1 Introduction
In this paper, all graphs are finite, simple and undirected. Let and be the vertex set and the edge set of , respectively. We denote by and the maximum and minimum degree of , respectively.
A signed graph is a graph in which each edge has a positive or negative signature. Precisely, a signed graph is a graph with a signature . Edges with signature are positive edges while edges with signature are negative edges. A graph with no signature is usually called an unsigned graph. Signed graphs appeared first in a mathematical paper of Harary [4] in 1955, and have been rediscovered many times because they come up naturally in many unrelated areas [7, 8]. For example, signed graphs have been used in social psychology [1] to model social situations, with positive edges representing friendships and negative edges representing enmities between nodes, which represent people.
This paper focuses in the coloring problems of sighed graphs. First, let us look at the proper coloring of a signed graph. Actully, there are two definitions on this in the history.
In an early paper, Zaslavsky [6] defines a (signed) coloring of a signed graph with colors or with signed colors to be a mapping such that for every edge , if , and if . Recently in 2016, Máčajová, Raspaud and Škoviera [3] pointed out that Zaslavsky’s definition has disadvantage that this definition for signed graphs does not directly transfer from the definition for unsigned graphs. Thereby, they diverged from the above definition adopted by Zaslavsky and proposed the following definition, which aligns the definitions for both unsigned and signed versions.
Let , where is if , and if . A mapping such that for every edge is a (signed) -coloring of a signed graph . The signed chromatic number of , denoted by , is the smallest integer such that admits an -coloring.
Define the all-positive signed graph as having all positive edges. Since the signed coloring rules for such a graph are equivalent to the ones for an unsigned graph, its signed chromatic number is the same with its (unsigned) chromatic number. Hence the proper coloring of signed graphs generalizes the one of unsigned graphs. In view of this, it is natural to generalize the improper coloring of unsigned graphs to its signed version.
A tree--coloring of an unsigned graph is a function from the vertex set to so that the graph induced by is a union of trees for every . The minimum integer so that admits a tree--coloring is the vertex arboricity of , denoted by . The notion of vertex arboricity was introduced by Chartrand, Kronk and Wall [2] in 1968.
The aim of this paper is to investigate the vertex arboricity of signed graphs, i.e., investigate the signed version of the vertex arboricity. Before doing this, we shall properly define the tree-coloring of a signed graph. In order to complete this work, we first introduce an equivalent definition for the (signed) -coloring of a signed graph .
Mapping the vertex set of a signed graph into , we then obtain a (signed) -coloring of .
Let be the subgraph of whose vertex set is the set of vertices colored by or and edge set is the set of positive edges with two end-vertices colored both by or both by , along with the set of negative edges with one end-vertex colored by and the other colored by . Clearly, .
By the definitions of , one can easily conclude that is a proper coloring of if and only if is an empty graph (i.e., graph with no edge) for every . Follow this idea, we can naturally define one another kind of coloring of the signed graph as follows.
Definition 1.
A signed tree-coloring of is a vertex coloring so that is a forest for every and . If is a function from to , then we call a (signed) tree--coloring of . The minimum integer such that admits a signed tree--coloring is the signed vertex arboricity of , denoted by .
2 Main results and their proofs
Switching a vertex in a signed graph means negating the signs of all the edges incident with that vertex. Switching a set of vertices means negating all the edges that have one end in that set and one end in the complementary set. Switching a series of vertices, once each, is the same as switching the whole set at once.
If a signed graph is obtained from a signed graph by a series of switchings, then we say that and are switching equivalent.
Theorem 2.
If and are switching equivalent, then is tree--colorable iff is tree--colorable, hence .
Proof.
We just need prove this result if is obtained from by switching one vertex . Let be a tree--coloring of and let . In what follows we claim that if we recolor by then we would get a tree--coloring of
If , then is a forest, since any edge incident with is not contained in and thus .
If , then we suppose, to the contrary, that contains a cycle . If , then any edge incident with is not an edge of and thus is cycle in , a contradiction. Hence we assume that . Let be path derived from by deleting and . Since for every and for every , is a subgraph of . By the definition of , we have and . Note that . Since and , and . This implies that and are two edges in . Therefore, becomes a cycle in , a contradiction. Hence is a forest.
