1 Introduction
In a proper graph coloring, we want to assign to each vertex of a graph one of a fixed number of colors in such a way that adjacent vertices receive distinct colors. Proper graph coloring models a number of realworld problems related to scheduling or distributing limited resources in a way that avoids conflicts (e.g., scheduling classes into time slots so that any two classes taught by the same teacher occur at different times, or the compiler assigning variables to registers so that any two variables which are used at the same time reside in different registers).
In such applications, it is common for the vertices to prefer to be colored by certain colors (e.g., teachers may prefer to teach or not to teach at certain times, the variables may be more profitably kept in specific registers in case some assembler instructions can only be applied to those registers). Usually, it is not possible to satisfy all such preferences. This motivates the following definitions (which we present in a more general list coloring setting, for a reason we discuss below).
A list assignment for a graph is a function that to each vertex assigns a set of colors, and an coloring is a proper coloring such that for all . A graph is choosable if is colorable from every assignment of lists of size at least . A weighted request is a function that to each pair with and assigns a nonnegative real number. Let . For , we say that is satisfiable if there exists an coloring of such that
An important special case is when at most one color can be requested at each vertex and all such colors have the same weight (say for at most one color , and for any other color ): A request for a graph with a list assignment is a function with such that for all . For , a request is satisfiable if there exists an coloring of such that for at least vertices .
In particular, a request is satisfiable if and only if the precoloring given by extends to an coloring of . The corresponding precoloring extension problem has been studied in a number of contexts: as a tool to deal with small cuts in the considered graph by coloring one part of the graph recursively and then extending the corresponding precoloring of the cut vertices to the other part [2, 5, 12], as a way to show that a graph has many different colorings[6, 11], or from the algorithmic complexity perspective [3, 8]. In planar graphs, it is known that precoloring of any set of vertices at distance at least three from one another extends when at least 5 colors are used [1] and for sufficiently distant vertices this holds also in the list coloring setting [7]; on the other hand, a precoloring of two arbitrarily distant vertices of a planar graph does not necessarily extend to a coloring [9].
Dvořák, Norin and Postle [4] asked a related question: In a given class of graphs, is it always possible to satisfy at least a constant proportion of the requests? We say that a graph with the list assignment is flexible if every request is satisfiable, and it is weighted flexible if every weighted request is satisfiable (of course, weighted flexibility implies flexibility). Dvořák, Norin and Postle [4] established some basic properties of the concept and proved that several interesting graph classes are flexible:

For every , there exists such that degenerate graphs with assignments of lists of size are weighted flexible.

There exists such that every planar graph with assignment of lists of size is flexible.

There exists such that every planar graph of girth at least five with assignment of lists of size is flexible.
They also raised a number of interesting questions, including the following one.
Problem 1.
Does there exists such that every planar graph and assignment of lists of size

five in general,

four if is trianglefree,

three if has girth at least five
is (weighted) flexible?
Let us remark that planar graphs are choosable [13] but not necessarily choosable [16], trianglefree planar graphs are choosable by a simple degeneracy argument but not necessarily choosable [17], and planar graphs of girth at least are choosable [15]. Also, let us remark that the analogous questions in the ordinary proper coloring setting are trivial: If all vertices of a colorable graph are assigned the same list of length , then with this uniform list assignment is weighted flexible, as is easy to see by considering the colorings of arising from a fixed coloring by permuting the colors [4].
We answer the part (b) of Problem 1 in positive.
Theorem 2.
There exists such that each planar trianglefree graph with assignment of lists of size four is weighted flexible.
Let us remark that the underlying choosability result for Theorem 2 is a trivial average degree argument. While the proof of the flexibility result also exploits bounded average degree of the trianglefree planar graphs, it somewhat unexpectedly turns out to require much more involved reducibility and discharging arguments.
2 Flexibility and reducible configurations
To prove weighted flexibility, we use the following observation made by Dvořák et al. [4].
Lemma 3.
Let be a graph and let be a list assignment for . Suppose is
colorable and there exists a probability distribution on
colorings of such that for every and , . Then with is weighted flexible.Let be a graph. For a positive integer , a set is independent if the distance between any distinct vertices of in is greater than . Let
denote the characteristic function of
, i.e., if and otherwise. For functions that assign integers to vertices of , we define addition and subtraction in the natural way, adding/subtracting their values at each vertex independently. For a function and a vertex , let denote the function such that for and . A list assignment is an assignment if for all .Suppose is an induced subgraph of another graph . For an integer , let be defined by for each . For another integer , we say that is a reducible induced subgraph of if

