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Determining the flow numbers of signed eulerian graphs
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Electronic Notes in Discrete Mathematics 38 (2011) 585–590 www.elsevier.com/locate/endm
Determining the flow numbers of signed eulerian graphs 2,4 ˇ Edita M´aˇcajov´a 1,3 Martin Skoviera
Department of Computer Science Comenius University 842 48 Bratislava, Slovakia
Abstract We determine the flow numbers of signed eulerian graphs. Keywords: eulerian graph, signed graph, nowhere-zero flow
1
Introduction
A signed graph is a graph with each edge labelled with a sign, + or −. In this paper we study flows on signed eulerian graphs. Our aim is to generalise the well-known fact that every eulerian graph has a nowhere-zero 2-flow. As in the unsigned case, a flow on a signed graph follows a certain orientation. An orientation of a signed graph is obtained by dividing each edge into two half-edges each of which receives its own direction. A positive edge 1 2 3 4
Supported in part by APVV-0223-10 and by VEGA, Grant 1/0406/09 Supported in part by APVV-0223-10 and by VEGA, Grant 1/0634/09 Email: [email protected] Email: [email protected]
1571-0653/$ – see front matter © 2011 Elsevier B.V. All rights reserved. doi:10.1016/j.endm.2011.09.096
586
E. Máˇcajová, M. Škoviera / Electronic Notes in Discrete Mathematics 38 (2011) 585–590
has one half-edge directed from and the other half-edge directed to its endvertex; a negative edge has both half-edges directed either towards or from their respective end-vertices. Let A be an abelian group. A nowhere-zero A-flow on a signed graph G is an assignment of an orientation and a value from A − 0 to each edge in such a way that for each vertex of G the sum of incoming values equals the sum of outgoing values (Kirchhoff ’s law ). A nowhere-zero k-flow is a Z-flow with values in the set {±1, ±2, . . . , ±(k − 1)} where k ≥ 2. In contrast to unsigned graphs, there is no simple characterisation of signed graphs that admit a nowhere-zero integer flow. We call such graphs flow-admissible. The study of nowhere-zero integer flows on signed graphs was initiated by Bouchet [1] and was motivated by a duality relationship between tensions and flows on nonorientable surfaces. The fundamental problem is to determine, for a given signed graph G, the smallest integer k such that G has a nowhere-zero k-flow. This value is called the flow number of G and is denoted by Φ(G). In 1983, Bouchet [1] conjectured that every flow-admissible signed graph has a nowhere-zero 6-flow. This conjecture constitutes, in fact, a signed counterpart to the celebrated 5-Flow Conjecture of Tutte. Its best general approximation to date is a theorem of Z´ yka [6] stating that the constant 6 can be replaced with 30. Much better estimates were later obtained for signed graphs with sufficiently high connectivity by Xu and Zhang [5] and very recently by Raspaud and Zhu [6]. Nevertheless, very little is known about precise flow numbers of signed graphs from particular classes. The purpose of the present paper is to resolve this problem for signed eulerian graphs.