If , then is a forest by the above result.
In conclusion, we prove that is a forest for any . Therefore, is a tree--coloring of . ∎
Actually, Theorem 2 grantees that Definition 1 is well-defined.
A cycle of a signed graph is positive if it has an even number of negative edges. A singed graph is said to be balanced if all of its cycles are positive.
By (resp. ), we denote the graph with vertex set (resp. ) and edge set (resp. ).
Lemma 3.
Let and be balanced signed graphs such that if , and let be a graph with signature such that if and if . Let and be signed tree-colorings of and , respectively, such that for any . If or , then combining with we obtain a signed tree-coloring of .
Proof.
Fist, we consider the case that . One can image that is drawn on one side of the edge while is drawn on the other side of the edge , so is non-crossed in this drawing.
If is not a signed tree-coloring of , then contains a cycle for some color . Clearly, , , and the graph and induced by and are cycles, respectively. Otherwise, or contains a cycle, a contradiction.
By symmetry, we consider the following two cases.
If , then , because otherwise is a cycle in , a contradiction. On the other hand, since is a path with , the edge of which is positive iff its end-vertices are colored both by or both by , and is negative iff one of its end-vertices is colored by and the other is colored by , there is an even number of negative edges in . Hence when
, there is an odd number of negative edges in the cycle
, contradicting the fact that every cycle of is positive.If and , then , because otherwise is a cycle in , a contradiction. On the other hand, since is a path with and , the edge of which is positive iff its end-vertices are colored both by or both by , and is negative iff one of its end-vertices is colored by and the other is colored by , there is an odd number of negative edges in . Hence there is an odd number of negative edges in the cycle , contradicting the fact that every cycle of is positive.
Second, we consider the case that . One can image that is drawn in the interior of the triangle while is drawn in the exterior of the triangle , so are non-crossed in this drawing.
Actually we can prove that is a forest for every and . Suppose, to the contrary, that contains a cycle for some color . One can easily conclude that there is an edge among , say , and then a path on initiated from and ended with , so that is a cycle in , and meanwhile, , because otherwise or contains a cycle, a contradiction. At this stage, using the same arguments as the ones applied for the first major case, we would obtain a contradiction to the fact that is a balanced signed graph. ∎
A plane graph is a near-triangulation if the boundary of every face, except possibly the outer face, is a cycle on three vertices, and is triangulation if the boundary of every face is a cycle on three vertices. Clearly, every triangulation is near-triangulation.
Theorem 4.
Let be a balanced signed graph and let be a near-triangulation with outer face . If there is a list of colors to every vertex in so that , , for every and for every , then there is a signed tree-coloring of so that for every .
Proof.
We prove it by the induction on . If there is an edge with , then let be the signed graph induced by the vertices on and inside the cycle and let be the signed graph induced by the vertices on and inside the cycle . Without loss of generality, assume that . Clearly, and are near-triangulations with outer face and , respectively.
Since , , for every and for every , there is a signed tree-coloring of so that for every by the induction hypothesis. Now and have been colored with and , respectively.
Define a list of colors to every vertex in so that , and for every . Since for every and for every , there is a signed tree-coloring of so that for every .
Combining with , we obtain a signed tree-coloring of so that for every by Lemma 3.
Hence in the following we assume that there is no such an edge with . Let be neighbors of in cyclic order around . By the previous assumption, we have , since is a near-triangulation. Hence is a near-triangulation with outer face . Let be a color in that is different from . Define a list of colors to every vertex in so that for every and for every . Since for every , for every . Hence we can apply the induction hypothesis to and conclude that there is a signed tree-coloring of so that for every . Now we complete the coloring of by coloring with . Since and for every , is a forest. Note that we would not mind whether is equal to or not. Hence is a signed tree-coloring of so that for every . ∎
Actually, Theorem 4 directly implies the following
Corollary 5.
If is a balanced signed triangulation, then .
