Restricted cycle factors and arc-decompositions of digraphs. J. Bang-Jensen and C. J. Casselgren

Transcription

1 Restricted cycle factors and arc-decompositions of digraphs J. Bang-Jensen and C. J. Casselgren REPORT No. 0, 0/04, spring ISSN 0-467X ISRN IML-R- -0-/4- -SE+spring

2 Restricted cycle factors and arc-decompositions of digraphs Jørgen Bang-Jensen Department of Mathematics Uniersity of Sothern Denmark DK-50 Odense, Denmark Carl Johan Casselgren Department of Mathematics Linköping Uniersity SE-58 8 Linköping, Sweden Noember 4, 04 Abstract. We stdy the complexity of finding -factors with arios restrictions as well as edge-decompositions in (the nderlying graphs of) digraphs. In particlar we show that it is N P-complete to decide whether the nderlying ndirected graph of a digraph D has a -factor with cycles C, C,..., C k sch that at least one of the cycles C i is a directed cycle in D (while the others may iolate the orientation back in D). This soles an open problem from []. Or other main reslt is that it is also N P-complete to decide whether a -edge-colored bipartite graph has two edge-disjoint perfect matchings sch that one of these is monochromatic (while the other does not hae to be). We also stdy the complexity of a nmber of related problems. In particlar we proe that for eery een k, the problem of deciding whether a bipartite digraph of girth k has a k-cycle free cycle factor is N P-complete. Keywords: Cycle factor, -factor, mixed problem, NP-complete, Complexity, cycle factors with no short cycles Introdction Notation not introdced here follows [, 5]. We distingish between (non-directed) cycles and directed cycles in digraphs, where the former is a sbgraph that corresponds to a cycle in the nderlying graph of a digraph. The notions of a (non-directed) path and a directed path are defined in a similar way. A cycle factor in a digraph is a spanning sbgraph consisting of directed cycles, and a -factor in a digraph (or graph) is a spanning sbgraph consisting of cycles. We denote by UG(D) the nderlying graph of a directed graph D. address: jbj@imada.sd.dk. Parts of this work was done while the athor attended the program Graphs, hypergraphs and compting at Institte Mittag-Leffler, spring 04. The research of Bang-Jensen was also spported by the Danish research concil nder grant nmber -0078B address: carl.johan.casselgren@li.se. Part of the work done while the athor was a postdoc at Mittag-Leffler Institte. Research spported by Mittag-Leffler Institte.

3 In this paper we consider seeral ariations on the problem of finding cycle factors of digraphs. The problems of deciding if a gien graph has a -factor and if a gien digraph has a cycle factor are fndamental problems in combinatorial optimization, and both these problems are well-known to be solable in polynomial time, see e.g. [, 5]. Here, we are particlarly interested in problems concerning the complexity of deciding existence of spanning sbgraphs that in a sense lies inbetween -factors and cycle factors. In particlar, we answer the qestion of the complexity of the following two problems by the first athor: Problem.. [, Problem ] -factor with at least one directed cycle. Instance: A digraph D. Qestion: Does D hae a -factor F sch that at least one cycle in F is a directed cycle, while the rest of the cycles do not hae to respect the orientations of arcs in D? Problem.. [7] Disjoint perfect matchings one of which is monochromatic Instance: A -edge colored bipartite graph B = (U, V ; E). Qestion: Does B hae two edge-disjoint perfect matchings M, M so that eery edge of M has color, while M may se edges of both colors? This problem is eqialent to the following problem (see [, Section 6.7]). Problem.. Semi-directed -factors of bipartite digraphs Instance: A bipartite digraph B = (X, Y ; A). Qestion: Does UG(B) hae a -factor which is the nion of a perfect mathching from X to Y in D (respecting the orientation) and a perfect matching in U G(B)? Ths we are asking for a collection of cycles coering all ertices of B sch that eery second edge (starting from X) is oriented from X to Y in D, whereas the remaining edges do not hae to respect the orientation of D. The motiation for stdying sch mixed problems for digraphs, that is, problems concerning strctres in a digraph D where only part of the strctre has to respect the orientation of the arcs of D, is that this way one can obtain new insight into the complexity of arios problems which hae natral analoges for graphs and digraphs. As an example, in [, ] the problem of deciding for a digraph D the existence of a directed cycle C in D and a cycle C in UG(D) which are ertex disjoint was stdied. It was shown that this problem is polynomially decidable for the class of digraphs with a bonded nmber of cycle transersals of size (ertices whose remoal eliminates all directed cycles) and N P-complete if we allow arbitrarily many transersal ertices. For (di)graphs one can decide the existence of two disjoint (directed) cycles in polynomial time [, ]. Note that the ariant of Problem. where we ask if D has a -factor F sch that at most one cycle in F is not directed is N P-complete. This can easily be proed as follows: It is N P-complete to decide if a gien graph G is Hamiltonian. Let D be a digraph with a cycle factor, and let D be an acyclic orientation of the gien graph G. Next, let D be the disjoint nion of D and D (or add an arbitrary arc between a ertex of D and a ertex of D if one wants D to be connected). Then G is Hamiltonian if and only if D has a -factor where at most one cycle is not a directed cycle. We show that Problem. and. are both N P-complete. Theorem.4. -factor with at least one directed cycle is N P-complete.

