Maximum cycle packing in three families of Petersen graphs

Transcription

1 Technische Universität Dortmund Wirtschafts- und Sozialwissenschaftliche Fakultät Diskussionsbeitrge des Fachgebietes Operations Research und Wirtschaftsinformatik Maximum cycle packing in three families of Petersen graphs Peter Recht 1 Eva-Maria Schulte-Loh Technische Universität Dortmund, Operations Research und Wirtschaftsinformatik, Vogelpothsweg 87, D 221 Dortmund, Germany, peter.recht@udo.edu 1 1 corresponding author

2 Maximum cycle packing in three families of Petersen graphs Abstract: For k {1, 2, 3} let P(k) be the family of the generalized Petersen graphs. The paper addresses to the packing number ν of a graph (P (n, k) P(k), i.e. the maximal number of edge-disjoint cycles in P (n, k) and a corresponding cycle packing Z (n, k). For graphs in these families the packing number can be given in closed form: it will be shown that ν(p (n, 1)) = n 2, if n, ν(p (n, 2)) = n 3, if n 9 and ν(p (n, 3)) = n, if n 28, respectively. Keywords: cycle packing, Petersen graphs, extremal problems in graph theory 1 Introduction 1.1 Cycle packing and cycle packing number of a graph Let G = (V, E) be a graph with vertex set V and edge set E. By V and E, we denote the cardinalities of V and E, respectively. Sometimes we will write V (G) and E(G), when we focus on the vertices and edges of G, respectively. A graph G = (V, E ) is a subgraph of G, if V V and E E. In this case we write G G. Two subgraphs G = (V, E ), G = (V, E ) G are called edge-disjoint if E E =. Let v i1, e 1, v i2, e 2,..., e r 1, v ir be a finite sequence of vertices v ij and pairwise distinct edges e j = (v ij, v ij+1 ) of G. Then the subgraph W with vertices V (W ) = {v i1, v i2,..., v ir } and edges E(W ) = {e 1, e 2,..., e r 1 } is called a walk in G. The number E(W ) denotes the length of the walk. The edge e 1 is the starting edge, the edge e r 1 is the end edge of W. If W is 2

3 closed (i. e. v i1 = v ir ) we call it a circuit in G. A path is a walk W in which all vertices v V (W ) have degree d(v) 2. A closed path C will be called a cycle. The minimal length of a cycle in G is called the girth of G, and is denoted by girth(g). A cycle packing Z = {C 1,..., C l } of G is a collection of l pairwise edge-disjoint cycles in G. The cycle packing problem consists of finding a maximum cycle packing Z, i. e. a cycle packing of cardinality ν(g) = max{l {C 1,..., C l } is a cycle packing of G}. The quantity ν(g) is called the cycle packing number of G. Although a maximum cycle packing relates to a very natural question in graph theory, there are few papers that have focused attention on this topic systematically (see (1), (3), (), (7), (6)),(8), (9), (10), (11)) This paper deals with properties of the cycle packing number for some special families of graphs. 2 Cycle packing in P(1), P(2) and P(3) 2.1 The generalized Petersen graph P (n, k) Let n, k be natural numbers such that n 5 and 1 k < n 2. A graph P (n, k) = (V 1 V 2, E 1 E 2 E 3 ) is called a generalized Petersen graph (2), if V 1 = {v 0, v 1, v 2,..., v n 1 }, V 2 = {u 0, u 1, u 2, u 3,..., u n 1 }, E 1 = {(v 0, v 1 ), (v 1, v 2 ),..., (v n 2, v n 1 ), (v n 1, v 0 )}, E 2 = {(u i, u i+k ) 0 i n 1} 2 and E 3 = {(v i, u i ) 0 i n 1}. For fixed k 2 we denote by P(k) := {P (n, k) n > 2k} the family of Petersen graphs of density k. A graph P (n, k) P(k) obviously has 2n vertices and 3n edges. We will call the sets V 1 and E 1 outer vertices and outer edges, respectively. The sets V 2 and E 2 will be inner vertices and inner edges, respectively, and an element of E 3 is called a joining edge. The subgraph G out := (V 1, E 1 ) P (n, k) is a cycle, and the subgraph G int := (V 2, E 2 ) P (n, k) sometimes is called a (n, k) star-polygon (5). Let Z = {C 1,..., C q } be a cycle packing in P (n, k). For v V ( e E) we say v is blocked (or e is blocked, respectively) by Z, if there is no cycle 2 subscripts are to read modulo n 3

