1 Introduction
A coloring (or simply a coloring) of a graph is a surjective function . We refer to the values assigned to vertices as their colors. A proper coloring is such that for any edge . A proper coloring can also be given as a partition of the vertex set of into disjoint stable sets , . In this case, is the color class, formed by the vertices such that . The graph is colorable if it admits a proper coloring. In what follows, by a coloring we always mean a proper coloring.
Graph colorings are a natural model for problems in which a set of objects is to be partitioned according to some prescribed rules. Usually, the rules are related to conflicts between the objects to be partitioned. This model is remarkably useful for scheduling problems [Werra.85], such as frequency assignment [Gamst.86], register allocation [Chow.Hennessy.84, Chow.Hennessy.90], and the finite element method [Saad.96].
While it is easy to find a coloring when no bound is imposed on the number of color classes, for most of these applications the challenge consists in finding one that minimizes the number of colors. The chromatic number of a graph is the minimum number of colors in a coloring of ; it is denoted by and we say that is chromatic if . An optimal coloring is any coloring with colors. To decide if a given graph is colorable is an complete problem, even if is not part of the input [Hol81]. The chromatic number is even hard to approximate: for all , there is no algorithm that approximates the chromatic number within a factor of unless [Has96, Zuc07].
Because of the hardness results and the existing practical applications, the study of coloring heuristics is motivated. In most coloring heuristics, such as the well studied DSATUR
[Bre79], the greedy algorithm is present. The greedy algorithm works on an input graph and an ordering of its vertices. For each from up to , we color with the smallest color such that no vertex in is already colored . We call a coloring obtained by the greedy algorithm a greedy coloring.A nice property is that there always exists an ordering that produces an optimal coloring of . To see this, let be a coloring of graph . The order obtained by ordering every vertex of color 1 first, then the vertices of color 2, and so on, clearly produces a coloring that uses at most colors. Hence, minimizing the number of colors used by the greedy algorithm equals to finding the chromatic number of . This is why the parameter studied is related to the worstcase scenario, i.e., the maximum value for which the greedy algorithm produces a coloring with colors. This is called the Grundy number and is denoted by . It is a very well studied parameter, but we refrain from citing any work on it here because we will focus on a special type of greedy colorings.
Connected greedy colorings. In this paper we consider a variant of the greedy algorithm called the connected greedy algorithm. A connected ordering of is an ordering of the vertices with the property that has at least one neighbour in {} for every . The connected greedy algorithm works similar as the greedy algorithm, but only takes as input connected orderings. A connected greedy coloring is one obtained by the connected greedy algorithm. Up to our knowledge, this has been first introduced in [hertz1989connected], where the authors call the connected greedy algorithm SCORE (Sequential coloring based on a Connected ORdEr).
The concept of Connected Grundy Number is naturally defined as being the maximum for which has a connected greedy coloring. But observe that, contrary to the traditional greedy algorithm, it is not obvious that there always exists a connected ordering that produces an optimal coloring. Indeed, in [babel1994hard], the authors present a graph on 18 vertices for which this does not hold. They believed that it was the smallest such graph, but up to our knowledge a formal proof has not been given yet. Also, we mention that there is an infinite number of such graphs, as shown in [BCD+14]. Therefore, it makes sense to define the minimization parameter related to connected greedy colorings. The connected chromatic number of a graph is defined as the minimum integer for which admits a connected greedy coloring; it is denoted by . Interestingly enough, this cannot be larger than the chromatic number plus one [BCD+14]:
(1) 
Given graph parameters and , a graph is called perfect if for every induced subgraph of . In [ChSe79], the authors prove that the perfect graphs are exactly the cographs. Following this result, concerning connected greedy colorings, in [hertz1989connected] the authors characterize a subclass of perfect graphs (they make other constraints on the order), while in [KT.18] the authors characterize the clawfree perfect graphs. We mention that the complexity of recognizing perfect, perfect, and perfect are all open.
Concerning complexity results, in [BCD+14] it is proved that deciding if is hard, while in [BFKS.15] the authors prove that it is complete to decide whether has a connected greedy coloring with colors, for every fixed . The latter result has been generalized in this article, as will be seeing forward.
Here, we are interested about whether there exists a hard dichotomy for the free graphs, as the one below for the traditional coloring problem, proved by Král et. al. in 2001 (a graph is free if it does not contain a copy of as induced subgrah). In what follows, denotes the graph obtained from the union of and . Also, given a graph , we denote by the class of free graphs.
Theorem 1.1 ([kral2001complexity])
Let be fixed. Given and a positive integer , deciding whether can be done in polynomial time if is an induced subgraph of or , and is an complete problem otherwise.
Even before this result was presented, it was already known that deciding whether is complete for line graphs for every fixed [Hol81]. Because line graphs are clawfree, it follows that deciding is complete for free graphs, when contains a claw. Also, in 2007 Kaminski and Lozin [KL.07] proved that for every fixed and , given a graph with girth at least , deciding whether is complete. This gives us that deciding for is complete for every fixed and every fixed . Therefore, if the problem is polynomial in , then must be a linear forest (forest of paths). This is why much work has been done on the problem of deciding, for fixed values of , whether for graphs with no certain induced paths. It has been proved that it is polynomialtime solvable for free graphs and every positive integer [HKLSS.10], and when for free graphs [RS.04] and, more recently, free graphs [B.etal.17]. Also, it is complete for free graphs and , and for free graphs and [huang2016improved]. Therefore, the only open cases are: 4coloring free graphs, and 3coloring free graphs, for . These results are summarized in Table 1.
free  


