Independent even cycles in the Pancake graph and greedy Prefix reversal Gray codes.

Transcription

1 Graphs and Combinatorics manuscript No. (will be inserted by the editor) Independent even cycles in the Pancake graph and greedy Prefix reversal Gray codes. Elena Konstantinova Alexey Medvedev Received: date / Accepted: date Abstract Any l cycle, 6 l n!, can be embedded in the Pancake graph P n, n 3. In this paper we prove the existence of a family of maximal sets of even independent cycles with length bounded by O(n ). We introduce the new concept of Prefix reversal Gray codes based on the independent cycles which extends the known greedy Prefix reversal Gray code constructions considered by Zaks and Williams. We derive the cases of non-existence of such codes using the fastening cycle approach. Along with that we give necessary condition for the existence of greedy Prefix reversal Gray codes in the graph. Keywords Pancake graph Pancake sorting even cycles prefix reversal Gray codes fastening cycle Mathematics Subject Classification (000) 05C15 05C5 05C38 90B10 1 Introduction The Pancake graph P n = (Sym n,pr),n, is the Cayley graph on the symmetric group Sym n of permutations π = [π 1 π...π n ] written as strings in one line notation, The research was supported by Grant of the Russian Foundation of Basic Research and by Grant NSh of President of Russia for Leading Scientific Schools. E. Konstantinova Sobolev Institute of Mathematics, Acad. Koptyug av. 4, Novosibirsk, , Russia Novosibirsk State University, Pirogova str., Novosibirsk, , Russia Tel.: e konsta@math.nsc.ru A. Medvedev Sobolev Institute of Mathematics, Acad. Koptyug av. 4, Novosibirsk, , Russia Central European University, Nador ut. 9, Budapest, 1051, Hungary Tel.: an medvedev@yahoo.com

2 Elena Konstantinova, Alexey Medvedev where π i = π(i) for any 1 i n, with the generating set PR = {r i Sym n : i n} of all prefix reversals r i reversing the order of any substring [1,i], i n, of a permutation π when multiplied on the right, i.e. [π 1 π...π i π i+1...π n ]r i = [π i...π π 1 π i+1...π n ]. An edge corresponding to a prefix reversal r i is called an r i edge. This graph is a connected vertex transitive (n 1)-regular graph of order n!. It is Hamiltonian [10], moreover, it has the structure such that all cycles of length l, where 6 l n!, can be embedded in the Pancake graph P n,n 3, but there are no cycles of length 3,4 or 5 [3,7,8]. A cycle of length l is also called an l cycle. The following results are known for 6 and 8 cycles in the Pancake graph. We shortly write C l = (r a r b ) k for a cycle form C l = r a r b...r a r b, where l = k,a b, and r a r b appears exactly k times. The form C l = (r a r b ) k is canonical if a > b. Vertexdisjoint cycles covering all vertices of the graph form a maximal set of independent cycles. Theorem 1 [4,5] A maximal set of independent 6 cycles in P n,n 3, contains n! 6 cycles presented by the only canonical form C 6 = ( ) 3. A maximal set of independent 8 cycles in P n,n 4, contains n! 8 cycles. In particular, there are only two canonical forms, namely C8 1 = ( ) 4 and C8 = ( ) 4, giving a maximal set in the Pancake graph P 4. The study of Gray codes was started after the introduction of a Binary Reflected Gray code on a set of binary strings [1]. The code itself can be interpreted as a Hamiltonian cycle in the n cube, and there is a bijective correspondence between the set of n bit cyclic Gray codes and the set of Hamiltonian cycles in the n cube [9]. Further, the definition of combinatorial Gray codes has been introduced as a method of generating combinatorial objects so that successive objects differ in some pre specified small way [,6]. Thus, by setting the Cayley graph with the vertex set of permutations and by describing Hamiltonian cycles in this graph we shall refer to Gray codes implicitly. The notion of the Prefix reversal Gray code as an outstanding class was first introduced in 013 by A. Williams and J. Sawada [1], and was suggested as a Hamiltonian cycle algorithm in the Pancake graph. The hamiltonicity of the Pancake graph was known since 1984, when S. Zaks [10] has introduced the algorithm of successive generation of permutations by suffix reversals, which are obviously isomorphic to prefix reversals. This algorithm gives another example of the Prefix reversal Gray code and relies on the hierarchical structure of the graph. The graph P n, n 3, is constructed from n copies P n 1 (i), 1 i n, with the vertex set {[π 1 π...π n 1 i]}, and vertices between copies are connected with external edges from the set {{[ jπ...π n 1 i],[iπ n 1...π j]}, 1 i < j n}, where π k {1,...,n}\{i, j}, k n 1. For example, P 3 is isomorphic to a 6 cycle, and P 4 is constructed from 4 copies of P 3 = C6. Williams [11] gave an algorithmic interpretation of two above mentioned Hamiltonian cycles constructions called the greedy approach. Consider the ordered set of prefix reversals GR = {r m1, r m,...,r mk } PR, where k n 1. Then the greedy Hamiltonian cycle is called a Hamiltonian cycle formed by consecutive application of the leftmost suitable prefix reversal from the set GR. In this setting, we call GR