2
Results
Consider a signed graph G carrying an A-flow φ and let P = e1 e2 . . . er be an u-v trail in G. By sending a value b ∈ A from u to v along P we mean reversing the the orientation of the edge e1 so that it leaves u, adding b to φ(e1 ), and adding ±b to φ(ei ) for all other edges of P in such a way that Kirchhoff’s law is fulfilled at each inner vertex of P . The next theorem is due to Xu and Zhang [5]. Sending 1 along an eulerian trail proves one implication; the converse follows from the fact that the values on negative edges, each with the extroverted orientation, sum to 0. Theorem 2.1 [5] A connected signed graph has a nowhere-zero 2-flow if and only if it is eulerian and has even number of negative edges. Before proceeding to eulerian graphs with odd number of negative edges
E. Máˇcajová, M. Škoviera / Electronic Notes in Discrete Mathematics 38 (2011) 585–590
587
we need to explain the concept of switching equivalence of signed graphs and of flows. Let G be a signed graph and let X be a set of vertices of G. Switching the signature of G at X means changing the sign of each edge with exactly one end in X; this is equivalent to switching at each vertex of X is a succession. Note that switching does not change the total sign of any circuit. A circuit is said to be balanced if its total sign is +. A graph is balanced if it contains no unbalanced circuit, that is, if it can be switched to an all-positive graph. Consider a signed graph G that carries a nowhere-zero flow. If we reverse the direction of an edge and replace the flow value by its opposite, the flow is preserved. Similarly, if we change the direction of each half-edge incident with a given vertex v, switch the signature at v and leave the flow values intact, the flow also remains preserved. It follows that the existence of a nowhere-zero k-flow is invariant under switching and changing the orientation. The following result is proved in [3]. Theorem 2.2 A 2-edge-connected signed graph is flow-admissible if and only if it can be switched to a graph with exactly one negative edge. The following two theorems deal with the existence of nowhere-zero 3-flows and 4-flows on signed eulerian graphs. Theorem 2.3 Every flow-admissible signed eulerian graph with odd number of negative edges has a nowhere-zero 4-flow. Sketch of proof. Let G be a flow-admissible eulerian graph with odd number of negative edges. If G contains two edge-disjoint unbalanced circuits, then there is a third unbalanced circuit, edge-disjoint from the previous two. With some effort these circuits can be extended into three edge-disjoint eulerian subgraphs G1 , G2 , and G3 with odd number of negative edges each such that G = G1 ∪ G2 ∪ G3 . Since G is connected, one of them, say G2 , shares a vertex with G1 and a vertex with G3 . By Theorem 2.1, G1 ∪ G2 and G2 ∪ G3 have nowhere-zero 2-flows φ1 and φ2 , respectively, whose combination φ1 +2φ2 is the required nowhere-zero 4-flow on G. The rest of the proof consists in showing that G indeed contains two edge-disjoint unbalanced circuits. Choose any unbalanced circuit N of G, and set G0 = G − E(N ). If G0 is unbalanced, then it contains an unbalanced circuit, which together with N provides two required edge-disjoint unbalanced circuits of G. We can therefore assume that G0 is balanced. Two cases occur depending on whether G0 is connected or not. Assume that G0 is connected. Harary’s Balance Theorem [2] implies that the vertex-set of G0 decomposes into two subsets such that all negative edges
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are between the sets and all positive edges are within the sets. The vertices of G0 divide the circuit N into segments, pairwise edge-disjoint paths whose first and last vertex is in G0 and all inner vertices lie outside G0 . A multisegment is a path formed by a chain of segments. A segment or a multisegment is essential if it is positive with end-vertices in different partite sets, or is negative with end-vertices in the same partite set. In other words, a segment J is essential if and only if the graph G0 ∪ J is unbalanced. It is not difficult to show that N has an odd number of essential segments. Let S1 , S2 , . . . , Sk be all the segments of N . If S1 is the only essential segment, then each G0 ∪ Si with i ≥ 2 is balanced, and by a repeated use of Harary’s Balance Theorem, G0 ∪S2 ∪S3 . . .∪Sk is also balanced. Since G0 ∪S1 is unbalanced, every unbalanced circuit of G traverses S1 . But then for every edge e of S1 the graph G − e is balanced, contradicting Theorem 2.2. Hence at least three segments of N are essential. We pick two of them, say Si and Sj . We show that Si and Sj can be extended to essential multisegments Si+ and Sj+ , respectively, sharing a vertex. Clearly, Si and Sj have at most one common vertex. If Si and Sj do have a common vertex, we set Si+ = Si and Sj+ = Sj . Otherwise N = Si U1 Sj U2 for suitable multisegments U1 and U2 . If, say, U1 is essential, then we set Si+ = Si and Sj+ = U1 . Otherwise both U1 and U2 are inessential, and therefore either positive with ends in the same partite set or negative with ends in different partite set. If both U1 and U2 are negative, then either both Si and Sj are positive with ends in