International Journal of Combinatorial Optimization Problems and Informatics. E-ISSN:

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1 International Journal of Combinatorial Optimization Problems and Informatics E-ISSN: International Journal of Combinatorial Optimization Problems and Informatics México Karim, Lutful; Khan, Nargis Simple permutations of the classes Av(321, 3412) and Av(321, 4123) have polynomial growth International Journal of Combinatorial Optimization Problems and Informatics, vol. 2, núm. 2, mayoagosto, 2011, pp International Journal of Combinatorial Optimization Problems and Informatics Morelos, México Available in: How to cite Complete issue More information about this article Journal's homepage in redalyc.org Scientific Information System Network of Scientific Journals from Latin America, the Caribbean, Spain and Portugal Non-profit academic project, developed under the open access initiative

2 International Journal of Combinatorial Optimization Problems and Informatics, Vol. 2, No. 2, May-Aug 2011, pp ISSN: Simple permutations of the classes Av(321, 3412) and Av(321, 4123) have polynomial growth Lutful Karim and Nargis Khan* Department of Computing and Information Science University of Guelph, Canada *Department of Computer Science Ryerson University, Canada Abstract. A permutation is called simple if its only blocks i.e. subsets of the permutation consist of singleton and the permutation itself. For example, 2134 is not a simple permutation since it consists of a block 213 but 3142 is a simple permutation. The basis of a class of permutations is a set of patterns, which is minimal under involvement and do not belong to the permutation. In this paper we prove that the number of simple permutations a n of the pattern class Av(321, 3412) follows the recurrence a n = a n 1 +a n 2 for n 4 and the pattern class Av(321, 4123) follows the recurrence a n = a n 2 +a n 3 for n 7. Thus, these pattern classes have polynomial growth. 1 Introduction Permutation is a bijection f : [n] [n] of the set [n]. Total number of permutations of length n = n!. For example, 312 is permutation of length 3. Here, f:[3] [3] is defined by f(1)=3, f(2)=1, f(3)=2. Sn= set of all permutations of length n. For example, S2 = 12, 21 Let α and β be two permutations. β is involved in α (denoted by β α) if β is order isomorphic to a subsequence of α. For example, the permutation is involved in as is order isomorphic to the subsequence of the permutation A set of permutations A is called a pattern class or simply a class if, whenever α is a permutation in A and β is involved in α then β A. For example, the class of all increasing permutations is a pattern class. The basis of a class A is the set of permutations which are minimal under involvement and do not belong to A. For example, the pattern class whose basis is 21 consists of only the increasing permutations I = 1, 12, 123, 1234 etc. Thus, I is the class of permutations restricted by 21 and is also denoted I = Av(21). Enumerating the permutations of a pattern class with a finite basis is one of the major research issues in this area. A block of a permutation is a set of consecutive positions that also has consecutive values. A permutation is simple if its only blocks are singletons and the permutation itself. For example, the permutation is not simple as it has a nontrivial block On the other hand, 2413, 3142 and are simple permutations. A pattern class Av(α) has polynomial growth if the number of permutations a n of this class follows bound: a n (α) An m where A and m are some constants. Simple permutations play an important role in the study of permutation pattern classes where, permutation classes are used in pattern recognition, machine learning, search and optimization. Hence, studying the structure of simple permutations has a great importance. Albert and Atkinson have shown that a pattern class containing only finitely many simple permutations have algebraic generating function. [2,8,9].The work done by Brignall et al [10] prove that a sufficiently long simple permutation in fact contains two nearly disjoint simple sub-permutations. The work done by Robert Brignall [7] presents properties such as enumeration, partially well-ordered pattern classes containing finitely many simple permutations. In this paper, we investigated the structures of the pattern classes Av(α, β) with two basis α and β of length 3 and 4 where, α = 321 and β= 3412 or 4123 and it is found that simple permutations of Av(321, 3412) and Av(321, 4123) have polynomial growth. The remainder of this paper is organized as follows.