4 Figure 1: The generalized Petersen graph P (17, 3) C P (n, k), different from those in Z, containing v (or e, respectively) such that {C, C 1,..., C q } are edge disjoint. Obviously, all edges and vertices of cycles in Z are blocked. Figure 2: Yellow vertices and edges are blocked by the black bold cycle 2.2 The family P(1) The family P(1) of generalized Petersen graphs also appears as the family of prism graphs. The following statement on the packing numbers is easy to prove. Theorem 1 Let n. Then ν(p (n, 1)) = n 2

5 Proof: Observe, that for n ν(p (n, 1)) n 2, since there is packing Z = {C 1, C 2,...C n 2 }, in which each of the edge sets E(C i ) consist of exactly one inner edge, one outer edge and two joining edges. But, since girth(p (n, 1)) =, the packing number satisfies ν(g) 2n, i.e. ν(p (n, 1)) = n 2. Remark:1 Obviously, ν(p (3, 1) = The family P(2) For the family P(2) of generalized Petersen graphs the following theorem holds Theorem 2 Let n 9. Then ν(p (n, 2)) = n 3 To prove the theorem, we will formulate some lemmas. The first lemma is straight-forward and its proof is omitted. Lemma 3 (i) If n 9, then ν(p (n, 2)) 2n 5 (ii) If n 9, then n 3 ν(p (n, 2)) (iii) 0 n mod 2 G int decompose into two cycles each of length n 2. (iv) 1 n mod 2 G int is a cycle. Lemma Let n 9. Assume ν(p (n, 2)) > n 3. Let k 0 be the number of cycles that are contained in a maximum cycle packing Z (n, 2) of P (n, 2), having no vertex in V 1. Then k 0 = 0. Proof: First note, that k 0 2. The case k 0 = 2 is impossible, since in this case ν(p (n, 2)) = 3 n 3 would hold. Assume k 0 = 1 and let C (0) 1 Z (n, 2) be the corresponding cycle with only inner edges. Obviously E(C (0) 1 ) E 2, since otherwise ν(p (n, 2)) = 2 < n 3. Therefore, 0 n mod 2, i.e. G int decompose into two cycles of length n 2 5

6 each. One of these cycles is C (0) 1. For every cycle C Z (n, 2) \ C (0) 1 as well V (C) V 1 as V (C) V 2 holds. Note, that all edges e E 3 that are adjacent to V (C (0) 1 ) are blocked by C(0) 1. Moreover V (C) V 1 2, since otherwise 1 n mod 2 would hold. Hence V (C) V 1 3. Therefore, ν(p (n, 2)) 1 + n 3 holds. Using the assumption, the equality ν(p (n, 2)) = 1 + n 3 follows. Next we show, that at least four vertices in V 1 are blocked by {C, C (0) 1 }. Obviously, C cannot block more than five vertices in V 1, otherwise it is impossible to built the n 3 cycles using the n vertices in V 1. V (C) may not contain vertices from V 1, since otherwise E(C) would contain at least one edge in E 3, that is adjacent to a vertex in V (C (0) 1 ). If V (C) V 1 = {v i, v i+1, v i+2 } then E(C) E 3 contains the two joining edges (v i, u i ) and (v i+2, u i+2 ). Obviously, the edges (v i+2, v i+3 ) and (v i, v i 1 ) E 1 are blocked by C, and either (v i+3, u i+3 ) E 3 or (v i 1, u i 1 ) E 3 is blocked by C (0) 1. Hence {C, C(0) 1 } block four vertices in V 1. If we now consider two arbitrary but different cycles C s, C t Z (n, 2)\C (0) at least seven outer vertices are blocked by {C s, C t, C (0) 1 }. But this implies that Z (n, 2) \ C (0) 1 2n 7 Therefore: n 3 < ν(p (n, 2)) = 1 + n n 7, i.e. n 3 < 1 + 2n 7. But for n 9, the inequality n n 7 holds. We conclude k 0 = 1 is impossible. 1, Lemma 5 Let n 9. Assume ν(p (n, 2)) > n 3. Let k 2 the number of cycles that are contained in a maximum cycle decomposition Z (n, 2) of P (n, 2), having exactly two outer-vertices. Then k 2 = 0 Proof: Let C (2) Z (n, 2) be a cycle such that C (2) V 1 = 2. Hence 1 n mod 2 and E(C (2) ) consists either of n 2 or n edges in E 2. Therefore, k 2 2, but indeed k 2 1, since ν(p (n, 2)) > 3. Assume k 2 = 1 and consider the case that C (2) uses n 2 edges in E 2. I.e. the remaining cycles in Z (n, 2)\{C (2) } will use at most n 2 +1 of the remaining edges in E 2. Obviously, there is a path P G int with E(P ) = E 2 \ E(C (2) ). The starting edge and end edge of P are both incident with vertices from V (C (2) ), hence they are blocked by C (2). 6