4 Theorem.5. Disjoint perfect matchings one of which is monochromatic is N P-complete. In fact, in the latter case, we shall proe that this problem is N P-complete already for bipartite graphs with maximm degree. Gien an n n array A where each cell contains a (possibly empty) sbset of {,,..., n}, we say that A is aoidable if there is an n n Latin sqare L sch that each cell of L does not contain a symbol that appears in the corresponding cell of A. We also say that L aoids A. If a cell in A is empty, then we also say that this cell contains entry. In [6] it was proed that determining whether a gien array where each cell contains a sbset of {, } is aoidable is N P-complete. Using Theorem.5 we can proe the following strengthening of that reslt. Corollary.6. The problem of determining whether an array where each cell contains either the symbol, the set {, } or is aoidable is N P-complete. Proof. We redce Problem. to the problem of aoiding an array where each cell contains symbol, the set {, } or is empty. Let G be a balanced bipartite graph on n + n ertices with a edge coloring f sing colors and. Let {x,..., x n } and {y,..., y n } be the parts of G. We form an n n array A from G as follows: If x i y j / E(G), then we set A(i, j) = {, }. If x i y j E(G) and f(x i y j ) =, then we set A(i, j) =. If x i y j E(G) and f(x i y j ) =, then the cell (i, j) of A is empty. It is straightforward to erify that there are two disjoint generalized diagonals D and D in A, where D has no cell with symbol and D has no cell with symbol (and therefore a Latin sqare aoiding A) if and only if G has disjoint perfect matchings one of which does not contain any edge with color. Next, we consider a ariant of a problem stdied by Hartigsen: in [0] he proed that the problem of deciding if a bipartite graph has a 4-cycle-free -factor (i.e. a -factor with no 4-cycle) can be soled in polynomial time. The problem of determining if a general graph has a -factor withot -cycles is also solable in polynomial time []. The analogos qestion of the complexity of existence of -factors in general graphs where no cycle has length 4 or less is open to the best of or knowledge, while the problem of determining if a graph G has a -factor where all cycles hae length at least 6 is N P-complete if G is bipartite, and ths also in the general case (see e.g. [4, 0]). For digraphs, the problem of determining whether a general digraph has a cycle factor where eery cycle has length at least is N P-complete, see e.g. [4]. Here we consider the analogos qestion for bipartite digraphs. We strengthen the reslt of [4] as follows: Proposition.7. The problem of determining if a gien bipartite digraph D has a -cycle-free cycle factor is N P-complete. Using this reslt we show how to proe the following: Theorem.8. For each een k, the problem of determining if a bipartite digraph D with no directed cycles of length at most k has a (k + )-cycle-free cycle factor is N P-complete. A generalized diagonal of an n n matrix A is a collection of elements a,π(),..., a n,π(n) where π is a permtation of [n].

5 In particlar, the problem of determining if a gien oriented bipartite graph has a cycle factor with no 4-cycle is N P-complete; so the natral analoge of Hartigsen s positie reslt on 4-cyclefree -factors in bipartite graphs does not hold in the digraph setting. Finally, we consider decompositions of digraphs. It is well-known that the problem of determining if the edge set of a gien graph has a decomposition into edge-disjoint cycles is solable in polynomial time, as is also the analogos problem of deciding if the arc set of a gien digraph has a decomposition into arc-disjoint directed cycles. Here we shall proe the following: Theorem.9. It is N P-complete to decide for a gien digraph D = (V, A) whether there is a decomposition of A into arc-disjoint cycles C,..., C k of D with the property that at most one of these cycles is not directed. Note that the opposite problem, where we ask for a decomposition of A(D) into arc-disjoint cycles C,..., C k of D sch that C is a directed cycle bt all the other cycles do not hae to respect the orientation of the arcs, is triial: As we want a decomposition into arc-disjoint cycles of D it follows that UG(D) mst be Elerian. Hence if D contains any directed cycle C, this can play the role of C aboe, showing that the answer is no if and only if either D is acyclic or UG(D) is not Elerian. The rest of the paper is organized as follows: In section we first proe Proposition.7 and sing this reslt, we proe Theorem.8 and Theorem.4. Section is conclded by the proof of Theorem.5. Section contains the proof of Theorem.9 and in Section 4 we gie an instance of Problem. which can be soled in polynomial time. Restricted cycle factors For the proof of Proposition.7 we will se a reslt on aoiding arrays. As mentioned aboe, in [6] it was proed that the following problem is N P-complete: Problem.. Aoiding mltiple-entry arrays with symbols Instance: An n n array A, sch that each cell is either empty or contains a sbset of the symbols in {, }. Qestion: Is A aoidable? Proof of Proposition.7. We shall redce Problem. to the problem of deciding if a gien bipartite digraph has a -cycle-free cycle factor. Let A be an n n array, where each cell is either empty or contains a sbset of {, }. We form an n n bipartite digraph D with parts X = {x,..., x n } and Y = {y,..., y n } by for all i, j inclding the arc (x i, y j ) in A(D) if and only if the symbol does not appear in cell (i, j) of A, and inclding the arc (y j, x i ) if and only if the symbol does not appear in the cell (i, j) of A. We shall proe that A is aoidable if and only if D has a -cycle-free cycle factor. Sppose that A is aoidable, which means that there is a Latin sqare L aoiding A. Consider the generalized diagonals B, B in L, containing all cells with entries and, respectiely. Since L aoids A, any cell in A corresponding to a cell in B does not contain symbol ; and any cell in A corresponding to a cell in B does not contain the symbol. Hence, if (i, j) B, then (x i, y j ) D; and if (i, j) B, then (y j, x i ) D. Ths the set of arcs {(x i, y j ) : (i, j) B } {(y j, x i ) : (i, j) B } 4

6 indces a cycle factor F in D, and since B and B are disjoint, F is -cycle-free. Conersely, sppose that D has a -cycle-free cycle factor F. Let F be the set of arcs going from X to Y in F, and let F be the arcs going from Y to X. Define an array P by ptting the symbol in cell (i, j) if and only if (x i, y j ) F, and ptting the symbol in cell (i, j) if and only if (y j, x i ) F. Since F is -cycle-free, each cell in P has at most one entry. In fact, P is an n n partial Latin sqare where both symbols and occr exactly n times. Ths we may partition the nfilled cells of P into n generalized diagonals B,..., B n and then assign symbol i to all cells of B i, i =,..., n. So P is completable to a Latin sqare L. Moreoer, since (x i, y j ) A(D) if and only if the symbol does not appear in cell (i, j) of A, and (y j, x i ) A(D) if and only if the symbol does not appear in the cell (i, j) of A, L aoids A. D D x x x 5 x 6 x x 4 a a y y y 5 b y 6 y y 4 b Figre : Constrcting D from D in the proof of Theorem.8. Proof of Theorem.8. We shall proe the theorem in the case when k =, by redcing the problem of existence of -cycle-free cycle factors in bipartite digraphs to the problem of existence of 4-cycle-free cycle factors in oriented bipartite graphs. For k >, there is a similar redction. Let D be a bipartite digraph. From D we shall form a bipartite digraph D that does not hae any -cycles as follows: (i) For each (directed) -cycle of D we do the following: Remoe the arcs (, ) and (, ), add 6 new ertices x,..., x 6, let Q = x x x x 4 x 5 x 6 x be a directed 6-cycle joining these ertices, and add arcs (, x ), (x, ), (, x 4 ), (x 6, ). 5