for every , is colorable for every assignment , and

for every independent set in of size at most , is colorable for every assignment .
Note that (FORB) in particular implies that for all . Before we proceed, let us give an intuition behind these definitions. Consider any assignment of lists of size to vertices of . The function describes how many more (or fewer) available colors each vertex has compared to its degree. Suppose we color , and let be the list assignment for obtained from by removing from the list of each vertex the colors of its neighbors in . In , each vertex has at least available colors, since each color in corresponds to a neighbor of in . Hence, (FIX) requires that is colorable even if we prescribe the color of any single vertex of , and (FORB) requires that is colorable even if we forbid to use one of the colors on the independent set .
The following lemma is implicit in Dvořák et al. [4]; we include a proof for completeness.
Lemma 4.
For all integers and , there exists as follows. Let be a graph of girth at least . If for every , the graph contains an induced reducible subgraph with at most vertices, then with any assignment of lists of size is weighted flexible.
Proof.
Let and . For a graph satisfying the assumptions and an assignment of lists of size , we prove the following claim by induction on the number of vertices; the part (i) implies that with is weighted flexible by Lemma 3:
There exists a probability distribution on colorings of such that

for every and a color , the probability that is at least , and

for every color and every independent set in of size at most , the probability that for all is at least .
The claim clearly holds for a graph with no vertices, the basic case of the induction. Hence, suppose that . By the assumptions, there exists of size at most such that is reducible. By the induction hypothesis, there exists a probability distribution on colorings of satisfying (i) and (ii). Choose an coloring from this distribution and let be the list assignment for defined by . Note that for all , and thus has an coloring by (FORB) applied with . Among all colorings of , choose one uniformly at random, extending to an coloring of .
Let us first argue that (ii) holds. Let and . By the induction hypothesis, we have for all with probability at least . If , this implies (ii). Hence, suppose that . For , let , and for , let . Note that for all , and by (FORB), has an coloring. Since has at most colorings, we conclude that the probability that for all is at least . Hence, the probability that for all is at least , implying (ii).
Next, let us argue that (i) holds. For , this is true by the induction hypothesis. Hence, suppose that , and let be the set of neighbors of in . Since has girth at least and all vertices in have a common neighbor, the set is independent in . Furthermore, (FORB) implies , and thus . Hence, by the induction hypothesis we have for all with probability at least . Assuming this is the case, (FIX) implies there exists an coloring of which gives the color . Since has at most colorings, we conclude that the probability that is at least . Hence, (i) holds. ∎
The argument used to prove Lemma 4 also implies the following fact.
Lemma 5.
Let and be integers and let be a graph. If for every , the graph contains an induced reducible subgraph with at most vertices, then has at least colorings from any assignment of lists of size .
Note that in subgraphs with at most vertices, reducibility means that in the (FORB) property, we only care about sets of size . In particular, the arguments from Sections 3 imply that planar trianglefree graphs have exponentially many colorings from lists of size four. Of course, there exist simpler proofs of this fact, see e.g. [10] for the trianglefree case (even in a more general setting of graphs on surfaces).
3 Trianglefree planar graphs
In this section, we prove Theorem 2. The proof is by the discharging method: We first describe a number of configurations ensuring the existence of a small reducible subgraph, then perform a doublecounting argument to show that one of these configurations appears in any trianglefree planar graph, so that Lemma 4 applies.
3.1 List coloring preliminaries
We use the following wellknown fact.
Lemma 6 (Thomassen [14]).
Let be a connected graph and a list assignment such that for all . If either there exists a vertex such that , or some connected component of
is neither complete nor an odd cycle, then
is colorable.This has the following consequence.
Corollary 7.
Let be a connected graph and let be a vertex of . Let be an assignment of nonempty lists to vertices of such that is colorable. Let , …, be the vertex sets of the components of , and for , let be the number of neighbors of in . Let if for all and has a connected component that is neither complete nor an odd cycle, and otherwise. If , then is colorable.
Proof.
For , let and let consist of the colors such that does not have an coloring that assigns the color to . We claim that ; if we prove this to be the case, the claim follows, since then there exists a color , by the definition of the sets there exists an coloring of assigning to the color for , and the combination of these colorings gives an coloring of .
Let be the list assignment for such that for and . By the definition of , the graph is not colorable. If for all and has a connected component that is neither complete nor an odd cycle, then Lemma 6 implies . Otherwise, by the assumptions has an coloring, and since this coloring cannot be extended to an coloring of , we conclude that . ∎
3.2 Reducible configurations
When coloring from lists of size four, vertices of degree at most three can be colored greedily. This argument no longer works in the flexibility setting, as this greedy coloring does not have any freedom to satisfy requests. However, the following weaker claim holds.
Lemma 8.
If is a trianglefree graph, then a vertex of degree at most two, or two adjacent vertices of degree three, form a reducible subgraph.
Proof.
Suppose is a vertex of of degree at most two; then . Let be the subgraph of formed by . Then , and and . Hence, is colorable whenever is a assignment or assignment. Consequently, is reducible.
Suppose now and are adjacent vertices of of degree three; then for . Let be the subgraph of induced by . The function satisfies and . Note that the only nonempty independent sets in are and , and for . Clearly, is colorable whenever is an assignment or assignment, and thus is reducible. ∎
We now describe a quite general class of reducible configurations. Let be a trianglefree graph and a vertex of . A stalk is one of the following subgraphs (see Figure 1):