different partite sets or are both negative with ends in the same partite set. But then N = Si U1 Sj U2 is balanced – a contradiction. Therefore at least one of U1 or U2 , say U1 , is positive (with ends in the same partite set). Then we may simply set Si+ = Si and Sj+ = U1 Sj . In each case we have found essential multisegments Si+ and Sj+ sharing precisely one vertex. Let a and b be the end-vertices of Si+ and let b and c the end-vertices of Sj+ . Since G0 is eulerian, it has an eulerian trail T which traverses each of vertices a, b, and c at least once. It follows that T = T1 T2 T3 where T1 is an a-b-subtrail and T2 is an b-c-subtrail of T . Then Si+ T1−1 and Sj+ T2−1 are two edge-disjoint unbalanced closed trails. Each of them contains an unbalanced circuit, so G contains two edge-disjoint unbalanced circuits. The case with G0 disconnected is more involved and is deferred to [3]. 2 A signed eulerian graph G is triply odd if it has a decomposition into three eulerian subgraphs G1 , G2 , and G3 , with odd number of negative edges each, that share a vertex. Theorem 2.4 Let G be a connected signed eulerian graph. Then Φ(G) = 3
E. Máˇcajová, M. Škoviera / Electronic Notes in Discrete Mathematics 38 (2011) 585–590
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if and only if G is triply odd. Sketch of proof. Let G1 , G2 , and G3 form a triply odd decomposition of G with common vertex v. For i ∈ {1, 2, 3} let Ti be an eulerian trail in Gi starting at v. If we send from v the value 1 along T1 and T2 , and the value −2 along T3 , the resulting valuation will clearly be a nowhere-zero 3-flow on G. Since each of the trails is unbalanced, G has odd number of negative edges, and from Theorem 2.1 we get that Φ(G) ≥ 3. Hence Φ(G) = 3. To prove the converse we first show that the proof reduces to 6-regular allnegative graphs. Let G be an arbitrary signed eulerian graph with a nowherezero 3-flow φ. We choose the orientation of G with each edge carrying a positive value and fix a vertex v. If there is a pair of edges e and f at v with φ(e) = φ(f ) such that e is directed to and f is directed from v, we remove the path ef and add a new edge g joining the other ends of ef whose sign equals the product of the signs of e and f . If the new graph has a triply odd component, then G is also triply odd. Hence we may assume that all the edges at v carrying the same value 1 or 2 are directed either to or from v. By Kirchhoff’s law, the valency of v is a multiple of 6. If it is greater than 6, we replace v with two new vertices v and v and join the edges originally incident with v to v and v in such a way that v becomes 6-valent and Kirchhoff’s law is satisfied at both vertices. Again, if a component of the new graph is triply odd, so is G itself. Therefore we may assume that G is 6-regular. Since each edge of G carries a positive value, the situation at each vertex v is that either four incident edges with value 1 are directed into v and two edges with value 2 are directed from v (Type 1) or vice versa (Type 2). Observe that an edge joining vertices of the same type is negative whereas an edge joining vertices of different type is positive. By Harary’s Balance Theorem, G is switching equivalent to an all-negative 6-valent graph, as claimed. We now use induction to show that if G is an all-negative 6-valent graph with odd number of edges, then G is triply odd. As in the proof of Theorem 2.3, G is an edge-disjoint union G1 ∪ G2 ∪ G3 , where each Gi has odd number of edges, G1 ∩ G2 contains a vertex u, and G2 ∩ G3 contains a vertex v. If G has a 2-edge-cut, then the result easily follows from the induction hypothesis. Assume that G is 4-edge-connected. If G2 contains four edge4 disjoint u-v-paths P1 , P2 , P3 , and P4 , we observe that G2 − ( i=1 E(Pi )) is an eulerian graph each component of which intersects P1 ∪ · · · ∪ P4 . Let H2 be the subgraph formed from two of the paths plus incident components, and let K2 have the rest of G2 . One of these two subgraphs has odd number of edges, say H2 . Then {G1 , H2 , K2 ∪ G3 } is a triply odd decomposition of G. If G2 does not have four edge-disjoint u-v paths, then G1 and G3 have a
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common vertex, and the only case to consider is that each Gi has a 2-edge-cut. This case is more involved and is therefore deferred to [3]. 2 The following theorem summarises Theorems 2.1-2.4. Theorem 2.5 Let G be a connected signed eulerian graph. Then (i) G has no nowhere-zero flow if and only if G is unbalanced and G − e is balanced for some edge e; (ii) Φ(G) = 2 if and only if G has even number of negative edges; (iii) Φ(G) = 3 if and only if G is triply odd; (iv) Φ(G) = 4 otherwise. We conclude with a result that characterises the existence of nowhere-zero A-flows on a signed eulerian graph for every abelian group A. Theorem 2.6 Let G be a signed eulerian graph and let A be a nontrivial abelian group. The following statements hold true. (i) If A contains an involution, then G admits a nowhere-zero A-flow. (ii) If A ∼ = Z3 , then G admits a nowhere-zero A-flow if and only if G is triply odd. (iii) For all other groups, G has a nowhere-zero A-flow unless G is unbalanced and G − e is balanced for some edge e.