7 (ii) For each arc (a, b) of D which is not in any -cycle we do the following: Remoe the arc (a, b), add 6 new ertices y,..., y 6, let Q = y y y y 4 y 5 y 6 y be a directed 6-cycle joining these ertices, and add arcs (a, y ), (y 6, b). The graph obtained by repeating the procedre (i) for each -cycle of D, and the procedre (ii) for each arc of D which is not in any -cycle, we denote by D (see Figre ). It is easily erified that D is bipartite and contains no -cycles. For the constrction (i) we say that the arcs (, x ), (x, ), (, x 4 ), (x 6, ) are the arcs of D corresponding to (, ) and (, ). We also say that the 6-cycle Q is the 6-cycle of D associated with. For the constrction (ii) we say that the arcs (a, y ), (y 6, b) are the arcs of D corresponding to (a, b). We also say that the directed 6-cycle Q is the directed 6-cycle of D associated with (a, b). Let s now proe that D has a -cycle-free cycle factor if and only if D has a 4-cycle-free cycle factor. Sppose first that D has a -cycle-free cycle factor F, and let C be a directed cycle of F. Then C defines a directed cycle C in D in the following way: Let (a, b) be an arc of C that is not in any -cycle of D, and let y y... y 6 y be its associated directed 6-cycle in D along with the arcs (a, y ), (y 6, b) of D corresponding to (a, b); we define a directed path P = ay y y y 4 y 5 y 6 b in D corresponding to (a, b). Since F does not contain any directed -cycles, we can proceed similarly for any arc in C that is in a -cycle: sppose that (, ) is in the -cycle and that (, ) is in C. Assme frther that x x... x 6 x is the associated directed 6-cycle in D and that (, x ), (x, ), (, x 4 ), (x 6, ) are the arcs of D corresponding to (, ) and (, ); we define a directed path P = x x x x 4 x 5 x 6 in D corresponding to (, ). By concatenating all directed paths in D corresponding to arcs of C in D, we obtain a directed cycle C in D. Moreoer, if C and C are disjoint directed cycles in F, then the corresponding cycles C and C clearly traerses different ertices of D in D, so they are disjoint. Let F be a sbgraph of D consisting of the collection of cycles in D arising from cycles in F ia the aboe constrction. Then F coers all ertices in V (D ) V (D) and each cycle of F clearly has length at least 6. Moreoer, the sbgraph of D indced by V (D ) \ V (D) is a collection of disjoint directed 6-cycles, and if some ertex of sch a directed 6-cycle is in F, then all ertices of this directed 6-cycle is in F. Therefore, F together with the sbgraph of D indced by V (D ) \ V (F ) forms a cycle factor of D. Moreoer, each cycle in this cycle factor has length at least 6. Sppose now that D has a 4-cycle-free cycle factor F. Let (a, b) be an arc of D that is not in any -cycle, y y... y 6 y be the associated directed 6-cycle in D, and (a, y ) and (y 6, b) be the corresponding arcs in D. We need the following easy claim, the proof of which is omitted. Claim.. It holds that (a, y ) A(F ) if and only if (y 6, b) A(F ). We define E to be the set of all arcs (a, b) in D that are not in any -cycles and satisfying that there are ertices y, y 6 V (D ) \ V (D) sch that (a, y ) A(F ) and (y 6, b) A(F ). Let E be the set of all corresponding arcs (a, y ) and (y 6, b) in D that are in F. It follows from Claim. that E indces a sbgraph with maximm in- and otdegree at most in D, and that a ertex in D has the same in- and otdegree in D[E ] as in D [E ]. 6

8 Next, let be a directed -cycle in D, x x... x 6 x be the associated 6-cycle in D, and (, x ), (x, ), (x 6, ), (, x 4 ) be the corresponding arcs in D. We will se the following claim which follows easily from the fact that F is a 4-cycle-free cycle factor. Claim.. (I) If (, x ) A(F ), then (x 6, ) A(F ), and {(x, ), (, x 4 )} A(F ) =. (II) If (x 6, ) A(F ), then (, x ) A(F ) and {(x, ), (, x 4 )} A(F ) =. (III) If (x, ) A(F ), then (, x 4 ) A(F ) and {(x 6, ), (, x )} A(F ) =. (IV) If (, x 4 ) A(F ), then (x, ) A(F ) and {(x 6, ), (, x )} A(F ) =. We define E to be the set of all arcs (, ) in D sch that (, ) is in some -cycle, and there are ertices x, x V (D ) \ V (D) satisfying that (, x) A(F ) and (x, ) A(F ). Let E be the set of all corresponding arcs (, x) and (x, ) in D that are in F. It follows from Claim. that E indces a sbgraph with maximm in- and otdegree at most in D, and that a ertex in D has the same in- and otdegree in D[E ] as in D [E ]. Moreoer, there is no -cycle in D[E ]. Frthermore, the sets E and E are disjoint, and E E contains eery arc of A(F ) that is incident with a ertex of D; so it follows that the sbgraph of D indced by E E is a cycle factor with no cycle of length. D i H i i i a a c c 4 b c b c Figre : Constrcting H from D in the proof of Theorem.4. Step. Proof of Theorem.4. We shall redce the problem of deciding existence of -cycle-free cycle factors in bipartite digraphs (the problem in Theorem.7) to the problem -factor with at least one directed cycle. Gien a bipartite digraph D, we shall constrct a digraph D from D and then show that D has a -cycle-free cycle factor F if and only if D has a -factor F sch that at least one cycle of F is a directed cycle of D. 7

