Graph Coloring
CS 594
Stephen Grady
Overview
-Terminology
-History
4 Color Theorem
Kempe’s flawed proof
-Simple Bounds on chromatic number
-Coloring on Digraphs
-Applications
Terminology
-Color: Any labeling assigned to an element of a
graph.
-Proper coloring: No color may be adjacent to
itself.
Terminology
-Chromatic number (χ(G)): Minimum colors
needed to properly color
graph.
-Chromatic Equivalency: Two graphs are
chromatically equivalent if
they have the same χ(G)
Chromatic Polynomial
-The possible number of ways to color a graph
using k colors.
Example K3+1 can be colored 12 different ways with 3
colors
Types of Coloring
-Vertex-Vertices are labeled
-Edge-Edges are labeled
-Total-Both vertices and edges are labeled
-Many more...
Vertex coloring
-Coloring most concerned with
-Other colorings can be transformed to vertex
coloring
Edge: color vertices of line graph
Planer: color vertices of dual
History
-First studied as a map coloring problem.
-What is the minimum number of colors needed
to color a map. Or stated another way, what is
the minimum number of colors needed to color a
planar graph?
4 Color Theorem
-Francis Guthrie first noticed he could color the
map of the counties of England using only 4
colors.
-Naturally the question of if this holds true for all
maps (planar graphs) arose.
4 Color Theorem
-The question was passed along in 1852
Guthrie's brother->Augustus de Morgan->William
Hamilton
-Eventually it was brought up to the London
Mathematical Society in 1879
-That same year Alfred Kempe published a paper
claiming to contain a proof that 4 colors suffice.
Sir Alfred Kempe
-1849-1922
-Trinity College, Cambridge
-22nd wrangler
-1877: Flawed “straight line linkage”
proof, though ideas basis
of proof in 2002
-1879: Flawed 4 color theorem proof though ideas
basis of proof in 1976
Percy John Heawood
-1861-1955
-Exeter College, Oxford
-Much of life dedicated to
4 color theorem
-1890:Disproved Kempe's
4 color theorem proof
-1890:Established 5 color theorem
based on Kempe's work
Kemp's Flawed 4 Color Theorem
-Kempe's proof rested on the properties of what
are known as Kempe chains.
-A Kempe chain is a bicolored path between any two
non-adjacent vertices.
Kempe's Argument
-Assume Δ(G) ≤5 (There is a proof but I’ll leave
that to whomever covers planarity)
-Consider the smallest planar graph G in terms of
|V| that requires 5 colors to color properly.
-Delete any vertex v in G creating G'. Since G' is
smaller than G, G' is 4 colorable.
-Color all vertices in G' using 4 colors.
Kempe's Argument
-Now add back v to G' recreating G.
-There are three cases to consider when deciding
how to color v with one of the four colors used.
deg(v)=1,2 or 3
deg(v)=4
deg(v)=5
Case 1
-v has degree 1,2 or 3.
-Since the neighbors of v only take up to 3 of the
possible colors in this case coloring v is trivial.
Case 2
-v has degree 4 and neighbors a, b, c and d.
-If the neighbors of v do not use all 4 colors, then
coloring v is trivial.
-If all 4 colors are used by the neighbors of v then
choose any two neighbors a,c of v and consider
the subgraph H of G induced by the colors of a
and c.
-This creates two possibilities
Case 2 Subcase i
-There does not exist a Kempe chain between a
and c
-If this is the case then consider the subgraph J of
H induced by taking all vertices that have a path
to a.
-Perform a color swap on J and now a and c share
the same color and v may be colored with the
remaining color.
Case 2 Subcase ii
-There does exist a Kempe chain between a and
c.
-Now consider the subgraph H' of G induced by
the colors of b and d.
-Because G is planar there cannot exist a Kempe
chain from b to d.
-Simply perform a color swap on b as in
subcase i.
Case 3
-v has degree 5 with neighbors a,b,c,d and e.
-Add edges to all neighbors of v while staying
planar
-Assume all 4 colors are used by neighbors of v.
-Just like in case 2 there exist 2 subcases.
Case 3 Subcase i
-Consider three neighbors b, e and d.
-As in case 2 consider the subgraph H of G
induced by the colors of b and e.
