A CAT Algorithm for Generating Permutations with a Fixed Number of Inversions

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Edmund Hensley
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1 A CAT Algorithm for Geeratig Permutatios with a Fixed Number of Iversios Scott Effler Frak Ruskey Departmet of Computer Sciece, Uiversity of Victoria, Caada Abstract We develop a costat amortized time (CAT) algorithm for geeratig permutatios with a give umber of iversios. We also develop a algorithm for the geeratio of permutatios with give idex. Key words: Permutatio, exhaustive listig, iversio, idex. 1 Itroductio Amog all statistics of permutatios, the umber of iversios are of paramout importace to computer scietists because of their coectio with sortig ad searchig algorithms. Furthermore, they have iterestig combiatorial properties ad are itesely studied i the mathematical literature (e.g., Margolis [7], Clark [3]). Both the computatioal ad mathematical properties of permutatios with a give umber of iversios are icely summarized by Kuth [4]. However, we kow of o published algorithms for geeratig all -permutatios with a give umber of iversios (or with give idex). There are may published algorithms for geeratig various types of permutatios ad may of these ru i costat time per permutatio, i a amortized sese. For example, there are algorithms for deragemets [1], ivolutios [10], up-dow permutatios [], ad liear extesios of posets [9],[5]. Research supported i part by NSERC grat OGP addresses: seffler@csc.uvic.ca (Scott Effler), fruskey@cs.uvic.ca (Frak Ruskey ). Preprit submitted to Iformatio Processig Letters 14 April 00

2 A algorithm rus i costat amortized time (CAT) if the amout of computatio, after a small amout of preprocessig, is proportioal to the umber of objects that are geerated [8]. A iversio of a permutatio π is a pair (π i,π j ) where i < j ad π i > π j [4]. Let I(π) deote the umber of iversios i π. Clearly, 0 I(π) ( ), with the extremes occurrig for 1 ad 1. Followig Kuth [4] we use I (k) to deote the umber of permutatios of [] with k iversios. By [] we deote the set {1,,..., }. This paper presets a CAT algorithm for geeratig permutatios of [] with a give umber of iversios. The algorithm is outlied i Sectio. The idex of a permutatio π is the sum of idices j such that π j > π j+1, deoted J(π) [4]. A theorem of MacMaho [8] establishes a bijectio betwee classes of permutatios couted by the two statistics. Theorem 1.1 The umber of permutatios of [] with k iversios is the same as the umber of permutatios of [] with idex k; i.e., both sets have cardiality I (k). Eve though there is a bijectio betwee two sets, a fast algorithm for geeratig oe set will ot ecessarily imply the existece of a fast algorithm for geeratig the other set. A algorithm for geeratig permutatios with fixed idex k that experimetally appears to be CAT i the rage < k < ( ) is give i Sectio 3. Geeratig Permutatios with a Give Number of Iversios I this sectio, we develop a CAT algorithm for geeratig all permutatios π S where I(π) is equal to some give value k..1 Algorithm The algorithm works from right to left i the oe lie otatio of the permutatio by placig elemets oe at a time. The key to the algorithm is the followig lemma. Lemma.1 Give atural umbers, k with 0 k ( ), ad x [], there is a permutatio π S of the form π = π 1 π π 1 x such that I(π) = k if

