Constant Time Generation of Set Partitions

Transcription

1 930 PAPER Special Section on Selected Papers from the 17th Workshop on Circuits and Systems in Karuizawa Constant Time Generation of Set Partitions Shin-ichiro KAWANO a) and Shin-ichi NAKANO b), Members SUMMARY In this paper we give a simple algorithm to generate all partitions of {1, 2,, n} into k non-empty subsets. The number of such partitions is known as the Stirling number of the second kind. The algorithm generates each partition in constant time without repetition. By choosing k = 1, 2,, n we can also generate all partitions of {1, 2,, n} into subsets. The number of such partitions is known as the Bell number. key words: algorithm, enumeration, the Stirling number of the second kind, the Bell number, Gray code 1. Introduction It is useful to have the complete list of objects for a particular class. One can use such a list to search for a counterexample to some conjecture, to find the best object among all candidates, or to experimentally measure an average performance of an algorithm over all possible inputs. Many algorithms to generate a particular class of objects, without repetition, are already known [1], [6] [9], [11], [16]. Many excellent textbooks have been published on the subject [2], [5], [15]. In this paper we consider the following generation problem. For a positive integer n and k < n, lets (n, k) denote the set of all partitions of {1, 2,, n} into k non-empty subsets. For instance, for n = 4andk = 2thereareseven such partitions: {1, 2, 3} {4}, {1, 2, 4} {3}, {1, 3, 4} {2}, {2, 3, 4} {1}, {1, 2} {3, 4}, {1, 3} {2, 4}, {1, 4} {2, 3}. Thus S (4, 2) = 7. The cardinality S (n, k) of such partitions is called the Stirling number of the second kind, and we can compute the number by the recurrence: S (n, k) = S (n 1, k 1) + k S (n 1, k). by a very small amount, and output each object as the difference from the preceding one. Such orderings of objects are known as Gray codes [4], [12] [14]. Let G be a graph, where each vertex corresponds to each object and each edge connects two similar objects. Then the Gray code corresponds to a Hamiltonian path of G. A Gray code for S (n, k) is known in [12]. The main idea is reversing sublists technique based on the recurrence above. The paper also gives a generation algorithm to generate each partition in S (n, k) in constant time in amortized sense. In this paper we give a simple algorithm to generate all partitions in S (n, k), base on a new simple tree structure on the partitions in S (n, k). The algorithm generates each partition in constant time (in ordinary sense). Our algorithm outputs each partition as the difference from the preceding one. The main idea of the algorithm, so called family tree method, is as follows. We first define a rooted tree (See Fig. 1) such that each vertex corresponds to a partition in S (n, k), and each edge corresponds to a relation between two partitions. Then with traversing the tree we generate all partitions in S (n, k). With a similar technique we have already solved some generation problems for graphs [6], [9]. In this paper for the first time we apply the technique for non-graph objects. The rest of the paper is organized as follows. Section 2 introduces the family tree. Section 3 presents our first algorithm. The algorithm generates each partition in S (n, k) in O(1) time on average. In Sect. 4 we improve the algo- The recurrence means that we either put the last object into a class by itself, or we put it together with some non-empty subset of the first n 1 objects [3, p259]. Generally, generating algorithms produce huge outputs, and the outputs dominate the running time of the generating algorithms. So if we can compress the outputs, then it considerably improves the efficiency of the algorithms. Therefore many generating algorithms output objects in an order such that each object differs from the preceding one Manuscript received June 23, Manuscript revised October 3, Final manuscript received December 10, The authors are with the Department of Computer Science, Gunma University, Kiryu-shi, Japan. a) kawano@msc.cs.gunma-u.ac.jp b) nakano@msc.cs.gunma-u.ac.jp DOI: /ietfec/e88 a Fig. 1 The family tree T 5,3. Copyright c 2005 The Institute of Electronics, Information and Communication Engineers