References [1] Bouchet, A., Nowhere-zero integral flows on a bidirected graph, J. Combin. Theory Ser. B 34 (1983) 279–292. [2] Harary, F., On the notion of balance of a signed graph, Michigan Math. J. 2 (1953-1954), 143–146; Addendum, ibid., preceding p. 1. ˇ [3] M´ aˇcajov´ a, E., and M. Skoviera, Nowhere-zero flows on signed eulerian graphs, manuscript. [4] Raspaud, A., and X. Zhu, Circular flow on signed graphs, J. Combin. THeory Ser. B (2011), doi: 10.1016/j.jctb.2011.02.007 [5] Xu, R., and C.-Q. Zhang, On flows in bidirected graphs, Discrete Math. 299 (2005), 335–343. [6] Z´ yka, O., Nowhere-zero 30-flows on bidirected graphs, KAM Series No. 87-26, Charles University, Prague, 1987.
Electronic Notes in Discrete Mathematics 38 (2011) 585–590 www.elsevier.com/locate/endm
Determining the flow numbers of signed eulerian graphs 2,4 ˇ Edita M´aˇcajov´a 1,3 Martin Skoviera
Department of Computer Science Comenius University 842 48 Bratislava, Slovakia
Abstract We determine the flow numbers of signed eulerian graphs. Keywords: eulerian graph, signed graph, nowhere-zero flow
1
Introduction
A signed graph is a graph with each edge labelled with a sign, + or −. In this paper we study flows on signed eulerian graphs. Our aim is to generalise the well-known fact that every eulerian graph has a nowhere-zero 2-flow. As in the unsigned case, a flow on a signed graph follows a certain orientation. An orientation of a signed graph is obtained by dividing each edge into two half-edges each of which receives its own direction. A positive edge 1 2 3 4
Supported in part by APVV-0223-10 and by VEGA, Grant 1/0406/09 Supported in part by APVV-0223-10 and by VEGA, Grant 1/0634/09 Email: [email protected] Email: [email protected]
1571-0653/$ – see front matter © 2011 Elsevier B.V. All rights reserved. doi:10.1016/j.endm.2011.09.096
586
E. Máˇcajová, M. Škoviera / Electronic Notes in Discrete Mathematics 38 (2011) 585–590
has one half-edge directed from and the other half-edge directed to its endvertex; a negative edge has both half-edges directed either towards or from their respective end-vertices. Let A be an abelian group. A nowhere-zero A-flow on a signed graph G is an assignment of an orientation and a value from A − 0 to each edge in such a way that for each vertex of G the sum of incoming values equals the sum of outgoing values (Kirchhoff ’s law ). A nowhere-zero k-flow is a Z-flow with values in the set {±1, ±2, . . . , ±(k − 1)} where k ≥ 2. In contrast to unsigned graphs, there is no simple characterisation of signed graphs that admit a nowhere-zero integer flow. We call such graphs flow-admissible. The study of nowhere-zero integer flows on signed graphs was initiated by Bouchet [1] and was motivated by a duality relationship between tensions and flows on nonorientable surfaces. The fundamental problem is to determine, for a given signed graph G, the smallest integer k such that G has a nowhere-zero k-flow. This value is called the flow number of G and is denoted by Φ(G). In 1983, Bouchet [1] conjectured that every flow-admissible signed graph has a nowhere-zero 6-flow. This conjecture constitutes, in fact, a signed counterpart to the celebrated 5-Flow Conjecture of Tutte. Its best general approximation to date is a theorem of Z´ yka [6] stating that the constant 6 can be replaced with 30. Much better estimates were later obtained for signed graphs with sufficiently high connectivity by Xu and Zhang [5] and very recently by Raspaud and Zhu [6]. Nevertheless, very little is known about precise flow numbers of signed graphs from particular classes. The purpose of the present paper is to resolve this problem for signed eulerian graphs.