3  4  5  
¶  ¶  ¶  ¶  
6  ¶  ?  
7  ¶  
8  ? 
Now, coming back to our problem, observe that asking whether is not the same as asking whether for a given . Indeed, the former question is not in , while the latter is. Below, we formally define these problems.
Concercing Problem CGCDecision, part of our results follow directly from previous ones. This is because of the following easy proposition and the fact that some of the classes are closed under the addition of an universal vertex. To see that the proposition holds, just observe that , where is obtained from by adding a universal vertex.
Proposition 1
Let be a graph class and suppose that deciding is complete if for fixed . If is closed under the addition of a universal vertex, then deciding is complete on
Observe however that, when is the cycle on 3 vertices or is a claw, then the class is not closed under the addition of universal vertices. Without getting much ahead of ourselves, we mention that we actually investigate the complexity of deciding for fixed , when is a chromatic free graph. This gives us hardness results for both problems. However, because we were not able to obtain such hardness results for every possible configuration of , we do not have a dichotomy for the Problem CGCEquality, while Problem CGCDecision have the same dichotomy as in Theorem 1.1.
Theorem 1.2
Let be a fixed graph, and be a positive integer. If is an induced subgraph of or , then Problem CGCDecision can be solved in polynomial time. Otherwise, the problem is complete. Furthermore, if is considered to be fixed and is at least 7, then it remains complete when is not a linear forest or contains a as induced subgraph.
Theorem 1.3
Let be a fixed graph and . If is not a linear forest or contains a as induced subgraph, then CGCEquality is hard. Also, if is an induced subgraph of or of , then .
Observe that the polynomial case of CGCEquality in the theorem above consists of a very simple algorithm: it always says “yes” if is free, for an induced subgraph of or . We ask whether this is always the case when is a path:
Question 1
Does there exist such that for every , while deciding whether is hard for every ?
Also, as already mentioned, we actually prove that deciding is hard even if is a chromatic graph, for some fixed values of . The only proof where we did not succeed in fixing was for the free graphs. Therefore, we ask:
Question 2
For fixed , given a free chromatic graph , can one decide in polynomial time whether ?
Now, concerning only free graphs, from Table 1 and Proposition 1 we get the situation depicted in Table 2. Position of this table tells us the complexity of deciding for . The completeness results propagate along the rows because of the proposition above, and along the columns because . The row related to is entirely polynomial because of the result in [HKLSS.10] previously mentioned and by Theorem 1.2. The question marks are open problems and they propagate along the rows in a column.
free  


