

2 96 JAMES EAST AND PETER J MCNAMARA apointi of its domain to a (possibly new point j in its image If the elements of n are thought of as points, equally spaced along a line, then the point i has been moved a distance of i j units Summing these values, as i varies over the domain of α, gives the (total wor performed by α, denoted by w(α We may also consider the total and average wor performed by a collection S of partial transformations, being the quantities w(s = 1 α S w(α andw(s = w(s respectively It is the S purpose of the current article to calculate w(s andw(s when S is either PT n itself, or one of its six ey subsemigroups: T n = { α PT n dom(α =n }, the (full transformation semigroup; I n = { α PT n α is injective }, the symmetric inverse semigroup; S n = T n I n, the symmetric group; PO n = { α PT n α is order-preserving } ; O n = T n PO n ; and POI n = I n PO n For the above definitions, recall that a partial transformation α PT n is orderpreserving if iα jα whenever i, j dom(α andi j 1 Figure 1 illustrates the various inclusions; for a more comprehensive picture, see [2] Figure 1: The relevant part of the subsemigroup lattice of PT n Our interest in this topic began after attending a tal by Tim Lavers at a Semigroups Special Interest meeting in Sydney 2004, in which the formula w(o n =(n 12 2n 1 We write iα for the image of i n under α PT n, although the semigroup operation on PT n (composition as binary relations does not feature in the current wor

3 WORK PERFORMED BY A TRANSFORMATION SEMIGROUP 97 was conjectured We also note that the quantity 1 w(s n n= n2 1 has been calculated n previously in relation to turbo coding [1] although, in the absence of such external considerations, we feel that w(s andw(s are the more intrinsic quantities Our results are summarized in Tables 1 and 2 below, where the reader will notice some interesting relationships such as w(o n =w(poi n andw(s n =w(t n Table catalogues the calculated values of w(s forn =1,,10 S Formula for w(s S n n!(n 2 1 T n n n (n 2 1 PT n I n POI n n n (n+1 n (n 2 n n 1 ( n 1 2! (n 12 2n O n PO n n i,j=1 (n 12 2n n,l=0 i j ( i 1 ( j+ 1 ( n i l ( n j+l l Table 1: Formulae for the total wor w(s performed by a semigroup S PT n S S n T n PT n I n n n Formula for w(s ( n ( n l=0 l n 2 1 n 2 1 n 2 n 2l! 1 n 1 ( n 1 2! POI n ( 2n n 1(n 12 2n O n PO n ( n m=0 ( n m ( n+m 1 m ( 2n 1 1(n n 12 2n 1 n i,j=1 n,l=0 i j ( i 1 ( j+ 1 ( n i l ( n j+l l Table 2: Formulae for the average wor w(s performed by an element of a semigroup S PT n

4 98 JAMES EAST AND PETER J MCNAMARA n w(s n w(t n w(pt n w(i n w(poi n w(o n w(po n Table : Calculated values of w(s for small values of n The article is organized as follows In Section 2 we obtain a general formula for w(s involving the cardinalities of certain subsets M i,j (S ofs In Section we consider separately the seven semigroups described above, calculating the cardinalities M i,j (S, and thereby obtaining explicit formulae for w(s ineachcase The formulae obtained in this way are in a closed form when S is one of S n, T n, PT n, but expressed as a sum involving binomial coefficients in the remaining four cases In Section 4 we prove Lavers conjecture, which essentially boils down to a proof of the identity ( ( p + q 2n p q p q = n2 2n 1, p n p p,q=0 giving rise to the postulated closed form for w(o n =w(poi n By contrast, the expression w(i n = n n I n 1 may not be simplified further, since no closed form is nown for I n = n ( n 2! It is not nown to the authors whether a closed form for w(po n exists, but the presence of large prime factors suggests that the situation could not be as simple as that of w(o n ; for example, w(po 9 = Unless specified otherwise, all numbers we consider are integers, so a statement such as let 1 i 5 should be read as let i be an integer such that 1 i 5 It will also be convenient to interpret a binomial coefficient ( p q to be 0 if p<q 2 General Calculations We now mae precise our definitions and notation The wor performed by a partial transformation α PT n in moving a point i n is defined to be w i (α = { i iα if i dom(α 0 otherwise,