2 KAWANO and NAKANO: CONSTANT TIME GENERATION OF SET PARTITIONS 931 rithm so that it generates each partition in O(1) time. Finally Sect. 5 is a conclusion. 2. The Family Tree In this section we define a tree structure among partitions in S (n, k). For positive integers n and k < n, leta S (n, k) bea partition of {1, 2,, n} into k non-empty subsets B 1 B 2 B k. Assume that the B 1, B 2,, B k are ordered by their smallest elements. For each i = 1, 2,, n, definea i = j so that i B j. Then for each A S (n, k) a unique sequence a 1 a 2, a n is defined. For example, {1, 2, 4} {3} {5} defines We can observe that any sequence derived as above has the following property. (a) If i = 1thena i = 1. (b) If i > 1thena i 1 + max{a 1, a 2,, a i 1 }. (c) k = max{a 1, a 2,, a n }. We call such a sequence a k-restricted growth sequence. We can also observe the following. Let R(n, k) bethesetofallk-restricted growth sequences satisfying (a) (c) above. Let A = a 1 a 2, a n be a k-restricted growth sequence in R(n, k). For each j = 1, 2,, k, defineb j = {i a i = j}. Then each A R(n, k) defines a unique partition B 1 B 2 B k of {1, 2,, n}. For example, defines {1, 2, 4} {3} {5}. Thus there is a simple bijection between S (n, k) and R(n, k), and we can easily translate one into the other. We designate one special sequence k, which corresponds to a special partition {1, 2,, n k + 1} {n k + 2} {n k + 3} {n}, and call it the root sequence. Then we define the parent sequence P(A) for each sequence A in R(n, k) except for the root sequence as follows. Let A = a 1 a 2 a n be a sequence in R(n, k), and assume that A is not the root sequence. We have the following two cases. Case 1: a 1 a 2 a 3 a n. Let b be the maximum i such that a i k (n i) and n k + 1 i n. This means a b is the rightmost disagree between A and the root sequence k. Note that, b 2 holds since a 1 = 1forA and a 1 matches to the root sequence by (a). (Note that we have assumed k < n.) Also b n holds since a n = k by (c) and the condition a 1 a 2 a 3 a n.nowa i = k (n i) holds for each i = b + 1, b + 2,, n, by the choice of b. Properties (b) and (c) mean that A must contain every 1, 2,, k, so the condition a 1 a 2 a 3 a n means a b 2, a b = k (n b)+1 = a b+1,anda b 1 is equal to either a b or a b 1. In this case we subtract one from a b, and define P(A) = a 1 a 2 a b 1 (a b 1)a b+1 a b+2 a n. Then the former rightmost disagree integer a b now matches the corresponding integer in the root sequence. Now P(A) S (n, k) holds, because (a) holds since b 1, (b) holds since a b = a b+1 and a b 1 is either a b or a b 1, (c) holds since a n = k in A and b n. For instance, P(112223) = with b = 4, and P(111233) = with b = 5, where the rightmost disagree a b is underlined. Case 2: Otherwise. Let b be the maximum i such that 1 i < n and a i > a i+1. Note that b 1sincea 1 = 1by(a). In this case we swap a b and a b+1, and define P(A) = a 1 a 2 a b 1 a b+1 a b a b+2 a b+3 a n. The swapped a b and a b+1 are underlined. Now P(A) R(n, k) holds, because (a) holds since b 1, (b) holds since the swap a b and a b+1 (< a b )never destroys the property (b), and (c) holds since we only swap two integers. For instance, P(122131) = with b = 5, and P(122113) = with b = 3, where the pair (a b, a b+1 ) is underlined. If P(A) is the parent sequence of A then we say A is a child sequence of P(A). Note that A has the unique parent P(A), while P(A) may have many child sequences. We have the following lemma by the case analysis above. Lemma 2.1: If A R(n, k) anda is not the root sequence, then P(A) R(n, k). By the lemma above, given a sequence A in R(n, k), where A is not the root sequence, by repeatedly finding the parent sequence of the derived sequence, we