A proper coloring of a graph is an assignment of colors to vertices of a graph such that no two adjacent vertices receive the same color. A graph is k-colorable if it can be properly colored with k colors. For example, the famous Four Color Theorem states that “Evey planar graph is 4-colorable“. This is tight, since a complete graph on four vertices is 4-colorable but not 3-colorable. Deciding if a graph is 3-colorable is NP-hard. It is natural to ask which planar graphs are 3-colorable.

Grotzsch’s Theorem :Every triangle-tree planar graph is 3-colorable.

Dvorak, Kawarabayashi and Thomas [DKT '08] presented a very short proof of Grotzsch’s theorem and a linear-time algorithm for 3-coloring such graphs.

Given a graph and given a set L(v) of colors for each vertex v, a list coloring is a proper coloring such that every vertex v is assigned a color from the list L(v). A graph is k-list-colorable (or k-choosable) if it has a proper list coloring no matter how one assigns a list of k colors to each vertex.

If a graph is k-choosable then it is k-colorable (set each L(v) = {1,…k}). But the converse is not true. Following is a bipartite graph (2-colorable) that is not 2-choosable (corresponding lists are shown).

A graph is k-degenerate if each non-empty subgraph contains a vertex of degree at most k. The following fact is easy to prove by induction :

Exercise :A k-degenerate graph is (k+1)-choosable

Are there k-degenerate graphs that are k-choosable ? Following are some known results and open problems :

- Every bipartite planar graph is 3-choosable [Alon & Tarsi '92]. It is easy to prove that every bipartite planar graph is 3-degenerate.
- Every planar is 5-choosable [Thomassen '94]. Note that every planar graph is 5-degenerate. There are planar graphs which are not 4-choosable [Voigt '93].
- Every planar graph of girth at least 5 is 3-choosable. This implies grotzsch’s theorem in a very cute way [Thomassen '03]. There are planar graphs of girth 4 which are not 3-choosable [Voigt '95].

Open Problems :

- There is a conjecture stating
“Every 3-colorable planar graph is 4-choosable”. Prove or disprove this conjecture. A positive answer implies four color theorem !!- Another open problem (I learnt this problem from Robin Thomas‘s course on Graph Minors in Spring’2008) is
“Find a linear-time algorithm to 3-list-color planar graphs of girth 5″. Thomassen’s Proof [Thomassen '03] gives a quadratic algorithm. I could not convert his proof into a linear-time algorithm. Probably it requires a new proof. It might be possible to get an algorithm using the data structures of [KK'03].

**References :**

**[Alon & Tarsi '92]**N. Alon, M. Tarsi:**Colorings and orientations of graphs.***Combinatorica*12(2): 125-134 (1992)**[Thomassen '94]**C. Thomassen:**Every Planar Graph Is 5-Choosable.***J. Comb. Theory, Ser. B*62(1): 180-181 (1994)**[Voigt '93]**M. Voigt:**List colourings of planar graphs.***Discrete Mathematics*120(1-3): 215-219 (1993)**[Thomassen '03]**C. Thomassen:**A short list color proof of Grötzsch’s theorem.***J. Comb. Theory, Ser. B*88(1): 189-192 (2003)**[Voigt '95]**M. Voigt :**A not 3-choosable planar graph without 3-cycles.***Discrete Mathematics*146(1-3): 325-328 (1995)**[DKT '08]**Z. Dvorak and K. Kawarabayashi and R. Thomas :**Three-coloring triangle-free planar graphs in linear time.***SODA 2009*.**[KK'03]**Lukasz Kowalik, Maciej Kurowski:**Short path queries in planar graphs in constant time.***STOC 2003*: 143-148