SAMPLES OF HIGHER RATED WRITING: LAB 5
1. Task Description
Lab report 5. (April 14) Graph Coloring. Following the guidelines, and keeping in mind the
assessment criteria, using where appropriate experimental examples, graphics, or mathematical
proofs, introduce the reader to graph coloring problems by a real-world scenario different
from the one described in class. Give special attention to Questions 2, 3, 5 (H, I, J, K),
10, 12. Question 13: The vertex coloting game is described at http://en.wikipedia.org/wiki/Graph
coloring game. (a) For the graphs in Fig. 5.9 H, J and K, determine for each of the following
values of k (the number of colors the players may use) which player has a winning strategy
- k = 2, 3, 4, 5, 6. (b) How does this value of k compare with the chromatic number of the
graph? (c) If a finite graph G has chromatic number k, does TWO then have a winning strategy
in the vertex coloring game allowing k colors?
2. Introduction
Example 1
Imagine that you are working as a security consultant for a very prestigious art gallery. The gallery
owner asks you to design a layout for placing new security cameras that will ensure that every part of the
gallery is visible by camera at all times with the smallest possible amount of cameras. How would you proceed?
It turns out that the security camera puzzle can be modelled as a graph theory problem. To see how this is,
let us first establish a few definitions.
A graph, in this case, is a set of vertices, or nodes, that are connected by a set of edges; to be precise, the
graph is the larger set that is the union of the set of vertices and the set of edges. For our purposes, we are
assuming that every graph is finite, that is, there are a finite amount of vertices and edges in any graph.
Then, a colouring of a graph is an arrangement that assigns a colour to each node such that, if two nodes
are connected by an edge, they cannot have the same colour. We can ask many questions about colouring
graphs; for example, we could ask "What is the minimum number of colours needed in order to colour a
specific graph?" or "How many different ways can we colour a graph with k colours?"
As it turns out, these are the type of questions that would be considered in the art gallery scenario. In this
paper, some additional definitions and theoretical results about graph colouring will first be presented, and
then the application of these results to the art gallery situation, along with other examples, will be discussed
and presented alongside some numerical results.
3. Theory
First, a graph colouring that uses at most k colours for some positive integer k is called a (proper) k colouring.
The first question we can ask about k colouring is whether one exists for any given graph. With the exception
of graphs which contain a loop, or a vertex that is connected to itself, the answer seems to be an obvious
"yes," because we can always pick a value of k greater than or equal to the number of vertices on a graph.
However, this line of thought is neither interesting nor useful in the majority of applications that use graph
colouring. So, to make the question more interesting, we should ask if, given certain restrictions on what
value k takes, there still exists a k colouring for a graph (for our purposes, we will assume that none of the
graphs in question have a loop).
Now, the first question we will examine is "What is the minimum value of k for which a graph has a k
colouring?" For example, suppose we have a graph G and we find a k colouring of G with 3 colours. Then,
when we try to find a k colouring that uses only 2 colours, we find that this task is impossible. Then, for G, the
minimum value of k is 3. Then, k = 3 is referred to as the chromatic number of G, which is sometimes represented with the symbol χ(G). Then, if χ(G) = k for a graph G and positive integer k, G is called k-chromatic.
The chromatic number has a couple of simple properties associated with it. First, let G be a graph with n
vertices. Then:
1. χ(G) ≤ n
This follows from the remark at the beginning of this section, that a k colouring can always be achieved by
assigning a unique colour to every vertex.
2. χ(G) = 1 if and only if G has no edges.
Date: April 14, 2015.
1
This follows from a simple contradiction. If χ(G) = 1, then every vertex is the same colour. However, if G
has at least one edge, then at least two vertices with the same colour are connected, and we do not have a k
colouring of G.
In addition to knowing the minimum value of k needed to have any k-colouring of a graph, it is worthwhile to
ask "How many different k-colourings does a graph have for a specific k?" Assuming that k is larger than the
chromatic number of the graph being examined (so the amount of possible k-colourings is non-zero), there are
a couple of interesting results that come from this question. As it turns out, we can define a function, PG (k),
called the chromatic polynomial which counts the number of colourings as a function of k, the number of
colours being used.