9 So let D be a bipartite digraph. First we constrct the axiliary digraph H from D. Sppose that D has r directed -cycles and denote them by T i = i i i, i =,..., r. (i) For each arc (a, b) of D that is not in any -cycle we do the following: remoe the arc (a, b), and let Q ab be an orientation of a non-directed 4-cycle with 4 new ertices c, c, c, c 4 diided into partite sets {c, c } and {c, c 4 } where we orient all edges towards c or c 4, and add the arcs (a, c ) and (c, b) (see Figre ). (ii) For each -cycle i i i of D we do the following: remoe the arcs ( i, i ) and ( i, i ) and add two disjoint directed 6-ertex paths and L (x) i L (y) i = x(i) x(i) x(i) 4 x(i) 5 x(i) 6 = y(i) y(i) y(i) 4 y(i) 5 y(i) 6 on new ertices, along with the additional arcs ( the arcs ( i, 4 ), (y(i), i), ( 5, i), ( i, (iii) For each i =,..., r, let Z i = {z (i), z(i) (indices taken modlo r) add the arcs in (see Figre )., z(i), x(i) 6 ) (see Figre ). {(z (i), z(i) ), (z(i), z(i) ), (x(i) 6, z(i) ), (y(i) 6, z(i) ) and (y(i), y(i) 6 ). Moreoer, add } be a set of three new ertices, and for i =,..., r ), (z(i), x(i+) ), (z (i), y(i+) )} Denote the reslting graph by H. For the constrction (i), we say that Q ab is the 4-cycle associated with (a, b). Next, we shall constrct the graph D from H by proceeding as follows for each ertex of V (H) V (D): let {q,..., q l } be the in-neighbors of in H. For i =,..., l, add a set {w (i), w(i), w(i) } of new ertices and replace the arc (q i, ) with the directed path q i w (i) w(i) w(i) ; these are called connecting paths between N H () and ; moreoer, add l additional directed -ertex paths p () p() p(),..., p(l ) p (l ) p (l ), on altogether (l ) new ertices {p (), p(), p(),..., p(l ), p (l ), p (l ) }; these are called non-connecting paths between between N H () and. Next, for each i =,..., l we add the arcs (p (i), w(i) ), (p(i), w(i+) ), (w (i+), p (i) ), (w(i), p(i) ). By repeating this process for eery ertex of V (H) V (D) we obtain the digraph D (see Figre 4 for an example). For a ertex of V (H) V (D), denote by J, the sbgraph of D indced by NH (), and all connecting and non-connecting paths between and N H (). 8

10 . z (i ) z (i ) z (i ) z (i) z (i) z (i) x (i+) 4 x (i+) x (i+) 6 y (i+) y (i+) 5 y (i+) x (i+) 5 x (i+) x (i+) y (i+) y (i+) 4 y (i+) 6 z (i+) z (i+) z (i+). Figre : Constrcting H from D in the proof of Theorem.4. Step. 9

11 q q q H D q q q w () p () w () p () w () w () p () w () p () w () w () p () w () p () w () Figre 4: Constrcting D from H in the proof of Theorem.4. Sppose now that D has a -cycle-free cycle factor F. We shall form the reqired -factor F in D, by first showing how F indces a -factor F H in H, and then demonstrating how F H yields the reqired factor F in D. Consider a cycle C in F. Then C defines a corresponding cycle C H in H in the following way: Sppose that (a, b) is an arc of D that lies on C and is not in any -cycle, and let {c, c, c, c 4 } be the ertices of the associated 4-cycle of (a, b) as in Figre. Then we define a corresponding (non-directed) path P = ac c c 4 c b in H. Sppose that ( i, i ) is an arc of a -cycle that lies on C. Then ( i, i ) / A(C), and we define a corresponding (non-directed) path P = i 4 x(i) x(i) x(i) x(i) 6 x(i) 5 i in H. (The case when ( i, i ) is in C is analogos.) By concatenating all paths in H corresponding to arcs of C in D, we obtain a cycle C H in H coering the same ertices of V (D) as C. It shold be clear that the cycles in F in the way described aboe defines a sbgraph ˆF H of H that consists of a collection of disjoint cycles, sch that none of the cycles in ˆF H is a directed cycle in H, and each ertex in D has in- and otdegree in ˆF H. We shall now define a directed cycle C dir in H. Note that for each j, if some ertex of L (x) j is in ˆF H, then all ertices of L (x) j are in ˆF H and no ertex of L (y) j is in ˆF H. Let I be a maximal sbset of indices i of {,..., r} sch that no ertex of L (x) i is in ˆF H. Now, we define Ĉdir to be the sbgraph of H indced by i I V (L (x) i ) i {,...,r}\i V (L (y) i ) Z Z r. 0