3  4  5  
¶  ¶  ¶  ¶  
6  ?  ?  ?  
7  ?  ?  
8  ?  ? 
We mention that to prove Theorems 1.2 and 1.3, we actually investigate the edge version of the problems. Since every line graph is clawfree, we get that the problem is hard for clawfree graphs as well, which means that if is a nonlinear forest, then the problem is hard on free graphs. Let denote the connected chromatic index of (which equals the connected chromatic number of , the line graph of ). Observe that, by Vizing’s Theorem and Equation 1, we get , for every graph . However, we were not able to find a graph with ; note that such a graph would necessarily be a Class 2 graph (a graph is Class 2 if ). Hence, we pose the following question.
Question 3
Does there exist a Class 2 graph such that ?
Another aspect that relates to our investigation is the notion of hardtocolor graphs. In [babel1994hard], a connected graph is called globally hardtocolor if, for every and every , if is greedily colored using a connected order starting at with color , then the produced coloring uses more than colors. To prove the hardness results for free graphs, we construct a globally hardtocolor free graph. This leads us to the following question (by our results, we know that the answer is in ):
Question 4
What is the smallest such that there exists a globally hardtocolor free graph?
Finally, although not directly related to the studied problem, we would like to expose another interesting and nontrivial aspect of greedy colorings in order to pose one last question. In [ChSe79], the authors prove that for every integer , there exists a greedy coloring of with colors. We ask whether the same holds for connected greedy colorings.
Question 5
Let be any graph. Does there always exist a connected greedy coloring with colors, for every ?
Many other questions can be posed on these colorings since, up to our knowledge, very little is known about both parameters, with only the aforementioned articles having been published on the subject.
2 Outline of the proof
First, observe that, for , the graph classes and are closed under the addition of a universal vertex. Therefore, by Proposition 1 and Theorem 1.1, we get that CGCDecision is complete on and on , for and . In fact, we get stronger results since these problems are complete even for fixed , as we mentioned before. However, these results do not imply that CGCEquality is also hard on these graph classes. Here, we actually prove the following results, which imply hardness of both CGCEquality and CGCDecision.
Lemma 1
For every and , deciding if is complete even when restricted to chromatic graphs in .
We mention that the hardness for follows from a result in [BCD+14].
Lemma 2
For every , deciding if is complete for free edgechromatic graphs.
These lemmas finish the proof of Theorem 1.2. As for Theorem 1.3, it remains to study the case where is a linear forest, which is done in the following lemmas.
Lemma 3
Deciding if is hard for free graphs, for every .
Lemma 4
If is an induced subgraph of or of , then , for every free graph .
Before, we begin, we need one last definition. Given a vertex , an integer and a coloring of , we say that is a connected greedy coloring of (or CGC for short) if there exists a connected order of , , such that starts with and is obtained by coloring with and applying the greedy algorithm on . The similar is used for edgecoloring. Also, from now on we call a connected greedy coloring and a connected greedy edgecoloring by CGC and ECGC, for short .
3 free graphs
Here, we prove Lemma 1. It is known that deciding whether is complete for planar graphs [GJS76]. Observe that this and the Four Color Theorem imply that the following problem is also complete for every fixed . It suffices to add universal vertices to a planar graph to obtain such that if and only if .
Given an instance of the problem above, we construct a chromatic free graph such that if and only if . Lemma 1 follows because Problem COL COLORABLE is complete for every . We first construct a gadget that will admit cycles of undesired length, and afterwards we replace some of the edges in order to get rid of such cycles. Start with three disjoint cliques each of size , and let be a vertex of . Obtain by adding vertices and making complete to , complete to , complete to , and complete to . See Figure 1 for the construction of .
Now, given a colorable graph that contains a dominant vertex, denote by the graph obtained from by appending a copy of on each vertex of , identifying on vertex of . We first prove that if and only if , and then we show how to get rid of the undesired cycles. This is actually the proof presented in [BCD+14]. Note that, because deciding for a free graph is complete for every fixed and [KL.07] and by previously mentioned aspects, this gives us that CGCDecision is complete even when restricted to free chromatic graphs, for every ( or ) and . For the first part of the proof, it is essential to understand the following properties of graph .
Lemma 5
Let be the graph obtained as above, where is a positive integer, .

;

In every coloring of , vertices receive the same color;

For every and every , we get that vertices receive color at most in every CGC of with colors.
Proof
The first and second properties can be easily verified: it suffices to see that the colors used in cannot be used in any of the cliques . For the last property, consider any connected order starting with and suppose, without loss of generality, that . Then, no matter which one between and gets colored first, we get that it will have at most colored neighbors and, therefore, cannot have color bigger than . The property then follows from Property 2.
Now, observe that if , then Property 3 trivially implies that . On the other hand, if , because has an universal vertex, we get that . A CGC of with colors can be easily constructed from a CGC of with colors. This proves that if and only if , as we wanted to show.
Now, we construct gadgets that will replace the edges of so as to ensure that the obtained graph has no cycles of length , for and . In fact, we need a different gadget for each case because, when avoiding cycles of length 3, we end up creating cycles of length 5, and viceversa. First, we show the gadget necessary for the case .
Let be obtained as follows. Start with a , , a clique of size , and a clique of size . Let every vertex in be complete to , and let be complete to . Finally, add vertex and make it complete to . Figure 2 depicts .
Now, let be obtained from by replacing every edge with a copy of , except the edges of incident to the copies of in the gadgets of type . Observe that has no induced cycles of length 5; also, because the distance between and in is 4, we get that also does not have cycles of length 5. Now, we prove that if and only if , thus finishing the proof of the case . For this, we need the following properties to hold.
Lemma 6
Let be obtained as above, where is a positive integer, .