3 Independent even cycles in the Pancake graph and greedy Prefix reversal Gray codes. 3 to be the greedy sequence. For instance, Williams cycle is formed with greedy sequence GR W = {r n, r n 1,...,, } and Zaks cycle is formed with greedy sequence GR Z = {,,...,r n 1, r n }. We show below that both above Prefix reversal codes follow the certain type of graph hierarchy in their construction. For instance, Zaks construction is based on taking the maximal set of independent 6 cycles of the form C 6 = ( ) 3 and connecting them hierarchically together such that: 1) we start with the Hamiltonian cycle H4 Z in P 4 removing one edge from each of four vertex disjoint 6 cycles C 6 = ( ) 3, getting vertex disjoint 6 paths, and then connecting them by four edges; ) we get the Hamiltonian cycle H5 Z in P 5 removing one edge from each of five Hamiltonian cycles H4 Z in copies P 4(i), 1 i 5, having five vertex disjoint 4 paths, and then connecting them by five r 5 edges; 3) in general, we get the Hamiltonian cycle Hn Z in P n removing one r n 1 edge from each of n Hamiltonian cycles Hn 1 Z in copies P n 1(i), 1 i n, having n vertex disjoint (n 1)! paths, and then connecting them by n external r n edges. Williams construction gives the Hamiltonian cycle Hn W in P n, n 4, based on taking the maximal set of independent n cycles of the form C n = (r n r n 1 ) n, and connecting them together in the following way: 1) we start with the Hamiltonian cycle H4 W in P 4 removing one edge from each of three vertex disjoint 8 cycles C 8 = ( ) 3, getting vertex disjoint 8 paths, and then connecting them by three edges; ) we get the Hamiltonian cycle H5 W in P 5 removing one edge from each of twelve vertex disjoint 10 cycles C 10 = (r 5 ) 4, getting vertex disjoint 10 paths, and then connecting them by or edges such that these edges together with the removed edges constitute the Hamiltonian cycle H4 W ; 3) in general, we get a Hamiltonian cycle Hn W in P n removing one r n 1 edge from each of (n 1)! vertex disjoint n cycles C n = (r n r n 1 ) n, getting (n 1)! vertex disjoint n paths, and then connecting them by r i edges, i n, such that these edges together with the removed r n 1 edges form the Hamiltonian cycle Hn 1 W of previous index, which is based on the maximal set of independent cycles of the form C (n 1) = (r n 1 r n ) n 1. Thus, both algorithms are based on a maximal set of independent cycles of the same form, getting corresponding independent paths and then connecting them in a prescribed way. In this paper we consider a general approach for constructing Prefix reversal Gray codes based on a maximal set of independent cycles in the Pancake graph. Further we omit maximal set for shortness, and call the edges not belonging to any of independent cycles as external edges to these cycles. We say that the Hamiltonian cycle H n in P n,n 3, is based on the independent cycles of length l if it consists of their parts as l paths obtained in the same way as above and connected together by external edges to these cycles. The edges of independent cycles not belonging to the Hamiltonian cycle H n are called unused edges. The external edges along with the unused edges form the fastening cycle H n related to the Hamiltonian cycle H n. For example, Zaks and Williams Hamiltonian cycles in the Pancake graph P 4 are presented schematically in Figure 1 where unrelated edges of the graph are omit-