12 Let s now define C dir = Ĉdir {(x (), x() 6 ),..., (x(r), x(r) 6 )} {(y(), y() 6 ),..., (y(r), y(r) 6 )}. C dir is a directed cycle of H disjoint from ˆF H. Now consider the graph H = H V (C dir ) V ( ˆF H ). Since each ertex of D has in- and ot degree in ˆF H and all ertices of Z Z r are in V (C dir ), H is a collection of disjoint (non-directed) 4- and 6-cycles. Define F H = ˆF H C dir H. Then F H is a spanning sbgraph of H where each component is a cycle, and exactly one cycle of F H is directed. Moreoer, each ertex in V (D) has in- and otdegree in F H. Let s now constrct F from F H. A cycle C in ˆF H translates into a cycle C in D in the following way: If (s, t) is in C and (s, t) A(D ), then t / V (D), becase any in-neighbor in D of a ertex in V (D) is not in H (by the constrction of D from H); and we inclde (s, t) in C ; otherwise, if (s, t) is in C and (s, t) / A(D ), then t V (D) and there is a connecting path P in J t with origin s and termins t. We inclde the path P in C. It shold be clear that the cycles in ˆF H in this way defines a sbgraph ˆF D of D which is a collection of disjoint cycles, and which coers all ertices of D. Note frther that the cycle C dir is in D as are also all the cycles in H. So it follows from the constrction of D from H that ˆD = D V ( ˆF D ) V (C dir ) V (H ) is a sbgraph of D consisting of eqally many connecting and non-nonnecting paths between ertices V (D) and their in-neighbors in H. It is easy to see that ˆD contains a -factor M where eery cycle has length 6, and each cycle contains exactly one nonconnecting path and some ertices of exactly one connecting path. The graph M C dir H ˆF D is the reqired -factor of D. Sppose now conersely that F is a -factor of D sch that at least one cycle of F is a directed cycle in D. Denote by C dir the directed cycle of D that is in F. We proe a series of claims concerning F. Claim.4. Let V (D) and consider the sbgraph J of D. Let {q,..., q l } be the in-neighbors of in H. Then exactly one of the arcs in J that are incident with a ertex in {q,..., q l } is in F ; and exactly one of the arcs in J that are incident with is in F. The proof of this claim is omitted. It is easily dedced by doing some case analysis sing e.g. Figre 4 and the fact that F is a -factor. It follows from Claim.4 that F indces a -factor F H in H in the following way: For each arc of F that is in H, we inclde this arc in F H ; for each ertex V (D), inclde the arc (q i, ) in F H, where q i is the niqe in-neighbor of in H that has otdegree in F. Obiosly, we hae that each ertex of V (D) has in- and otdegree in F H. Moreoer, the directed cycle C dir clearly corresponds to a directed cycle C (H) dir in F H. Consider an arc (a, b) in D that is not in any -cycle and its associated 4-cycle Q ab in H and label the ertices of Q ab according to Figre. It is easy to see that if (a, c ) is in F H, then (c, c ) is in F H, and ths (a, c ) is not in C (H) dir. Now consider an arc ( i, i ) that is in a -cycle of D. In H, this -cycle is replaced by the directed paths L (x) i and L (y) i and some additional arcs (see Figre ). It is easy to see that if ( i, 4 ) is in F H, then (, x(i) 4 ) is in F H, and ths ( i, 4 ) is not in C(H) dir. Since each ertex of V (D) has in- and otdegree in F H we hae the following: Claim.5. No ertex of V (D) is in C (H) dir. Claim.5 implies that V (C (H) dir ) V (Z ) V (Z r ) V (L (x) ) V (L(x) r ) V (L (y) ) V (L(y) r ).

13 Claim.6. (i) Z i V (C (H) dir ), for some i {,..., r}. (ii) If V (C (H) dir ), then {x(i),..., x(i) 6 } V (C(H) dir ). (iii) If V (C (H) dir ), then {y(i),..., y(i) 6 } V (C(H) dir ). (i) If Z i V (C (H) dir ), then Z i+ V (C (H) dir ), where indices are taken modlo r. Proof. Since L (x) L (x) r L (y) L (y) r is a collection of disjoint (non-directed) 6-cycles, (i) is tre. The statements (ii) and (iii) are straightforward to erify e.g. from Figre, and statement (i) follows easily from (ii) and (iii). Since C (H) dir is a directed cycle of H it follows from Claim.6 that for each i =,..., r, C (H) dir contains all ertices of L (x) i or L (y) i. Moreoer, C (H) dir contains all the ertices Z Z r, and no ertices of V (D). Now consider the graph Ĥ = H V (C(H) dir ). Clearly, ˆFH = F H V (C (H) dir ) is a -factor of Ĥ sch that each ertex of V (D) has in and ot-degree in ˆF H. Consider an arc (a, b) of D that is not in any -cycle and the corresponding associated 4-cycle Q ab in H, and label the ertices of Q ab according to Figre. We need the following easy claim, the proof of which is omitted. Claim.7. It holds that (a, c ) is in ˆF H if and only if (c, b) is in ˆF H. Now consider an arc ( i, i ) that is in a -cycle of D. We shall also need the following claim which follows easily from the facts that ˆF H is a -factor of Ĥ and that for each i =,..., r, C (H) dir contains all ertices of L (x) i or L (y) i. Claim.8. (I) If ( i, 4 ) A( ˆF H ), then ( 5, i) A( ˆF H ), and {(, i), ( i, )} A( ˆF H ) =. (II) If ( 5, i) A( ˆF H ), then ( i, 4 ) A( ˆF H ) and {(, i), ( i, )} A( ˆF H ) =. (III) If (, i) A( ˆF H ), then ( i, ) A( ˆF H ) and {( 5, i), ( i, 4 )} A( ˆF H ) =. (IV) If ( i, ) A( ˆF H ), then (, i) A( ˆF H ) and {( 5, i), ( i, 4 )} A( ˆF H ) =. Now, sing the two claims aboe and the fact that each ertex of V (D) has in and ot-degree in ˆF H, it is straightforward to erify that ˆF H yields a -cycle-free cycle factor in D as in the last part of the proof of Theorem.8. This completes the proof of the theorem. For the proof of Theorem.5, we shall se the fact that the following problem is N P-complete [8]. An (edge) precoloring of a graph G is a coloring of some of the edges of G. Problem.9. Edge precoloring extension. Instance: A -reglar bipartite graph G, a precoloring f of E E(G). Qestion: Can f be extended to a proper edge coloring of G sing precisely distinct colors?