3 ge(, k ) (1) if = 0 k = 0 the output π 1 π π N () for each x [N] \ {π +1,..., π N } (a) if rak(x) k ( ) 1 + rak(x) the (i) π := x (ii) ge( 1, k + rak(x) ) ad oly if Fig. 1. Geeratig Permutatios With Give Number of Iversios ( ) 1 x k + x. (1) Proof. We must show both directios of the lemma. : Suppose π = π 1 π 1 x ad I(π) = k. The umber of iversios i π 1 π 1 satisfies 0 I(π 1 π 1 ) ( ) 1. There are exactly x iversios of the form (π i, x). Thus, we have x I(π) ( ) 1 + x. : Suppose the iequality i the lemma is satisfied for, k, ad x. There is a permutatio π 1 π π 1 of [] \ {x} such that I(π 1 π π 1 ) = k + x sice 0 k + x ( ) 1 by (1). But the π1 π π 1 x S ad I(π 1 π π 1 x) = k. We oly recurse if it is possible to exted the partial permutatio to a permutatio with the give umber of iversios. That is to say, o brach i the computatio tree is a dead-ed. Such algorithms are said [8] to be BEST (backtrackig esurig success at termials). BEST algorithms are ofte fast i practice ad are geerally easier to aalyze. Our algorithm geerates permutatios of [N]. Defie rak(x) as the positio of x i the ordered list X = [N]\{π +1,..., π N } of remaiig elemets. Notice that the rak(x) is the umber of iversios created by placig elemet x at the curret positio,. The method is outlied i Figure 1. Lemma. Give that π +1 π N is a (N )-permutatio of {1,,..., N}, the call ge(, k) geerates all permutatios π 1 π π N of {1,,..., N} such that I(π 1 π ) = k. Proof. Our proof is by iductio o. Sice the empty permutatio has o ivolutios, the lemma is true whe =0. There are precisely rak(π ) iversios of the form (π i, π ). By Lemma.1 we kow that π ca be ay value such that rak(x) k ( ) 1 + rak(x). Iductively, for each possible value of x = π, the recursive call ge( 1, k +rak(π )) geerates all permutatios for which I(π 1 π 1 ) = k +rak(π ). Together with the rak(π ) iversios created by elemet π, we have geerated all -permutatios with k iversios. 3

4 Corollary.1 The call ge(n, K) geerates all permutatios π 1 π π N of {1,,..., N} such that I(π 1 π π ) = K. The followig result is importat i determiig the order to traverse the list X of remaiig elemets of the permutatio. Lemma.3 If a 1 < a < < a where {a 1, a,..., a } = [N]\{π +1,..., π N } the those a i that satisfy the coditio i lemma.1 form a cotiguous subsequece a i, a i+1,..., a j where i = 1 or j =. Proof. It is clear from (1) that the values occur i a cotiguous subsequece. We will show, uder the assumptio that 0 k ( ), that (1) is satisfied for x = 1 or x = (or both), which will suffice to prove the lemma. So suppose that (1) fails for x =. The either k < 0 or k > ( ) 1. Sice the former caot occur we have ( ) ( ) 1 < k. Hece which is (1) whe x = 1. ( ) 1 1 < k ( ) = ( ) 1 + 1, If 0 < k ( ) 1 the we traverse the list i reverse order from the largest to the smallest elemet. Otherwise, we traverse the list i order from the smallest to the largest elemet. I either case, if the test i step (a) of the algorithm of Figure 1 fails, we stop traversig the list. Implemeted i this way, the algorithm has the crucial property that its ruig time is proportioal to the total umber of odes i its computatio tree. I order to maitai the list we use the techique dubbed dacig liks by Kuth [6].. Example As a example, cosider geeratig all permutatios of legth 4 with iversios. The five such permutatios are {314, 314, 143, 143, 134}. The computatio tree is give i Figure. The list of remaiig permutatio elemets at each ode is give i square brackets followed by the curret values of ad k, alog with the curret partial permutatio. Cosider the leftmost child of the root as a example. We already kow that ay child permutatio eds i a 4 ad has two iversios to the left of the curret positio. The ext step is to fid a value for the third positio. This 4