4 4 Elena Konstantinova, Alexey Medvedev (a) Zaks Hamiltonian cycle H Z 4 (b) Williams Hamiltonian cycle H W 4 Fig. 1: Zaks and Williams Hamiltonian cycles in P 4 ted. The fastening cycle of Zaks Hamiltonian cycle H4 Z is an 8 cycle presented by the canonical form H4 Z = ( ) 4, and the fastening cycle of Williams Hamiltonian cycle H4 W is a 6 cycle presented by the canonical form HW 4 = ( ) 3. In this setting the problem of identifying the maximal sets of independent cycles of the same form constitutes the problem on its own. In this paper we define a family of independent even cycles which covers both of the above mentioned constructions and study its properties in Section. In Section 3 we provide the analysis of the forms of the fastening cycles and study the cases of fastening cycles with which Hamiltonian cycles based on the independent cycles of Lemma 1 do not exist. In Section 4 we prove the necessary condition for the existence of greedy Hamiltonian cycles based on the fastening cycle s approach. Finally, we conclude the paper with Section 5 containing the discussion of open questions. A family of independent even cycles We use a segmented representation of permutations with the following notation. Let π = [π 1...π i...π j...π n ], then a segment α = π i...π j of a permutation π consists of all elements between π i and π j inclusive. If π = [αβ], where α = π 1 π...π i, β = π i+1...π n, then α is the length of the corresponding segment. The segment α with the reversed order of elements is denoted α. By orientation of a segment α, α, we mean the order of its elements: α or α. We consider cycles of a form C l = (r n r m ) k, where l = k, n m, and give conditions on m and k for these cycles to state a family of independent cycles. Lemma 1 The Pancake graph P n,n 4, contains a maximal set of n! l independent l cycles of the canonical form C l = (r n r m ) k, (1)

5 Independent even cycles in the Pancake graph and greedy Prefix reversal Gray codes. 5 with l = k and with one of the following conditions: 1) if m n, then k = 4; ) if n < m < n 1, then { if q, t, p(p + 1), { if q = 1, t, p is even, or q, t = 1, p is odd; k = if q = 1, t, p is odd, or q, t = 1, p is even, p(p + 1), if q = 1, t = 1; p, if q = 0. () where n m, p = n n m, q = n (mod n m) and t = n m q; 3) if m = n 1, then k = n. Proof It is evident that applying an alternating sequence of prefix-reversals to a permutation we eventually arrive at a cycle. The cycles obtained in this way will be independent since every vertex is incident to exactly one r n edge and one r m edge in P n. Moreover, every permutation belongs to a cycle presented by (1). Hence, we have a maximal set of n! l independent cycles in P n,n 4. The length l of cycles (1) is obtained as follows. Since a cycle can be traversed both ways, for the sake of simplicity we shall traverse it backwards, i.e. apply the reverse sequence (r m r n ) k. Let m n. Applying (rm r n ) to the identity permutation we get: ( n ) [1...(m 1)m r m [m(m 1) ( n r m...(n 1)n] ) r (n 1)n] n ( r n n ) [n(n 1) (m 1)m], so that, the sequence (r m r n ) changes the order of elements in the first half, and then inverts the permutation. The halves behave independently under this action, thus it takes four applications of the sequence (r m r n ) to reach the identity permutation, therefore k = 4, and hence l = 8. Let n < m < n 1, then applying (rm r n ) to the identity permutation in this case we get: [13...(m 1)m(m + 1)...(n 1)n] r m r m r [m(m 1)...31(m + 1)...(n 1)n] n r n [n(n 1)...(m + 1)13...(m 1)m], so that, the sequence (r m r n ) reverses the order of the last (n m) segment of the identity permutation and moves it to the beginning. Let us consider a segmented permutation [β 0 α 1 β 1 α β...α p β p ], where p = n n m, αi β i = n m for any 1 i p, and β j = q, where q = n (mod n m), for any 0 j p, thus α i = t = n m q.