5 WORK PERFORMED BY A TRANSFORMATION SEMIGROUP 99 and the (total wor performed by α is w(α = i n w i (α For S PT n, we write w(s = w(α and w(s = 1 S w(s α S for the total and average wor performed by the elements of S, respectively For the remainder of this section, we fix a subset S PT n Fori n, put w i (S = α S w i (α, which may be interpreted as the total wor performed by S in moving just the point i Rearranging the defining sum for w(s gives w(s = w(α = w i (α = w i (α = w i (S α S α S i n i n α S i n For i, j n, we consider the set M i,j (S = { α S } i dom(α and iα = j, and write m i,j (S = Mi,j (S for the cardinality of M i,j (S Note that w i (α = i j for all α M i,j (S, so that w i (S = i j m i,j (S j n These preliminaries give the following result Lemma 1 Let S PT n Thenw(S = i,j n i j m i,j (S Specific Calculations We now use Lemma 1 as the starting point to derive explicit formulae for w(s for each of the semigroups S defined in Section 1 We consider each case separately, covering them roughly in order of difficulty When S is one of S n, T n, PT n,ori n,we will see that m i,j (S is independent of i, j n, andsow(s turns out to be rather easy to calculate, relying only on Lemma 1 and the well-nown identity ( n +1 i j =2 i,j n = n n

6 100 JAMES EAST AND PETER J MCNAMARA (The reader is reminded of the convention that ( n+1 =0ifn = 1 In each of the remaining three cases, the formulae we derive for the quantities m i,j (S yields an expression for w(s as a sum involving binomial coefficients We defer further investigation of the O n and POI n cases until Section 4, where we show that this so-obtained expression may be simplified It may be that some of the intermediate results of this section are already nown (for example Lemmas 2, 6 and 9 but the proofs, which are believed to be original, are included for completeness; the reader is referred to the introduction of [4] for a review of related studies Note that the proofs we give are largely geometrically motivated A partial transformation α PT n may be represented diagrammatically by drawing an upper and lower row of n dots, representing the elements of n (in increasing order from left to right, and drawing a line from upper vertex i to lower vertex j whenever i dom(α and iα = j In this way, the quantity m i,j (S may be interpreted as the number of ways to extend the partial map π i,j, pictured in Figure 2, to an element of S Figure 2: The partial map π i,j PT n with domain {i} and image {j} 1 The Symmetric Group S n To extend the partial map π i,j to a permutation of n, we must add n 1 lines, ensuring that they correspond to a bijection from n \{i} to n \{j} It follows then that m i,j (S n =(n 1! for all i, j n, and so Lemma 1 gives w(s n = i,j n The average wor is given by i j (n 1! = n n w(s n = w(s n n! (n 1! = n!(n2 1 = n This result may be found in [1], in a slightly different form

7 WORK PERFORMED BY A TRANSFORMATION SEMIGROUP The Transformation Semigroup T n To extend π i,j to a full transformation of n, each upper vertex must be connected by a line to a lower vertex Since the lower vertex of such a line is not constrained in any way, we see that m i,j (T n =n n 1 for all i, j n By Lemma 1, we therefore have w(t n = i j n n 1 = n n n n 1 = nn (n 2 1, i,j n and w(t n = w(t n = n2 1, n n giving rise to the first interesting (and coincidental, as far as we can tell identity w(t n =w(s n The Partial Transformation Semigroup PT n To extend π i,j to a partial transformation, each upper vertex may be connected to any lower vertex or else left unconnected It follows that m i,j (PT n =(n +1 n 1 for all i, j n and so, by Lemma 1, we have w(pt n = i,j n i j (n +1 n 1 = n n (n +1 n 1 = (n +1n (n 2 n, and w(pt n = w(pt n (n +1 = n2 n n Although w(pt n w(s n =w(t n, all three sequences are of course asymptotic to n 2 / 4 The Symmetric Inverse Semigroup I n To extend π i,j to an injective partial transformation, we must add at most n 1more lines, ensuring that they correspond to an injective partially defined map from n \{i} to n \{j} Since such a partial map obviously corresponds to an injective partial transformation on {1,,n 1}, weseethatm i,j (PT n = I n 1 for all i, j n Using the well-nown formula I m = m ( m 2!, it then follows that and w(i n = i,j n i j I n 1 = n n w(i n = w(i n I n I n 1 = n n = n n I n 1 I n n 1 ( 2 n 1!,