5 [4,3,,1] (k=, =4) _,_,_,_ [3,,1] (k=, =3) _,_,_,4 [4,,1] (k=1, =3) _,_,_,3 [4,3,1] (k=0, =3) _,_,_, [3,1] (k=1, =) _,_,,4 [3,] (k=0, =) _,_,1,4 [,1] (k=1, =) _,_,4,3 [4,1] (k=1, =) _,_,,3 [3,1] (k=0, =) _,_,4, [3] (k=0, =1) _,1,,4 [] (k=0, =1) _,3,1,4 [] (k=0, =1) _,1,4,3 [1] (k=0, =1) _,4,,3 [1] (k=0, =1) _,3,4, [] (k=0, =0) 3,1,,4 [] (k=0, =0),3,1,4 [] (k=0, =0),1,4,3 [] (k=0, =0) 1,4,,3 [] (k=0, =0) 1,3,4, Fig.. Example of the Algorithm of Figure 1 positio caot be 3 sice this leaves oly 1 ad, which ca oly produce oe iversio. Therefore, this ode has two childre, as show i the computatio tree. This process is cotiued recursively util all positios are filled i..3 Path Elimiatio We otice that there may be may successive odes i the computatio tree with oly oe child. This leads to iefficiecy sice, for example, to geerate the oe permutatio with k = 0 takes time Θ(), ad to geerate the 1 permutatios with k = 1 takes time Θ( ). To combat this problem, we will apply a path elimiatio techique (PET) from [8] ad stop the recursio early. This will allow us to elimiate most, but ot all of the odes i the recursive computatio tree that have oe child. Details of this process follow. First, if k = 0, there is always just oe valid permutatio of X; the oe with all elemets ascedig. We do ot elimiate such paths, uless they arise from a k = 1 ode higher i the tree. If k = 1, we ca easily geerate all valid permutatios of X i costat amortized time by startig with all elemets i ascedig order ad swappig oe pair of adjacet elemets at each step. For example, if = 4 ad X = [4], the permutatios are {134, 134, 143}. So, if k = 1, we stop the recursio ad geerate the 1 permutatios directly i time Θ(). Similarly, if k = ( ), the there is exactly oe valid permutatio; the the oe with all elemets descedig. To elimiate may of these 1-paths we otice that if k = ( ) 1 we ca easily geerate all valid permutatios i costat amortized time by startig with all elemets i descedig order ad swappig oe pair of adjacet elemets at each step. So, if k = ( ) 1 we agai stop the recursio ad geerate the 1 permutatios directly i time Θ(). 5

6 I Figure the two odes with darker borders are the roots of subtrees with k = 1 ad k = ( ) (with = 3). The subtrees at these odes are replaced with o-recursive calls. Oe 1-path remais ad it is idicated by the odes with the black dots i their lower right corers. The cost of computig this path is assiged to the subtree rooted at the highlighted k = 1 ode. Now, we state ad prove the mai theorem of this paper. The argumet that our algorithm rus i costat amortized time is depedet upo the list traversal ad path elimiatio techiques discussed earlier. Theorem.1 The algorithm of Figure 1 with path elimiatio techiques geerates permutatios with a give umber of iversios i costat amortized time. Proof. By our earlier discussio the computatio time of the algorithm is proportioal to the umber of odes i its recursive computatio tree. If call with k = 1 or k = ( ) 1 occurs, the the 1 permutatios are geerated i Θ() time by a separate o-recursive routie ad we regard the correspodig subtrees as beig replaced by a sigle ode i the recursive computatio tree. Now every root of a 1-path i the computatio tree has k = 0 or k = ( ). A root r of a 1-path with k = 0 ca occur oly oce amogst its sibligs, ad it must have a siblig q with k = 1. The Θ() cost of geeratig the oe path is charged to the 1 leaves of the subtree represeted by q. A root of a 1-path with k = ( ) is hadled i a aalogous maer. A subtree ca oly be charged oce by this scheme (uless = 3 i which case it ca be charged twice). If 1 < k < ( ) 1, the a iteral ode of the computatio tree has at least two childre. Thus the total umber of such iteral odes is less tha the umber of leaves. 3 Geeratig Permutatios with Give Idex I this sectio, we are cocered with the geeratio of all π such that J(π) is equal to some give value. The algorithm for this problem has a similar flavor to the algorithm of Figure 1 for geeratig permutatios with a give umber of iversios, but there are sigificat ad subtle differeces. For fixed k i the rage < k < ( ) the algorithm appears to be CAT i the sese that the amortized time per permutatio is observed to be decreasig for large eough. 6