3 Section 2 present and proves different lemmas to show that the generating function of Av(321, 3412) is polynomial. Similarly, Section 3 proves with lemmas, conjectures and diagrams that the number of simple permutations Av(321, 4123) has polynomial growth. Finally, section 4 concludes the paper with some future research directions. 2 Av(321, 3412) Let σ Av(321, 3412), σ = n. Consider the position where n lies in σ. The elements to the right of n are increasing (to avoid 321). The structure of the permutations avoiding 321 and 3412 is shown in figure 17. When at least 2 elements x 1, x 2 exist to the right of n the shaded region Z would be empty otherwise, ynx 1 x 2 would form 3412 where, y Z. Lemma 1. When 2 elements x 1, x 2 exist to the right of a simple permutation of Av(321, 3412) then the region X is empty. Proof. Assume the region X is not empty. Therefore, n 1 X. Then n 1 must be separated by position from x 2 by an element y and y < x 2. Thus, n, x 2, y would form a subsequence order isomorphic to 321 (see figure 2). Fig. 1. Av(321, 3412) Fig. 2. Fig. 3. Therefore, n 1 cannot exist in the region X. So, X is empty (see figure 3). Thus, there are maximum 2 increasing elements x 1, x 2 to the right of n. But there should be at least one element to the right of n as n cannot be the last element of a simple permutation (see Figure 3). First, we consider only one element x 1 exists to the right of n. The elements in the region Z must be increasing as in the figure 4 otherwise, y 1, y 2, x 1 forms a subsequence order isomorphic to 321 for two decreasing elements y 1, y 2 Z (see Figure 5). x 1 n 1 (to be simple). Now assume that x 1 < n 2. To satisfy the simplicity y separates n 1 and n by position and y < x 1. Therefore, n 2n 1yx 1 (black circles) would form a subsequence order isomorphic to 3412 (see Figure 6). For example, let x 1 = n 3. Then n 2, n 1, y, n 3 forms a subsequence order isomorphic to 3412 (see Figure 7). Hence x 1 = n 2. Therefore, when n is at the (n 1)th position the element at nth position is n 2. When x 1 = n 2, the element y that separates n 1 and n cannot be n 6. Otherwise, n 5, n 4, z, y would form a subsequence order isomorphic to 3412 where, z separates n 4 and n 3 (see figure 8). Therefore, y = n 5 (see figure 9), y = n 4 (see figure 10) or y = n 3. When y = n 3 (see figure 11), to avoid block composition (with n 1, n 3, n, n 2), n 1 and y should be separated by z. If z n 7 then a subsequence order isomorphic to 321 or 3412 would form (can be checked using a similar structure). Therefore, z = n 5 or n 6. Now, we consider 2 increasing elements x 1, x 2 exist to the right of n (see figure 12).

