Signed cycle double covers

The cycle double cover conjecture states that every bridgeless graph has a collection of cycles which together cover every edge of the graph exactly twice. A signed graph is a graph with each edge assigned by a positive or a negative sign. In this article, we prove a weak version of this conjecture that is the existence of a signed cycle double cover for all bridgeless graphs. We also show the relationships of the signed cycle double cover and other famous conjectures such as the Tutte flow conjectures and the shortest cycle cover conjecture etc. Mathematics Subject Classifications: 05C21, 05C22, 05C38


Introduction
The Cycle Double Cover Conjecture is a famous conjecture in graph theory.It states that every bridgeless graph has a collection of cycles which together cover every edge of the graph exactly twice.Let G be a graph.A collection of cycles of G is called a cycle cover if it covers each edge of G.A cycle double cover of G is such a cycle cover of G that each edge lies on exactly two cycles.A graph is called even if its vertices are all of even degree.A k-cycle double cover of G consists of k even subgraphs of G.This conjecture is a folklore (see [3]) and it was independently formulated by Szekeres [23], Itai and Rodeh [9], and Seymour [22].In fact, Szekeres raised this conjecture just for cubic graphs, while Seymour put it in a general case.
The cycle cover problem is closely related to the flow theory.Let G = (V, E) be a graph.An integer flow of G is an ordered pair (D, f ) satisfying s(e)=v f (e) = t(e)=v f (e) ∀v ∈ V, where D is an orientation of G and f is a function E → Z, s(e) denotes the source of e and t(e) denotes the tail of e.A nowhere-zero k-flow of G is an integer flow such that 0 < |f (e)| < k for every edge e of G.A group flow is defined similarly by taking values in an Abelian group instead of Z.For instance, if a flow takes values in the finite cyclic group Z k , then we call it a Z k -flow.The support of a flow f , denoted by supp(f ), is the set of all the edges e of G with f (e) = 0. Tutte [24] found the following relationship of the integer flow and the group flow.

Theorem 3. [24]
A bridgeless graph admits a nowhere-zero k-flow if and only if it admits a nowhere-zero group flow with group order k.
Generalizing the coloring problem of planar graphs, Tutte proposed the following three famous conjectures on integer flows.Conjecture 4. (The Five-Flow Conjecture [24]) Every bridgeless graph admits a nowherezero 5-flow.
Jaeger [12] and Seymour [20] derived two fundamental theorems on integer flows, the Eight-Flow Theorem and the Six-Flow Theorem.
The following relationship of cycle double covers and flows can be found in [11].

Theorem 9. [11]
A graph has a 4-cycle double cover if and only if it admits a nowherezero 4-flow.
For more backgrounds on cycle covers and flows we refer to the monographs [26,27].An ordered pair (G, ω) is called a weighted graph if ω : E(G) → Z is an integer-valued function which is called the weight.Thus a graph G can be viewed as a special weighted graph (G, ω) with the constant weight ω = 1.Let G i = (G, ω i ) be a weighted graph for i = 1, 2. The sum of G 1 and G 2 is defined by where (ω 1 + ω 2 )(e) = ω 1 (e) + ω 2 (e) for each e ∈ E(G).The difference of G 1 and G 2 is defined by where (ω 1 − ω 2 )(e) = ω 1 (e) − ω 2 (e) for each e ∈ E(G).
A general signed graph (G, ω) is such a weighted graph that ω(e) ∈ {0, ±1} for all e ∈ E(G).This generalizes the notion of signed graph which edges are allowed to take only two values ±1.Since the support of a general signed graph is just its signed subgraph, we simply call it a signed graph without confusion.Thus a subgraph H of G can be viewed as a signed graph (G, ω H ), where ω H (e) = 1 for e ∈ E(H) and ω A k-signed cycle double cover consists of k even signed graphs.Let Γ be a signed cycle double cover.The total number of negative edges of Γ, denoted by ν(Γ), is defined by Though Conjectures 1 and 2 are widely open and it is well known that the Petersen graph has no 4-cycle double cover, we establish the following weak version by replacing cycles with signed cycles.This is analogous in some sense to a result of Seymour [21] who proved the weak signed versions of the Berge-Fulkerson Conjecture [8,21] about double covers by perfect matchings and the Tutte Four-Flow Conjecture in its dual coloring form for cubic graphs.
Theorem 10.Every bridgeless graph of order n and size m has a 4-signed cycle double cover Γ with ν(Γ) min{n/3, m/9}.[18,Theorem 1.1] proved that every bridgeless cubic graph has a Fano coloring which uses at most six different lines.Moreover, if at most six points are used in the Fano coloring, then there is a nowhere-zero 4-flow.This is a consequence of the following theorem which is stated in the flow form instead of coloring.