8 102 JAMES EAST AND PETER J MCNAMARA 5 The Semigroup POI n From this point onward, calculation of the quantities m i,j (S is not as straightforward For 0 p, q n let POI p,q denote the set of all order-preserving injective partial maps from p to q (Note that we interpret = {1,,} to be empty if =0 Lemma 2 Let 0 p, q n Then POI p,q = ( p+q p ( = p+q Proof Let q = {1,,q } be a set in one-one correspondence with q Denote also by : q q the inverse bijection, so that we write i = i for all i q Consider the set Σ= { A p q } A = q For A Σ, define φ A POI p,q by dom(φ A =A p and im(φ A =q\(a q, noting that A p = q \ (A q, and that an element of POIp,q is completely determined by its domain and image It is then easy to chec that the maps determined by A φ A and φ dom(φ ( q \ im(φ are mutually inverse bijections between Σ and POI p,q The result follows since we clearly have Σ = ( p+q q q Remar Geometrically, this proof corresponds to the fact that, given p upper vertices and q lower vertices, an element of POI p,q is determined by choosing q vertices, and then joining the selected upper vertices to the unselected lower vertices An alternative proof begins by noting that POI p,q contains ( p q ( maps of ran, and then concludes by applying the identity ( p q ( ( = p+q q Lemma 4 Let i, j n Thenm i,j (POI n = ( i+j 2 i 1 ( 2n i j n i Proof Let α M i,j (POI n Then since iα = j and α is order-preserving, we see that α < j whenever dom(α and<i Thus, we may define a map λ α POI i 1,j 1 by dom(λ α =dom(α {1,,i 1} and im(λ α =im(α {1,,j 1} Similarly, we have α > j whenever dom(α and > i, andsowemayalso define a map ρ α POI n i,n j by dom(ρ α = { i dom(α, >i } and im(ρ α = { j im(α, >j } It is then easy to chec that the map α (λ α,ρ α defines a bijection from M i,j (POI n topoi i 1,j 1 POI n i,n j The result now follows from Lemma 2

9 WORK PERFORMED BY A TRANSFORMATION SEMIGROUP 10 Remar 5 The idea of the above proof is summed up in the schematic picture of a typical element of M i,j (POI n illustrated in Figure Figure : A schematic picture of an element of M i,j (POI n It follows by Lemmas 1 and 4 that the total wor performed by POI n is given by w(poi n = ( ( i + j 2 2n i j i j i 1 n i i,j n In Section 4 we revisit this formula, and show that in fact w(poi n = (n 12 2n An expression for w(poi n may be found by dividing through by POI n = POI n,n = ( 2n n 6 The Semigroup O n For 0 p n and q n let O p,q denote the set of all order-preserving maps from p to q Lemma 6 Let 0 p n and q n Then O p,q = ( ( p+q 1 p = p+q 1 q 1 Proof Consider the set Ω= { α POI p,q p dom(α } There is an obvious bijection O p,q Ω determined geometrically by removing all but the right-most lines from the connected components in the picture of α O p,q ; see Figure 4 For i q, put Ω i = {α Ω pα = i}, so that we have the disjoint union Ω = Ω 1 Ω q Clearly, the operation of removing the right-most line gives a bijection between Ω i and POI p 1,i 1 for each i q so that, by Lemma 2, we have O p,q = Ω = ( p + i 2 p 1 i q

