On coloring of fractional powers of graphs
Abstract
For , the fractional power of a graph is the th power of the subdivision of , where the subdivision is obtained by replacing each edge in with a path of length . It was conjectured by Iradmusa that if is a connected graph with and , then . Here we show that the conjecture does not hold in full generality by presenting a graph for which . However, we prove that the conjecture is true if is even. We also study the case when is odd, obtaining a general upper bound for graphs with .
Keywords: fractional power, chromatic number, regular graphs.
Math. Subj. Class.: 05C15
1 Introduction
Let be a simple finite graph, and let and be positive integers. The subdivision of , denoted by , is the graph formed from by replacing each edge with a path of length . The power of , denoted by , is the graph constructed from by joining every two distinct vertices with distance at most . The fractional power is then defined to be the power of the subdivision of ; that is, . Here we study the relation of and .
A total coloring of is a coloring of its vertices and edges such that no adjacent vertices, no adjacent edges, and no incident edge and vertex have the same color. The total chromatic number of is the least number of colors in such a coloring. The famous Total Coloring Conjecture, formulated independently by Behzad [1] and Vizing [6], states that for any simple graph . Since , and for every graph with , we can rewrite the Total Coloring Conjecture as follows:
If , then maximum cliques of are somewhat separated, so less colors may be needed, as conjectured by Iradumusa [2].
Conjecture 1 ([2]).
If is a connected graph with and , then .
Iradmusa showed that Conjecture 1 is true for . We refer the reader to [2] for the proof and several other results for .
The purpose of this paper is to further investigate Conjecture 1. Our first results is that Conjecture 1 is true if is even.
Theorem 1.
If is a graph with and with even, then .
We also study the conjecture when is odd. We show that the conjecture does not hold in full generality. In particular, it is not true for the cartesian product of and (triangular prism), , and . However, we give the following general bound.
Theorem 2.
If is a graph with and with odd, then .
The paper is organized as follows. In Section 2, we introduce notation and some known results that are used later in the paper. The proof of Theorem 1 is given in Section 3 () and Section 4 (). In Section 5 we prove Theorem 2 for graphs that are not complete, and we show that Conjecture 1 holds for infinitely many (about “half” of the values) if is odd. Section 6 deals with complete graphs, where it is shown Conjecture 1 is true for all complete graphs. Finally, Section 7 discusses the connection between dynamic coloring and Conjecture 1.
We make the following conjecture.
Conjecture 1.1.
Conjecture 1 holds except when .
2 Notation and Preliminaries
In this paper we only consider finite simple graphs. Next we always assume that and are positive integers such that . The graph is constructed from in two steps. First, every edge is replaced by a path (called a superedge) of length , forming . Second, edges joining vertices of distance at most in are added, forming .
Let () be a vertex of that lies on and has distance from in . If or , then is a branch vertex, otherwise is an internal vertex.
For an edge in , the ordered tupple of vertices of defined by
is called a bubble (at ). If is odd, then we say that the set of vertices