The chromatic polynomial also has a number of properties, and provides at least as much information about
a graph as the chromatic number. In fact, the chromatic number is the smallest positive integer that is not a
root of the chromatic polynomial; that is, χ(G) is the smallest positive integer for which PG (k) 6= 0. It follows
directly from this that every positive integer which is less than the chromatic number must be a factor of the
chromatic polynomial (in other words, any positive integer less than the chromatic number is a root of the
chromatic polynomial). Additionally, a graph with n vertices will have a chromatic polynomial of degree n,
that is, the leading power of k will be k n . It should be noted that, in general, finding the chromatic number
and chromatic polynomial for a graph is by no means a simple task (although, it is interesting to note that
checking if an approximated value for the chromatic number is correct is much less computationally complex
than directly computing the chromatic number). However, if we focus only on special types of graphs, we can
bypass this general difficulty and see that certain patterns emerge for these properties in special types of
graphs.
Applications and Numerical Results
Now, returning to the original scenario, it should be easier to see how one might begin examining different
camera placement schemes in an art gallery by using graph colouring. As you may have already guessed,
the floor plan of the gallery can be modelled using a graph by defining nodes at all corner locations. If we
let these nodes represent potential camera placements, then we can use edges to connect any two cameras
that are able to view the same region of the gallery, using this as our graph for which we want to find one or
more k colourings. If we assume that, from the defined corner positions, each camera can see three directions,
then we can look for k colourings using three colours, where each colour represents one possible setup. Then,
whichever colour occurs the least amount of times out of all k colourings found will represent the optimal
scheme that should be recommended to the museum director to ensure that the entire gallery can be seen by
the minimum amount of cameras.
In addition to having solved the security camera puzzle, though, the problem of graph colouring can be
applied to many other puzzles as well. For example, in a scheduling problem, one could model schedule blocks
as nodes and draw lines connecting blocks which cannot happen at the same time, and then look for a k
colouring in that graph which will tell them how to formulate the schedule being examined. Graphs can also
be used to model networks, geographical maps, and even games!
In order to visualise all of this information about graphs and graph colouring, we can use some of the various
tools available with Mathematica to create and work with graphs. For example, the following string of
commands will give us an image, the chromatic number and polynomial, and a minimal colouring for complete
graphs on 3,4,5, and 6 nodes:
g1 = CompleteGraph[3];
g2 = CompleteGraph[4];
g3 = CompleteGraph[5];
g4 = CompleteGraph[6];
ShowGraph[g1]
ChromaticNumber[g1]
ChromaticPolynomial[g1, k]
MinimumVertexColoring[g1]
ShowGraph[g2]
ChromaticNumber[g2]
ChromaticPolynomial[g2, k]
MinimumVertexColoring[g2]
ShowGraph[g3]
ChromaticNumber[g3]
ChromaticPolynomial[g3, k]
MinimumVertexColoring[g3]
ShowGraph[g4]
ChromaticNumber[g4]
ChromaticPolynomial[g4, k]
MinimumVertexColoring[g4]
Sample 2
2
In this article, we explore the proper coloring of the vertices of a finite graph. This notion of graph theory
allows for one to color each state of the United States a different color from those of bordering states, while
only using four colors (this is from the “Four Color Theorem”). Graph coloring has also been applied to many
scheduling problems.
The Movie Theater Problem
To illustrate the usefulness of proper graph coloring, we examine a particular scheduling problem:
Suppose you work for a movie theater. You are assigned to minimize the number of show-times while
still meeting the demand for each movie. Movies are classified into three categories: low-brow (LB), high-brow
(HB), and middle-brow (MB). Demand comes from two demographics: one demographic would like to watch
each LB and MB movie, and the other would like to watch each MB and HB movie. For the following movie
lineups, how many show-times are needed? (a) 1 LB, 1 MB, 1 HB; (b) 2 LB, 1 MB, 2 HB.