6 6 Elena Konstantinova, Alexey Medvedev Then, applying the sequence (r m r n ) to the segmented permutation above, we again get a segmented permutation such that: [β 0 α 1 β 1...α p β p ] (r m r n ) [β p α p β 0 α 1 β 1...α p 1 β p 1 ], where α i is always a starting part and β j is always an ending part of any moving segment α i β j. The action of the sequence (r m r n ) is periodic, namely after a certain number of these actions segments α s and β s return to their initial positions with the initial order of its elements. Therefore, to get the initial permutation by sequential application of (r m r n ), the periods of α s and β s must coincide. The number of α s is p and if they have orientation, i.e. when α i, then the period of α s is p. Similarly, if q 0, then the number of β s is (p + 1) and when β j, then the period of β s is (p + 1). When α s or β s consist of one symbol, the corresponding periods decrease by half. When q = 0 then length of β s is zero, and since n 3, α and only α segments determine the length of a cycle. Simple calculations provide us with the following cases: LCM ( p,(p + 1)), if q, t ; LCM (p,(p + 1)), if q, t = 1; k = LCM ( p, p + 1), if q = 1, t ; LCM (p, p + 1), if q = 1, t = 1; p, if q = 0, where LCM(a,b) stands for the Least Common Multiplier of numbers a and b. Using the fact that for all a,b N the LCM(a,b) GCD(a,b) = a b, where GCD(a,b) stands for the Greatest Common Divisor, we can rewrite cases in the form of (). Let m = n 1 and let π = [π 1 β], where β = π...π n, then π r n r n 1 = [π 1 β]r n r n 1 = [β π 1 ], and hence π (r n r n 1 ) n = [π 1 β](r n r n 1 ) n = [π 1 β], which means the initial vertex is reached after using exactly n prefix reversals. This completes the proof. An edge not belonging to a cycle but connecting two vertices of the cycle is called a chord. Corollary 1 The cycles presented by (1) have no chords. Proof Suppose there exists a chord in a cycle presented by (1) connecting two vertices π and π. Hence there exists a prefix-reversal r i, i {n,m}, such that π (r n r m ) s = πr i = π, s < k. Since the translation φ(π) = π r n r m performs a rotation of vertices along a cycle, the same holds for vertices φ(π) and φ(π ). Therefore, there should appear a 6-cycle of the form C 6 = (r n r m r i ) in P n, n 4, which contradicts Theorem 1 saying that there is the only form C 6 = ( ) 3 for 6 cycles in P n. 3 Fastening cycles We consider Hamiltonian cycles H n based on the independent cycles presented by (1). By Corollary 1 these cycles are non-chordal, therefore any external edge to them is adjacent to another independent cycle. In this case the related to H n fastening cycle

7 Independent even cycles in the Pancake graph and greedy Prefix reversal Gray codes. 7 H n has to have an alternating form with one edge from the cycle C l = (r n r m ) k and one external edge to these cycles, thus H n has form H n = H n (η,ξ ) = (r η r ξ ) t, (3) where r η {r n,r m } and r ξ PR\{r n,r m }, i.e. on every application r η and r ξ are permitted to take any value from the domain. The power t is hypothetical and is not specified directly. Since we restrict the form of independent cycles to C l = (r n r m ) k, then we can infer the structure of the fastening cycle from the structure of the Hamiltonian cycle and vice versa, therefore, we refer to a pair (H n,h n ) as an explicit description of a Hamiltonian cycle based on the independent cycles of the prescribed form C l = (r n r m ) k. In this section we investigate non-existence of Hamiltonian cycles (H n,h n ) due to the structure of fastening cycles. First we consider the fastening cycles of forms H n = (r m r j ) t and H n = (r n r j ) t for j n,m and prove that Hamiltonian cycles (H n,h n ) only exist in the Pancake graph P 4. We present the complete description of these cycles in Theorem 3. Then we consider the richer form H n = (r m r ξ ) t and prove the non-existence of Hamiltonian cycles (H n,h n ) for almost all m. We start with a technical lemma that gives a length of H n. Lemma In the Pancake graph P n,n 4, let H n be a Hamiltonian cycle based on the independent cycles of length l. Then the length of a fastening cycle H n is given by H n = n! l. (4) Proof By the definition of a maximal set of independent cycles the number of independent cycles is n! l. The number of unused edges equals to the number of external edges and equals to the number of independent cycles, thus the length is given by (4). 3.1 Fastening cycles H n (m, j) and H n (n, j) Let us fix the value of r ξ = r j and r η = r n or r η = r m in (3), i.e. we restrict ourselves to use only one type of the reversals on every application, and let us consider the fastening cycles of the forms H n (n, j) = (r n r j ) t and H n (m, j) = (r m r j ) t. Theorem In the Pancake graph P n, n 5, there are no Hamiltonian cycles (H n,h n ) with fastening cycles H n (n, j) or H n (m, j). Proof Assume there is a Hamiltonian cycle (H n,h n ) in P n with the fastening cycle of the form H n (n, j) = (r n r j ) t. Since H n (n, j) and independent cycles are defined by forms (3) and (1), then the length of the fastening cycle H n must coincide with one of the cycle lengths from Lemma 1. Let L max be the maximal length of cycles of the form C l = (r n r m ) k, then the following inequality should hold: n! L max H n (n, j) L max. (5)