and in every coloring of , vertices and receive distinct colors;

For every , , there exists a CGC of in which is colored with , and a CGC in which is colored with .
Proof
To see that just observe that and that a coloring of can be obtained by giving colors to , color 2 to , color 1 to , and colors to . Now, let be a coloring of . Because and are complete to of size , they receive the same color; similarly, and also receive the same color, and since is adjacent to , Property (i) follows. Now, given , , we construct the desired CGC as follows: color with and enough vertices of so that colors through appear in ; then, color with . If , color with , with , and finish coloring . Otherwise, color with , some vertices of with , with , and finish coloring . It is easy to see that this can be extended to as desired, and also that we could have started in and made our way to (this is because vertices and are symmetric with relation to ).
Observe that Property (ii) makes it possible to construct a CGC of , given a CGC of ; hence, if , then . Also, note that Property (i) and the fact that the edges incidents to the copies of vertices were not replaced ensures that Properties (1)(3) still hold on the copies of . Therefore, if , then Property (3) ensures us that , and the previous argument gives us that .
Now, we present the gadget needed for the case . We prove that Lemma 7 also holds for the consutrcted gadget, thus finishing our proof since the same arguments can be applied. For this, we introduce an operation on graphs. Let be any graph. The double myscielskian of is the graph obtained from as follows (observe Figure 3 to see the operation applied to the ): add two copies and of , and denote the copy of in by ; for every edge , add edges and , for and ; finally, add two adjacent vertices and and make complete to , for and . It is well known that the Mycieslki of a trianglefree chromatic graph produces a trianglefree chromatic graph [M55]. Also, if , an coloring of can be constructed from an optimal coloring of by coloring with and with , ; hence . Now, let be obtained by applying times the double mycielskian operation, starting with the . By what was said before, we get that and that is trianglefree. Also, since and are not adjacent, the graph constructed as before is also trianglefree. It remains to prove Properties (i) and (ii).
Lemma 7
Let be obtained as above, where is a positive integer, .

and in every coloring of , vertices and receive distinct colors;

For every , , there exists a CGC of in which is colored with , and a CGC in which is colored with .
Proof
Let be the additional vertices added at the last application of the double Myscielski operation. To see that , recall that the Mycielski operation increases the chromatic number by one. Hence, the double Mycielski operation increases by at least one; the fact follows because a coloring can be obtained from a coloring of by giving a new color to and using colors 1 and 2 in . Now, we prove the second part of Property (i) by induction on . Note that it trivially holds for the initial . So, suppose by contradiction that is a coloring of such that . Because , we can suppose that . Note that, by switching the color of every such that to , we obtain a coloring of : the color of certainly does not appear in , and we only change the colors of vertices contained in a stable set. But since , this coloring is a coloring of in which and receive the same color, contradicting the induction hypothesis.
Finally, consider , ; we prove Property (ii) also by induction on . Observe that, because and are symmetric in , we only need to prove the existence of a CGC in which is colored with . If , then and the property trivially holds. Now, suppose it holds for and consider the following cases:

: let be a CGC of in which . Then, let be obtained from by coloring with , for every , with , with 1 and with , for every (observe that this is at most );

: if , let and be a CGC in which is colored with ; and if , let and be a CGC in which is colored with . Also, let be any neighbor of in . Give color to , color to , color to , and color to . Let be the order that produces , and denote by the partial coloring of . Now, for each , if and , then let ; otherwise, let . Because for every , we know that if and , then color 1 does not appear in and color 1 is allowed for . Otherwise, because for each the neighbor of in is now colored with (including if ), we just need to prove that also has neighbors of colors 1 and 2 in in order to prove that is greedily colored. Let . If , then and has a neighbor in of color 2; otherwise, and has a neighbor of color 1 in . To finish coloring , give color , for every (observe that this is at most ), and color 3 to . Because , we know that as desired.