14 G H Figre 5: An edge precolored in G and the corresponding sbgraph of H. Sppose now that G is a cbic bipartite graph with a precoloring sing colors. By replacing eery edge precolored with the gadget in Figre 5, we obtain the graph H. It is easy to check that the precoloring of G can be extended to a proper -edge coloring if and only if the precoloring of H can be extended to a proper -edge coloring. Hence, the following problem is also N P-complete. Problem.0. Edge precoloring extension with only two colors in the precoloring. Instance: A -reglar bipartite graph G, a precoloring f of E E(G) sing only two distinct colors. Qestion: Can f be extended to a proper edge coloring of G sing precisely distinct colors? Proof of Theorem.5. We shall redce Problem.0 aboe to the problem of determining if a -edge-colored bipartite graph has two disjoint perfect matchings M and M so that eery edge of M has color. Let G be a -reglar bipartite graph with some edges colored and some edges colored. Denote this precoloring by f. We shall constrct a bipartite graph H with maximm degree from G, where eery edge in H is colored or. Then we will arge that the precoloring f can be extended to a proper -edge coloring of G if and only if there are edge-disjoint perfect matchings M and M in H, sch that M only contains edges colored. To this end, we define two -edge colored bipartite graphs B and C depicted in Figre 6, together with their compact notation. We say that w and z, and x and y are endpoints of the graphs B and C respectiely.

15 B Compact notation for B w z w z B C Compact notation for C x y x y C Figre 6: The graphs B and C and their compact notation. Now we define the graph H from G by replacing all edges of G according to the following procedre (see Figre 7). Let e = be an edge of G. If (a) e is colored, then is replaced by a sbgraph isomorphic to C by identifying the ertices x and y of C (see Figre 6) with the ertices and, respectiely; (b) e is colored, then the edge is in H and is colored in H as well; (c) e is ncolored and not adjacent to any colored edge of G, then is in H and we color it with ; (d) e is ncolored and adjacent to some edge colored bt not adjacent to any edge colored, then is replaced by a sbgraph isomorphic to C by identifying the ertices x and y of C (see Figre 6) with the ertices and, respectiely; (e) e is ncolored and adjacent to some edge colored bt not adjacent to any edge colored, then is in H and it is colored ; (f) e is ncolored and adjacent to some edge colored and also adjacent to some edge colored, then is replaced by a sbgraph isomorphic to B by identifying w and z of B (see Figre 6) with the ertices and, respectiely. 4

16 G H (a) C (b) (c) (d) C () (e) () (f) B Figre 7: Constrcting H from G. 5

17 The graph reslting from this process we denote by H and we denote its edge coloring by g. We say that a sbgraph isomorphic to B (C) is a B-sbgraph (C-sbgraph) of H if it arises in H by replacing an edge of G. Sppose first that the precoloring f of G can be extended to a proper -edge coloring f of G. We shall define the reqired matchings M and M in H. We first define a set ˆM. Let e = be an edge of G with f ( ) =. It follows from the constrction of H, that either E(H) and is colored, or and are endpoints of a C-sbgraph C in H. In the first case we inclde in ˆM, and in the second case we inclde the edges of C incident with and in ˆM ; additionally, a third edge of C is inclded in ˆM, so that these three edges form a perfect matching of C. By repeating this process for each e in G with f (e) =, we obtain the matching ˆM of H that coers all ertices of G. It follows that the ertices of H not coered by ˆM lies on B- or C-sbgraphs of H. For each sch sbgraph with ertices ncoered by ˆM we inclde two edges with color in the set M, so that ˆM M is a perfect matching in H. We set M = ˆM M. Let s now constrct M. Sppose that e = is colored nder f. It follows from the constrction of H, that E(H) and is colored or nder g. We inclde all edges e E(G) E(H) with f (e) = in M, and for each C- and B-sbgraph of H we also inclde two edges with color in M so that M is a perfect matching in H. Note that M M =, becase M only contains edges colored (nder g) from B- and C-sbgraphs and edges from E(G) with color nder f. So M and M are the reqired perfect matchings of H. Sppose now conersely that H has disjoint perfect matchings M and M sch that M contains no edges colored nder g. We shall proe that there are disjoint perfect matchings M and M in G sch that M contains all edges colored nder f, and M contains all edges colored nder f, which yields the desired conclsion. Consider a B-sbgraph B of H. Denote by s and t, respectiely, the ertices of B of degree that are not endpoints of B. Since M is a perfect matching of H that contains no edges colored, the two edges colored in B are both in M. Frthermore, since M is perfect, the two edges colored that are incident with s or t are in M. Ths, we hae the following: Claim.. If is an endpoint of a B-sbgraph B in H, then no edge of (M M ) E(B ) is incident with. Now consider a C-sbgraph C of H. Denote by a and b the ertices of degree in C that are not endpoints of C ; and by a and b the two ertices of degree three in C, where a and a are adjacent. Since M is a perfect matching of H that contains no edges colored, ab M. Moreoer, since M is perfect, the two edges in C colored are in M. Now consider the three edges of C that are incident with a or b. Since M is perfect, it follows that either exactly two or one of these edges are in M. Ths, we hae the following: Claim.. Let and be endpoints of one C-sbgraph C in H. Then no edge of E(C ) M is incident with or, and, moreoer if is incident with an edge from E(C ) M then is incident with an edge from E(C ) M. We now constrct M from M as follows: let V (G) and sppose that e = M. It follows from Claims. and. that is not in any B- or C-sbgraph of H, so E(G). Hence, by the constrction of H, either is ncolored nder f, or has color nder f. Let ˆM be the set of all edges e of M that are in B- or C-sbgraphs of H. We simply define M by setting M = M \ ˆM. Let s erify that M contains all edges precolored in G. If f(e) =, then g(e) =, and any edge adjacent to e in G is replaced by a B- or C-sbgraph in H. So Claims. and. imply that e M, and ths e M. 6