8 8 Elena Konstantinova, Alexey Medvedev From Lemma 1, we have L max 4p(p + 1) which gives us the following estimation: ( ( )) n n L max n ( n ) n m n m + 1 = n(n + ), (6) and from (5) we have: n! Lmax. Finally, substituting the inequality (6) we get: n! 1 n (n + ). The factorial function grows faster than any polynomial, so the inequality does not hold for n 7. The inequality does not also hold for n = 5,6, which is verified using the exact lengths of independent cycles and the fastening cycle H n (n, j). Hence, we have a contradiction and there is no Hamiltonian cycle (H n,h n ) in P n with the fastening cycle of the form H n (n, j). The same type of arguments are used to show that there is no a Hamiltonian cycle (H n,h n ) with the fastening cycle of the form H n (m, j). Theorem holds for P n with n 5. The case of n = 4 is given as follows. Theorem 3 In the Pancake graph P 4 there are only four Hamiltonian cycles constructions based on the independent cycles. Proof The collection of all possible maximal sets of independent cycles of the same form in P 4 is presented below in the Table 1. Table 1: Maximal sets of independent cycles in P 4 6 cycles 8 cycles 1 cycles C 6 = ( ) 3 C8 1 = ( ) 4 C1 1 = ( ) C8 = ( ) 4 C1 = ( ) In the case of independent 6 cycles, the unused edges correspond to the prefix reversal. Therefore, in order to obtain a Hamiltonian cycle one should fasten them with an 8 cycle every second reversal of which is. By Theorem 1, there are only two canonical forms of 8 cycles in P 4, namely, C8 1 and C 8. Hence, the corresponding Hamiltonian cycles are: 1. Zaks Hamiltonian cycle H4 Z in P 4, presented in Figure 1a;. Hamiltonian cycle H4 in P 4, presented in Figure a. In the case of independent 8 cycles, the fastening cycle must have length 6, thus it is unique by Theorem 1. Therefore, we have: 1. Williams Hamiltonian cycle H4 W in P 4, presented in Figure 1b;. Hamiltonian cycle H4 4 in P 4, presented in Figure b. In the case of independent 1 cycles, the fastening cycle must be of length 4. Since there are no 4 cycles in the Pancake graph, hence Hamiltonian cycles based on these independent cycles don t exist. Thus, all possible cases are considered and all Hamiltonian cycles based on the independent cycles in P 4 are presented in the Table.

9 Independent even cycles in the Pancake graph and greedy Prefix reversal Gray codes. 9 Table : Hamiltonian cycles based on the independent cycles in P 4 H4 i H4 i Description H4 1 = (( ) ) 4 H4 1 = ( ) 4 Zaks Hamiltonian cycle; H4 = (( ) ) 4 H4 = ( ) 4 based on independent cycles C 6 ; H4 3 = (( ) 3 ) 3 H4 3 = ( ) 3 Williams Hamiltonian cycle; H4 4 = (( ) 3 ) 3 H4 4 = ( ) 3 based on independent cycles C 8. (a) Hamiltonian cycle (H 4,H 4 ) in P 4 (b) Hamiltonian cycle (H 4 4,H4 4 ) in P 4 Fig. : Hamiltonian cycles (H 4,H 4 ) and (H4 4,H4 4 ) in P Fastening cycles of type H n (m,ξ ) Let us relax the condition on r ξ and fix r η = r m and consider the fastening cycle H n of the form H n (m,ξ ) = (r m r ξ ) t. For instance, the Williams prefix reversal Gray code corresponds to the Hamiltonian cycle (H n,h n (m,ξ )) in P n,n 4, when m = n 1. In general we prove the following non-existence result. Theorem 4 In the Pancake graph P n, n 5, there are no Hamiltonian cycles (H n,h n ) with the fastening cycle H n (m,ξ ) for m {,...,n 3}. Proof The fastening cycle H n (m,ξ ) in this case may consist of prefix-reversals up to r n 1, hence the length of H n (m,ξ ) is at most (n 1)!, which coincides with the length of a Hamiltonian cycle in P n 1. This provides us with a key estimate we use further. The case 1) of Lemma 1 gives the length l = 8 of independent cycles, and then H n (m,ξ ) should be of the following length (n 1)! H n (m,ξ ) = n! l = n! 4, which does not hold for n 5. The case of n = 4 is given by Theorem 3.

