A Note on Not-4-List Colorable Planar Graphs

The Four Color Theorem states that every planar graph is properly 4-colorable. Moreover, it is well known that there are planar graphs that are non-$4$-list colorable. In this paper we investigate a problem combining proper colorings and list colorings. We ask whether the vertex set of every planar graph can be partitioned into two subsets where one subset induces a bipartite graph and the other subset induces a $2$-list colorable graph. We answer this question in the negative strengthening the result on non-$4$-list colorable planar graphs.

Problem 1. Is every planar graph L-colorable for every 2-common 4-assignment L?
Since every proper coloring of the vertices gives a partition of the vertex set we may look for the problem from another point of view.
Problem 2. Let G be a planar graph. Is it possible to partition the vertex set of G into two sets in such a way that one partition set induces a bipartite graph and the other one induces a 2-list colorable graph?
If such a partition would always exist for planar graphs, then it would strengthen the Four Color Theorem. Moreover, we have the following relationship to Problem 1.
Claim 3. If the vertex set V of a planar graph G can be partitioned into V 1 and V 2 such that V 1 induces a bipartite graph and V 2 induces a 2-list colorable graph then G is L-colorable for every 2-common 4-assignment L.
Proof. Let G be a planar graph and L be a 2-common 4-assignment for the vertices of G with {α, β} ⊆ L(v) for all v ∈ V (G). Properly color the subgraph induced by V 1 with α and β and set L (v) = L(v) \ {α, β} for all v ∈ V 2 . Since the subgraph induced by V 2 is 2-list colorable it can be colored from the remaining lists L (v).
Since every acyclic graph is 2-list colorable we may put a stronger question for the partition of G.
Problem 4. Let G be a planar graph. Is it possible to partition the vertex set V into V 1 and V 2 such that the subgraph induced by V 1 is a bipartite graph and the subgraph induced by V 2 is a forest? Unfortunately, this is not possible for every planar graph as shown by Wegner in 1973 [8].
Theorem 5. There is a planar graph G such that in every proper 4-coloring of G the vertices of every two color classes induce a subgraph containing a cycle.
The construction of Wegner does not give an answer to Problem 1 but based on his construction, we were able to find our construction.
the electronic journal of combinatorics 25(2) (2018), #P2.46 Theorem 6. There is a planar graph G and a 2-common 4-assignment L such that G is not L-colorable.
Proof. We will construct a planar graph G and a 2-common 4-assignment L in two steps. In the first step we consider the subgraph G 1 of G, which is shown in Figure 1. The structures inside the triangles D 1 and D 2 are depicted separately outside. Let v 1 be precolored by 1 and v 2 be precolored by 2, and consider the list assignment for the other vertices of G 1 given in Figure 2. Assume that there is a proper coloring c that assigns every vertex v a color c(v) ∈ L(v) such that adjacent vertices get different colors.
At first, let v 10 be colored by α. Clearly, one of the vertices v 4 and v 5 must be colored by 3 since otherwise v 3 would not be colorable. Secondly, let v 10 be colored by β. This can be handled using analogous arguments by interchanging the roles of α and β.
Therefore, the subgraph G 1 of G with given precoloring and list assignment as in Figure 2 is not L-list colorable. Next, consider the complete graph K 4 where the list of all vertices is {1, 2, α, β}. Construct the graph G as follows. For every edge xy in K 4 add two copies of the graph G 1 identifying its edge v 1 v 2 with the edge xy, once with x = v 1 and y = v 2 and the other time with y = v 1 and x = v 2 (see Figure 3) and then identify the vertices u, v, and w.
Clearly, two vertices of of K 4 must be colored by 1 and 2 giving exactly the above precoloring for one of the corresponding subgraphs G 1 . Hence, G is planar and not L-list colorable, where all lists of the list assignment L have length 4 and contain the elements α and β.
Since the vertices u, v, and w in Figure 3 are identified, the graph G constructed above has 8 + 12 · 13 = 164 vertices.
Considering Claim 3 and Theorem 6 we obtain the answer to Problem 2, which also improves the above mentioned result of Wegner. Moreover, in some sense this conclusion is a sharpness result for the Four Color Theorem.

Corollary 7.
There is a planar graph G such that in every proper 4-coloring of G the vertices of every two color classes induce a subgraph that is non-2-list colorable.
Finally, let us mention a related concept introduced by Kratochvíl et al. in [4]. A list assignment L for a graph G = (V, E) is called a (k, c)-assignment if L(v) = k for all v ∈ V (G) and |L(v) ∩ L(w)| c for all edges vw ∈ E(G). In [3] it is mentioned that every the electronic journal of combinatorics 25(2) (2018), #P2.46 planar graph is L-list colorable for every (4, 1)-assignment L. Moreover, there exist planar graphs G and corresponding (4, 3)-assignments L such that G is not L-list colorable. So far, there is no result for (4, 2)-assignments and the following problem remains open. Problem 8. Is every planar graph L-list colorable for every (4, 2)-assignment L?