is the crust at . Lastly, , called a middle part, is the tuple of vertices between the two bubbles (or two crusts if is odd) on the edge defined by
For a tuple we define . Also, we write and . So, we have and . If is another tuple, then means (i.e. ) for all for some function . To shorten notation, we often write instead of . Lastly, we use the symbol to denote the merging of two tuples as follows: .
We are ready to state results of Iradmusa that we use later in this paper. The first theorem gives the exact values of .
Theorem 2.1 ([2]).
If is a graph and such that , then
Since for any graph , it suffices to show to prove Conjecture 1. By Theorem 2.1, we thus need to construct a coloring of using colors if is even, and colors if is odd.
By [2, Lemma1], if , then . This result can be generalized as follows.
Lemma 2.2.
Let be a graph, and such that . If for some , then .
Theorem 2.3 ([2]).
If is a connected graph with and , then .
The next lemma follows by inductively applying Lemma 2.2 and Theorem 2.3. We will use it repeatedly.
Lemma 2.4.
If for all and some , then for all with . In Particular, if Conjecture 1 holds for , then it holds for all with .
Throughout this paper, we will also use the following wellknown result by König [4].
Lemma 2.5 ([4]).
For every graph with maximum degree there exists a regular graph containing as an induced subgraph.
Lemma 2.5 enables us to restrict our attention to regular graphs. Indeed, if is not regular, we can find a regular graph , color , and use the coloring of on . The following result is proven in [2, Theorem 3] using a special type of vertex ordering and induction. Here we give another simple proof.
Lemma 2.6.
If is a connected graph with maximum degree , then there exists a proper coloring using the color exactly on branch vertices.
Proof.
First, we color all branch vertices with . It remains to color the internal vertices with colors . Let be a graph that arises if we delete all branch vertices of . Then , and is neither an odd cycle nor a complete graph. Thus, by Brooks’ theorem. ∎
Observe that the graph that arises if we delete all branch vertices of is isomorphic to the line graph of . Therefore, we can think of coloring of internal vertices of as of coloring of edges of . So, Lemma 2.6 equivalently states that for a graph with degree , there is a proper edge coloring . We refer to such a coloring as a halfedge coloring of , and to the ’halves’ of edges of that correspond to the edges in simply as halfedges of . For an edge in , denotes the halfedge on that is adjacent to .
Lemma 2.7.
If is a graph with maximum degree , then there exists a (proper) halfedge coloring .
3 Theorem 1 for noncubic graphs
In this section we show that Conjecture 1 is true if is even and . We sketch the basic idea of the technique used here. Suppose that we have a halfedge coloring of using colors. For each color we introduce new colors , and use them on each bubble whose corresponding halfedge has color . So, we use colors to color vertices of bubbles. As a next step we show that it is possible to color the middle vertices from the same set of colors. So, if we use an additional color for branch vertices of , then we can conclude that , as needed.
Theorem 3.1.
If is even and is a connected graph with , then .
Proof.
By Lemma 2.5, we can assume that is regular graph, where . Also, it is sufficient to prove the claim for (Lemma 2.4). By Theorem 2.1, . Since trivially , we only need to construct a proper vertex coloring that uses colors.
Let , where , be a halfedge coloring of whose existence is ensured by Lemma 2.7. Then we define a vertex coloring as follows.