7 It follows, that at most n = n 2 1 of the edges in E 2 can be used for the cycles in Z (n, 2) \ {C (2) }. But this leads to the contradiction: ν(p (n, 2)) = Z (n, 2) 1 + ( n 2 )/2 n 3, where the last inequality holds for n 9. In the same way the case that C (2) uses n 2 +1 of the edges in E 2 is treated. Lemma 3 and Lemma now complete the proof of Theorem 1: we conclude for n 9 that, if ν(p (n, 2)) > n 3 would hold, all the cycles in Z (n, 2) must have at least 3 outer vertices. But in this case P (n, 2) admits at most n 3 of such cycles. A contradiction. Remark 2: A straight-forward consideration shows that for the packing numbers ν(p (n, 2)), 5 n 8 we get: ν(p (5, 2)) = 2, ν(p (6, 2)) = 3, ν(p (7, 2)) = 2, ν(p (8, 2)) = 3 2. The family P(n, 3) For the family P(n, 3) of generalized Petersen graph the following theorem holds Theorem 6 Let n 28. Then ν(p (n, 3)) = n As before, we will formulate some lemmas in order to prove the theorem. Lemma 7 (i) If n 16, then ν(p (n, 3)) n 3 (ii) If n 7, then ν(p (n, 3)) n (iii) If n 20, then G out / Z (n, 3) (iv) 0 n mod 3 G int decompose into 3 cycles each of length n 3. (v) 1 n mod 3 or 2 n mod 3, respectively G int is a cycle. Lemma 8 Let n 28. Assume ν(p (n, 3)) > n. Let k 0 be the number of cycles that are contained in a maximum cycle packing Z (n, 3) of P (n, 3), 7

