List coloring, an intriguing extension of traditional graph coloring, introduces dynamic color assignment by associating each vertex with a list of permissible colors. The list chromatic number, denoted as χl(G), signifies the minimum colors needed to properly color a graph under list constraints. Brooks' Theorem characterizes when a graph is optimally list colorable. Algorithmic techniques, including greedy approaches and backtracking, are pivotal in solving list coloring problems. Applications span various domains such as resource allocation, wireless network frequency assignment, and register allocation. List coloring's link to the classic chromatic number, its variations, computational complexity, and ongoing developments highlight its versatile role in computer science and mathematics, making it a captivating and practical field of study and application.

1. Date
List Coloring
Submitted by:
Amit Kumar
2019MT60744
Supervised by:
Dr. Bhawani Sankar Panda,
Professor
Department of Mathematics

2. What is a proper colouring of a graph?
Painting all the vertices of a graph with colors such that no two
adjacent vertices have the same color is called the proper coloring.
Example:
Improper Coloring Proper colorings

3. What is chromatic number?
Chromatic number of a graph or χ(G) is the minimum unique
colors needed to provide a proper coloring of graph.
χ(K4,2) = 2
Chromatic number
of Any bipartite
graph is 2.

4. What is list colouring?
For a graph G, a list coloring of G is a proper coloring of the
vertices of G from color lists L(v) available at each vertex.
Let A be a set of colors, let L(v) be some subset of A assigned to
the vertex v for each vertex in G. A list coloring is a function f from
v to L(v) such that f(v) ≠ f(u) for any two adjacent vertices u, v.
A={1,2,3}
A valid colouring f: v->L(v) is as
follows:
f(y1)=1
f(x1)=3
f(x2)=2

5. K-choosable and List chromatic number
We call a graph k-choosable if there exists a proper coloring for
any list of k colors assigned to each vertex
The list chromatic number (or choice number) of a graph G,
denoted χl(G), is the minimum k for which G is k-choosable
Consider the example of K4,2, we know that that the chromatic
number or χ(K4,2) is 2. In fact χ of any bipartite graph is 2.
What about χl(K4,2)?

6. K-choosable and List chromatic number
K4,2 is not 2-choosable, here is a counter example:
Consider the following 2-lists for each node:
It is impossible to provide an appropriate coloring of the above
graph, hence it is not 2 colorable.

7. K-choosable and List chromatic number
From the previous example, we can get an intuition of the
following result:
χ(G)<=χl(G)
Since, if we assign {1,2,3,…, χ(G)-1} to each vertex, we know the
graph is not colorable using the definition of χ(G).
Also, χ(G)<=Δ(G) + 1, since all the vertices can be colored
differently than the neighbors.

8. K-choosability of Trees.
We know χ(T)= 2, so χl(T)>=2.
In fact χl(T)=2.
Proof:
Let T be a tree, where each vertex v has |L(v)|=2,
Let T’ be the tree obtained after removing all the
leaves from T.
T is colorable if T’ is colorable, Since for any leaf
in T we can color it exactly different from its
parent’s color in T’.
Apply the same process on T’ again and again till
we’re left with the root of the tree, Which is
colourable.

9. K-choosability of certain Bipartite graphs
For complete bipartite graphs G of the form Kn^n,n , χl(G) ≥ n + 1.
Proof:
Let G=XUY, |X|=nn, and |Y|=n.
Let L(yi)={i1,i2,..,in}, and let L(xi) be one possible set formed by
taking one color from each L(yi). Then all possible sets will be
covered in X with one color from each L(yi). Now consider any
possible coloring f of G.
Then there exists an xi with L(xi)={f(y1), f(y2), f(y3),… ,f(yn)}, (since
all possible sets were covered), this vertex will hence be uncolorable.

10. K-choosability of certain Bipartite graphs
For all complete bipartite graphs G=XUY of the form Kn,n,
where n=2k-1Ck, χl(G) ≥ k + 1
Say we use a total of 2k-1 colors, i.e. |A|=2k-1.
Consider n unique subsets of A, of size K, assign each one of these
subsets to n vertices in X and Y each.
Notice: In coloring n vertices of X, we use at least k colors.
Say we use k-1 colours out of 2k-1 not using rest k colors, then there exists a
vertex in x with L(x) exactly equal to these k colors.(since all possible subsets
are covered)
There exists a vertex in Y, y such that L(y) is exactly these k
colors.(all possible sets), hence this vertex will be not colorable.

11. K-choosability of planar graphs
Proof:
Suppose G is maximally planar (i.e. all bounded faces are triangles,
and the outer face is a cycle C). We show that if G has two adjacent
external vertices with different lists of size 1, all other external
vertices have lists of size 3, and all internal vertices have lists of size
5, then G can be properly colored.
Since adding edges to a graph cannot reduce its list chromatic
number, it is sufficient to prove this claim for maximal G.
We perform the proof using induction.