18 Now we constrct M from M as follows: let V (G) and sppose that e = M. Let ˆM be the set of all edges e of M that are in B- or C-sbgraphs, and which are not incident to endpoints of sch sbgraphs. It follows from Claim. that e is not in any B-sbgraph of H, and if is in a C-sbgraph, and ths is an endpoint of a C-sbgraph C, then the other endpoint w of C is also incident with an edge from M E(C ). For each sch edge in M we inclde the edge w in M. Note frther that any edge of M \ ˆM that is not in any B- or C-sbgraph of H is in G. Ths, we inclde all sch edges in M. Clearly, M coers all ertices of G and is disjoint from M. Let s erify that all edges precolored in G are in M. Let e E(G) with f(e) =. Then e is replaced by a C-sbgraph C in H, and any edge that is adjacent to e in G is replaced by an edge colored or a B-sbgraph in H. Since M only contains edges colored nder g and contains no edge of a B-sbgraph B that is incident with an endpoint of B, it follows that the endpoints of C are coered by edges from M E(C ). Ths e M. Restricted decompositions of digraphs In this section we proe Theorem.9. The following is a well-known fact, see e.g. [, Excercise 4.8]. Proposition.. The arc set of a digraph D = (V, A) can be decomposed into arc-disjoint directed cycles C,..., C p for some p if and only if we hae d + D () = d D () for all ertices V. Proof of Theorem.9. Obsere that if the arc set A can be decomposed into arc-disjoint directed cycles, then the problem is triial, so we can assme that is not the case. Similarly, if any ertex has d + () d () >, there can be no soltion as all directed cycles contribte the same to the in- and ot-degree of any ertex. So by Proposition., D is a yes-instance if and only if there is a cycle C which coers all the ertices in V + V where V + = { V : d + () = d () + } and V = { V ; d () = d + () + } and satisfying that eery ertex V + has otdegree in C, eery ertex V has indegree in C, and eery ertex V (C) \ (V + V ) has in- and otdegree in C. Note that after remoing the arcs of sch a cycle C the reslting digraph D will satisfy the condition in Proposition.. Now we show how to redce the hamiltonian cycle problem for cbic bipartite graphs to or problem (this problem is well-known to be NP-complete [9]) Let B = (X, Y ; E) be a cbic bipartite graph (note that X = Y ) and form the directed graph D(B) as follows: Orient all edges of E from X to Y. Add one new ertex s and the following arcs: all arcs from s to X as well as all arcs from Y to s. The digraph D(B) satisfies that d + (s) = d (s), all ertices x X hae d + (x) = d (x) + and all ertices y Y satisfy d (y) = d + (y) +. By the remarks aboe, if D has the desired decomposition into arc-disjoint cycles C, C,..., C k sch that C is the only non-directed cycle, then the arcs of C mst correspond to a hamiltonian cycle back in B. Conersely, if C is a hamiltonian cycle of B, then by remoing the corresponding arcs in D(B) the reslting digraph D has a decomposition into arc-disjoint -cycles all of which contain s and one arc of a perfect matching from X to Y. 4 A polynomial instance of Problem. An easy conseqence of Hall s marriage theorem (see e.g. [, Theorem 4..]) is that eery balanced bipartite graph on n+n ertices with minimm degree at least n has a perfect matching. It is worth noting that this implies that Problem. is solable in polynomial time if the balanced bipartite graph B on n + n ertices has minimm degree at least n + : if this holds, then Problem. has 7

19 a positie answer if and only if there is a matching M containing only edges colored, becase if there is sch a matching M, then B M has a perfect matching by applying Hall s theorem to B M. With a little more effort we can proe the following strengthening of this reslt: Proposition 4.. Problem. is solable in polynomial time for balanced bipartite graphs on n + n ertices with minimm degree at least n. Proof of Proposition 4.. Let B be a bipartite graph with parts V and V both of which hae size n. We may assme that n 6, since smaller instances can srely be checked in polynomial time. We will deal with the cases that n is een and n is odd separately. By the aboe remark we may assme that δ(b) = n/. In the case when n is odd we will proe that if Problem. has a negatie answer for B, then either B has no perfect matching with only edges of color, or if B has sch a matching, then there are sbsets W V and W V of size n sch that (A.) B[W W ] is a one-factor, (A.) B[W V \ W ] is complete bipartite, (A.) B[W V \ W ] is complete bipartite, and (A.4) for eery perfect matching M containing only edges of color in B, eery edge of B[W W ] is contained in M. In the case when n is een we will proe that if Problem. has a negatie answer for B, then either B has no perfect matching M with only edges of color, or if B has sch a matching, then, by renaming V and V if necessary, there is a sbset W of V of size n +, and a maximal sbset X of V of size n/ or n/ + containing all ertices x of V with exactly one edge joining x with a ertex of W. Moreoer, W and X satisfy the following: (B.) the edge set of B[W X] is a matching of size n/ or n/ +, (B.) The sbgraph of B indced by V \ W and X is complete bipartite, (B.) if X has size n/ +, then the sbgraph of B indced by W and V \ X is complete bipartite, (B.4) if X has size n/, then eery perfect matching M with only edges colored contains each edge of B[W X], (B.5) if X has size n/ +, then eery perfect matching M with only edges colored contains at least n/ edges from B[W X]. We first deal with the case when n is odd. If B has no perfect matching with only edges of color, then there is nothing to proe, so sppose that B has sch a matching M. Consider the graph B = B M, and sppose that Problem. has a negatie answer for B, in particlar, there is no perfect matching in B. Then, by Hall s theorem there is a sbset W of V sch that N B (W ) < W. Since δ(b ) = n/, we hae that W n/. It is easy to see that if we choose a minimal set W with this property, then W n/. Hence, there is sch a set W of size exactly n/. We set W = V \ N B (W ). Then W = n/, and it is easy to see that N B (W ) = n/. Since δ(b ) = n/, this implies that the sbgraphs B [W N B (W )] and B [W N B (W )] of B are complete bipartite; and ths each ertex of W is matched to a ertex 8