To solve these problems, we apply a graphical framework. Represent each movie as a node and demand as
edges. For (a), we get:
Figure 1
Now apply colors to the nodes such that no two nodes share the same color for any edge, that is, ensure that
the edges are properly colored. The number of colors used represents the number of show-times.
3
Figure 2
We repeat the process for (b).
Figure 3
Here, the movie theater would need three show-times to meet the demand for the five movies.
Counting Colors
As counting colors has been shown to be very useful, we aim for an algorithm for doing so. Therefore, we
need to define the number of colorings.
Definition: P (G; x), the number of proper colorings of a graph G using x colors, is the chromatic polynomial
of G.
4
Counting the number of the proper colorings may be thought of as subtracting the number of improper
colorings from the total number of colorings. So, focusing on edge e of Graph H, we remove the edge and
examine the resulting graphs.
Figure 4
This decomposition is to be repeated until only empty graphs (graphs without edges) remain. To illustrate,
we decompose a simpler, three-edge graph.
Figure 5
An empty graph with k vertices can be colored in xk ways. So the chromatic polynomial of the above graph
is (x3 − x2 ) − (x2 − x), or x3 − 2x2 + x. The smallest nonnegative x for which P (G; x) > 0 is the chromatic
number (finding this number would be ideal for the movie theater problem). The algorithm used is known as
the Birkhoff-Lewis Algorithm.
Sample 3
Graphs are useful tools for representing numerous problems. By developing tools for analyzing graphs,
mathematicians can greatly impact a variety of other fields. Examples of problems for which graphs are good
representations are making road trips, planning future offerings for flights, and scheduling class offerings.
There are several real-world scenarios that graph coloring can be applied to. For example, one of these is
figuring out the best sites on the internet. If we make every page on the internet a vertex and draw a line
connecting two vertices when the pages are linked, it creates a graph. Google uses this technique to find out
which pages are better than others when people use their search engine. A page that is classified as good
has many other pages linked to it. Pages that do not receive very many views will not have many edges
connected to other vertices. This way, the success of pages can be ranked. This is called PageRank. [1 ]
In this report, we will explore how to find chromatic numbers and polynomials of graphs, how to play graph
coloring games, and how players can win a vertex coloring game based on a graph’s chromatic number and
the number of colors that are used to color it.
TERMINOLOGY
Graph: A graph is a collection of vertices and edges between the vertices.
Connected: A graph is connected if for any two vertices there is an edge-path from one to the other.
Component: Vertices are a component if they are connected to another vertex.
Proper Vertex Coloring: A vertex coloring of a graph is proper if no two vertices sharing an edge have
5
the same color.
Path: A path is an order pairing of edges such that each sequential edge is connected to the next edge in the
pairing.
Hamiltonian path: A Hamiltonian path is a path that visits every vertex once.
Euler path: An Euler path is a path that visits every edge exactly once.
Polynomial time problem: A polynomial run time problem is a problem for which the number of operations required to find a solution can be represented using only polynomial terms. Many difficult problems
have no known polynomial time solutions, but such solutions are highly desirable.
Chromatic number: The smallest number of colors needed to color a graph with no two adjacent vertices
sharing the same color.
Chromatic Polynomial: A chromatic polynomial is a polynomial generated from a graph G such that the
polynomial will produce the number of proper colorings for an inputted number of colors x. This is denoted
p(G, x).
TECHNICAL FINDINGS
Color-Assigning Method
We found that for most graphs the chromatic polynomial could be found by choosing a starting vertex, and
designating this vertex to any of the available colors. Then we choose an adjacent vertex that has not yet
been assigned, and assign it to any of the available colors where an available color is a choice that keeps the
coloring of the graph proper. If a vertex is connected to k vertices that have already been assigned a color,
this vertex is assigned the value (x − k). If there are no adjacent vertices that have not been assigned a
value, jump to another vertex that is connected to a vertex that has already been assigned. If no unassigned
vertices are adjacent to previously assigned vertices, choose one of the unassigned vertices, and start the
assignment process over again. Once all vertices have been assigned a value, the chromatic polynomial is all
of the values of the vertices multiplied together.