Since all-positive signed graph is balanced and the signed tree-coloring for such a graph is equivalent to the tree-coloring for the unsigned graph , the signed vertex arboricity of is the same with the (unsigned) vertex arboricity of . Hence we have
Corollary 6.
If is a triangulation, then .
Specially, since every planar graph is a subgraph of a triangulation and if , we deduce a well-known result contributed by Chartrand, Kronk and Wall [2].
Corollary 7.
If is a planar graph, then .
Now let us focus on the class of -minor-free graphs, which is a larger class than the one of planar graphs. Before showing the second main result of this paper, we present some additional useful lemmas.
Lemma 8.
Lemma 9.
Let be a balanced signed graph and let be a near-triangulation. If there is a triangle in and a list of colors to every vertex in so that , , , for every and has a signed tree-coloring by coloring with , respectively, then there is a signed tree-coloring of so that for every .
Proof.
We prove it by the induction on the order of . If both the interior and exterior of the triangle contains vertices of , then let denote the graph induced by the vertices inside or on , and let denote the graph induced by the vertices outside or on . Clearly, we can apply the induction hypothesis to both and , and then obtain signed tree-colorings and of and so that for every and , respectively.
Combining with , we obtain a signed tree-coloring of so that for every by Lemma 3.
Hence we assume, without loss of generality, that is the outer face of . Let be neighbors of in cyclic order around . Since is a near-triangulation, . Color with and define a list of colors to every vertex in so that for every and for every . Since , for every and for every , there is a signed tree-coloring of so that for every by Theorem 4. Combining with the color on , we obtain a coloring of so that for every . Next, we show that is a signed tree-coloring of .
In fact, is for any , and thus is a forest, because is a signed tree-coloring. Hence we just claim that is a forest.
If or , then is a forest, since for every . If , then by symmetry we consider three cases according to the fact that is a positive cycle. If , then , and . If and , then , and . If and , then , and . In each case does not have the signed tree-coloring as mentioned in the lemma, a contradiction. ∎
Lemma 10.
Let be a signed Wagner graph (see Figure 1). If there is a list of colors to every vertex in so that and for en edge , and for every , then there is a signed tree-coloring of so that for every .
Proof.
By symmetry, considering two speical cases here is enough. First, if and , then let , , and choose , , , , , . It is easy to check that the resulting coloring is a signed tree-coloring as required. Note that we may have or by the above choice of coloring but it does not matter at all. Second, if and , then we can similarly solve it, so omit the proof. ∎
Theorem 11.
If is an edge-maximal -minor-free graph with balanced signature and a list of colors is given to every vertex in so that , then there is a signed tree-coloring of so that for every . Specially, if we choose for every , then the previous coloring is a signed tree-3-coloring of and thus the signed vertex arboricity of is at most 3, i.e., .
Proof.
We prove it by the induction on the order of . By Lemma 8, , where is an edge-maximal -minor-free graph, is an edge-maximal planar graph (i.e., triangulation) or the Wagner graph, and or .
If , then let . Since is an edge-maximal -minor-free graph with smaller order, by the induction hypothesis, has a signed tree-coloring such that for every . Now, the vertices and have been colored, and then by Theorem 4, there is a signed tree-coloring of such that for every and , . Combining with , we obtain a signed tree-coloring of such that for every by Lemma 3.
If , then is an edge-maximal planar graph (i.e., triangulation), because Wagner graph does not contain triangles. Let . Since is an edge-maximal -minor-free graph with smaller order, by the induction hypothesis, has a signed tree-coloring such that for every . Now, the vertices and have been colored so that admits a signed tree-coloring, thus by Lemma 9, there is a signed tree-coloring of such that for every and , , . Combining with , we obtain a signed tree-coloring of such that for every by Lemma 3. ∎
Note that all-positive signed graph is balanced and its signed vertex arboricity is the same with its (unsigned) vertex arboricity, and that if is a subgraph of an unsigned graph . Hence Theorem 11 directly deduces the following corollary.
Corollary 12.
If is a -minor-free graph, then .
References
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