3SAT23Colorability - Theorem1. Proof:...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Theorem 1.  3–Colorability is NP–Complete. Proof:  Given a YES instance of 3–Colorability and a 3-coloring of the vertices is  easy, in polynomial time, to verify the instance is YES. Thus 3-Coloribility is in the  set NP. In order to transform 3SAT to 3-colorability, we have to construct a  graph  of  node s and  edge s which is  3-colorable  if and only if the given instance of  3SAT   can be satisfied. This means we need a  graph  that can be 3-colored only when a  corresponding expression evaluates to true. For this example, the three colors  are  TRUE FALSE  and  OTHER . To construct the graph, start with a node named  root  that is colored  OTHER . Also, for every  variable   x  in the expression, create a  pair of nodes  x  and  ¬x . This pair is connected to each other and to the  root , to  create a  triangle  for each variable:
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 07/14/2011 for the course COT 4610 taught by Professor Dutton during the Fall '10 term at University of Central Florida.

Page1 / 3

3SAT23Colorability - Theorem1. Proof:...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online