8 having no outer-vertices. Then k 0 = 0. Proof: First note, that k 0 3, but the case k 0 = 3 is impossible, since in this case ν(p (n, 3)) = < n would hold. The the situation k 0 = 2 can occur if and only if 0 n mod 3. The two corresponding cycles C (0) 1, C(0) 2 use 2n 3 of the inner edges of P (n, 3). I.e. the cycles in Z (n, 3) \ {C (0) } must have inner edges and they can use at 1, C(0) 2 most n 3 of the remaining ones. If n is even, at most n 6 cycles are possible in Z (n, 3) \ {C (0) 1, C(0) 2 }, if n is odd at most ( n 3 1)/2 cycles are possible in Z (n, 3) \ {C (0) }. Hence, this situations lead to 1, C(0) 2 n < ν(p (n, 3)) = Z (n, 3) 2 + n 6, a contradiction for n 20. Now, assume k 0 = 1 and let C (0) 1 Z (n, 3) be the corresponding cycle. Obviously E(C (0) 1 ) E 2, since otherwise ν(p (n, 3)) = 2 < n. Therefore, 0 n mod 3, i.e. G int decomposes into three cycles of length n 3 each. One of these cycles must be C (0) 1. Every cycle C Z (n, 3) \ C (0) 1 must have as well inner as outer vertices. Note, that all joining edges that are adjacent to V (C (0) 1 ) are blocked by C(0) 1. The vertex set V (C) cannot have exactly two or exactly three outer vertices, since in those cases 1 n mod 3 or 2 n mod 3, respectively, would hold. Hence V (C) V 1. Therefore, ν(p (n, 3)) 1 + n holds. The assumption then implies the equality ν(p (n, 3)) = 1 + n. We now show, that at least five vertices of V 1 are blocked by {C, C (0) 1 }. Obviously, C cannot block more than seven outer vertices (otherwise it would be impossible to get n cycles in Z (n, 3) \ C (0) 1, using the n outer vertices. If C allows exactly four vertices in V 1, then it either has two or four joining edges. In case of two joining edges, let V (C) V 1 = {v i, v i+1, v i+2, v i+3 }. Obviously, both outer edges (v i+3, v i+ ) and (v i, v i 1 ) are blocked by C, and at least one of the joining edges (v i+, u i+ ) or (v i 1, u i 1 ) is blocked by. Hence {C, C(0) 1 } at least block five outer vertices. By the same reason, in the second case (four joining edges) the outer vertices C (0) 1 {v i, v i+1, v j, v j+1 } of V (C) must have the property, that at least one of the vertices {v i 1, v i+2, v j+2 } must be blocked by C. Again, {C, C (0) 1 } block five 8

9 outer vertices. Hence, if we consider two arbitrary but different cycles C s, C t Z (n, 3)\ C (0) 1, at least nine outer vertices are blocked by them. But this implies that Z (n, 3) \ C (0) 1 2n 9 and n < ν(p (n, 3)) = 1 + n 1 + 2n 9, i.e. n < 1 + 2n 9. But for n 28 we just have n 1 + 2n 9. Therefore, we conclude k 0 = 1 is impossible. Lemma 9 Let n 20. Assume ν(p (n, 3)) > n. Let k 2 the number of cycles that are contained in a maximum cycle packing Z (n, 3) of P (n, 3), having exactly two outer-vertices. Then k 2 = 0 Proof: Let C (2) Z (n, 3) with V (C (2) ) V 1 = 2. Hence, either 1 n mod 3 or 2 mod 3. In the first case E(C (2) ) E 2 = n 3, in the second case E(C (2) ) E 2 = n Therefore, k 2 3, but k 2 = 3 is impossible, because ν(p (n, 3)) > 5. Hence k 2 2. Let k 2 = 2 and assume 1 n mod 3. The corresponding cycles C (2) 1, C(2) 2 use 2 n 3 of the inner edge. I.e. the remaining cycles in Z (n, 3)\{C (2) 1, C(2) 2 } will use at most n 2 n 3 = n 3 +1 of the remaining inner edges. G int \ {C (2) } decompose into at most 1, C(2) 2 two paths P 1 and P 2. At least one of these two paths, say P 1, consists of more than one inner edge. But the starting edge and end edge of P 1 are incident with vertices from V (C (2) 1 ) and V (C(2) 2 ), respectively. Therefore, they are blocked by {C (2) 1, C(2) 2 }. Moreover, at least one edge of P 2 is blocked by {C (2) 1, C(2) 2 } (one is blocked if E(P 2) consists of only one inner edge, otherwise two edges are blocked). We conclude, that at most n = n 3 2 of the edges in E 2 can be used for the cycles in Z (n, 3) \ {C (2) 1, C(2) 2 }. Let n 1 and n 2 be the numbers of edges in P 1 and P 2, that can be used for the cycles in Z (n, 3) \ {C (2) 1, C(2) 2 }, i.e. n 1 + n 2 n 3 2. Three possible situations on n 1 and n 2 can occur. If n 1, n 2 are both even numbers, we can have at most n n 2 2 ( n 3 2)/2 n 3 /2 cycles in Z (n, 3) \ {C (2) 1, C(2) 2 }. 9