Máčajová and Škoviera
Theorem 11.A graph admits a nowhere-zero 4-flow if and only if it admits a nonsurjective nowhere-zero Z 3 2 -flow.Though the bound in Theorem 10 is hardly improved (see Corollary 18), it may not be sharp because of the following result.
Theorem 12.There exist infinitely many graphs of size m such that all 4-signed cycle double covers have at least m/15 negative edges.
In view of Theorems 10 and 12, we propose the following conjecture.
Conjecture 13.Every bridgeless graph of size m has a 4-signed cycle double cover with at most m/15 negative edges.
Note that Theorem 10 implies Conjecture 13 for all bridgeless graphs of order n and size m 5n.We prove Theorems 10, 11, and 12 in Section 2 and finish with a discussion in Section 3 on the relationship of the signed cycle double cover and other conjectures, such as the Shortest Cycle Cover Conjecture and the Tarsi Conjecture on the shortest 3-cycle covers etc.

Proofs of main results
The proof of Theorem 10 is based on the Six-Flow Theorem and the Eight-Flow Theorem.It is to construct directly a signed cycle double cover which differs from the approach of Seymour [21] who made signed covers by perfect matchings indirectly from multicolorings.The idea of our proof is as follows.
A 6-flow induces a Z 3 × Z 2 -flow.The support C of the Z 2 -flow can be taken as a disjoint union of circuits with size at most n.The Z 3 -flow induces a 3-cycle double cover of its support.All even subgraphs form an Abelian group under the operation of symmetric difference, and all even weighted graphs also form an additive group, though all even signed graphs do not.Applying the three operations of addition, subtraction, and symmetric difference to C and the three cycles, and taking care of the weights of all edges, we get two 4-signed cycle double covers with disjoint sets of negative edges E 1 and E 2 appearing only on the support C. The three disjoint sets E 1 , E 2 , and E(C) \ E 1 ∪ E 2 form a partition of all edges of C. But both E 1 and E 2 might be of size greater than n/3.In this case, we make a new Z 3 -flow by pivoting the Z 2 -flow so that one of the two sets E 1 and E 2 changes to the set E(C) \ E 1 ∪ E 2 which is of size at most n/3.
An 8-flow induces a Z 3 2 -flow that in turn induces a partition of all edges into 7 subsets.The three supports of the Z 2 -flows all induce even subgraphs.Taking symmetric differences of the three supports yields four new even subgraphs.The key observation is that every one of the 7 subsets of edges is covered exactly 4 times by the 7 even subgraphs and the corresponding 4 even subgraphs form a signed cycle double cover.Bermond et al. [2] observed that the Z 3 2 -flow can be taken so that one of the supports is large with at least 2m/3 edges.Three of the 7 subsets of edges are disjoint with the largest support, so we can take the smallest one as the set of negative edges which size turns out be at most m/9.
Proof of Theorem 10.Let G be a bridgeless graph of order n and size m.We first show that G has a 4-signed cycle double cover with ν n/3.By Theorem 3 and the Six-Flow Theorem, the graph G admits a nowhere-zero Z 3 × Z 2 -flow, say (f, g).Let C be the even subgraph of G induced by supp(g).By the proof of the Six-Flow Theorem [20], the flow g can be selected such that C is a union of disjoint circuits with |E(C)| n.Note that f is a nowhere-zero Z 3 -flow on supp(f ) which can be turned to a nowhere-zero 3-flow by Theorem 3. Since a 3-flow can be viewed as a 4-flow, the support of f has a 3-cycle double cover, denoted by {C 01 , C 02 , C 03 }.Indeed, if the pair (g 1 , g 2 ) is a nowhere-zero Z 2 2 -flow on supp(f ), then let C 01 and C 02 be the even subgraphs induced by supp(g 1 ) and supp(g 2 ) respectively, and let C 03 = C 01 C 02 be the subgraph of G induced by the symmetric difference E(C 01 ) E(C 02 ). Let As for the flow f , both supp(f 1 ) and supp(f 2 ) also have 3-cycle double covers, denoted by {C 11 , C 12 , C 13 } and {C 21 , C 22 , C 23 }, respectively.One can easily verify that is a signed even graph of G. Indeed, let e be an edge of G and we go through the following three cases.Note that only in Case 3 can negative edges occur in C 1 or C 2 and all such negative edges are on C. Indeed, an edge e in C 1 is negative if and only if f (e) = g(e) = 1 and an edge e in C 2 is negative if and only if f (e ) = 2 and g(e ) = 1. Let ).Note that f is still a Z 3 -flow and the difference between f and f can only occur on C. Thus (f , g) is also a nowherezero , and E 2 = E 0 .As above, G has a 4-signed cycle double cover with ).As the above argument, (f , g) is also a nowhere-zero Next we show that G has a 4-signed cycle double cover with ν m/9.By Theorem 3 and the Eight-Flow Theorem, the graph G admits a nowhere-zero Z 3  2 -flow, say f = (f 1 , f 2 , f 3 ).Let C i be the even subgraph of G induced by supp(f i ) for i = 1, 2, 3.As observed by Bermond et al. [2], the flow f can be selected such that We define seven subsets of E(G) as follows, 2 -flow of G, the seven subsets defined above form a partition of E(G).