have the unique sequence A, P(A), P(P(A)), of sequences in R(n, k), which eventually s with the root sequence. (If Case 2 occurs then the number of general reverse pair, which is a pair (a i, a j ) such that a i > a j and i < j, decrease, and if the number of general reverse pair reaches to zero, then Case 1 occurs and then the sum of the sequence decrease by one. Thus A, P(A), P(P(A)), never lead into a cycle.) By merging these sequences we have the family tree of R(n, k), denoted by T n,k, such that the vertices of T n,k correspond to the sequences in R(n, k), and each edge corresponds to each relation between some A and P(A). For instance, T 5,3 is shown in Fig. 1, where each dashed line corresponds to the relation with Case 1, and each solid line corresponds to Case 2. This proves that all partitions are in the tree. 3. Algorithm In this section we give an algorithm to construct T n,k and generate all partitions. If we can generate all child sequences of a given sequence in R(n, k), then in a recursive manner we can construct T n,k, and generate all sequences in R(n, k). How can we generate all child sequences of a given sequence? For any sequence P = p 1 p 2 p 3 p n in R(n, k), each child sequence C = c 1 c 2 c 3 c n of P is one of the following two types. Type 1: C satisfies Case 1 in Sect. 2, that is c 1 c 2 c 3 c n holds. Type 2: C satisfies Case 2 in Sect. 2, that is c i > c i+1 holds

3 932 for some i. We need to observe those types more. For a sequence A = a 1 a 2 a n, a consecutive pair (a i, a i+1 )ona is called a reverse pair if a i > a i+1. Type 1: C satisfies Case 1 in Sect. 2. In this case c 1 c 2 c 3 c n holds. Let b be the maximum i such that c i k (n i). Then P = P(C) = c 1 c 2 c b 1 (c b 1)c b+1 c b+2 c n. Note that, since C has no reverse pair, P has either one or zero reverse pair. The only possible reverse pair of P is (p b 1, p b ) = (c b 1, c b 1), since we have subtracted one from c b. Thus if P has two or more reverse pairs, then P has no child sequence with Type 1. We have the following two subtypes. Type 1(a): P has exactly one reverse pair. Let (p x, p x+1 ) be the reverse pair. If p y = p y+1 for some y>x then C never satisfies Case 1. For instance, p 5 = p 6 for P = with x = 2and y = 5 > 2, then P has no child sequence with Type 1, since by adding one to p 3 we have a sequence C = with c 1 c 2 c 3 c n,howeverp(c) = P(122233) = = P. If p x > p x+1 + 1thenalsoC never satisfies Case 1. For instance, p 3 > p 4 + 1forP = with x = 3, then P has no child sequence with Type 1, since after adding one to p 4 the resulting sequence C = still has a reverse pair, a contradiction. Otherwise (1) p n = k, p n 1 = k 1,, p x+1 = k (n (x + 1)), and (2) p x = p x hold. Then P has a child sequence p 1 p 2 p x (p x+1 + 1)p x+2 p x+3 p n with Type 1. For instance, P = has a child sequence , P = has a child sequence , and P = has a child sequence , where each reverse pair is underlined. Type 1(b): P has no reverse pair. Since n > k there is some integer i such that p i = p i+1. Let x be the maximum such integer i. Nowp n = k, p n 1 = k 1,, p x+1 = k (n (x + 1)) hold. In this case we have two subcases. If x n 1, then P has a child sequence C = p 1 p 2 p x (p x+1 + 1)p x+2 p x+3 p n. Note that now c x+1 = c x+2 holds. Otherwise x = n 1 holds, then P has no child sequence with Type 1. For instance, P = has a child sequence , P = has a child sequence , where p x and p x+1 are underlined. On the other hand, P = has no child sequence with Type 1. Thus P has at most one child sequence with Type 1. We have two cases. If P is not the root sequence, then the child sequence can be derived by adding one to p x+1 for both Type 1(a) and (b), where p x is the rightmost disagree between P and the root sequence. Especially if P is the root sequence, then the child sequence can be derived by adding one to rightmost 1 in P. For example if P = then we have a child sequence Type 2: C satisfies Case 2 in Sect. 2. In this case c i > c i+1 holds for some i. Let b