10 If n 1, n 2 are both odd numbers, we can have at most (n 1+1) 2 + (n 2+1) 2 ( n 3 2)/2 + 1 = n 3 /2 cycles in Z (n, 3) \ {C (2) 1, C(2) 2 }. If n 1 +n 2 is odd, we have have n 1 +n 2 +1 n = n 3 1 and at most (n 1 +n 2 +1) 2 = ( n 3 1)/2 n 3 /2 cycles are possible in Z (n, 3)\{C (2) 1, C(2) 2 }. For all three situations we get the contradiction: n < ν(p (n, 3)) = Z (n, 3) 2 + n 3 /2 n, where the last inequality holds for n 20. Now, let k 2 = 2 and assume 2 n mod 3. In this case the two cycles C (2) 1, C(2) 2 use 2( n 3 +1) of the inner edges of P (n, 3), i.e. n 2 n 3 2 = n 3 inner edges remain for cycles in Z (n, 3) \ {C (2) }. Similarly as before, 1, C(2) 2 these remaining inner edges decompose into at most two paths in G int. At least one of these two components consists of more than one inner edge. In just this component at least two edges (starting edge and end edge of the path) are incident with vertices from C (2) 1 and C (2) 2, respectively. Therefore, these special edges are blocked. Moreover, at least one edge of the second path is blocked (one is blocked if it consists of only one edge. Otherwise two edges are blocked). Hence, we conclude that, beside C (2) 1 and C (2) 2, at most ( n 3 2 1)/2 of the inner edges can be used for the cycles in Z (n, 3) \ {C (2) }. We conclude 1, C(2) 2 ν(p (n, 3)) = Z (n, 3) 2 + n 3 /2 n, getting a contradiction as before. Assume k 2 = 1. Since every cycle C Z (n, 3) \ C (2) has at least four outer vertices, we get n (n 2) + 1 ν(p (n, 3)) + 1 n + 1, hence ν(p (n, 3)) = n + 1 = (n 2) + 1, i.e. n = (n 2). But then either (n 2) N or (n 3) N. Therefore, every cycle C Z (n, 3) \ C (2) must have as well inner as outer vertices. At most one of the cycles in Z (n, 3) \ C (2) can block more than four vertices in V 1 (otherwise it is impossible to built the n cycles using the n 2 outer vertices.) Let {v i, v i+1 } be the outer vertices of C (2). If V (C) V 1 =, then C has either two or four joining edges. In the first case (two joining edges), let V (C) V 1 = {v j, v j+1, v j+2, v j+3 }. Obviously, both outer edges (v j+3, v j+ ) and (v j, v j 1 ) are blocked by C. If v j 1, v j+ / V (C (2) ), at least one of the joining edges (v j+, u j+ ) or (v j 1, u j 1 ) is blocked by C (2). Hence, C blocks at least five outer vertices. But there exist at most two distinct cycles 10