4 Fig. 4. Fig. 5. Fig. 6. Fig. 7. Fig. 8. Fig. 9. Fig. 10. Fig. 11. Fig. 12.

5 x 1 n 1 otherwise, n, x 1 would form a block. Hence, x 2 = n 1 and x 1 n 2. Let x 1 n 5. Elements in the region Y cannot be decreasing otherwise, y 1, y 2, x 1 would form a subsequence order isomorphic to 321 where, y 1, y 2 are 2 decreasing elements of Y (See Figure 13). Fig. 13. Fig. 14. Fig. 15. In figure 14, the element y separates n 3 and n 2 by position. As y < x 1 (otherwise would form 321), n 4, n 3, y, x 1 (black circles) would form a subsequence order isomorphic to Hence, at most two elements n 3 and n 2 exist in the area Y. Therefore, x 1 = n 4 (when Y contains n 3 and n 2) or x 1 = n 3 (when Y contains n 2). When x 1 = n 3, the element y that immediately precedes n cannot be n 7 otherwise, n 6, n 5, z 2, y would form a subsequence order isomorphic to 3412 where, z 2 separates n 5 and n 4 and y n 7 (see figure 15). Therefore, y = n 4, n 5 or n 6 when x 1 = n 3. When x 1 = n 4, the element y just to the left of n cannot be n 5 otherwise, n 3, n 2, y, n 4 would form a subsequence order isomorphic to 3412 (see figure 16). Therefore, y = n 2 when x 1 = n 4. Let a n counts the simple permutations that have n in the position n 1 or n 2. Suppose, type A is the simple permutations where n is in the position n 2 and type B is the simple permutations where n is in the position n 1. These two types are inverse of one another and so there are 1 2 a n of each type. Put b n = 1 2 a n and consider b n counts the type A permutations (in this case, the position n 1 contains either n 3 or n 4). b n 1 of these permutations with positions n 1 containing n 3 and b n 2 of these permutations with positions n 1 containing n 4. Summary 1. n lies in the position n 1 or n b n counts the type A simple permutations α 1 α 2 nα n 2 α n 1 where α n 2 = n 3 or n 4 and α n 1 = n Type B simple permutations are the inverse of type A simple permutations. Thus, b n = 1 2 a n counts both type of simple permutations. Conjecture There are b n 1 simple permutations of the form α 1, α 2, n, n 3, n 1 and there are b n 2 simple permutations of the form α 1, α 2, n, n 4, n 1. Lemma 2. If βnxn 1 is simple of length n and belongs to Av(321, 3412) then βn 1x is also simple of length n 1 and belongs to Av(321, 3412) (Here x = n 3) Proof. Let βn 1x is not simple and thus, there exists a block δ in βn 1x. As βnxn 1 is simple β does not have any block. Let δ contains x = n 3 and so also n 1. Then δ x (n 1) must also be a block of β and so, δ x (n 1) = 1 and δ = 3. Thus, δ includes the last element y of β and y should be equal to n 2. But y n 2 otherwise, ynxn 1 would be a block of βnxn 1. Therefore, no block exists in βn 1x and hence, βn 1x is simple of length n 1 and belongs to Av(321, 3412).

6 Lemma 3. If βn 1x is simple of length n 1 and belongs to Av(321, 3412) then βnxn 1 is also simple of length n and belongs to Av(321, 3412) (Here x = n 3) Proof. Suppose βnxn 1 is not simple. Therefore, βnxn 1 contains a block δ. Let δ includes n 1 and x = n 3. As βn 1x is simple, n 1x is not a block. So, δ also includes n and z where, z is the last element of β. δ to be a block, z should be equal to n 2. But δ n 2 (figure 15). As δ includes three elements n 1, x and z of βn 1x and βn 1x is simple, δ should include all the elements of βn 1x (If a block X contains two or more elements of a simple permutation π then X contains all the elements of π). Therefore, βnxn 1 is simple and belongs to the class Av(321, 3412). From Lemma 8 and 9, we see that there is a one-one correspondence between the simple permutation βn 1n 3 of length n 1 and simple permutations βnn 3n 1 of length n. Corollary 1. The number of simple permutations in Av(321, 3412) of the form βnn 3n 1 is b n 1. Lemma 4. If βnyn 1 is simple of length n and belongs to Av(321, 3412) then βy is also simple of length n 2 and belongs to Av(321, 3412) (y = n 4) Proof. Let βy is not simple and thus, there exists a block in βy. As βnyn 1 is simple β does not have any block. Let contains y = n 4. Then y must also be a block of β and so, y = 1 and = 2. Thus, includes the last element z of β and z should be equal to n 3 or n 5. But we have already shown that when y = n 4 then z = n 2 (see figure 16). Therefore, no block exists in βy and hence, βy is simple of length n 2 and belongs to Av(321, 3412). Lemma 5. If βy is simple of length n 2 and belongs to Av(321, 3412) then βnyn 1 is also simple of length n and belongs to Av(321, 3412) (Here y = n 4) Proof. Suppose βnyn 1 is not simple. Therefore, βnyn 1 contains a block. Let includes n 1 and y = n 4. y, n 1 is not a block and so, also includes n and z where, z is the last element of β. z = n 2 (figure 16). As includes two elements x and z of βy and βy is simple, should include all the elements of βy (If a block X contains two elements of a simple permutation π then X contains all the elements of π). Therefore, βnyn 1 is simple and belongs to the class Av(321, 3412). From Lemma 10 and 5, we see that there is a one-one correspondence between the simple permutation βn 4 of length n 2 and simple permutations βnn 4n 1 of length n. Corollary 2. The number of simple permutations in Av(321, 3412) of the form βnn 4n 1 is b n 2. From corollary 3 and 4, we can conclude that b n = b n 1 + b n 2 where, b n counts the type A simple permutations and b n = 1 2 a n (as type A is the inverse of type B simple permutations). Considering the same number of simple permutations of type B as of type A, we get the following recurrence for the number of simple permutations a n of Av(321, 3412): a n = a n 1 + a n 2 a n a n 1 a n 2 = 0 f(x) xf(x) x 2 f(x) = a 0 + (a 1 a 0 x + (a 2 a 1 a 0 )x 2 + (a 3 a 2 a 1 )x 3 + α n=4 (a n a n 1 a n 2 )x n f(x)(1 x x 2 ) = (x + x 2 3x 3 ) f(x) = x+x2 3x 3 1 x x 2 3 Av(321, 4123) Let σ Av(321, 4123), σ = n. Consider the position where n lies in σ. The elements to the right of n are increasing (to avoid 321). The structure of the permutations avoiding 321 and 4123 is shown in figure 17.