2
Results
Consider a signed graph G carrying an A-flow φ and let P = e1 e2 . . . er be an u-v trail in G. By sending a value b ∈ A from u to v along P we mean reversing the the orientation of the edge e1 so that it leaves u, adding b to φ(e1 ), and adding ±b to φ(ei ) for all other edges of P in such a way that Kirchhoff’s law is fulfilled at each inner vertex of P . The next theorem is due to Xu and Zhang [5]. Sending 1 along an eulerian trail proves one implication; the converse follows from the fact that the values on negative edges, each with the extroverted orientation, sum to 0. Theorem 2.1 [5] A connected signed graph has a nowhere-zero 2-flow if and only if it is eulerian and has even number of negative edges. Before proceeding to eulerian graphs with odd number of negative edges
E. Máˇcajová, M. Škoviera / Electronic Notes in Discrete Mathematics 38 (2011) 585–590
587
we need to explain the concept of switching equivalence of signed graphs and of flows. Let G be a signed graph and let X be a set of vertices of G. Switching the signature of G at X means changing the sign of each edge with exactly one end in X; this is equivalent to switching at each vertex of X is a succession. Note that switching does not change the total sign of any circuit. A circuit is said to be balanced if its total sign is +. A graph is balanced if it contains no unbalanced circuit, that is, if it can be switched to an all-positive graph. Consider a signed graph G that carries a nowhere-zero flow. If we reverse the direction of an edge and replace the flow value by its opposite, the flow is preserved. Similarly, if we change the direction of each half-edge incident with a given vertex v, switch the signature at v and leave the flow values intact, the flow also remains preserved. It follows that the existence of a nowhere-zero k-flow is invariant under switching and changing the orientation. The following result is proved in [3]. Theorem 2.2 A 2-edge-connected signed graph is flow-admissible if and only if it can be switched to a graph with exactly one negative edge. The following two theorems deal with the existence of nowhere-zero 3-flows and 4-flows on signed eulerian graphs. Theorem 2.3 Every flow-admissible signed eulerian graph with odd number of negative edges has a nowhere-zero 4-flow. Sketch of proof. Let G be a flow-admissible eulerian graph with odd number of negative edges. If G contains two edge-disjoint unbalanced circuits, then there is a third unbalanced circuit, edge-disjoint from the previous two. With some effort these circuits can be extended into three edge-disjoint eulerian subgraphs G1 , G2 , and G3 with odd number of negative edges each such that G = G1 ∪ G2 ∪ G3 . Since G is connected, one of them, say G2 , shares a vertex with G1 and a vertex with G3 . By Theorem 2.1, G1 ∪ G2 and G2 ∪ G3 have nowhere-zero 2-flows φ1 and φ2 , respectively, whose combination φ1 +2φ2 is the required nowhere-zero 4-flow on G. The rest of the proof consists in showing that G indeed contains two edge-disjoint unbalanced circuits. Choose any unbalanced circuit N of G, and set G0 = G − E(N ). If G0 is unbalanced, then it contains an unbalanced circuit, which together with N provides two required edge-disjoint unbalanced circuits of G. We can therefore assume that G0 is balanced. Two cases occur depending on whether G0 is connected or not. Assume that G0 is connected. Harary’s Balance Theorem [2] implies that the vertex-set of G0 decomposes into two subsets such that all negative edges
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E. Máˇcajová, M. Škoviera / Electronic Notes in Discrete Mathematics 38 (2011) 585–590