11 C, C Z (n, 3) \ {C (2) with v i 1 V (C ) and v i+2 V (C ). Because of ν(p (n, 3)) > 7, the packing Z (n, 3) \ {C (2), C, C } contains at least four cycles, with two of them blocking at least nine outer vertices. That is not possible. In the second case (four joining edges) the outer vertices {v j, v j+1, v l, v l+1 } V (C) must have the property, that at least one of the vertices {v j 1, v j+2, v l+2 } must be blocked by C. Hence, C blocks five outer vertices. Because of ν(p (n, 3)) > 7, the packing Z (n, 3) \ C (2) contains at least six cycles, with two of them blocking at least nine outer vertices. Therefore, such a situation cannot occur and we can conclude k 2 = 0. Lemma 10 Let n 28. Assume ν(p (n, 3)) > n. Let k 3 the number of cycles that are contained in a maximum cycle packing Z (n, 3) of P (n, 3), having exactly three outer-vertices. Then k 3 = 0 Proof: Let C (3) Z (n, 3) be a cycle with exactly three outer vertices. Hence either 1 n mod 3 or 2 mod 3. In the first case E(C (2) ) E 2 = n 3 + 1, in the second case E(C(2) ) E 2 = n 3. Therefore, k 3 3, but k 2 = 3 is impossible, because ν(p (n, 3)) > 7. Hence k 3 2. Assume that k 2 = 2 and 2 n mod 3. In this case the two cycles C (3) 1, C(3) 2 use 2 n 3 of the inner edges of P (n, 3), i.e. n 2 n 3 = n inner edges remain for cycles in Z (n, 3) \ {C (3) 1, C(3) 2 }. G int \ {C (3) 1, C(3) 2 } decompose into two paths P 1 andp 2. It is important to note, that both paths P 1 and P 2 consist of more than one inner edge. (If P 1 would consist of only one inner edge, then at least one of the two cycles C (3) 1, C(3) 2 have n inner edges, that only happens if 1 n mod 3.) Therefore, the starting edges and the end edges of P 1 and P 2 are incident with vertices in V (C (3) 1 ) and V (C (3) 2 ), respectively. These special edges are blocked. Hence, at most n = n 3 2 edges in E 2 can be used for the cycles in Z (n, 3) \ {C (3) 1, C(3) 2 }. Now, consider the second case, i.e. k 3 = 2 and 1 n mod 3. In this case the two cycles C (3) 1, C(3) 2 use 2( n 3 + 1) of the inner edges of P (n, 3), i.e. n 2( n 3 + 1) = n 3 1 inner edges remain for the cycles in Z (n, 3) \ {C (3) 1, C(3) 2 }. As in the previous lemma, G int \ {C (3) 1, C(3) 2 } decompose into at most two paths in. At least one path consists of more than one inner edge and least two of its edges (starting edge and end edge) 11

12 are incident with vertices in V (C (3) 1 ) and V (C(3) 2 ), respectively. Therefore, these special edges are blocked. As before, at least one edge of the second path is blocked. We get that at most n = n 3 edges in E 2 can be used for the cycles in Z (n, 3) \ {C (3) 1, C(3) 2 }. Now, we can argue as above for getting the contradiction for n 28. ν(p (n, 3)) = Z (n, 3) 2 + ( n 3 )/2 n. Assume k 3 = 1. Since every cycle C Z (n, 3) \ C (3) has at least four outer vertices, we get n (n 3) + 1 ν(p (n, 3)) + 1 n + 1, hence ν(p (n, 3)) = n + 1 = (n 3) + 1. From this follows n = (n 3), that implies (n 3) N. Every cycle C Z (n, 3) \ C (3) must have as well inner as outer vertices, but cannot block more than four outer vertices (otherwise it is impossible to built the n cycles in C Z (n, 3) \ C (3), using the n 3 vertices in V 1 ). Therefore, C allows exactly four outer vertices and either has two or four joining edges. Let {v i, v i+1, v i+2 } be the outer vertices of C (3). If 1 n mod 3, then v i 1 and v i+3 are blocked by C (3). In this case it would be impossible to built the n further cycles in C Z (n, 3) \ C (3). The case 2 n mod 3 remains. If, in this case, C has two joining edges, let v j, v j+1, v j+2, v j+3 be its outer vertices. Obviously, both outer edges (v j+3, v j+ ) and (v j, v j 1 ) are blocked by C. If v j 1, v j+ / C (3), at least one of the joining edges (v j+, u j+ ) or (v j 1, u j 1 ) is blocked by C (3). Hence, C blocks at least five outer vertices. But, there only exist at most two distinct cycles C C in in C Z (n, 3) \ C (3) with v i 1 C and v i+3 C. Because of ν(p (n, 3)) > 7, Z (n, 3) \ C (3) contains at least one cycle C, which blocks five outer vertices. But this is impossible. If C has outer vertices {v j, v j+1, v l, v l+1 } and four joining edges, at least one of the vertices {v j 1, v j+2, v l+2 } must be blocked by C. I.e., C blocks at least 5 outer vertices, a situation that cannot occur. conclude C (3) / Z (n, 3), i.e. k 3 = 0. We in summary Proof of Theorem 5 From the last lemmas we conclude for n 28 that, if ν(p (n, 3)) > n would hold, all the cycles in Z (n, 3) must have at least four outer vertices. But in such a case P (n, 3) admits at most n of such cycles, a contradiction. Therefore, Theorem 5 holds. 12