7 ge(, k) (1) if = 0 k = 0 the output π 1 π π N () for each x [N] \ {π +1,..., π N } (a) if x > π +1 the k := k else k := k; (b) if rak(x) k ( ) rak(x) + 1 the (i) π := x; (ii) ge( 1, k ); Fig. 3. Geeratig Permutatios With Give Idex (assumes π N+1 = N + 1). 3.1 Algorithm As was the case for iversios, this algorithm works from right to left. We assume that we are geeratig permutatios of [N]. I order to treat all positios i a uiform maer we will assume that π N+1 = N + 1. Some simple results will help us costruct a algorithm i much the same way as was doe for iversios. Cosider a permutatio of [] of the form π 1 π π 1 x. For x fixed i positio, the permutatio of smallest idex is (x+1)(x+) (x + ( x)) 1 (x 1) x ad the permutatio of greatest idex is (x 1) 1 (x+( x)) (x+)(x+1) x. The idex of those permutatios is x ad ( ) x + 1, respectively. Based o this observatio ad reasoig similar to that used to prove Lemma 1, we are led to the followig lemma. Lemma 3.1 Give a triple (, x, k) of itegers, there is a permutatio π of the form π = π 1 π 1 x such that J(π) = k if ad oly if x k ( ) x + 1. () The algorithm of Figure 3 geerates all -permutatios with idex k. It is based o Lemma 3.1, except that we eed to use rak(x) istead of x ad if x > π +1 the the idex target eeds to be adjusted by. Lemma 3. Give that π +1 π N is a (N )-permutatio of [N], the call ge(, k) will geerate all permutatios π 1 π π N of [N] such that J(π 1 π π ) = k. 7

8 Proof. We will prove this result by iductio o. Sice the empty permutatio has idex 0. the lemma is true whe =0. If = N or π < π +1 the π does ot cotribute to the idex. By Lemma 3.1 we kow that π ca be ay value such that rak(x) k ( ) rak(x)+ 1. For each possible value of x = π, we recurse with call ge( 1, k). By our iductive assumptio this call geerates all permutatios such that J(π 1 π 1 ) = k. If π > π +1 the π cotributes to the idex. By Lemma 3.1 we kow that π ca be ay value such that rak(x) k ( ) rak(x) + 1. For each possible value of x = π, we recurse with call ge( 1, k ). By our iductive assumptio this call geerates all permutatios such that J(π 1 π 1 ) = k. Together with the added to the idex by π, we have geerated all -permutatios with idex k. Corollary 3.1 The call ge(n, K) will geerate all permutatios π 1 π π N of {1,,..., N} such that J(π 1 π ) = K. A similar result to Lemma.3 is ecessary to determie which order to traverse the list X = [N] \ {π +1,..., π N } of remaiig elemets of the permutatio. Lemma 3.3 If a 1 < < a p 1 < π +1 < a p < < a where {a 1, a,..., a } = [N] \ {π +1,..., π N } the, for 1 i < p, those a i that satisfy the coditio i Lemma 3.1 for (, a i, k) form a cotiguous subsequece a i, a i+1,..., a j where i = 1 or j = p 1, ad for p i <, those a i that satisfy the coditio i Lemma 3.1 for (, a i, k ) form a cotiguous subsequece a i, a i+1,..., a j where i = p or j =. Proof. It is clear from () that the values occur i a cotiguous subsequece. For 1 i < p, we will show that if () is satisfied for a i the it is satisfied for a i+1,..., a p 1 or a 1,..., a i 1 (or both), which will suffice to prove the first half of the lemma. If () is satisfied for a i the we have rak(a i ) k ( ) rak(a i ) + 1. If rak(a i ) k the x = rak(a j ) satisfies () for 1 j < i. If ( ) rak(a i ) + 1 k the x = rak(a j ) satisfies () for i < j < p. For p i, the argumet is idetical except k is replaced by k. At Step () of the algorithm, we will traverse the list X up to four times. Oce from largest to smallest, oce from smallest to largest ad i both directios 8