Conjecture 1. The Color-Assigning Method produces the chromatic polynomial when applied to a graph G.
Proof For a vertex connected to k vertices that have already been assigned a color, assigning the value
(x − k) is representative of the number of colors that vertex could be colored. In order for the coloring to
remain proper, no connected vertices may be the same color. Thus any color may be assigned to that vertex
other than the color of the two other vertices, leaving the coloring choices (x − k). The chromatic polynomial
of the graph is represented by the product of all of the assignments because the number of combinations of
colors is equal to the product of the number of colors available at each vertex.
Birkhoff-Lewis Counting Algorithm
The Birkhoff-Lewis Counting Algorithm is an algorithm that, given the graph g, produces the chromatic
polynomial for g.
The algorithm works by creating two new graphs from the original graph. One of the graphs has an edge
from connected vertices v1 and v2 removed, while the other has combined v1 and v2 into a new vertex v3 such
that v3 has all of the connections that v1 and v2 previously had. The chromatic polynomial for the original
graph is equal to the chromatic polynomial of the graph with the edge between v1 and v2 removed minus
the chromatic polynomial of the graph with the vertices merged. By repeating this process, the chromatic
polynomial can be found as a combination of the chromatic polynomials for graphs with no connections, and
graphs with a single vertex.
Conjecture 2. The chromatic polynomial for a graph which contains a single unconnected vertex is x.
Proof With a single vertex and x colors, each of the colors may be used to color the vertex, thus the number
of proper colors is equal to the number of colors available. Thus the chromatic polynomial is x.
Conjecture 3. The chromatic polynomial for a graph with n vertices and no connections is xn .
Proof The number of color choices available for a vertex is dependent on the number of edges that vertex
has to other vertices. If a vertex has no connections, there are no possible color conflicts, so all of the x
considered colors are valid choices. With n coloring spots and x valid colors for each spot, the chromatic
polynomial becomes xn .
6
4. Technical Findings
Since a complete graph will have every node connected to every other node, it would only make sense that the
chromatic number for a complete graph on 2, 3, 4, 5, and 6 vertices will be 2, 3, 4, 5, and 6 respectively. This
is the case because no two vertices can have the same color in a complete graph. For example, the following
is a complete graph on 6 vertices:
2
1
3
6
4
5
Clearly, the graph shows that the chromatic number has to be 6 because no two vertices can share the same
color.
4.1. Question 2.
Sample 1
to every other vertex. A simple example to imagine would be to think of a triangle, all 3 vertexes connect to
the remaining 2 vertexes.
Now consider a graph of n nodes that is complete. We can ask how many colors does it take to color that
graph. We can begin by looking at 1, 2, 3 and 4 These correspond to a point, a line, a triangle, and a square
seen below.
Since each of the points connects to each of the other points so a 1 node graph needs 1 color, 2 nodes needs 2
all the way to 4 nodes needs 4. This gives us a hint that we need to do a proof by induction to prove how
many colors is needed for a completed graph of n nodes. First lets prove that 1 ∈ S since that is a graph of 1
node it only needs 1 color to be colored. Next we need to prove that if an n node graph needs n colors then
an n+1 node graph needs n+1 colors. Assume that n ∈ S then the n node graph has an n chromatic color.
Now consider that graph with an added node. To properly color this node it has to be separate from each
other node. Since each other node has a unique color since the graph is complete it must be a new color
making the chromatic color of the n + 1 node graph n + 1, so n + 1 ∈ S so S = N.
4.2. Question 5.
Sample 1
H: A graph with 6 vertices and 9 edges.
Chromatic polynomial: x6 − 9x5 + 36x4 − 75x3 + 78x2 − 31x = (x − 1)·x·(x4 − 8x3 + 28x2 − 47x + 31)
Chromatic number: 2
2
1
3
6
4
5
I: A graph with 6 vertices and 5 edges.