the electronic journal of combinatorics 25(4) (2018), #P4.63 We claim that G has a 4-signed cycle double cover with ν To see this, we define an incidence matrix M in the following way.The rows of M correspond to the seven subsets of edges listed in the following order: The columns of M correspond to the seven even subgraphs of G listed in the following order: Let M = (a ij ) 7×7 , where a ij = 1 if the ith subset is a subset of edges in the jth even subgraph, otherwise a ij = 0 for i, j = 1, 2, . . ., 7. Explicitly, It is easily verified that M has the following property: Each row of M has exactly 4 non-zero entries and of the other six rows has exactly 2 non-zero entries at the 4 columns corresponding to such 4 non-zero entries of that row.
Let E * be one of the seven subsets of edges with minimum size.Without loss of generality, we may assume that E * = E 001 , i.e., the 7th subset.Select the four even subgraphs corresponding to the four non-zero entries of the 7th row, i.e., C 3 , Proof of Theorem 11.Theorem 3 implies that a graph admitting a nowhere-zero 4-flow has a nowhere-zero Z 2  2 -flow which can obviously be viewed as a nonsurjective Z 3 2 -flow.Conversely, if G admits a nonsurjective nowhere-zero Z 3  2 -flow, then G has a 4-signed cycle double cover without negative edges by the proof of Theorem 10, i.e., G has a 4-cycle double cover, which is equivalent to the fact that G admits a nowhere-zero 4-flow by Theorem 9.
Proof of Theorem 12. Consider the Petersen graph P of order 10 and size 15.Since P fails to admit a nowhere-zero 4-flow, it has no 4-cycle double cover by Theorem 9. Thus every 4-signed cycle double cover of P has at least one negative edge.In fact, the lower bound can be achieved, see Fig. 1 for instance: The dotted edge is negative and the first   even signed graph together with other three even subgraphs shown in Fig. 1 form a 4signed cycle double cover of P with only one negative edge.For a natural number k, a k-subdivision of a graph is obtained from the graph by subdividing each edge exactly k times.It is easy to see that a k-subdivision of the Petersen graph also has a corresponding 4-signed cycle double cover with only k negative edges.As in [14], one can construct a more general graph of size m such that its 4-signed cycle double covers all have at least m/15 negative edges.Denote by P the graph obtained from the Petersen graph P in the following way: Delete a vertex v of P and the three edges vv 1 , vv 2 , and vv 3 which are incident with v, and add three new vertices u 1 , u 2 , u 3 which are called the arms of P and three new edges u i v i for i = 1, 2, 3, see Fig. 2. Take a bridgeless cubic graph G of order n.A graph denoted by G ⊗ P , is constructed from G and P as follows: Replace every vertex v of G by a copy of P , denoted by P v .For every edge uv of G, select an arm of P u which is not used and such a arm of P v , and identify these two arms forming a new vertex of degree two.It is clear that G ⊗ P is also bridgeless.Select a 4-signed cycle double cover of G ⊗ P arbitrarily (the existence of such a cover is ensured by Theorem 10), say Γ = {C 1 , C 2 , C 3 , C 4 }.For every subgraph P v of G ⊗ P , identifying its three arms results in a graph isomorphic to P , and the collection Γ v = {C 1 ∩ P v , . . ., C 4 ∩ P v } corresponds to a 4-signed cycle double cover of the graph.Thus ν(Γ v ) 1 and since the graph G ⊗ P has exactly 15n edges.
(1) Let Γ be a 4-signed cycle double cover of G with ν negative edges.Replacing all the negative edges in Γ by normal edges, we get a cycle cover, denoted by {C 1 , C 2 , C 3 , C 4 }, which covers all the negative edges of Γ exactly 4 times and the other edges exactly twice.The total length of this cycle cover is 4ν + 2(m − ν) = 2m + 2ν.Without loss of generality, we assume that C 1 has the maximum length among the cycle cover, and then the other three even subgraphs C 2 , C 3 , and C 4 make up a cycle cover with length no more than 3(m + ν)/2.
(2) Let {C 1 , C 2 , C 3 } be a 3-cycle cover of G with length l.Clearly, G admits a nowherezero Z 3  2 -flow, say f = (f 1 , f 2 , f 3 ), such that C i is induced by supp(f i ) for i = 1, 2, 3. Following the notation such as the edge subsets E i 1 i 2 i 3 in the proof of Theorem 10, we have l = Corollary 17, compared with the result of Jamshy et al. [13], indicates that it might be hard to prove Conjecture 13, and Corollary 18 shows that the bound in Theorem 10 might also be hardly improved since the bound 7m/75 is slightly better than m/9.