10 104 JAMES EAST AND PETER J MCNAMARA Figure 4: The bijection O p,q Ω; see the proof of Lemma 6 for an explanation of the notation The result now follows from the identity s ( ( r + r + s +1 = r s Remar 7 An argument similar to that used in the proof of Lemma 2 may also be used here An element α Ω is completely determined by the sets dom(α \{p} {1,,p 1} and q \ im(α q This gives rise to a bijection between Ω and the set { A {1,,p 1, 1,,q } A = q 1 }, which has cardinality ( p+q 1 q 1 Lemma 8 Let i, j n Thenm i,j (O n = ( i+j 2 i 1 ( 2n i j n i Proof The proof follows a similar pattern to the proof of Lemma 4 Rather than include all the details, we simply refer to Figure 5 which gives a schematic picture of an element of M i,j (O n, indicating a bijection between M i,j (O n and O i 1,j O n i,n j+1 Figure 5: A schematic picture of an element of M i,j (O n In particular, we have m i,j (O n =m i,j (POI n foralli, j n, sothatw(o n = w(poi n The differing cardinalities of O n and POI n mean that w(o n w(poi n However, since POI n = ( ( 2n n =2 2n 1 =2 On,n =2 O n n,wedo have the relationship w(o n =2w(POI n

11 WORK PERFORMED BY A TRANSFORMATION SEMIGROUP The Semigroup PO n For 0 p n and q n let PO p,q denote the set of all order-preserving partial maps from p to q Lemma 9 Let 0 p n and q n Then PO p,q = ( p ( q + 1 Proof For A p write PO A p,q = { α PO p,q dom(α =A } Wethenhavethe disjoint union PO p,q = PO A p,q A p Now, for any 0 p, thereare ( p subsets A p for which A = and, for each such subset A, wehave PO A p,q = O,q = ( q+ 1, the last equality following by Lemma 6 This shows that PO p,q = p ( PO A p p,q = A p ( q + 1 The upper limit may be changed to n, in light of the convention that ( p =0if >p Lemma 10 Let i, j n Then m i,j (PO n = ( ( ( ( i 1 j + 1 n i n j + l l l,l=0 Proof Again we find that there is a bijection between M i,j (PO n and PO i 1,j PO n i,n j+1 The result now follows from Lemma 9 It follows, by Lemmas 1 and 10, that w(po n = i,j=1,l=0 ( ( ( ( i 1 j + 1 n i n j + l i j l l An expression for w(po n is then found by dividing through by PO n = PO n,n = ( n ( n + 1

12 106 JAMES EAST AND PETER J MCNAMARA 4 The Proof of Lavers Conjecture We now turn to the tas of proving the conjectured result of Lavers that w(o n = (n 12 2n In light of Section 6, this amounts to a proof of the identity ( ( i + j 2 2n i j i j =(n 12 2n i 1 n i i,j n Replacing n by n + 1, and introducing the new parameters p = i 1andq = j 1, the identity taes on the more pleasing form: ( ( p + q 2n p q p q = n2 2n 1 p n p p,q=0 The remainder of this section is devoted to a proof of this identity and, hence, a proof of the conjecture For 0 m 2n, define f(n, m = m ( m m 2p p p=0 ( 2n m n p noting first that f(n, 0 = f(n, 2n = 0 We now find a closed form for the remaining values of m In what follows, for a real number x and non-negative integer m, we use the standard falling factorial notation, writing (x m = x(x 1 (x m +1, Lemma 11 We have the following identities: f(n, 2 +1= 2 (2n 2 1! (2 +1! n (n 1!(n 1!!! f(n, 2 = 2 (2n 2! (2! n (n!(n 1!!( 1! for 0 n 1 (12 for 1 n 1 (1 Proof We apply two different methods of proof, one to each identity, and each of which may be adapted to treat the other case We first present a purely human-discovered proof of (12 Let 0 n 1 Consider the degree polynomial P (x = ( p x (p (x 1 ( p p p=0 in an indeterminate x In the defining sum for f(n, m, the terms with p = i and p = m i are equal In this way, we calculate f(n, 2 +1= 2(2n 2 1! n!(n 1! P (n