9 startig from the largest elemet a p 1 less tha π +1. Our data structure allows us to idetify a p 1 i the list i costat time. I all of these cases, if the test of step (b) of the algorithm of Figure 3 fails, we stop traversig the list. Also ote that we must keep track of where each traversal eds, makig sure that we do ot recurse o the same elemet twice. Implemeted i this way, the algorithm has the crucial property that its ruig time is proportioal to the total umber of odes i its computatio tree. Similar improvemets ca be made to this algorithm as were made to the iversio geeratio algorithm so as to produce a costat amortized time algorithm, but oly if < k < ( ), ad this has oly bee observed experimetally. To be precise, for > 4, the ratio of recursive calls to permutatios geerated is strictly decreasig. To get this CAT behavior we must use the path elimiatio techiques described i the followig paragraphs. Let {a 1, a,..., a } = [N] \ {π +1,..., π N } where a 1 < a < < a. If k = 0 ad a < π +1 the there is oly oe valid permutatio; the oe with all elemets ascedig. We do ot elimiate such paths, uless they arise from a k = 1 ode, where a < π +1, higher i the computatio tree. If k = 1 ad a < π +1, we ca easily geerate all valid permutatios i costat amortized time by startig with the idetity ad doig swaps (π 1, π ), (π 1, π 3 ),..., (π 1, π ), pritig each itermediate permutatio. Similarly, if k = ( ) ad a < π +1, the there is exactly oe valid permutatio; the oe with all elemets descedig. To elimiate may of these 1-paths we otice that if k = ( ) 1 ad a < π +1 we ca easily geerate all valid permutatios i costat amortized time by startig with all elemets i descedig order ad swappig as doe above for k = 1. 4 Fial Remarks The algorithms described i this paper are used i the secod author s Combiatorial Object Server at httpd:// i the permutatios sectio. Java implemetatios of the algorithms ca also be dowloaded there. Refereces [1] Selim G. Akl, A New Algorithm for Geeratig Deragemets, BIT 0 (1980) 7. 9

10 [] Bruce Bauslaugh ad Frak Ruskey, Geeratig Alteratig Permutatios Lexicographically, BIT, 30 (1990) [3] L.A. Clark, A Asymptotic Expasio for the Number of Permutatios with a Certai Number of Iversios, Electroic Joural of Combiatorics, 7 (000) #R50. [4] Doald E. Kuth, The Art of Computer Programmig, Volume 3: Sortig ad Searchig. Secod Editio, Addiso-Wesley, [5] Doald E. Kuth, The Art of Computer Programmig, Volume 4: Prefascicle B, Jauary 00. [6] Doald E. Kuth, Dacig Liks, Milleial Perspectives i Computer Sciece, (Houdmills, Basigstoke, Hampshire: Palgrave, 000), [7] B.H. Margolis, Permutatios with iversios, Joural of Iteger Sequeces, 4 (001) Article [8] Frak Ruskey, Combiatorial Geeratio. Workig versio of book i progress, [9] G. Pruesse ad F. Ruskey, Geeratig Liear Extesios Fast, SIAM J. Computig, 3 (1994) [10] D. Roelats va Baroaigie, Costat time geeratio of ivolutios, Proceedigs of the Twety-third Southeaster Iteratioal Coferece o Combiatorics, Graph Theory, ad Computig (Boca Rato, FL, 199). Cogr. Numer. 90 (199),

4.3 Growth Rates of Solutions to Recurrences

4.3 Growth Rates of Solutions to Recurrences 4.3. GROWTH RATES OF SOLUTIONS TO RECURRENCES 81 4.3 Growth Rates of Solutios to Recurreces 4.3.1 Divide ad Coquer Algorithms Oe of the most basic ad powerful algorithmic techiques is divide ad coquer.