Case 1 .
f (e) = 0, g(e) = 0.In this case, we have e ∈ E(C) ∩ supp(f 1 ) \ supp(f ), and then e is in exactly two of {C 11 , C 12 , C 13 } and none of {C 01 , C 02 , C 03 }.Thus e is in only one of {C C 11 , C C 12 , C C 13 } and ω C 1 (e) = 1.Case 2. f (e) = 0, g(e) = 0.In this case, we have e ∈ supp(f ) ∩ supp(f 1 ) \ E(C), and then e is in exactly two of {C i1 , C i2 , C i3 } for i = 0, 1.Thus e is in exactly two of {C C 11 , C C 12 , C C 13 } and ω C 1 (e) = 0. Case 3. f (e) = 0, g(e) = 0.If f (e) = 1, then e ∈ E(C) ∩ supp(f ) ∩ supp(f 1 ) and thus e is in exactly two of {C i1 , C i2 , C i3 } for i = 0, 1 and in only one of {C C 11 , C C 12 , C C 13 } which implies that ω C 1 (e) = −1.If f (e) = 2, then e ∈ E(C) ∩ supp(f ) \ supp(f 1 ) and thus e is in exactly two of {C 01 , C 02 , C 03 } and in all of {C C 11 , C C 12 , C C 13 } which implies that ω C 1 (e) = 1.The even signed graph C 1 together with the three even subgraphs C C 21 , C C 22 , C C 23 forms a 4-signed cycle double cover of G, denoted by Γ 1 , which is shown and E 2 = E 1 .Thus G also has a 4-signed cycle double cover with ν = min{|E 1 |, |E 2 |} = min{|E 0 |, |E 1 |} n/3.This completes the proof of the first part.
and C 1 C 2 C 3 .By the above property of M , these four even subgraphs form a cycle cover of G which covers every edge of E 001 4 times and all the other edges twice.Replacing all edges of E 001 in C 3 by negative edges, we get a 4-signed cycle double cover with ν = |E * | min{|E 010 |, |E 011 |, |E 001 |} (m − |E(C 1 )|)/3 m/9.This completes the proof of the second part.

Figure 1 :
Figure 1: A 4-signed cycle double cover of the Petersen graph.

Figure 2 :
Figure 2: The graph P obtained from the Petersen graph.

Table 1 :
in Table 1.Similarly, let C 2 = C C 21 + C C 22 + C C 23 − C 01 − C 02 − C 03 , which together with the three even subgraphs C C 11 , C C 12 , C C 13 forms another 4-signed cycle double cover of G, denoted by Γ 2 , see Table 2. C (e) ω C 1 (e) ω C C 21 +C C 22 +C C 23 (e) The 4-signed cycle double cover Γ 1 .