13 WORK PERFORMED BY A TRANSFORMATION SEMIGROUP 107 So it suffices to prove the polynomial identity (2 +1! P (x = (x 1 ( (14!! We do this by induction on, noting first that when = 0 both sides of (14 are identically equal to 1 Suppose now that 1 n 1andthat0 l< We consider P ( + l + 1 In the defining sum, terms with p l 1 will be zero and so, replacing the index of summation by r = p + l, wehave P ( + l +1= l r=0 which is readily checed to be equal to ( l +1 2r ( + l +1 ( l+r l (l r, l + r (2 +1! l!! (2l +1! P l( + l +1 By an inductive hypothesis, (2l +1! P l (x = (x 1 (l, l!l! and it quicly follows that (14 holds for the distinct x-values x = +1,+2,,2 Since the identity (14 involves polynomials of degree, it suffices to verify it for one more value of x Noting that the defining sum for P (0 has only one non-zero term (namely the p = 0 term, it is easy to chec that both sides of (14 are equal to ( 1 (2+1! when x = 0 So (14 holds, and the proof of (12 is complete! We now present a computer-aided proof of (1 using the WZ method [5] Let 1 n 1 Define F (n,, p = 2n( p( ( 2 2n 2 p n p (n ( ( 2 2n 2, n noting that the desired result is equivalent to 1 F (n,, p =1 (15 p=0 A computer implementation of Gosper s algorithm [] gives us the identity F (n +1,,p F (n,, p =G(n,, p +1 G(n,, p, where the function G is defined by G(n,, p = p( +1 p F (n,, p 2n(n +1 p Available at

14 108 JAMES EAST AND PETER J MCNAMARA Summing over p {0,, 1}, and noting that G(n,, =G(n,, 0 = 0, we obtain the equality 1 1 F (n +1,,p= F (n,, p p=0 So it suffices to prove (15 in the case n = + 1 only This however is a triviality as only the p = 1 term in the sum is non-zero This completes the proof of (1 p=0 Proposition 16 The following identity holds: ( ( p + q 2n p q p q = n2 2n 1 (17 p n p p,q=0 Proof Consider the two generating functions (2 +1! ( +2 x =(1 4x /2 and 4 x =(1 4x!! 2 After squaring the first, and equating coefficients of x n 1, Lemma 11 gives n n 1 ( n +1 f(n, 2 +1= 4 n Similarly, starting from (2! (!( 1! x =2x(1 4x /2 and 4 1 x =4x 2 (1 4x, 2 =1 =2 squaring the first, and looing at the coefficient of x n,weobtain n n 1 ( n f(n, 2 = 4 n =1 Returning to the original sum in (17, we rewrite it to first sum over those p and q for which p + q = m is fixed, and get ( ( p + q 2n p q p q p n p p,q=0 = 2 m=0 f(n, m n 1 n 1 = f(n, 0 + f(n, f(n, 2+f(n, 2n [( n +1 = 2 n 2 = n 2 2n 1 4 n 1 + ( n 2 =1 4 n 1 ]

15 WORK PERFORMED BY A TRANSFORMATION SEMIGROUP 109 References [1] D Daly and P Vojtěchovsý, How Permutations Displace Points and Stretch Intervals, Ars Combin 90 (2009, [2] M Delgado and V H Fernandes, Abelian Kernels, Solvable Monoids and the Abelian Kernel Length of a Finite Monoid, Semigroups and Languages, eds I M Araújo, M J J Branco, V H Fernandes and G M S Gomes, World Scientific, 2004, pp [] R W Gosper, Decision Procedure for Indefinite Hypergeometric Summation, Proc Nat Acad Sci USA 75 (1978, [4] A Laradji and A Umar, Combinatorial Results for Semigroups of Order- Decreasing Partial Transformations, J Integer Seq 7( (2004, 14pp (electronic [5] H S Wilf and D Zeilberger, Rational Functions Certify Combinatorial Identities, J Amer Math Soc (1990, (Received Nov 2009; revised 11 Nov 2010