are between the sets and all positive edges are within the sets. The vertices of G0 divide the circuit N into segments, pairwise edge-disjoint paths whose first and last vertex is in G0 and all inner vertices lie outside G0 . A multisegment is a path formed by a chain of segments. A segment or a multisegment is essential if it is positive with end-vertices in different partite sets, or is negative with end-vertices in the same partite set. In other words, a segment J is essential if and only if the graph G0 ∪ J is unbalanced. It is not difficult to show that N has an odd number of essential segments. Let S1 , S2 , . . . , Sk be all the segments of N . If S1 is the only essential segment, then each G0 ∪ Si with i ≥ 2 is balanced, and by a repeated use of Harary’s Balance Theorem, G0 ∪S2 ∪S3 . . .∪Sk is also balanced. Since G0 ∪S1 is unbalanced, every unbalanced circuit of G traverses S1 . But then for every edge e of S1 the graph G − e is balanced, contradicting Theorem 2.2. Hence at least three segments of N are essential. We pick two of them, say Si and Sj . We show that Si and Sj can be extended to essential multisegments Si+ and Sj+ , respectively, sharing a vertex. Clearly, Si and Sj have at most one common vertex. If Si and Sj do have a common vertex, we set Si+ = Si and Sj+ = Sj . Otherwise N = Si U1 Sj U2 for suitable multisegments U1 and U2 . If, say, U1 is essential, then we set Si+ = Si and Sj+ = U1 . Otherwise both U1 and U2 are inessential, and therefore either positive with ends in the same partite set or negative with ends in different partite set. If both U1 and U2 are negative, then either both Si and Sj are positive with ends in different partite sets or are both negative with ends in the same partite set. But then N = Si U1 Sj U2 is balanced – a contradiction. Therefore at least one of U1 or U2 , say U1 , is positive (with ends in the same partite set). Then we may simply set Si+ = Si and Sj+ = U1 Sj . In each case we have found essential multisegments Si+ and Sj+ sharing precisely one vertex. Let a and b be the end-vertices of Si+ and let b and c the end-vertices of Sj+ . Since G0 is eulerian, it has an eulerian trail T which traverses each of vertices a, b, and c at least once. It follows that T = T1 T2 T3 where T1 is an a-b-subtrail and T2 is an b-c-subtrail of T . Then Si+ T1−1 and Sj+ T2−1 are two edge-disjoint unbalanced closed trails. Each of them contains an unbalanced circuit, so G contains two edge-disjoint unbalanced circuits. The case with G0 disconnected is more involved and is deferred to [3]. 2 A signed eulerian graph G is triply odd if it has a decomposition into three eulerian subgraphs G1 , G2 , and G3 , with odd number of negative edges each, that share a vertex. Theorem 2.4 Let G be a connected signed eulerian graph. Then Φ(G) = 3
E. Máˇcajová, M. Škoviera / Electronic Notes in Discrete Mathematics 38 (2011) 585–590
589
if and only if G is triply odd. Sketch of proof. Let G1 , G2 , and G3 form a triply odd decomposition of G with common vertex v. For i ∈ {1, 2, 3} let Ti be an eulerian trail in Gi starting at v. If we send from v the value 1 along T1 and T2 , and the value −2 along T3 , the resulting valuation will clearly be a nowhere-zero 3-flow on G. Since each of the trails is unbalanced, G has odd number of negative edges, and from Theorem 2.1 we get that Φ(G) ≥ 3. Hence Φ(G) = 3. To prove the converse we first show that the proof reduces to 6-regular allnegative graphs. Let G be an arbitrary signed eulerian graph with a nowherezero 3-flow φ. We choose the orientation of G with each edge carrying a positive value and fix a vertex v. If there is a pair of edges e and f at v with φ(e) = φ(f ) such that e is directed to and f is directed from v, we remove the path ef and add a new edge g joining the other ends of ef whose