Chromatic polynomial: x6 − 5x5 + 10x4 − 10x3 + 5x2 − x = x·(x − 1)5
Chromatic number: 2
7
2
1
6
3
5
4
J: A graph with 7 vertices and 11 edges.
Chromatic polynomial: x7 −11x6 +50x5 −121x4 +165x3 −120x2 +36x = (x−3)·(x−1)·x·(x−2)2 ·(x2 −3x+3)
Chromatic number: 4
2
1
3
7
4
5
6
K: A graph with 8 vertices and 10 edges.
Chromatic polynomial: x8 −10x7 +43x6 −103x5 +148x4 −127x3 +60x2 −12x = x·(x−2)2 ·(x−1)3 ·(x2 −3x+3)
Chromatic number: 3
2
1
3
8
4
7
5
6
Sample 2
Our next investigation was to use sagemath.org to find the chromatic polynomials for each of the graphs in
figure below. The error “Graph is not defined” suggest that the graph class is not part of the online library.
Therefore, the next step was to find it by hand. This involves picking a vertex, assigning x possible colors to
it, and then picking another. If it is connected to the first, that color can’t be used. If it is connected to two,
then we need to know if the two are different or the same to assign a subtraction value. The in class examples
showed that if a vertex’s color options can’t be determined, its corresponding factors are an unfactorable
polynomial of degree 2.
Graph H’s is x(x – 1)(x – 1)(unfactorable polynomial)(unfactorable polynomial)(unfactorable polynomial).
The last 3 terms are the result of not knowing how the non-adjacent vertices relate to teach other, so that we
cannot assign x – some number of color options removed. Since (x – 1) is the smallest known term, we tried
x = 2 first, which worked.
Graph I’s is x(x – 1) (x – 1) (x – 1) (x – 1) (x – 1). Clearly the center can have any value, and the outside
ones cannot have that value. Only 2 colors are needed.
Graph J’s is x(x – 1)(x – 1)(x – 2)(x – 2)(x - 3)(unfactorable), with the last factor depending on non-adjacent
vertices of unknown relationships. Therefore, x = 4 is a likely first guess. Note that J contains the 4 vertex
complete graph, a square with opposite vertices connected. See the figure, which uses 4 colors.
Graph K’s chromatic polynomial is x(x – 1)(x – 1)(x – 1)(x – 1)(x – 2)(unfactorable)(unfactorable). This
means at least 3 colors are needed, but maybe more are needed, based on the last term being unknown. Only
3 were indeed needed, in the figure.
8
Sample 3
Graph H:
A proper coloring that we found for this graph was [[2, 4, 6], [1, 3, 5]] where the each grouping is vertices of
the same color. The chromatic number for this graph is 2.
9
Graph I:
A proper coloring that we found for this graph was [[6], [1, 2, 3, 4, 5]] where the each grouping is vertices of
the same color. The chromatic number for this graph is 2.
Graph J:
A proper coloring that we found for this graph was [[6, 2], [7], [1, 5], [4, 3]] where the each grouping is vertices
of the same color. The chromatic number for this graph is 4.
Graph K:
10
A proper coloring that we found for this graph was [[5, 3], [1, 2, 6, 8], [4, 7]] where the each grouping is
vertices of the same color.The chromatic number for this graph is 3.
4.3. Question 10.
Sample 1
If the chromatic number of a graph G is k (for some positive integer k), what factors do you know must exist
for the chromatic polynomials of G?
If the chromatic number of graph G is k then the outcome of the smallest factor of the chromatic polynomial
of G must equal 1. These factors must be present: (x)(x − 1)(x − 2)...(x − (k − 1)).
Why is this so? Recall that the least number of needed colors to color a graph is k, by definition of the
chromatic number. Therefore, all numbers a such that 0 ≤ a < k must fail to create any valid color
combinations: In other words, the chromatic polynomial must equal 0 when x = a for the a values described
above. To ensure this happens, the factors (x)(x − 1)(x − 2)...(x − (k − 1)) must be present in the factorization
of the chromatic polynomial.