-If no Kempe chain, do a color swap.
-If Kempe chain, repeat for b and d.
Case 3 Subcase ii
-Both b to d and b to e have a kempe chain.
-Consider the subgraph H' of G induced by colors
on a and d. As with case 2 because G is planar
there cannot exist a Kempe chain from a to d so
a color swap can be performed on a.
-Repeat for c and e.
So, why is it flawed?
Haewood's Counter
What if the Kempe chain's b to d and b to e cross?
4 Color Theorem
-Finally proven in 1976 by Kenneth Appel and
Wolfgang Haken at University of Illinois.
-First proof using a computer as an aid.
4 Color Theorem
-Used the idea of an unavoidable set of
reducible configurations
-Had to check 1,936 graphs to prove
minimum counterexample to 4 color theorem
could not exist.
Complexity
-Decision: Given a graph G can it be colored with
k colors? NP-Complete
-Optimization: What is χ(G)? NP-Hard
Bounds on χ(G)
-Brooke's Theorem
-Clique number
Brooke's Theorem
χ(G) ≤ Δ(G)
Except for Kn and C2n+1
χ(G) ≤ Δ(G)+1
Clique Number
-χ(G) >= ω(G)
-A clique of size Kn must be colored with n colors.
Mycielski's Theorem
-There exists a triangle free graph with arbitrarily
high χ(G).
-Generalized with Mycielski graphs.
Coloring on Digraphs
-Gallai-Roy Theorem
-χ(G) =L+1 where L is the longest path in its
shortest orientation.
Applications
-Applications are generally those that
must satisfy some constraint
Scheduling
Register allocation
Determining if graph is bipartite
Sudoku
Scheduling
Used
to find minimum number of time slots needed
with no time conflicts.
Each time slot represented by a color.
Each edge represents time conflict
Register Allocation
In
an attempt to optimize code, compilers will
allocate multiple variables to the same register.
However, multiple variables allocated to the same
register cannot be called at the same time.
Naturally this becomes a coloring problem.
Register Allocation
Determine if Graph is Bipartite
Bipartite
Can
Graphs always have χ(G)=2
check if graph is 2 colorable in linear time
Sudoku
Sudoku Graph Transformation
Coloring a Sudoku Graph
Sudoku
Why
study Sudoku
Methods for solving Sudoku can be generalized to
solve problems like protein folding.
Open Problems
Erdos-Faber-Lovasz
Reed’s
upper bound
Conjecture
Erdos-Faber-Lovasz Conjecture
Can n kn graphs each sharing only one vertex be
colored with n colors.
Reed’s Upper Bound
χ(G) ≤ (1+Δ(G)+ω(G))/2
Homework
-Prove that for any planar graph 5 colors suffice
(Assume Δ(G) ≤5)
-Which two of these graphs are chromatically
equivalent?
k4,4, P7, k5, Peterson graph
-How many ways can a k6 be colored using 7
colors?
Email: [email protected]
References
1.
2.
3.
4.
https://en.wikipedia.org/wiki/Graph_coloring
https://en.wikipedia.org/wiki/Alfred_Kempe
https://en.wikipedia.org/wiki/Percy_John_Heawood
https://en.wikipedia.org/wiki/Wrangler_%28University_of_Ca
mbridge%29
5. https://en.wikipedia.org/wiki/Four_color_theorem
6. https://en.wikipedia.org/wiki/Mycielskian
7. https://en.wikipedia.org/wiki/Gallai%E2%80%93Hasse%E2%80
%93Roy%E2%80%93Vitaver_theorem
8. http://www.math.illinois.edu/~dwest/openp/
9. http://math.ucsb.edu/~padraic/ucsb_2014_15/math_honors
_f2014/math_honors_f2014_lecture4.pdf
10. https://en.wikipedia.org/wiki/Sudoku
References
11. https://en.wikipedia.org/wiki/Bipartite_graph
12.
http://www.math.rutgers.edu/~sk1233/courses/grap
htheory-F11/planar.pdf
13.
http://www.skidmore.edu/~adean/MC3021309/Print
Slides/MC302_131203_P.pdf
14. Ercsey-Ravasz, M. and Z. Toroczkai (2012). "The Chaos
Within Sudoku." Scientific Reports 2: 725.