sign equals the product of the signs of e and f . If the new graph has a triply odd component, then G is also triply odd. Hence we may assume that all the edges at v carrying the same value 1 or 2 are directed either to or from v. By Kirchhoff’s law, the valency of v is a multiple of 6. If it is greater than 6, we replace v with two new vertices v and v and join the edges originally incident with v to v and v in such a way that v becomes 6-valent and Kirchhoff’s law is satisfied at both vertices. Again, if a component of the new graph is triply odd, so is G itself. Therefore we may assume that G is 6-regular. Since each edge of G carries a positive value, the situation at each vertex v is that either four incident edges with value 1 are directed into v and two edges with value 2 are directed from v (Type 1) or vice versa (Type 2). Observe that an edge joining vertices of the same type is negative whereas an edge joining vertices of different type is positive. By Harary’s Balance Theorem, G is switching equivalent to an all-negative 6-valent graph, as claimed. We now use induction to show that if G is an all-negative 6-valent graph with odd number of edges, then G is triply odd. As in the proof of Theorem 2.3, G is an edge-disjoint union G1 ∪ G2 ∪ G3 , where each Gi has odd number of edges, G1 ∩ G2 contains a vertex u, and G2 ∩ G3 contains a vertex v. If G has a 2-edge-cut, then the result easily follows from the induction hypothesis. Assume that G is 4-edge-connected. If G2 contains four edge4 disjoint u-v-paths P1 , P2 , P3 , and P4 , we observe that G2 − ( i=1 E(Pi )) is an eulerian graph each component of which intersects P1 ∪ · · · ∪ P4 . Let H2 be the subgraph formed from two of the paths plus incident components, and let K2 have the rest of G2 . One of these two subgraphs has odd number of edges, say H2 . Then {G1 , H2 , K2 ∪ G3 } is a triply odd decomposition of G. If G2 does not have four edge-disjoint u-v paths, then G1 and G3 have a
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common vertex, and the only case to consider is that each Gi has a 2-edge-cut. This case is more involved and is therefore deferred to [3]. 2 The following theorem summarises Theorems 2.1-2.4. Theorem 2.5 Let G be a connected signed eulerian graph. Then (i) G has no nowhere-zero flow if and only if G is unbalanced and G − e is balanced for some edge e; (ii) Φ(G) = 2 if and only if G has even number of negative edges; (iii) Φ(G) = 3 if and only if G is triply odd; (iv) Φ(G) = 4 otherwise. We conclude with a result that characterises the existence of nowhere-zero A-flows on a signed eulerian graph for every abelian group A. Theorem 2.6 Let G be a signed eulerian graph and let A be a nontrivial abelian group. The following statements hold true. (i) If A contains an involution, then G admits a nowhere-zero A-flow. (ii) If A ∼ = Z3 , then G admits a nowhere-zero A-flow if and only if G is triply odd. (iii) For all other groups, G has a nowhere-zero A-flow unless G is unbalanced and G − e is balanced for some edge e.
References [1] Bouchet, A., Nowhere-zero integral flows on a bidirected graph, J. Combin. Theory Ser. B 34 (1983) 279–292. [2] Harary, F., On the notion of balance of a signed graph, Michigan Math. J. 2 (1953-1954), 143–146; Addendum, ibid., preceding p. 1. ˇ [3] M´ aˇcajov´ a, E., and M. Skoviera, Nowhere-zero flows on signed eulerian graphs, manuscript. [4] Raspaud, A., and X. Zhu, Circular flow on signed graphs, J. Combin. THeory Ser. B (2011), doi: 10.1016/j.jctb.2011.02.007 [5] Xu, R., and C.-Q. Zhang, On flows in bidirected graphs, Discrete Math. 299 (2005), 335–343. [6] Z´ yka, O., Nowhere-zero 30-flows on bidirected graphs, KAM Series No. 87-26, Charles University, Prague, 1987.