Further NP-Complete Problems

• Now that we have CIRCUIT-SAT in NP-Complete it is much easier to add other problems to this set.
• We can reduce all problems in NP to CIRCUIT-SAT in polynomial time
• and if we can reduce CIRCUIT-SAT to a problem in polynomial time,
• then we can reduce all problems to the new problem in polynomial time.
• SAT
  • Given a boolean expression, consisting of variables , and or, not implication, iff, and parenthesis, determine if the expression is satisfiable.
  • Given a set of values for the input, we can check this in polynomial time, thus it is in NP
  • It makes sense that we can reduce CIRCUIT-SAT to SAT, so it is in NP-Complete
  • The niave reduction is not polynomial, it can become exponential.
  • The reduction considers an equation of the "values" of each of the wires.
• 3CNF-SAT
  • Given an expression of the form (x+y+z)(a+b+c)(c+d+e) ...
  • juxtaposition is and
  • + is or
  • Can we find a set of inputs that satisfy this expression
  • Again, verification in polynomial time is easy
  • Reducing SAT to 3CNF-SAT is accomplished by
    • building a parse tree to represent the arbritrary expression
    • each subtree represents a binary expression
    • Which can be simplified into 3CNF
• CLIQUE
  • In an undirected Graph G=(V,E), a clique is a set V' in V suchthat for all u,v in V, (u,v) is in E
  • The size of a clique is the number of verticies in the clique
  • The clique problem (CLIQUE) is finding the clique of maximum size in a graph.
  • 
  • Checking to see if something is a clique can be done in O(|V|²)
  • So it is in NP
  • Reducing NP to CLIQUE is not too bad
    • Given an expression in 3CNF
    • For each term of each factor, create a node
    • Group all of the nodes from a factor togeather
    • Connect nodes of a factor to all nodes in other factors that are other than not of the node.
    • Let each factor be set of nodes, either X or not X
    • Find a clique of size k
    • If the size of the clique is the same as the number of factors in the original expression then the expression is satisfiable.
    • Consider (x+y+!z)(x+!y+z)(!x+y+z)