Side note: Recall that the chromatic polynomial of a complete graph of vertex count k has this as its exact
polynomial. The connection is this: A complete graph of vertex count k needs exactly k colors to abide by
the rules of graph coloring, no more and no less.
4.4. Question 12.
Sample 1
Looking at the complete graphs we can see that the expanded polynomial is interesting. For this we are going
to take a 10 node completed graph and remove one edge from the same node until it is separate from the
graph. When we did this we were returned the following chromatic polynomial
0 edged removed
x^10 - 45*x^9 + 870*x^8 - 9450*x^7 + 63273*x^6 - 269325*x^5
+ 723680*x^4 - 1172700*x^3 + 1026576*x^2 - 362880*x
1 edged removed
x^10 - 44*x^9 + 834*x^8 - 8904*x^7 + 58737*x^6 - 246876*x^5
+ 656396*x^4 - 1054576*x^3 + 916992*x^2 - 322560*x
2 edged removed
x^10 - 43*x^9 + 798*x^8 - 8358*x^7 + 54201*x^6 - 224427*x^5
+ 589112*x^4 - 936452*x^3 + 807408*x^2 - 282240*x
3 edged removed
x^10 - 42*x^9 + 762*x^8 - 7812*x^7 + 49665*x^6 - 201978*x^5
+ 521828*x^4 - 818328*x^3 + 697824*x^2 - 241920*x
4 edged removed
x^10 - 41*x^9 + 726*x^8 - 7266*x^7 + 45129*x^6 - 179529*x^5
+ 454544*x^4 - 700204*x^3 + 588240*x^2 - 201600*x
5 edged removed
x^10 - 40*x^9 + 690*x^8 - 6720*x^7 + 40593*x^6 - 157080*x^5
+ 387260*x^4 - 582080*x^3 + 478656*x^2 - 161280*x
6 edged removed
11
x^10 - 39*x^9 + 654*x^8 - 6174*x^7 + 36057*x^6 - 134631*x^5
+ 319976*x^4 - 463956*x^3 + 369072*x^2 - 120960*x
7 edged removed
x^10 - 38*x^9 + 618*x^8 - 5628*x^7 + 31521*x^6 - 112182*x^5
+ 252692*x^4 - 345832*x^3 + 259488*x^2 - 80640*x
8 edged removed
x^10 - 37*x^9 + 582*x^8 - 5082*x^7 + 26985*x^6 - 89733*x^5
+ 185408*x^4 - 227708*x^3 + 149904*x^2 - 40320*x
9 edged removed
x^10 - 36*x^9 + 546*x^8 - 4536*x^7 + 22449*x^6 - 67284*x^5
+ 118124*x^4 - 109584*x^3 + 40320*x^2
Looking at the coefficient associated with each polynomial we get a pattern of subtraction from them. The
table below shows the power of x and the difference each edge creates in that coefficient.
The completed graph
x power difference
10
0
9
1
8
36
7
546
6
4536
5
22449
4
67284
3
118124
2
109584
1
40320
This seems to indicate that removing one from each will eventually remove the x1 term which makes intuitive
sense because when that is completely separate it would be a free coloring.
Sample 2
For this question, we talk about chromatic polynomials of the form:
an xn + an−1 xn−1 + an−2 xn−2 . . . a2 x2 + a1 x
In all of the graphs we looked at we noted that:
• an = 1
• an−1 = −|E| where |E| is the number of edges
•
Pn
i=1
(ai ) = 0
We believe that there are more connections, yet due to the limited scope of our study, the above 3 were the
only few that arose.
Sample 3
For this question, we decided to look at n-cycle graphs, because they were the one exception to the trend
that we previously found for finding the chromatic number of a graph.
To begin we generated n-cycle graphs for n = 3 − 6, and found their chromatic polynomials.
Graph A:
12
The chromatic polynomial for this graph is x3 − 3x2 + 2x which factors into (x − 2)(x − 1)(x).