: a similar argument can be applied by getting a CGC of in which is colored with , when , or in which is colored with 1, otherwise.
4 Line graphs
Here we prove Lemma 2 by making a reduction from the problem of deciding whether is 3edgecolorable when is a trianglefree cubic graph, which is known to be complete [Hol81]. The idea follows the one applied for free graphs: for each , given an instance of the problem above, we append on each vertex some copies of a gadget that ensures that the obtained graph is such that if and only if . For this, we first construct what we call edge gadgets.
For each , let be obtained from a complete bipartite graph with parts and by adding , making complete to , complete to and adding edges and . Graph is depicted in Figure 4.
Now, we construct the gadget that will be appended on the vertices of . Observe Figure 5 to follow the construction. Consider copies of the edge gadget and denote them by and , . Let be obtained by identifying vertices into vertex , vertices into vertex , vertices into vertex , and finally adding a new vertex adjacent to .
Lemma 8
Let , and be an edgegadget obtained as above. Then the following hold:

In every edge coloring of , edges and receive the same color; and

For every , there exists a ECGC of .
Furthermore, let be constructed as before. Then, is trianglefree edgechromatic graph and the following holds.

has a CGC with colors if and only if .
Proof
Because is a bipartite graph, it has no triangles and, by König’s line coloring Theorem, we know that . In fact, observe that the only perfect matching containing must also contain , which gives us Property (1) below. Property (2) is also easy to be verified.
Now, by Property (1), the colors in are the same as the colors in . This means that, in every edge coloring of , edge must get the same color as edges and . Because edge has only 3 adjacent edges that are incident in , and edge has only edges that are incident in , we get that if we start to color in with color , we end up using colors only if (recall that ). Using Property (2), one can verify that a ECGC exists when .
We are finally ready to finish our proof. So, let be a trianglefree cubic graph, and let be obtained from by appending copies of on each vertex of . If , then a ECGC of with colors can be constructed as follows. Let be a connected order of and let be a 3edgecoloring of that uses colors . For each , color the copies of incident to using colors 1 through on the appended edges (this is possible by Property (3)), then color the uncolored edges incident to in ascending order of their colors. Now, suppose that . By Property (3) and the fact that for every , we get that the colors 1 through are all used in the copies of appended on . This means that the edges of can only use colors , which gives us a 3edgecoloring of .
5 free graphs
In this section, we prove Lemma 3. In [huang2016improved], it is proved that deciding is complete when is a free graph. Observe that if is a free graph and is obtained from by adding a universal vertex, then is also free and is such that if and only if . Therefore, deciding whether is complete even if is a free graph with a universal vertex. We reduce this problem to the problem of deciding whether for free graphs. For shortness, we call the former problem 5Col free and the latter CGC free.
The following lemma proved in [BCD+14] will be useful.
Lemma 9
Let be a connected graph, and be any positive integer. Then, there exists a CGC of with at most colors.
In our reduction, we use an auxiliary graph with the following properties.

is connected, and for every and every , there is no CGC of with 5 colors;

There exists such that there are no induced with extremity in ;

is a free graph.
We first assume that such a graph exists, and later we show how to construct it. Let be an instance of 5Col free and let be its universal vertex. Also, let be obtained from by identifying vertices and ; denote the resulting vertex by , the vertices corresponding to by and the vertices corresponding to by . Because of Properties (2) and (3), the fact that is free and that dominates in , we know that is also a free graph. We claim that if and only if . This finishes the proof.
First, let , and observe that, because , we get that . Let be an optimal coloring of . Since is universal in , we know that ; it suffices to start by coloring with 1 and using any optimal greedy ordering for (recall that it always exists). Also, because , by Lemma 9 we know that there exists a CGC of with at most colors. This means that can be extended to a connected greedy coloring of with colors.
Now, suppose that and let be an optimal connected greedy coloring of . Note that, since , we get that . Also, note that either starts in , in which case restricted to is a CGC of , or starts in , in which case restricted to is a CGC of . In both situations, we get that the number of used colors is bigger than 5 by Property (1).
It remains to construct the desired auxiliary graph. Observe Figure 6 to follow the construction. We start with a complete graph on vertices . Then, for each , add vertices and make each adjacent to two vertices of
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