12. K-choosability of planar graphs
Proof:
Base case: If n = 3, then G is a triangle. Let V (G) = {v1, v2, v3},
and L(v1) = {1} and L(v2) = {2}. Then, since |L(v3)| = 3, there
exists a color for v3 that is different from v1 and v2.
Induction step: We consider graphs on > 3 vertices. Let C =
v1v2v3...vpv1 be the external cycle of G. Suppose that v1 and vp
have different color lists both of size 1. We consider two cases:
1) Suppose G has a chord between vertices vi and vj in C. We consider the two smaller
cycles C’ = v1v2...vivj ...vpv1 and C’’ = vivi+1...vjvi . We first apply our induction
hypothesis to C’ and its interior. This provides a proper coloring where vi , vj are
adjacent and differently colored. We then apply our induction hypothesis to C’’ and
its interior. The resulting colorings give us a list coloring of G.
2) If G is chordless?

13. K-choosability of planar graphs
Proof:
If G is chordless: (say v1 and vp are adjacent)
Let {v1, u1, ..., um, v3} be the neighborhood of v2. Since G is maximally planar,
v1u1...umv3 must form a path from v1 to v3, since G is chordless, {u1, ..., um} are
all internal vertices of G.
Consider G’=G-v2 , Let v1 be colored with color c. Since |L(v2)| = 3, there exist
two colors x, y ∈ |L(v2)| such that x ≠ c and y ≠ c. For all ui ∈ {u1, ..., um}, let
L’(ui) = L(ui) {x, y}. This is possible since all such vertices are internal and
therefore have lists of size 5. Therefore, |L’(ui)| ≥ 3.
Now apply induction to G’ , now only v3 could be colored x or y, color v2
differently.

14. Characterisation of Graphs with Xl(G)=2
Theorem: G is 2-choosable if and only if the core of G(pruning
leaves) lies in T={K1, C2m+2, Θ2,2,2m}.
We call a graph G a Θ-graph if G consists of two vertices u and v
with three vertex-disjoint paths between them.

15. Characterisation of Graphs with Xl(G)=2
The proof for one way for K1 is clear. Consider Θ2,2,2m
Case 1: L(A1)=L(A2)=L(A3)=….=L(A2m+1)={x,y}, simply colour
them alternating and then colour of A1 and A2m+1 will be same, so B
and D are colourable.

16. Characterisation of Graphs with Xl(G)=2
Case 2: L(Ai)≠L(Ai+1) for some i,
 Tentatively choose xi ∈ Ai-Ai+1, Now choose xi-1 ∈ Ai-1-{xi} , keep going till
choosing x1, now if A2m+1 = {b,d} and {L(B),L(D)} ≠{{x1,b},{x1,d}}, then ∃ x ∈
A2m+1 s.t L(B)-{x,x1} ≠Φ and L(D)-{x,x1} ≠Φ, then choose x for A2m+1, now
keep going by choosing x2m ∈ A2m-{x}, till xi+1=Ai+1-xi+2. We know that xi+1 ≠xi,
since xi ∈ Ai-Ai+1

17. Characterisation of Graphs with Xl(G)=2
Case 2: L(Ai)≠L(Ai+1) for some i,
 If {L(B),L(D)} ={{x1,b},{x1,d}}, then go the other way, i.e take xi+1 ∈
Ai+1-Ai, and xi+2 ∈ Ai+2-{xi+1}, .. Till x2m+1 ∈ A2m+1={b,d}, now let B and D
be colored x1, so choose a color for A1 from A1-x1, keep going till we
complete circle.

18. Characterisation of Graphs with Xl(G)=2
Other side of the proof is done by exhausting all possibilities, by
showing that either G is in T or else G contains a subgraph of the
following type:
1. An odd cycle.
2. 2 node disjoint even cycles connected by a path.
3. Two even cycles having exactly one node in common .
4. Θa,b,c where a≠2 and b≠2 .
5. Θ2,2,2,2m+1
Then we prove the above 5 categories are not 2 choosable

19. Further Readings:
Josep D´ıaz, Oznur Ya¸sar Diner, Maria Serna, and Oriol Serra. On
list k-coloring ¨ convex bipartite graphs, 2020Josep D´ıaz, Oznur
Ya¸sar Diner, Maria Serna, and Oriol Serra. On list k-coloring ¨
convex bipartite graphs, 2020.
The above paper talks about polynomial time solving of List-k
Coloring (Each list at vertex is a subset of {1,2,..k}) for convex
bipartite graph.
Future goals expanding the above result for circular convex
bipartite graphs and studying cubic bipartite graphs.

20. References
1. Abel Romer. List colorings. 2020.
2. Paul Erdos, Arthur L Rubin, and Herbert Taylor. Choosability in
graphs. Congr. Numer, 26(4):125–157, 1979.
3. Noga Alon and Michael Tarsi. Colorings and orientations of
graphs. Combinatorica, 12:125–134, 1992.
4. Carsten Thomassen. Every planar graph is 5-choosable. Journal
of Combinatorial Theory Series B, 62(1):180–181, 1994.
5. Josep D´ıaz, Oznur Ya¸sar Diner, Maria Serna, and Oriol Serra.
On list k-coloring ¨ convex bipartite graphs, 2020