Applied Combinatorics, 4th Ed.
Alan Tucker
Section 2.4
Coloring Theorems
Prepared by Amanda Dargie
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Triangulation of a polygon
• The process of adding a set of straight-line chords
between pairs of vertices of a polygon so that all interior
regions of the graph are bounded by a triangle
• [these chords cannot cross each other nor can they cross
the sides of a polygon]
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Theorem 1
The vertices in a triangulation of a polygon can be 3-colored.
Proof
• by induction on n, the # of edges of the polygon
• for n=3, give each corner a separate color
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• Assume, any triangulated polygon with less than n boundary
edges, n  4, can be 3-colored.
•Consider a triangulated polygon T with n boundary edges.
• Pick a chord edge e.
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Note: T must have at least one chord edge since n  4, otherwise
T would not be triangulated.
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• This chord e splits T into two smaller triangulated polygons
which can each be 3-colored by the induction assumption.
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• The 3-colorings of the two subgraphs can be combined to yield
a 3-coloring of the original triangulated polygon by permuting
the names for the colors in one subgraph if necessary.
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Application – Art Gallery Problem
What is the smallest number of guards needed to watch
paintings along the n walls of an art gallery?
•The walls are assumed to form a polygon.
•The guards need to have a direct line of sight to every point on
the walls.
•A guard at a corner is assumed to be able to see the two walls
that end at that corner.
Note: [r] denotes the largest integer  r.
Corollary (Fisk, 1978)
The Art Gallery Problem with n walls requires at most n / 3 guards.
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Proof
• Triangulate the polygon formed by the walls of the art
gallery
• Observe that a guard at any corner of a triangle has all
sides of the triangle under surveillance
• Now obtain a 3-coloring of this triangulation.
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• Pick one of the colors (lets
pick red) and place a security
guard at every red vertex
• Now all of the sides of the
triangle are watched and
therefore all walls of the
gallery are watched since
every triangle has one red
vertex on it
• A polygon with n walls has n corners
• If there are n corners and 3 colors, then some color is used at
n / 3 or fewer corners.
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Definition
X(G) = 2 if and only if all circuits have even length. [Let X(G)
denote the chromatic number of the graph G.]
Recall: (Theorem 2, sec 1.3) a connected graph is bipartite if
and only if all circuits have even length
Therefore, being bipartite is the same as being 2-colorable.
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Theorem 2 (Brooks, 1941)
• If the graph G is not an odd circuit or a complete
graph, then X(G)  d, where d is the maximum
degree of a vertex of G.
In this case, the maximum degree of
each vertex, d, is 4. Simply, color the
non-adjacent vertices the same color
(in this case blue) and then the other
vertices their own color.
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Theorem 3 (Vizing, 1964)
• If the maximum degree of a vertex in a graph G is
d, then the edge chromatic number of G is either d
or d +1.
The edge chromatic number of G is defined as the number of colors it
takes to color each edge so no edge incident to the same vertex shares
the same color.
In this case, the maximum degree
of a vertex in G, d = 4 but we
need 5 colors for there not to be
any edges sharing colors that are
incident to the same vertex.
Therefore, the edge chromatic
number of G is d +1 = 4 +1 = 5.
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Theorem 4
It has already been proven that planar graphs can be 4-colored but since the proof
is extremely long and complicated we’re going to look at the next best thing..
Proving that a planar graphs can be 5-colored.
• Every planar graph can be 5-colored.
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Proof
Recall: (proved in exercise 18, sec. 1.4) any connected planar graph
has a vertex of degree at most 5
Consider only connected planar graphs (we can 5-color unconnected
planar graphs by 5-coloring each connected part)
Assume all graphs with n -1 vertices (n 2) can be 5-colored
G has a vertex x with at most degree 5
Delete x from G to obtain a graph with n -1 vertices (by
assumption, can be 5-colored)
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Proof con’t
Then reconnect x to the graph and try to color properly
If the degree of x is at most 4, then we can assign x a color
different from that of it’s neighbors.
If deg x = 5, but only 4 colors are used in coloring x
neighbors, then again, x can be colored a 5th color.
x
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Proof con’t
If deg x = 5 and all 5 colors are used on x’s neighbors, then pick two
colors not next to each other (for example, purple and blue) and look at
the connected subgraph with those two colors, starting at one of them
If they do not hook back to each other (if the purple does not hook
back to the blue by a path of alternating purple and blue) then switch
the colors on that piece (switch purple to blue and blue to purple)
Now color x the color not adjacent to it.
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Proof con’t
If the two colors hook back (if blue and purple hook back) then pick
two other colors not next to each other (let’s pick green and pink)
We know that green and pink cannot hook back to each other because
in order to they would need to cross the blue-purple path and the graph
would no longer be planar
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Proof con’t
Switch the green and pink on one section of the graph so that x now
becomes adjacent to two vertices of the same color
Now, since x’s 5 neighbors share only 4 different colors, color x the 5th
color.
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For the class to try:
Show that a planar graph G with 8 vertices
and 13 edges cannot be 2-colored.
Hint: Use results in Section 1.4 to show that G must
contain a triangle.
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Solution
By using Euler’s Formula, r = e – v + 2, we know
that any graph with 8 vertices and 13 edges must
have r = 13 – 8 + 2 = 7 regions and therefore must
contain a triangle. We know that a triangle can be
at smallest, 3-colored, and therefore this graph
cannot be 2-colored.
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