be the maximum integer i such that c i > c i+1. Now c b+1 c b+2 c n holds, and the rightmost reverse pair of C is (c b, c b+1 ). By swapping the pair (c b, c b+1 )inc, wederive P = p 1 p 2 p n = c 1 c 2 c b 1 c b+1 c b c b+2 c b+3 c n. The swapped pair is underlined. Note that c b+2 c b+3 c n still holds as it was. Let x be the maximum integer i such that p i > p i+1. We can observe that x b + 1, since otherwise x b + 2 holds which means (c x, c x+1 ) (c b, c b+1 )is the rightmost reverse pair of C, a contradiction. Similarly, x b holds since if x = b then (c b, c b+1 ) is not a reverse pair in P(C), a contradiction. Therefore, either x = b + 1or x b 1 holds. Thus either b = x 1orb x + 1 holds. Let P[i] be the sequence derived from P by swapping p i and p i+1. We have several cases. If p i = p i+1 then P[i] is not a child sequence of P,since P[i] = P. If p i > p i+1 then P[i] is not a child sequence of P,since then c i < c i+1 holds in P[i], and P(P[i]) P. Otherwise, now p i < p i+1. If both p i and p i+1 are the leftmost occurrences of the integers p i and p i+1 and p i + 1 = p i+1,thenp[i] is not a child sequence of P,sinceP[i] is not a k-restricted growth sequence. For instance, for P = 11123, P[4] = is not a k-restricted growth sequence. Otherwise, now p i < p i+1 and at least one of {p i, p i+1 } is not the leftmost occurrence of the integer. We have the following four cases. For each i, x + 1 i n 1, P[i] is a child sequence. In this case b = i holds. For i = x, P[i] is not a child sequence. For each i, 1 i x 2, P[i] is not a child sequence. Especially, for i = x 1, we have two subcases. If p x 1 p x+1 then P[x 1] is a child sequence, otherwise P[x 1] is not a child sequence. For instance, for P = with x = 4, P[3] = is a child sequence, since P(12312) = 12132, while for P = with x = 4, P[3] = is not a child sequence, since P(12321) = P. Procedure find-all-children(a = a 1 a 2 a n ) {A is the current sequence.} 01 Output H {Output the difference 02 from the preceding sequence.} 03 if A is the root sequence A r then 04 Let a x be the 2nd last 1 in A. 05 else Let a x be the rightmost disagree 06 between A and the root sequence A r. 07 if the number of reverse pairs 08 in A is at most one and 09 either a x = a x+1 + 1ora x = a x+1 then 10 find-all-children (a 1 a 2 11 a x (a x+1 + 1)a x+2 a n ){Type 1} 12 for each i, x + 1 i n 1suchthat 13 a i < a i+1 and at least one of 14 {a i, a i+1 } is not

4 KAWANO and NAKANO: CONSTANT TIME GENERATION OF SET PARTITIONS 933 Fig. 2 The family tree T 6,3. 15 the leftmost occurrence of the integer. 16 find-all-children (A = a 1 a 2 17 a i 1 a i+1 a i a i+2 a n ) {Type 2} 18 if a x 1 a x+1 then 19 find-all-children (a 1 a 2 20 a x 2 a x a x 1 a x+1 a x+2 a n ) {Type 2} Algorithm find-all-partitions(n, k) Output the root sequence A r find-all-children(a r ) Theorem 3.1: The algorithm uses O(n) space and runs in O( S (n, k) ) time. Proof. Since we traverse the family tree T n,k and output each sequence at each corresponding vertex of T n,k, we can generate all the sequences in S (n, k) without repetition. We maintain the following for the current sequence. (1) A list of reversed pairs, in the order of occurrences in the sequence, (2) the consecutively matched subsequences to the root sequence, and (3) consecutive occurrences of the same integers. To maintain those we need O(1) time for each output sequence. Other parts of the algorithm need only constant time for each edge of T n,k. Thus the algorithm runs in O( S (n, k) ) time. Some examples of the family trees are shown in Fig. 1 and Fig. 2, where each dashed line corresponds to Type 1, and each solid line corresponds to the relation with Type Modification The algorithm in Sect. 3 generates all sequences in S (n, k) in O( S (n, k) ) time. Thus the algorithm generates each sequence in O(1) time on average. However, after generating a sequence corresponding to the last vertex in a large subtree of T n,k, we have to merely return from the deep recursive call without outputting any sequence. This may take much time. Therefore, we cannot generate each sequence in O(1) time (in ordinary sense). However, a simple modification [10] improves the algorithm to generate each sequence in O(1) time. The algorithm is as follows. Procedure find-all-children2(a, depth) {A is the current sequence and depth is the depth of the recursive call.} 01 if depth is even 02 then Output A 03 {before outputting its child sequences.} 04 Generate child sequences A 1, A 2,, A x 05 by the method in Sect. 3, and