13 Remark 1: A straight-forward consideration shows that for the packing numbers ν(p (n, 3)), 7 n 27 we get: ν(p ( 7, 3)) = 2, ν(p ( 8, 3)) = 2, ν(p ( 9, 3)) =, ν(p (10, 3)) = 3, ν(p (11, 3)) = 3, ν(p (12, 3)) =, ν(p (13, 3)) = 3, ν(p (1, 3)) = ν(p (15, 3)) =, ν(p (16, 3)) =, ν(p (17, 3)) =, ν(p (18, 3)) = ν(p (19, 3)) = 5, ν(p (20, 3)) = 5, ν(p (21, 3)) = 5, ν(p (22, 3)) = 5 ν(p (23, 3)) = 5, ν(p (2, 3)) = 6, ν(p (25, 3)) = 6, ν(p (26, 3)) = 6 ν(p (27, 3)) = 7 The following figure shows the graph P (27, 3) together with a maximum cycle packing Z (27). Obviously it contains a cycle of type C (0), i.e. a cycle with no outer vertices. For n = 27 this situation can happen, since in this case the inequality n < 1 + 2n 9 is valid. Indeed, ν(p (27, 3)) = Figure 3: Maximum cycle packing Z (27, 3). ν(p (27, 3)) =

14 Remark2 : For k {1, 2, 3} the theorems above state, that there is n k N such that for all n n n k the equality ν(p (n, k)) = k+1 holds. This is not true for densities k. The reason is, that girth(p (n, k)) is not monotone in k, i.e. for k a graph P (n, k) contains cycles of length 8 if n is sufficiently large. Hence, for even k and 0 n mod 2k, we get ν(p (n, k)) 2n 8 for some r N. But n k+1 = 2kr k+1 < kr 2. = n = kr 2 References [1] L. Beineke, A Survey of Packings and Coverings of Graphs, in: G. Chartrand, S. Kapoor (Eds.), The Many Facets of Graph Theory, Vol. 110 of Lecture Notes in Mathematics, Springer-Verlag, Berlin, [2] M. E. Watkins, A Theorem on Tait Colorings with an Application to the Generalized Petersen Graphs, Journal of Combinatorial Theory [3] J. Moon, On Edge Disjoint Cycles in a Graph, Canadian Mathematical Bulletin [] H. Voss, Eigenschaften von Graphen, die eine bestimmte Anzahl kantenfremder Kreise enthalten und in denen der Grad der Knotenpunkte nach unten beschrnkt ist., in: P. Erdos, A. Renyi, V. Sos (Eds.), Combinatorial Theory and its Applications, Vol. of Colloquia Mathematica Societatis Janos Bolyai, North-Holland Publishing Company, Amsterdam, [5] H. s. M. Coxeter, Regular Polytopes, 3rd ed. New York: Dover, [6] H. Walther, H. Voss, ber Kreise in Graphen, VEB Deutscher Verlag der Wissenschaften, Berlin, 197. [7] H. Voss, Maximal Circuits and Paths in Graphs - Extreme Cases, in: A. Hajnal, V. Sos (Eds.), Combinatorics, Vol. 18 of Colloquia Mathematica Societatis Janos Bolyai, North-Holland Publishing Company, Amsterdam, 1978,

15 [8] A. Caprara, Sorting Permutations by Reversals and Eulerian Cycle Decompositions, SIAM Journal on Discrete Mathematics 12 (1) [9] A. Caprara, A. Panconesi, R. Rizzi, Packing Cycles in Undirected Graphs, Journal of Algorithms 8 (1) [10] J. Degenhardt and P. Recht. On Maximal Edge-disjoint Cycle Decompositions in Graphs. Diskussionsbeitrage des Fachgebiets Operations Research und Wirtschaftsinformatik, Nr. 28, Technische Universitat Dortmund, D-221 Dortmund, [11] J. Degenhardt and P. Recht. On a relation between the cycle packing number and the cyclomatic number of a graph, 2008, submitted 15