10 10 Elena Konstantinova, Alexey Medvedev Let us consider the case ) of Lemma 1, when q 0. Notice that the fastening cycle H n (m,ξ ) does not contain r n edges, hence goes through the vertices of the only copy of the graph P n 1. For the sake of simplicity, let us fix this copy to P n 1 (n). However, every independent cycle has vertices from p + 1 o(p + 1) copies, depending on the length q of β segments and the labels of these copies are determined by the first and the last elements of β s. Since α 0, there are vertices with element n in one of the α segments. Thus, there are independent cycles without vertices from the chosen copy P n 1 (n). Since H n (m,ξ ) must fasten all independent cycles, then in this case the Hamiltonian cycle H n with the above defined fastening cycle does not exist. Now let us consider the case ) of Lemma 1, when q = 0. Then independent cycles are presented by the permutations consisting only of α segments, and the length of independent cycles is l = k = 4 n n m, so the length of H n(m,ξ ) should be as follows: (n 1)! H n (m,ξ ) = n! l which does not hold for n m 3. m = (n 1)!n, 4 Necessary condition for the existence of the greedy Prefix reversal Gray codes The only two known Hamiltonian cycles that are based on greedy approach are Zaks and Williams. The challenge is to find other greedy sequences or to prove that they don t exist. In this Section we prove the necessary condition for the greedy sequences to exist. Suppose we have the greedy sequence GR = {r m1, r m,...,r mk }, where k n 1. Then by Lemma 1 we have maximal sets of independent cycles of all forms C li = (r mi r mi+1 ) k i, where 1 i k 1. Let us define a sunflower S(l 1,l,...,l i ), 1 i k 1, related to the greedy sequence GR, as an induced subgraph of P n, in the following recursive way: S(l 1 ) is the cycle of form C l1 = (r m1 r m ) k 1; S(l 1,l ) is defined as k disjoint l 1 paths obtained from cycles C l1 by removing one edge r m from each and connecting paths together by r m3 edges attached to the ends of different l 1 paths such that the removed r m edges and new r m3 edges form a cycle C l = (r m r m3 ) k ; S(l 1,l,...,l i ) is defined as k i disjoint paths obtained from sunflowers S(l 1,l,..., l i 1 ) by removing one edge r mi from each and connecting paths together by r mi+1 edges attached to the ends of different paths such that the removed r mi edges and new r mi+1 edges form a cycle C li = (r mi r mi+1 ) k i. Notice that the sunflower S(l 1,l,...,l i ) is a cycle for every 1 i k 1, hence it is natural to define its length recursively: S(l 1 ) is the cycle of length l 1, therefore S(l 1 ) = l 1 ;