7 Fig. 16. Fig. 17. Fig. 18. The shaded region X is empty otherwise, nx 1 x 2 y would form 4123 where, y X. Thus, there are maximum 2 increasing elements x 1, x 2 to the right of n. But there should be at least one element to the right of n as n cannot be the last element of a simple permutation. First, we consider only one element x 1 exists to the right of n. The elements in the region Z must be increasing otherwise, y 1, y 2, x 1 forms a subsequence order isomorphic to 321 for two decreasing elements y 1, y 2 Z (see Figure 18). x 1 n 1 (to be simple). Now assume that x 1 < n 2. Fig. 19. Fig. 20. Fig. 21. To satisfy the simplicity z 1 separates n 2 and n 1 by position and z 2 separates n 1 and n by position. If z 1 > z 2 then n 2, z 1, z 2 (black circles) would form a subsequence order isomorphic to 321 (see Figure 19). If z 1 < z 2 then n 2, z 1, z 2, x 1 (black circles) would form a subsequence order isomorphic to 4123 (see Figure 20). Hence x 1 = n 2. Therefore, when n is at the (n 1)th position the element at nth position is n 2. Now, we consider 2 increasing elements x 1, x 2 exist to the right of n (See Figure 21). Lemma 6. If 2 increasing elements x 1, x 2 exist to the right of n then the region Z is empty (Figure 24). Proof. Assume the region Z is not empty. Therefore, n 1 Z. Then n 1 must be separated by position from n by an element y. If y lies below x 1 then n 1yx 1 x 2 forms a subsequence order isomorphic to If y lies above x 1 then n 1yx 1 forms a subsequence order isomorphic to 321 (see Figure 22, 23). Therefore, no element lies in the region Z (see Figure 24). x 1 n 1 otherwise, n, x 1 would form a block. Hence, x 2 = n 1 and x 1 n 2. Let x 1 n 5. Elements in the region Y cannot be decreasing otherwise, y 1, y 2, x would form a subsequence order isomorphic to 321 where, y 1, y 2 are 2 decreasing elements of Y. In figures 25 and 26, the element z 1 separates n 4 and n 3 by position and z 2 separates n 3 and n 2 by position. If z 1 > z 2 then n 4, z 1, z 2 (black circles) would form a subsequence order isomorphic to 321 (see Figure 25).

8 Fig. 22. Fig. 23. Fig. 24. Fig. 25. Fig. 26. Fig. 27. If z 1 < z 2 then n 4, z 1, z 2, x 1 (black circles) would form a subsequence order isomorphic to 4123 ( see Figure 26). Hence, there at most two elements n 3 and n 2 exist in the area Y. Therefore, x 1 = n 4 (when Y contains n 3 and n 2) or x 1 = n 3 (when Y contains n 2). Fig. 28. Fig. 29. Fig. 30. Similarly, we can show that when only one element x 1 = n 2 is to the right of n then z 1 = n 4 or n 5 (by considering n 1 as n and two elements z 1, x 1 to the right of n 1, see Figure 27). Let a n counts the simple permutations that have n in the position n 1 or n 2. a n 2 of these permutations are with positions n 2 containing n and a n 3 of these permutations are with positions n 1 containing n. Summary 1. n lies in the position n 1 or n a n counts the simple permutations α 1 α 2 nα n 2 α n 1 where α n 2 = n 3 or n 4 and α n 1 = n a n counts the simple permutations α 1 α 2 α n 2 nα n 1 where α n 2 = n 4 or n 5 and α n 1 = n 2. Conjecture