20 in W nder M, and ths B[W W ] is a one-factor. Hence, we hae proed that (A.), (A.) and (A.) holds. Sppose now that there is some matching M in B that does not contain eery edge of B[W W ]. Then B = B M has some edge connecting a ertex w of W and a ertex w of W. Since B[W (V \ W )] and B[W (V \ W )] are complete bipartite it follows that B {w w } has a perfect matching, which contradicts that Problem. has a negatie answer. We conclde that (A.4) holds. Note that deciding whether B has a perfect matching M with only edges of color can be done in polynomial time e.g. sing flows (see [, Section 4.]). Assming that B has sch a matching, the problem of finding either a perfect matching in B M or a set W as in the argment aboe can be soled in polynomial time by well-known algorithms for constrcting maximm matchings in bipartite graphs. Moreoer, gien the set W, the problem of erifying whether there is a set W, sch that (A.)-(A.4) hold, is clearly solable in polynomial time. We now consider the case when n is een. If B has no matching with only edges of color, then there is nothing to proe, so sppose that B has sch a matching M. Consider the graph B = B M, and sppose that Problem. has a negatie answer, that is, there is no perfect matching in B. Then, by Hall s theorem there is a set A that is a sbset of V or V sch that N B (A) < A. Since δ(b ) = n/, we hae that A n/. On the other hand, if we choose a minimal set A with this property, then A n/. Hence, there is sch a set A of size exactly n/ with N B (A) = n/. Assme withot loss of generality that A V. Clearly B [A N B (A)] is complete bipartite. We set W = V \ N B (A). Then W has size n/ + and N B (W ) = n/. Since δ(b ) = n/, it follows that all ertices in A are matched to ertices in W nder M. Hence, the edge set of B[A W ] is a matching. Let X V be a maximal sbset of ertices sch that eery ertex of X has degree in B[X W ]. Clearly X contains A, and ths has size at least n/. On the other hand, since B has minimm degree n/, X n/ +. Sppose first that X has size n/, and ths X = A. Then (B.) holds, and (B.) as well, becase B[A N B (A)] is complete bipartite. Let s proe that eery perfect matching M with only edges colored in B contains all edges of B[X W ], i.e. that (B.4) holds. Sppose that there is a perfect matching M in B with only edges colored sch that wx / M, where w W and x X. Set B = B M. It follows that the bipartite graph B = B [W \ {w} V \ X] is balanced, and contains no isolated ertex. Moreoer, eery ertex of W has degree at least n/ in B, and ths B contains at least n /4 n edges. If B has no perfect matching, then by Hall s theorem there is a sbset W W of size n/, sch that N B (W ) = n/. Conseqently, there are two ertices, V \ X of degree in B. This implies that B has at most (n/ )(n/ ) + edges, a contradiction becase B has at least n /4 n edges and n 6. Ths B has a perfect matching ˆM. Now consider the balanced bipartite graph B = B [X \ {x} N B (X)]. Eery ertex of X has degree at least n/ in B. So proided that n 6, B has a perfect matching ˆM. This means that ˆM ˆM {xw} is a perfect matching in B, contradicting that Problem. has a negatie answer. Ths we conclde that (B.4) holds. Sppose now that X has size n/ +. Recall that the edge set of B[W A] is a matching, and we hae A X. Then (B.) holds, since sppose that x X \ A is adjacent to the same ertex in W as a ertex a A. Since W = n/ +, this means that there is some ertex w W that has no neighbor in X, which means that w has degree n/ in B, a contradiction. Hence, B[W X] is a one-factor. That (B.) and (B.) hold follows easily from the facts that δ(b) = n/ and V \W = V \X = n/. Let s now proe that (B.5) holds. 9

21 Sppose that there is a perfect matching M with only edges of color sch that w x / M and w x / M, where w i W and x i X, i =,. Set B = B M. It follows from (B.) and (B.) that B = B [W \ {w, w } V \ X] and B 4 = B [V \ W X \ {x, x }] are balanced bipartite graphs with parts of size n/ where each ertex has degree at least n/. Hence, B has a perfect matching ˆM and B 4 has a perfect matching ˆM 4. Then ˆM ˆM 4 {x w, x w } is a perfect matching in B, contradicting that Problem. has a negatie answer. Hence, (B.5) holds. As mentioned aboe, deciding whether B has a perfect matching with only edges of color can be done in polynomial time. Assming that B has sch a matching, the problem of finding the set W in the proof aboe can be soled in polynomial time by well-known algorithms for constrcting maximm perfect matchings in bipartite graphs. The set X can clearly also be fond in polynomial time. Moreoer, gien the sets W and X, the problem of erifying whether W and X satisfy that (B.)-(B.5) hold is clearly solable in polynomial time. We conclde that Problem. is solable in polynomial time for a balanced bipartite graph on n + n ertices with minimm degree at least n/. References [] J. Bang-Jensen and G. Gtin, Digraphs: Theory, Algorithms and Applications nd Ed., Springer Verlag, London 009. [] J. Bang-Jensen and M. Kriesell, On the problem of finding disjoint cycles and dicycles in a digraph, Combinatorica (6) (0) [] J. Bang-Jensen, M.Kriesell, A. Maddaloni and S. Simonsen, Vertex-disjoint directed and ndirected cycles in general digraphs, J. Combinatorial Theory, Ser B 06 (04) -4. [4] M. Bläser, B. Siebert, Compting Cycle coers withot short cycles, Proc. 9th Ann. Eropean Symp. on Algorithms (ESA), Lectre Notes in Compt. Sci. 6, pages 69-79, 00. [5] J.A. Bondy and U.S.R. Mrty, Graph Theory, Springer Gradate texts in Mathematics 44, Springer Verlag 008. [6] C. J. Casselgren, On aoiding some families of arrays, Discr. Math. (0), [7] Open problems collection from the program Graph, Hypergraphs and Compting at Mittag- Leffler Institte, [8] J. Fiala, NP-completeness of the edge precoloring extension problem on bipartite graphs, Jornal of Graph Theory 4 (00), pp [9] M.R. Garey, D.S. Johnson and R.E. Tarjan, The planar Hamiltonian circit problem is NPcomplete, Siam J. Compt. 5 (976) [0] Daid Hartigsen, Finding maximm sqare-free -matchings in bipartite graphs, Jornal of Combinatorial Theory, Series B Volme 96, Isse 5, 006, Pages [] D. Hartigsen, Extensions of Matching Theory, Doctoral thesis, Carnegie-Mellon Uniersity, 984. [] L. Loász, On graphs not containing independent circits (in Hngarian), Mat. Lapok bf 6 (965)

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