Group no: 09

NAME :Tarafder,Md. Shakibuzzaman Id:13-23384-1

Feroz,Adnan Ahmed 13-22916-1

In graph theory, graph coloring is a special case of graph labeling.

It is an assignment of labels traditionally called "colors" to elements of a graphsubject to certain constraints.

Graph Coloring is an assignment of colors (or any distinct marks) to the vertices of a graph. Strictly speaking, a coloring is a proper coloring if no two adjacent vertices have the same color.

Coloring theory started with the problem of coloring the countries of a map in such a way that no two countries that have a common border receive the same color.

If we denote the countries by points in the plane and connect each pair of points that correspond to countries with a common border by a curve, we obtain a planar graph.

Graphs are used to depict ”what is in conflict with what”, and colors are used to denote the state of a vertex.

So, more precisely, coloring theory is the theory of ”partitioning the sets having Internal unreconcilable conflicts.

Vertex Coloring: It is a way of coloring the vertices of a graph such that no two adjacent vertices share the same color.

Edge Coloring: An edge coloring assigns a color to each edge so that no two adjacent edges share the same color.

Face Coloring : A face coloring of a planar graph assigns a color to each face or region so that no two faces that share a boundary have the same color.

Chromatic Number: The chromatic number of a graph is the minimum number of colors in a proper coloring of that graph. If chromatic number is r then the graph is r-chromatic.

Chromatic number: 4

Polynomial which gives the number of ways of proper coloring a graph using a given number of colors

Ci = no. of ways to properly color a graph using exactly icolors

λ = total no of colors

λ Ci = selecting I colors out of λ colors

ΣCiλ Ci = total number of ways a graph canbe properly

colored using λ or lesser no. of colors

Pn(λ) of G = ΣCiλ Ci

P4 (λ) of G = C1(λ) + C2(λ) (λ-1)/2! + C3(λ) (λ-1) (λ-2) /3! + C4(λ) (λ-1) (λ-2) (λ- 3)/4!

Let G be a simple graph, and let PG(k) be the number of ways

of coloring the verticles of G with k colors in such a way that

no two adjacent vertices are assigned the same color. The

function PG(k) is called the chromatic polynomial of G.

As an example, consider complete graph K3 as shown in the

following figure.

Then the top vertex can be assigned any of the k colors, the left vertex can be assigned any k-1 colors, and right vertex can be assigned any of the k-2 colors.

The chromatic polynomial of K3 is therefore K(K -1)(K -2). The extension of this immediately gives us the following result.

If G is the complete graph Kn, then Pn(K) = K(K - 1)(K- 2) . . . (K - n +1).

Every non-trivial graph is atleast 2-chromatic.

If a graph has a triangle in it , then it is atleast 3-chromatic.

Chromatic Polynomial for a tree :

Pn(λ) of Tn = (λ) (λ-1)n-1 (tree is 2-chromatic)

This can be proved by Mathematical Induction.

Tree is 2-chromatic.

Theorem - the vertices of every finite planar graph can be coloured properly with five colours.

Proof-the proof is based on induction on vertices of a planar graph, since the vertices of all planar graph G with 1,2,3,4,5 can be properly coloured by 5 or less colours.

Let us assume that every planar graph with n-1 vertices is properly colourable with 5 colours or fewer. So we have to show that there is no graph of n-vertices which require more than 5-colours for proper colouring.

Scheduling

Mobile radio frequency assignment

Sudoku

Pattern matching

Register Allocation

Scheduling

Vertex coloring models to a number of scheduling problems .In the cleanest form, a given set of jobs need to be assigned to time slots, each job requires one such slot. Jobs can be scheduled in any order, but pairs of jobs may be in conflict in the sense that they may not be assigned to the same time slot, for example because they both rely on a shared resource. The corresponding graph contains a vertex for every job and an edge for every conflicting pair of jobs. The chromatic number of the graph is exactly the minimum make span, the optimal time to finish all jobs without conflicts.

When frequencies are assigned to towers, frequencies assigned to all towers at the same location must be different. How to assign frequencies with this constraint? What is the minimum number of frequencies needed? This problem is also an instance of graph coloring problem where every tower represents a vertex and an edge between two towers represents that they are in range of each other.

GSM (Global System for Mobile Communications, originally Group Special Mobile), was created in 1982 to provide a standard for a mobile telephone system.. Today,GSM is the most popular standard for mobile phones in the world, used by over 2 billion people across more than 212 countries.

GSM is a cellular network with its entire geographical range divided into hexagonal cells.

Each cell has a communication tower which connects with mobile phones within the cell.

All mobile phones connect to the GSM network by searching for cells in the immediate vicinity.

GSM networks operate in only four different frequency ranges. The reason why only four different frequencies suffice is clear: the map of the cellular regions can be properly colored by using only four different colors! So, the vertex coloring algorithm may be used for assigning at most four different frequencies for any GSM mobile phone network

Solving Sudoku puzzles can be expressed as a graph coloring problem. Consider the 4 × 4 = 22 × 22 case. The aim of the puzzle in its standard form is to construct a proper 9-coloring of a particular graph, given a partial 4-coloring. The graph in question has 16 vertices, one vertex for each cell of the grid. The vertices can be labeled with the ordered pairs (x, y), where x and y are integers between 1 and 4. In this case, two distinct vertices labeled by (x, y) and (x′, y′) are joined by an edge if and only if:

x = x′ (same column) or,

y = y′ (same row) or,

⌈ x/2 ⌉ = ⌈ x′/2 ⌉ and ⌈ y/2 ⌉ = ⌈ y′/2 ⌉ (same 2 × 2 cell)

The puzzle is then completed by assigning an integer between 1 and 4 to each vertex, in such a way that vertices that are joined by an edge do not have the same integer assigned to them