9 There are a n 2 simple permutations of the form α 1, α 2, n, n 3, n 1 and there are a n 3 simple permutations of the form α 1, α 2, n, n 4, n 1. Lemma 7. If βnxn 1 is simple and belongs to the class Av(321,4123) then βx is also simple and belongs to the class. Proof. There is no nontrivial block contained entirely in β since such a block would be a block of βnxn 1. Suppose βx contains a nontrivial block δ. Then δ x is a block of β. So δ x =1 and δ = 2. We have already shown that if a simple permutation of the form βnxn 1 belongs to Av(321, 4123) then x = n 3 or n 4. First, we consider x = n 4. As δ contains x, the other element of δ should be y = n 3 or n 5. From the figure 28, we see that a subsequence order isomorphic to 321 is formed when y = n 3. From the figures 29 and 30, we see that a subsequence order isomorphic to 4123 is formed when y = n 5. Therefore, y n 3 or y n 5. Therefore, βx does not have any nontrivial block and so, βx is simple and belongs to Av(321, 4123). Now, we consider x = n 3 then the other element of δ should be y = n 2 or n 4. If y = n 2 then y, n, n 3, n 1 would form a block of βnxn 1, a contradiction as βnxn 1 is simple. If y = n 4 then n 2 is to the left of n 4 but n 2 cannot be adjacent to y otherwise, n 2, y, n, x, n 1 would form a block of βnxn 1. Thus n 2 and y = n 4 must be separated by position by an element z and n 2zyx forms a subsequence order isomorphic to 4123 (see Figure 31). Therefore y n 2 or y n 4. Therefore, βx does not have any nontrivial block δ. Lemma 8. If βx is simple of length n 2 and belongs to the class Av(321,4123) then βnxn 1 is also simple of length n and belongs to Av(321,4123). Proof. Suppose βnxn 1 is not simple. Therefore, βnxn 1 contains a block δ. For βnxn 1 to be in Av(321, 4123), there are at most 2 increasing elements to the right of n. Therefore, x (< n 1) is a single element otherwise, there will be more that 2 elements to the right of n. Let δ includes n 1 and x. As βx is simple of length n 2 and belongs to Av(321, 4123), x = n 3 or n 4 (we already have shown). Therefore, x, n 1 is not a block and so, δ also includes n and y where, y is the last element of β. y = n 5 or n 6 when two elements y and x lie to the right of n 2 (figure 32) and y = n 2 when one element x lies to the right of n 2 (figure 33). Fig. 31. Fig. 32. As δ includes two elements x and y of βx and βx is simple, δ should include all the elements of βx (If a block X contains two elements of a simple permutation π then X contains all the elements of π). Therefore, βnxn 1 is simple and belongs to the class Av(321, 4123).