11 Independent even cycles in the Pancake graph and greedy Prefix reversal Gray codes. 11 S(l 1,l ) is the fastening of k = l cycles of length l 1, therefore S(l 1,l ) = l 1 l ; S(l 1,l,...,l i ) is the fastening of k i = l i sunflowers of previous index, therefore S(l 1,l,...,l i ) = l i S(l l 1,l,...,l i 1 ) = l 1... l i 1 l i. The sunflowers give us an approach to prove the necessary condition for the existence of the greedy Hamiltonian cycles. Theorem 5 Suppose Hn G is a greedy Hamiltonian cycle in the Pancake graph P n, n 4, with the greedy sequence GR = {r m1, r m,...,r mk }, where k n 1. Then for every initial greedy subsequence GR i = {r m1, r m,...,r mi }, where i k, there exists a maximal independent set of sunflowers S(l 1,l,...,l i ) and the length of Hn G satisfies Hn G = n! = 1 k 1 k i=1 l i, (7) where the l i is the length of a cycle of form C li = (r mi r mi+1 ) k i, 1 i k 1. Proof Prove first that having the greedy sequence {r m1, r m,...,r mk }, where k n 1, that produces the Hamiltonian cycle, every initial greedy subsequence {r m1, r m,..., r mi }, where 1 i k, produces a sunflower S(l 1,l,...,l i 1 ). By Lemma 1 the greedy subsequence {r m1, r m } gives the independent cycles or sunflowers S(l 1 ). Suppose we have the greedy subsequence {r m1, r m,...,r mi 1 } that forms the sunflower S(l 1,l,...,l i 1 ), where i k 1. Extend this greedy subsequence with r mi+1. Since this subsequence is a part of the greedy Hamiltonian cycle, thus it cannot create a self-intersecting cycle. Hence adding r mi+1 will produce a sunflower S(l 1,l,...,l i ). Prove now that these forms of sunflowers produce a maximal independent set. The sunflower S(l 1,l,...,l k 1 ) is the Hamiltonian cycle or one independent sunflower. By removing all r mk edges from the sunflower, we split the Hamiltonian l cycle into l k 1 / disjoint (l 1... l k ) paths which cover the vertex set. Closing these paths with r mk 1 edges we obtain the maximal set of independent sunflowers S(l 1,l,...,l k ). Since every sunflower now can be decomposed independently, then proceeding by analogy, we obtain the statement for every S(l 1,l,...,l i 1 ), where i k 1. 5 Concluding remarks and discussion One family of maximal sets of independent cycles has been presented in the paper. These cycles have even length. This result partly provides the answer on the question of existence of independent cycles in the Pancake graph. On the one hand, it is still unknown whether there exist maximal sets of independent cycles of odd length, in particular 7 or 9 cycles. On the other hand, one may consider cycles of fixed length but different forms to get maximal independent set.

12 1 Elena Konstantinova, Alexey Medvedev Open problem 1 Describe all maximal sets of independent cycles in the Pancake graph P n, n 4. The particular form of the Hamiltonian cycle (H n,h n (η,ξ )) has been proposed in this paper. Theorem 3 gives the full description of Hamiltonian cycles (H 4,H 4 (η,ξ )), therefore the following questions will be relative to the Pancake graph P n, n 5. Theorem 4 considers the form of the fastening cycle H n (m,ξ ) = (r m r ξ ) t, where m n 3 and the case m = n 1 corresponds to the Williams Hamiltonian cycle H W n. However, in the case when m = n the existence is still unresolved. Open problem Suppose the fastening cycle H n has form H n (n,ξ ) = (r n r ξ ) t for some t > 0. Is there a Hamiltonian cycle (H n,h n (n,ξ )) in the Pancake graph P n, n 5? The existence of the Hamiltonian cycles (H n,h n (η,ξ )) with the fastening cycle H n (η,ξ ) = (r η r ξ ) t, allowing different first prefix reversals r η {r n,r m } is not studied fully yet. This is the most general type of fastening cycles and is now solved partly only for H n (m,ξ ). Open problem 3 Suppose the fastening cycle H n (η,ξ ) has form H n (η,ξ ) = (r η r ξ ) t, where r η {r n,r m } and r ξ PR\{r n,r m }. Are there any Hamiltonian cycles (H n,h n (η,ξ )) in the Pancake graph P n, n 5, except of the known constructions? In the previous Section the necessary condition for the existence of greedy Hamiltonian cycles has been established. We have performed the computer experiment aimed to find new greedy sequences using the exhaustive search. Theorem 5 has narrowed the set of potential greedy sequences drastically and in the Table 3 we show the number of possible greedy sequences along with the number of those which satisfy the condition of Theorem 5 for n 11. Table 3: Results of the numerical experiment Index of P n P 5 P 6 P 7 P 8 P 9 P 10 P 11 # of all possible GR ! 10! # of GR with proper length However, none of the obtained sequences were Hamiltonian and in general the question of existence of new greedy sequences which produce Hamiltonian cycle is unresolved. Open problem 4 Are there any greedy sequences in the Pancake graph P n, n > 11, that produce Hamiltonian cycle, except of the known ones? References 1. E. N. Gilbert, Gray codes and paths on the n cube, Bell Systems Technical Journal, 37, (1958)

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