A path such that ; or

a path such that and ; or

a cycle such that and ; or

a path and an edge with such that and ; or

a path and a path with such that and ; or

a path , a path with , and an edge with such that and .
In all the cases, the root of the stalk is the vertex . In the case (c), we say that the vertex is the bud of the stalk; in the other cases, the stalk has no buds. For a subgraph of , a vertex is good if is the root of a stalk which is vertexdisjoint from ; in case the stalk has a bud , we say that is good using the bud .
Lemma 9.
Let be a plane trianglefree graph with the outer face bounded by a cycle such that each cycle in bounds a face. Let be a vertex of of degree , not contained in . If has good neighbors, no two of them using the same bud, then contains a reducible induced subgraph vertexdisjoint from with at most vertices.
Proof.
Let be a set of good neighbors of , no two of them using the same bud, and for , let be a stalk witnessing this is the case. Let be the subgraph of induced by . We clearly have , and thus it suffices to show that is reducible. This would be easy if the stalks were vertexdisjoint and there were no edges between them; however, we need to argue about such overlaps.
Since each cycle in bounds a face and is vertexdisjoint from , the following claim (which excludes many of the overlaps) holds.

If is a subgraph of and is not a cycle, then has a face that is not bounded by a cycle.
For and a vertex of denoted by or in the definition of a stalk, let us define . By (†) and the assumption that is trianglefree, we conclude that is equal to the distance between and in , with the following exceptions:

The vertex in case (c) is at distance from ,

the vertices and in case (d) may be at distance , , or from , and

the vertices and in case (f) may be at distance , , or from .
Suppose that are distinct and . If both and are equal to the distance between and , then , as otherwise we would have . If say is not equal to , then by the previous observation and the stalk is defined according to cases (c), (d), or (f). In the case (c), note furthermore that cannot be defined according to (c), as otherwise and would use the same bud. In conclusion, the following claim (which we will refer to as ()) holds.

is equal to the distance of from in , and is not adjacent to , or

and at least one of the stalks and is defined according to cases (d) or (f), or

, is adjacent to , one of and is defined according to (a) and the other one according to (c).
Note in particular that if a vertex has degree in and is adjacent to , then is the root of the stalk and does not belong to any other stalk.
Let . Let us first argue that satisfies (FIX).
Subproof.
Consider any vertex and a assignment for . Let be a color in and let be the list assignment for obtained from by removing from the lists of neighbors of . We need to argue that is colorable.
For all , we have . Hence, if then , and if , then . If , then note that by the definition of a stalk, each component of contains a vertex whose degree in is three, and we conclude that is colorable by applying Lemma 6 to each component. Hence, suppose that .
Note that the definition of a stalk ensures that has a neighbor in each component of that contains only vertices whose degree in is four. A neighbor of in is dangerous if either or belongs to a component of that contains only vertices of degree . If at most neighbors of are dangerous, then first greedily color the components of containing dangerous vertices (this is possible, since these dangerous vertices are adjacent to the vertex which has not been colored yet), then give a color from to (which is possible, since is greater than the number of dangerous neighbors of ), and finally extend the coloring to the remaining components of (which is possible, since each such component contains a vertex of degree three, which satisfies ). Thus, to prove that is colorable, it suffices to argue that has at most two dangerous neighbors.
Suppose for a contradiction that , , and are distinct dangerous neighbors of . Without loss of generality, , and thus the components of containing and only consist of vertices of degree . Since and and are adjacent to , () implies that and are the roots of stalks and . We conclude that and the removal of separates and from the vertices of degree in their stalks (possibly, is this vertex of degree ). If , this implies that neither of the stalks and is defined according to (d) or (f), and clearly neither is defined by (a). By (), we conclude that and are equal to the distance between and in , and is not a neighbor of . The same claim is implied by () when .
Consequently , and thus is the root of a stalk , , and . If , then contains three paths of length two between and , contradicting (†). Hence, . Since is nonadjacent to , together with the observation that if then the stalks , , and are not defined according to (d) and (f), this excludes the possibilities that the stalks are defined according to the cases (a), (b), (c), (d), and when also (f). Hence, either and the stalks , , and are defined according to (e), or , and the stalks are defined according to (f). However, it is easy to see that it is not possible to arrange the three stalks with distinct roots in the plane without exceeding the degree of or violating the condition (†). This is a contradiction, finishing the argument that (FIX) holds. ∎
Next, let us argue that satisfies (FORB).
Subproof.
Let be a independent set in of size at most , and let be a assignment for . We need to argue that is colorable. Let be the induced subgraph of obtained by initializing and repeatedly performing the following reductions as long as possible:

If satisfies , then let .