10 From Lemma 7 and 8, we see that there is a one-one correspondence between the simple permutation βx of length n 2 and simple permutations βnxn 1 of length n. Corollary 3. The number of simple permutations in Av(321, 4123) of the form βnxn 1 is a n 2. Lemma 9. If βn 1ynn 2 is simple and belongs to the class Av(321, 4123) then βy is also simple and belongs to the class. Proof. There is no nontrivial block contained entirely in β since such a block would be a block of βn 1ynn 2. Suppose βy contains a block. Then y would also be a block of β. So x =1 and = 2. Fig. 33. Fig. 34. Fig. 35. y n 3 otherwise, n 1ynn 2 would form a block and so, βn 1ynn 2 cannot be simple. If y n 6 then a subsequence order isomorphic to 321 or 4123 would form (see Figure 34 35). Therefore, y = n 4 or n 5. First, we consider y = n 5. As contains y, the other element of should be z = n 4 or n 6. If z = n 4 or n 6 then a subsequence order isomorphic to 321 or 4123 would form (Similar to Lemma 7). Therefore, z n 4 or z n 6. Therefore, βy does not have any nontrivial block and so, βx is simple and belongs to Av(321, 4123). Now, we consider y = n 4 then the other element of should be z = n 3 or n 5. If z = n 3 then z, n 1, n 4, n, n 2 would form a block of βn 1ynn 2, a contradiction as βn 1ynn 2 is simple. If z = n 5 then a subsequence order isomorphic to 4123 is formed (Similar to Lemma 7). Therefore, z n 3 or z n 5. Therefore, βy does not have any nontrivial block. Lemma 10. If βx is simple of length n 3 and belongs to the class Av(321,4123) then β, n 1, y, n, n 2 is also simple of length n and belongs to Av(321,4123). Proof. Suppose β, n 1, y, n, n 2 is not simple. Therefore, β, n 1, y, n, n 2 contains a block. Let includes n 2. Then must include n, y, n 1. As βy is simple and belongs to Av(321, 4123), y n 3 (it was shown that y = n 4 or n 5). Therefore, n 1, y, n, n 2 is not a block and so, also includes z where, z is the last element of β. As includes two elements y and z of βy and βy is simple, should include all the elements of βy. Therefore, β, n 1, y, n, n 2 is simple and belongs to the class Av(321, 4123). From Lemma 9 and 10, we see that there is a one-one correspondence between the simple permutation βy of length n 3 and simple permutations β, n 1, y, n, n 2 of length n. Corollary 4. The number of simple permutations in Av(321, 4123) of the form βn 1ynn 2 is a n 3. From Corollary 3 and 4, we can conclude that the enumeration of simple permutations of Av(321, 4123) follows the recurrence: a n = a n 2 + a n 3 a n a n 2 a n 3 = 0

11 4 Conclusion and Future Works Permutation pattern classes are used in machine learning, pattern recognition, search and optimizations. Hence, it is very important to study the structure of simple permutations and identify their properties. In this paper we prove that Av(321, 3412) and Av(321, 4123) have polynomial growth. More research can be done in this area to see whether any more non-isomorphic pattern classes with two basis of length 3 and 4 or both of length 4 with polynomial growth exists and to generalize the criteria of a pattern class to have polynomial growth. References 1. M. H. Albert and M. D. Atkinson. Simple permutations and pattern restricted permutations. Discrete Math., 300(1-3):1 15, M. H. Albert, M. D. Atkinson, and M. Klazar. The enumeration of simple permutations. J. Integer Seq., 6(4):Article , 18 pp. (electronic), M. Atkinson, M. Murphy, and N. Ruskuc. Partially well-ordered closed sets of permutations, M. D. Atkinson. Restricted permutations. Discrete Math., 195(1-3):27 38, M. D. Atkinson and T. Stitt. Restricted permutations and the wreath product. Discrete Math., 259(1-3):19 36, Miklos Bona. Combinatorics of Permutations. CHAPMAN HALL/CRC, Robert Brignall. A survey of simple permutations, Robert Brignall, Sophie Huczynska, and Vince Vatter. Decomposing simple permutations, with enumerative consequences, Jun Robert Brignall, Sophie Huczynska, and Vince Vatter. Simple permutations and algebraic generating functions, August Robert Brignall, Sophie Huczynska, and Vincent Vatter. Decomposing simple permutations, with enumerative consequences. Combinatorica, 28(4): , Adam Marcus and Gábor Tardos. Excluded permutation matrices and the Stanley-Wilf conjecture. J. Combin. Theory Ser. A, 107(1): , M.M Murphy. Restricted Permutations, Antichains, Atomic Classes and Stack Sorting. PhD thesis, University of St. Andrews, Daniel A. Spielman and Miklós Bóna. An infinite antichain of permutations. Electron. J. Combin., 7:Note 2, 4 pp. (electronic), 2000.