5 934 Fig. 3 A Gray code for S (5, 3). 06 recursively call find-all-children2 07 for each child sequence. 08 if depth is odd 09 then Output A 10 {after outputting its child sequences.} One can observe that the algorithm generates all sequences so that each sequence can be obtained from the preceding one by tracing at most three edges of T n,k. Note that if A corresponds to a vertex v in T n,k with odd depth, then we may need to trace three edges to generate the next sequence. Otherwise we need to trace at most two edges to generate the next sequence. Thus we can generate the next sequence by a constant number of operations. Therefore, we can generate each sequence in constant time. Note that each sequence is similar to the preceding one, since it can be obtained with at most three operations. See Fig. 3. Thus, we can regard the derived sequence of the sequences as a combinatorial Gray code [4], [12], [14], [15] for partitions. 5. Conclusion In this paper we gave a simple algorithm to generate all partitions in S (n, k). The algorithm generates each partition in constant time. By choosing k = 1, 2,, n we can also generate all partitions of {1, 2,, n} into subsets. The number of such partitions is known as the Bell number. [8] B.D. McKay, Isomorph-free exhaustive generation, J. Algorithms, vol.26, pp , [9] S. Nakano, Efficient generation of plane trees, Inf. Process. Lett., vol.84, pp , [10] S. Nakano and T. Uno, A simple constant time enumeration algorithm for free trees, IPSJ Technical Report, 2003-AL-91-2, [11] R.C. Read, How to avoid isomorphism search when cataloguing combinatorial configurations, Annals of Discrete Mathematics, vol.2, pp , [12] F. Ruskey, Simple combinatorial Gray codes constructed by reversing sublists, Proc. ISAAC93, LNCS 762, pp , [13] K.H. Rosen, ed., Handbook of Discrete and Combinatorial Mathematics, CRC Press, Boca Raton, [14] C. Savage, A survey of combinatorial Gray codes, SIAM Review, vol.39, pp , [15] H.S. Wilf, Combinatorial Algorithms: An Update, SIAM, [16] R.A. Wright, B. Richmond, A. Odlyzko, and B.D. McKay, Constant time generation of free trees, SIAM J. Comput., vol.15, pp , Shin-ichiro Kawano received B.E. and M.E. from Gunma University in 2001 and 2003, respectively. He is currently a doctoral student in the Graduate School of Engineering, Gunma University. His research interests include combinatorial algorithms and graph algorithms. Shin-ichi Nakano received B.E., M.E. and Ph.D. from Tohoku University in 1985,1987 and 1992, respectively. He is currently a Professor in the Department of Computer Science, Faculty of Engineering, Gunma University. His research interests include combinatorial algorithms and graph algorithms. References [1] T. Beyer and S.M. Hedetniemi, Constant time generation of rooted trees, SIAM J. Comput., vol.9, pp , [2] L.A. Goldberg, Efficient Algorithms for Listing Combinatorial Structures, Cambridge University Press, New York, [3] R. Graham, D.E. Knuth, and O. Patashnik, Concrete Mathematics, Addison-Wesley, [4] J.T. Joichi, D.E. White, and S.G. Williamson, Combinatorial Gray codes, SIAM J. Comput., vol.9, pp , [5] D.L. Kreher and D.R. Stinson, Combinatorial Algorithms, CRC Press, Boca Raton, [6] Z. Li and S. Nakano, Efficient generation of plane triangulations without repetitions, Proc. ICALP2001, LNCS 2076, pp , [7] G. Li and F. Ruskey, The advantage of forward thinking in generating rooted and free trees, Proc. 10th Annual ACM-SIAM Symp. on Discrete Algorithms, pp , 1999.