branch vertices: for every branch vertex .

bubbles: . (Recall that this means .)

middle parts: Consider a superedge . Since , there exist two colortuples , . Let and . Then we define and . See Figure 2.
We show that the coloring is proper. We need to check that the distance between any two vertices , with the same color is greater then in the subgraph of . This is true if since then and are branch vertices. Let now .

and : First, since the colors of vertices of each bubble are pairwise distinct. Next, neither nor since is proper and thus and cannot be at the same vertex. Also, and are not on the same superedge, so either or . If and , then and are two vertexdisjoint edges in and so . If and , then since by definition of . The case and is analogous.

and : As in the previous case, . But then since any two vertices from two different middle parts are in distance at least in .

and : If and are vertexdisjoint edges in , then again . So we may assume that . Then since and do not belong to the same superedge by definition of . Let be such that . Since , we have by definition of that . It follows that and , which together gives .
∎
4 Theorem 1 for cubic graphs
For a cubic graph , let be a proper halfedge coloring. A cycle in that uses only two of the colors on its halfedges is called a bad cycle in . The coloring is called a good halfedge coloring if there is no bad cycle in .
Lemma 4.1.
There exists a good halfedge coloring for every cubic graph .
Proof.
We consider an arbitrary halfedge coloring of . Let be bad cycles of . Observe that the cycles are pairwise edgedisjoint. We prove that we can eliminate the bad cycles by induction on . If , then there is nothing to prove. So suppose that .
Consider the cycle with its halfedges colored with colors . Then every halfedge incident with a vertex of has color . For an arbitrary vertex of , let be the neighbor of such that . So we have . Let be the neighbor of in such that . Then we define a halfedge coloring of by , , and for the remaining halfedges of (see Figure (a)a). Now uses all the three colors and the new coloring is again a proper halfedge coloring.
We show that no new bad cycle arises. Suppose it does. Let be a bad cycle in that was not a bad cycle in . Then either the edge or the edge is in . Suppose first. Then uses colors and on its halfedges, and thus contains the path . But the vertex of is adjacent to only one edge that uses colors and on its halfedges (since one of the colors was changed to on ), a contradiction. The case when also leads to a contradiction since no other edge than incident with has color on its halfedges. ∎
For a halfedge coloring , an halfedge is a halfedge colored with the color by , and an edge is an edge with the colors used on its halfedges.
An orientation of a graph is obtained by assigning a direction to each edge. We say that a halfedge is oriented inwards in if the head of the corresponding edge in the orientation of is a branch vertex of . Otherwise we say that is oriented outwards.
Lemma 4.2.
For every good halfedge coloring of there exists an orientation of such that for every edge , if and with , then one of the halfedges , is oriented inwards and the other one outwards.
Proof.
For , , let be the graph induced by edges of . Since there are no bad cycles in , the graph is a disjoint union of paths . Every halfedge adjacent to a vertex of has color . We orient all these neighboring halfedges subsequently for each . For , we orient every halfedge incident with with odd inwards and the rest outwards (so that the directions alternate along the path), as shown on Figure (b)b. We repeat the process for all graphs , , . Finally, we orient the remaining halfedges arbitrarily. ∎
Theorem 4.3.
If is even and is a graph with , then .
Proof.
By Lemma 2.5, we can assume that is cubic. By Theorem 2.1 we have that . So, we only need to find a coloring of that uses colors. Also, we only need to consider with by Lemma 2.4. We split the range into two parts.
Case 1: (first range). Let be a good halfedge coloring of , guaranteed by Lemma 4.1, and let be the orientation of given by Lemma 4.2 applied on and . Then we define a vertex coloring as follows.

branch vertices: for every branch vertex .

bubbles: Let be a bubble of . If the underlying halfedge of is oriented outwards, then let . Otherwise, let .

middle parts: Let . Since is in the first range, . Let . If the halfedge oriented inwards is adjacent to , then we define (as shown on Figure 4). Otherwise we let .
A routine check (similar as the one given in the proof of Theorem 3.1) shows that vertices of the same color are of distance at least in .
Case 2: (second range).
By Vizing’s theorem, is edgecolorable. Let be a proper edge coloring with the smallest number of edges colored . For , we call an edge that has color an edge. For , let be the subgraph of induced by edges and those edges whose both endpoints are incident to an edge. Observe that is a disjoint union of paths and cycles. So, we can orient the edges in each component in the same direction (i.e. so that the indegree as well as the outdegree of each vertex is at most in .) Note that there is no conflict in orienting the edge since , , are edgedisjoint. Indeed, suppose that two of them, say , , share an edge . Then is a edge, and by definition of , both ends of are adjacent to one edge and one edge. But then we can color with the remaining color , a contradicition with minimality of the number of edges. On the other hand, every edge is oriented since it is adjacent to four edges, out of which two (nonadjacent) edges have to have the same color.
The construction of a proper vertex coloring is more complicated that usual. Whereas both bubbles on an edge receive the same colortuple , the colortuples on the two bubbles on a edge differ. Also, the coloring of middle parts for edges and edges i not be the same. In the first case we split each into two (almost) equal parts, but in case of edges we split into three nonequal parts. The construction of is depicted on Figure 5.

branch vertices: for every branch vertex .

bubbles on edges:
Let that is oriented from to in . Then let and (so the bubbles on one edge are oriented in the same direction).