If induces a connected subgraph, neither a clique nor an odd cycle, and for all , then let .
Note that is uniquely determined, regardless of the choice of removed subgraphs, since the degrees in do not increase, and if a part of a set satisfying the assumptions of the second reduction gets removed by another reduction, then all vertices of will eventually be removed because of the first reduction. Using Lemma 6, it is easy to see that is colorable if and only if is colorable.
Note that for , we have if and , if , and if and . Consider a vertex and the stalk . A straightforward case analysis shows that if , then either , or satisfies one of the following conditions (see Figure 2).

consists of the edge , , and ; or,

is a path , , , , and ; or,

is a cycle , the stalk satisfies (c) and is its bud, , , , and either

, or

and ;
or,


consists of a path and an edge , , , , , , and ; or,

consists of a path and an edge , , , , , and ; or,

consists of an edge and a cycle , , , , , , , and .
The same case analysis also shows that if , then , and thus and are colorable. Hence, we can assume that . Analogously, we conclude that is colorable. Note also that is an induced subgraph of , except possibly for the case (iv), where or can be adjacent to (but not both by (†)).
Furthermore, if satisfies , then is the root of a stalk such that satisfies (iiib) or (vi), and in particular contains the other vertex of , and satisfies . Consequently, for at most one vertex .
Let , …, be the vertex sets of the components of , and let , …, be the corresponding integers defined in the statement Corollary 7. Each of the components contains a vertex of , and thus . We now argue that , implying that is colorable by Corollary 7, and thus is colorable as well. Let .
Let us first consider the case that consists of exactly one vertex . If for some , then satisfies (i), (ii), or (iiia). If is adjacent to , then all such stalks satisfy (i) or (iiia), and there is at most one such stalk satisfying (iiia) since no two stalks satisfying (c) use the same bud. Hence, is either an edge or a cycle, and . If is at distance from in , then the stalks intersecting satisfy (ii), and there are at most two such stalks sharing the vertex by (†). Hence, again. Note that , and thus if , then . If , then since and , and thus ; furthermore, the same argument shows , and consequently .
Therefore, we can assume that contains two vertices of , and since , we have and ; hence, , and it suffices to argue that .
If there exists a vertex with , then is the root of the stalk and satisfies (iiib) or (vi). The root is in a unique stalk by (), so there are no stalks satisfying (iv) or (v), or a stalk satisfying (iiib) or (vi) other than . The vertex of other than cannot belong to a stalk satisfying (ii) or (iiia) by the absence of triangles, (†), and the assumption that no two stalks use the same bud. Consequently, , and thus .
Hence, we can assume that for all , and in particular no stalk satisfies (iiib) or (vi). Let us now consider the case that satisfies (v) for some . We cannot have another stalk satisfying (iv) or (v), since then would contain a cycle that should be removed by the second reduction rule. Furthermore, the vertices of are not adjacent to in this case, and thus satisfies (ii) for all vertices . By (†), we conclude that . Note also that if , then . Hence, . Therefore, we can assume that no stalk satisfies (v).
Suppose now that satisfies (iv) for some . If contains another vertex with this property, then since does not contain a cycle (which would be removed by the second reduction rule), we conclude that consist of a cycle and vertices adjacent to , and by (†), neither nor is adjacent to . In this case, (†) implies that , and thus . Hence, we can assume that does not contain any vertex such that satisfies (iv). If neither of the vertices of is adjacent to , then satisfies (ii) for all , and by (†), and if . Hence, . Finally, let us consider the case that a vertex is adjacent to . Then the other vertex of is not contained in another stalk by (†), and can be contained in at most one stalk satisfying (iiia); hence .
Therefore, we can assume that satisfies (i), (ii), or (iiia) for every . Suppose that all vertices of are adjacent to , and thus satisfies (i) or (iiia) for every . By (†) and the assumption that no two stalks use the same bud, either for some , or there exist distinct such that . In the former case, we have . In the latter case, since is connected, we can also assume there is an edge between and ; however, this is not possible, since is trianglefree and satisfies (†).
Finally, suppose that there exists a vertex nonadjacent to , which necessarily is contained in a stalk such that satisfies (ii). Let and let . By (†), , and since is trianglefree, we have for all . Since , there exists
Comments
There are no comments yet.