13
Graph B:
The chromatic polynomial for this graph is x4 − 4x3 + 6x2 − 3x which factors into (x2 − 3x + 3)(x − 1)(x).
Graph C:
The chromatic polynomial for this graph is x5 − 5x4 + 10x3 − 10x2 + 4x which factors into (x − 2)(x −
1)(x)(x2 − 2x + 2).
Graph D:
14
The chromatic polynomial for this graph is x6 − 6x5 + 15x4 − 20x3 + 15x2 − 5x which factors into
(x − 1)(x)(x4 − 5x3 + 10x2 − 10x + 5).
From these graphs and their chromatic polynomials we noticed a fascinating pattern. The coefficients
on the chromatic expressions are the binomial coefficients but slightly altered.
Conjecture 4. For an n-cycle graph, the coefficients of the chromatic polynomial are equal to the coefficients
of the nth layer of Pascal’s triangle except the last coefficient of the chromatic polynomial is equal to one less
than the second to last number on that layer of Pascal’s triangle.
We found this pattern to be incredible because it suggests a connection between these types of graphs and
the pascals triangle. We were not able to find a proof to support our conjecture, but we included it because
we feel that it suggests something fundamentally interesting.
( It is amazing how everything ends up connected!)
4.5. Question 13. Here we record only the various ways in which the game was defined in the lab reports
submitted.
Sample 1
Two players, Q (P1) and M (P2) take turns colouring vertexes of a random graph. If the graph is successfully
(properly) coloured at the end of the game, then Q wins, but if the game reaches a point where a vertex v
cannot be properly coloured (it is connected to another vertex of every available colour), then M wins the game.
Sample 2
A game can be made from coloring graphs, the rules are as follows: First, this is a two player, perfect
information game. Player one goes first (referred to from now on as P1) and alternates turns with player two
(referred to as P2) to color the vertexes of a graph so that no vertexes sharing an edge have the same color.
A coloring of a graph that uses at most k − colorings is called a proper k coloring. The smallest number of
colors needed to color a graph is its chromatic number and is denoted X(G).
Sample 3
Rules: Draw any number of vertices and connect them any way you want. Player ONE and player TWO
color vertices of a graph with k colors. Player ONE will make a move by coloring a vertex with a certain
color. Player TWO then colors a different vertex with their own color, but there cannot be two vertices of the
same color that share an edge. A player wins when they arrive at a correct coloring of the graph by coloring
the last vertex, or if a player forces the other to violate the proper coloring condition by putting them in a
position where they must color a vertex whose adjacent vertices are the same color.
Sample 4
Two players may pick from a set of k colors to color a graph, taking turns. Player ONE wishes to color the
graph with no adjacent vertices having the same color, whereas Player TWO wishes to color vertices such
that this becomes impossible to do with only the k colors.
Sample 5
• PLAYER ONE and PLAYER TWO are coloring the vertices of a graph G with a set k of colors.
15
• PLAYER ONE and PLAYER TWO are taking turns, coloring properly a uncolored vertex (in the
standard version, PLAYER ONE begins).
• If a vertex v is impossible to color properly (for any color, v has a neighbor colored with it), then
PLAYER TWO wins.
• If the graph is completely colored, then PLAYER ONE wins.
5. Conclusion
Sample 1
To conclude, we have defined the basic notion of a graph, discussed the basic components of graph colouring,
and looked at some applications and results from the theoretical discussion. Graph colouring can be used in
numerous applications, from setting up security cameras to scheduling to modelling a network. Colouring
graphs is not a trivial task, but if the graph we want to colour is a special type of graph (like a tree), then we
can look to patterns in the chromatic polynomial to tell us more about the graph.
Continuing on, it seems worthwhile to begin seriously investigating computationally efficient methods for
finding the chromatic polynomial and chromatic number for large graphs. While it is always possible that the
end result will merely prove that there is no algorithm more computationally efficient than what we already
have for graph colouring, the enormous benefit of finding such an algorithm, if it does exist, certainly seems
worth the risk.
16