# 3-coloring is NP Complete

**Prerequisite: **NP-Completeness, Graph Coloring

** Graph K-coloring Problem:** A K-coloring problem for undirected graphs is an assignment of colors to the nodes of the graph such that no two adjacent vertices have the same color, and

**at most K**colors are used to complete color the graph.

** Problem Statement: **Given a graph

**G(V, E)**and an integer K = 3, the task is to determine if the graph can be colored using

**at most 3**colors such that no two adjacent vertices are given the same color.

** Explanation**:

An instance of the problem is an input specified to the problem. An instance of the

**3-coloring problem**is an undirected graph

**G (V, E)**, and the task is to check whether there is a possible assignment of colors for each of the vertices

**V**using only 3 different colors with each neighbor colored differently. Since an NP-Complete problem is a problem which is both in

**NP**and

**NP-hard**, the proof for the statement that a problem is NP-Complete consists of two parts:

- The problem itself is in
NP class.- All other problems in NP class can be polynomial-time reducible to that.(B is polynomial-time reducible to C is denoted as B ≤ P
^{C})

If the **2nd condition** is only satisfied then the problem is called **NP-Hard**.

But it is not possible to reduce every NP problem into another NP problem to show its NP-Completeness all the time. Therefore, to show a problem is NP-Complete, then proof that the problem is in **NP** and any **NP-Complete problem** is reducible to that i.e., if B is NP-Complete and B≤P^{C} then for C in NP, then C is NP-Complete. Thus, it can be concluded that the **Graph K-coloring Problem** is NP-Complete using the following two propositions:

**3-coloring problem is in NP:**

If any problem is in NP, then, given a certificate, which is a solution to the problem and an instance of the problem (A graph **G(V, E)** and an assignment of the colors **{c _{1}, c_{2}, c_{3}}** where each vertex is assigned a color from this three colors

**{c**), then it can be verified (Check whether the solution given is correct or not) that the certificate in polynomial time. This can be done in the following way:

_{1}, c_{2}, c_{3}}For each edge {u, v} in graph G verify that the color c(u) != c(v)

Hence, the assignment can be checked for correctness in the polynomial-time of the graph with respect to its edges O(V+E).

**3-coloring problem is NP-Hard:**

In order to prove that the 3-coloring problem is NP-Hard, perform a reduction from a known NP-Hard problem to this problem. Carry out a reduction from which the 3-SAT problem can be reduced to the 3-coloring problem. Let us assume that the 3-SAT problem has a 3-SAT formula of m clauses on n variables denoted by **x _{1}, x_{2}, …, x_{n}**. The graph can then be constructed from the formula in the following way:

- For every variable
**x**Construct a vertex_{i}**v**In the graph and a vertex_{i}**v**denoting the negation of the variable_{i’}**x**._{i} - For each clause c in m, add 5 vertices corresponding to values c1, c1, …, c5.
- Three vertices of different colors are additionally added to denote the values True, False, and Base (T, F, B) respectively.
- Edges are added among these three additional vertices
**T, F, B**to form a triangle. - Edges are added among the vertices
**v**and_{i}**v**and Base (B) to form a triangle._{i’}

**The following constraints are true for graph G: **

- For each of the pairs of vertices
**v**and_{i}**v**, either one is assigned a TRUE value and the other, FALSE._{i’} - For each clause c in m clauses, at least one of the literal has to hold TRUE value for the value to be true.

A small OR- gadget graph therefore can be constructed for each of the clause **c = (u V v V w)** in the formula by input nodes u, v, w, and connect output node of gadget to both False and Base special nodes.

Let us consider the formula **f = (u’ V v V w’)** AND **(u V v V w’)**

**Now the reduction can be proved by the following two propositions:**

Let us assume that the 3-SAT formula has a satisfying assignment, then in every clause, at least one of the literals **x _{i}** has to be true, therefore, the corresponding

**v**can be assigned to a TRUE color and

_{i}**v**to FALSE. Now, extending this, for each clause the corresponding OR-gadget graph can be 3-colored. Hence, the graph can be 3-colored.

_{i’}Let us consider that the graph G is 3-colorable, so if the vertex vi is assigned to the true color, correspondingly the variable x_{i} is assigned to true. This will form a legal truth assignment. Also, for any clause **C _{j} = (x V y V z)**, it cannot be that all the three literals x, y, z are False. Because in this case, the output of the OR-gadget graph for

**C**has to be colored False. This is a contradiction because the output is connected to Base and False. Hence, there exists a satisfying assignment to the 3-SAT clause.

_{j}** Conclusion: **Therefore, 3-coloring is an

**NP-Complete**problem.

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