3-Coloring is NP-Hard

3-Coloring is NP-Hard

Feliciano colella

December 4, 2014

Algorithm Design Homework 02

Outline of the presentation

1. De�nition of 3-Satis�ability.

2. De�nition of 3-Coloring.

3. Proof that 3-Sat ≤P 3-Color.

Feliciano colella 3-Coloring is NP-Hard December 4, 2014 2 / 8


The problems

3-Satis�ability

Input CNF formula Φ where each clause contains exactly 3 di�erent

literals.

Output Satisfying truth assignment for the formula Φ.

k-Coloring

Input A graph G = (V ,E ) with |V | = n vertices and |E | = medges.

Output A k-Coloring c : V → {1, . . . , k}, s.t.∀(x , y) ∈ E , C (x) 6= C (y).



The reduction

3-Coloring is in NP

Certi�cate A 3-Coloring c : V → {1, 2, 3}Certi�er Check if ∀(u, v) ∈ E , c(u) 6= c(v)

Hardness

I Show that the formula Φ is satis�able IFF exist a 3-Coloring for the

graph G .



The gadgets

Literals Gadget Clause gadget

x1 ¬x1

x2 ¬x2

xn ¬xn

R

T

F

ci1 ci2 ci3

R

T

F



The construction

x1 ¬x1

x2 ¬x2

xn ¬xn

R

T

F

c13

c12

c11

c23

c22

c21

c`3

c`2

c`1



The theorem

TheoremA CNF Formula Φ is satis�able ⇐⇒ exists a 3-Coloring for the graph GΦ.



The theorem


Φ is satis�able =⇒ GΦ is 3-Colorable

I Take a truth assignment φ for Φ and color the literals gadgets;

I Each clause have at least 1 literal set to True;

I All the clause gadget can be colored with respect to the constraint;

I We have a 3-Coloring.



The theorem


GΦ is 3-Colorable =⇒ Φ is satis�able

I Take a 3-Coloring for GΦ;

I If a literal is colored with True, set it to True in the formula Φ;

I For any clause, it cannot be that all the literals are True or False.

Otherwise we would have a clause gadget colored to False, but this is

impossible since they are connected to Base and False and we have

started from a 3-Coloring for GΦ;

I We have hence correct a truth assignment for Φ


Thank you for the attention.

3-Coloring is NP-Hard

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