Lecture 03-29-17

Today: Our 1st Reduction

CNF Conjunctive Normal Form

• Example:
  • (x₁ + x₂ + x₃)(x₁’ + x₂)(x₂’ + x₃’)(x₁’ + x₃)
• Special type of expression
• No negation over more than one literal
• Only negation over literals
• Each clause is just literals & OR
• Variable: X1
• Literal: X1bar

Cook’s Theorem

• States that the boolean satisifiability problem is NP-complete
• Significant breakthrough
• Prove NP-complete
  • Gary & Johnson has a list to reference
    • Can turn particular problem into that

CNF Satisifiability

• Whether or not there exists an interpretation that satisfies a given boolean formula

Clique Problem (Prove is NP-complete)

• Clique
  • All vertices in a clique are adjacent
• Particular graph: “Trampoline”
• Largest clique: 2-3-5-7
• Given a graph G and a number k, does this graph have a clique of size k?
  • K is not necessarily an integer, could be “half the nodes”
• Steps:
  1. Show clique is in NP
  2. Reduction
     • Turn known NP-complete in polynomial time into clique problem
     • Steps of a reduction:
       1. Make a clear reduction
       2. Prove that the reduction is correct
          • Ways:
            • If the answer to the new problem is yes, then the answer to the known NP-complete problem is yes & if the answer to the old problem is yes, then the answer to the new problem is yes
              • Almost always this way because of the definition of NP
                • NP there is a short proof that the answer is yes
            • If the answer to the known NP-complete problem is yes, then the answer to the new problem is yes as well. If the answer to the known NP-complete problem is no, then the answer to the new problem is no as well

CNF Satisifiability to Clique

1. Make a clear reduction
   • Given an boolean expression E, produce a graph G and a number K
     • E -> G,k
   • G has one vertex for every instance of a literal of E
   • An edge from x-y iff x and y are from different clauses and not the complements of each other
   • Now, k = number of clauses in E
   • Any expression E will be satisfiable if it has a clique of size K (part 2)
2. Prove that the reduction is correct
   • Suppose E is satisfiable
     • Take 1 true literal from each clause
     • Map to a set of k vertices which are each adjacent to each other
     • Since from separate clauses, none complement
   • Suppose G has a clique of size k
     • Take clique C of size k
       • k vertices
         • none of which are complements
         • all from different clauses (makes one literal true in each)
       • One thing true in each -> e is satisfiable
       • Fundamentally how all reductions will look like