middle parts on edges:
Case 1: is even. Let be a edge oriented from to in , and let . We split into three parts , , and .
Every edge is adjacent to at least one edge of each color , , (minimality). In particular, there is an edge adjacent to both and for some . Next, there is a edge and edge adjacent to and , respectively, for such that .
First, we color the central part by letting .
Second, we color the remaining parts and . Since , it suffices to show the coloring of , the coloring of is analogous. If the edge adjacent to is oriented inwards, then we set . Otherwise we define .
Case 2: is odd. We use the coloring defined in Case 1 on all but one vertex of , and then we color separately. So, is defined as above, but for and the three parts , , and . It remains to color on each .
For any , an edge is called a switching edge if is adjacent to edges and such that both and use the same colortuple . Then is a subgraph of induced by all switching edges and edges adjacent to at least one switching edge. Every component of is either a path or an even cycle, so we can orient the edges in in the same direction. We refer to the resulting oriented graph obtained from by orienting each component as .
Now we let for every edge . The whole graph is colored, but we have a conflict on vertices colored with . We do the following switching. Let be an oriented edge such that is an edge, where . Then we let and .

bubbles on edges:
We adopt the notation from the previous part. Let be the color missing around . If the edge adjacent to is oriented inwards, then , otherwise .

middle parts on edges:
We consider an edge , where . Let and be bubbles adjacent to and that use different colortuples. Then , and .
∎
5 Counterexample and Theorem 2
We begin this section by proving that Conjecture 1 is not true for if and . Before presenting the proof we need the following lemma.
Lemma 5.1.
Let be odd. If is a proper vertex coloring of with colors, then all vertices belonging to the same crust have the same color.
Proof.
By Theorem 2.1, . By the proof of Theorem 2.1 given in [2], maximal cliques in are induced by all vertices of distance and one vertex of distance from a branch vertex of degree in . In our terminology, a maximal clique consists of one branch vertex with , all bubbles at and one vertex of the crust . The coloring uses colors on and the bubbles at . Therefore, only one color remains for all the vertices of the crust . ∎
By writing “” for some coloring of we mean “ for all ”.
If , is odd, and is a proper vertex coloring of with colors, then for every edge since . So a necessary condition for to be proper is that the vertex coloring of defined by for every is proper.
Proposition 5.2.
If , then .
Proof.
By Lemma 2.1, . Thus, . By way of contradiction, suppose there exists a proper vertex coloring . Let be the vertices of as on Figure 6. By Lemma 5.1, all vertices of the same crust have to receive the same color. Since there are six crusts and only five colors, at least two crusts have to receive the same color, say . Also, no two adjacent crusts can have the same color. So, we can assume that (all other cases are symmetric).
Every maximal clique has to use all five colors . We try to color some vertex of each clique with . The color cannot be used for any other crust then on and and thus, it has to be used on some vertex in distance at most from for every .
For , the only vertex that can receive color is , because all of the others are too close to one of the two crusts , . Analogously for , the vertex has to receive color . Subsequently has to receive the color , because and are too close to the crust and is too close to . But then there is no vertex in distance at most from that can be colored with . Indeed, and are too close to , is too close to and is too close to . Hence, a contradiction.
∎
We now come to the two main theorems that support Conjecture 1 for odd . We first find a general upper bound for the chromatic number of for any noncomplete graph with maximum degree . Then we show that in about half of the possible choices for , the chromatic number of is in fact equal to . In the rest of this section, we assume that a graph is not a complete graph.
For a graph with , let be a proper vertex coloring and let be a halfedge coloring of . A halfedge is called an incompatible halfedge if . If otherwise , then is called a compatible halfedge. If every vertex of is adjacent to at most incompatible halfedges, then we say that and are incompatible. If , then we say that that and are compatible rather than incompatible.
Lemma 5.3.
For every connected noncomplete graph with maximum degree and for every proper vertex coloring there is a halfedge coloring such that and are incompatible.
Proof.
By contradiction. We consider a halfedge coloring that has the minimal number of noncompatible halfedges. Let be a vertex with three neighbors such that for all . Let , , . Observe that either or , otherwise we could switch colors on and , a contradiction with minimality. We can assume that . It follows that (since otherwise we could switch colors on and ), and next that (since otherwise we could switch colors on and ). But now we can recolor with , with , with , a contradiction with minimality. ∎
Theorem 5.4.
If is odd and is a connected noncomplete graph with maximum degree , then .
Proof.
Let and , where , be compatible colorings given by Lemma 5.3. Then we define an auxiliary (not necessarily proper) vertex coloring