lecture 30

lecture 30

• 1.
  CS 332: AlgorithmsNP Completeness Continued: Reductions
• 2.
  Review: P and NP What do we mean when we say a problem is in P ? What do we mean when we say a problem is in NP ? What is the relation between P and NP ?
• 3.
  Review: P and NP What do we mean when we say a problem is in P ? A: A solution can be found in polynomial time What do we mean when we say a problem is in NP ? A: A solution can be verified in polynomial time What is the relation between P and NP ? A: P  NP , but no one knows whether P = NP
• 4.
  Review: NP-Complete What,intuitively, does it mean if we can reduce problem P to problem Q? How do we reduce P to Q? What does it mean if Q is NP-Hard ? What does it mean if Q is NP-Complete ?
• 5.
  Review: NP-Complete What,intuitively, does it mean if we can reduce problem P to problem Q? P is “no harder than” Q How do we reduce P to Q? Transform instances of P to instances of Q in polynomial time s.t. Q: “yes” iff P: “yes” What does it mean if Q is NP-Hard ? Every problem P  NP  p Q What does it mean if Q is NP-Complete ? Q is NP-Hard and Q  NP
• 6.
  Review: ProvingProblems NP-Complete How do we usually prove that a problem R is NP-Complete? A: Show R  NP , and reduce a known NP-Complete problem Q to R
• 7.
  Review: Directed  Undirected Ham. Cycle Given: directed hamiltonian cycle is NP-Complete (draw the example) Transform graph G = (V, E) into G’ = (V’, E’): Every vertex v in V transforms into 3 vertices v 1 , v 2 , v 3 in V’ with edges ( v 1 , v 2 ) and ( v 2 , v 3 ) in E’ Every directed edge ( v , w ) in E transforms into the undirected edge ( v 3 , w 1 ) in E’ (draw it)
• 8.
  Review: Directed  Undirected Ham. Cycle Prove the transformation correct: If G has directed hamiltonian cycle, G’ will have undirected cycle (straightforward) If G’ has an undirected hamiltonian cycle, G will have a directed hamiltonian cycle The three vertices that correspond to a vertex v in G must be traversed in order v 1 , v 2 , v 3 or v 3 , v 2 , v 1 , since v 2 cannot be reached from any other vertex in G’ Since 1’s are connected to 3’s, the order is the same for all triples. Assume w.l.o.g. order is v 1 , v 2 , v 3 . Then G has a corresponding directed hamiltonian cycle
• 9.
  Review: Hamiltonian Cycle  TSP The well-known traveling salesman problem : Complete graph with cost c( i,j ) from city i to city j  a simple cycle over cities with cost < k ? How can we prove the TSP is NP-Complete? A: Prove TSP  NP ; reduce the undirected hamiltonian cycle problem to TSP TSP  NP : straightforward Reduction: need to show that if we can solve TSP we can solve ham. cycle problem
• 10.
  Review: Hamiltonian Cycle  TSP To transform ham. cycle problem on graph G = (V,E) to TSP, create graph G’ = (V,E’): G’ is a complete graph Edges in E’ also in E have weight 0 All other edges in E’ have weight 1 TSP: is there a TSP on G’ with weight 0? If G has a hamiltonian cycle, G’ has a cycle w/ weight 0 If G’ has cycle w/ weight 0, every edge of that cycle has weight 0 and is thus in G. Thus G has a ham. cycle
• 11.
  The SAT ProblemOne of the first problems to be proved NP-Complete was satisfiability (SAT): Given a Boolean expression on n variables, can we assign values such that the expression is TRUE? Ex: (( x 1  x 2 )   ((  x 1  x 3 )  x 4 ))  x 2 Cook’s Theorem: The satisfiability problem is NP-Complete Note: Argue from first principles, not reduction Proof: not here
• 12.
  Conjunctive Normal FormEven if the form of the Boolean expression is simplified, the problem may be NP-Complete Literal : an occurrence of a Boolean or its negation A Boolean formula is in conjunctive normal form , or CNF , if it is an AND of clauses, each of which is an OR of literals Ex: (x 1   x 2 )  (  x 1  x 3  x 4 )  (  x 5 ) 3-CNF : each clause has exactly 3 distinct literals Ex: (x 1   x 2   x 3 )  (  x 1  x 3  x 4 )  (  x 5  x 3  x 4 ) Notice: true if at least one literal in each clause is true
• 13.
  The 3-CNF ProblemThm 36.10: Satisfiability of Boolean formulas in 3-CNF form (the 3-CNF Problem ) is NP-Complete Proof: Nope The reason we care about the 3-CNF problem is that it is relatively easy to reduce to others Thus by proving 3-CNF NP-Complete we can prove many seemingly unrelated problems NP-Complete
• 14.
  3-CNF  Clique What is a clique of a graph G? A: a subset of vertices fully connected to each other, i.e. a complete subgraph of G The clique problem : how large is the maximum-size clique in a graph? Can we turn this into a decision problem? A: Yes, we call this the k-clique problem Is the k-clique problem within NP ?
• 15.
  3-CNF  Clique What should the reduction do? A: Transform a 3-CNF formula to a graph, for which a k -clique will exist (for some k ) iff the 3-CNF formula is satisfiable
• 16.
  3-CNF  Clique The reduction: Let B = C 1  C 2  …  C k be a 3-CNF formula with k clauses, each of which has 3 distinct literals For each clause put a triple of vertices in the graph, one for each literal Put an edge between two vertices if they are in different triples and their literals are consistent , meaning not each other’s negation Run an example: B = (x   y   z)  (  x  y  z )  (x  y  z )
• 17.
  3-CNF  Clique Prove the reduction works: If B has a satisfying assignment, then each clause has at least one literal (vertex) that evaluates to 1 Picking one such “true” literal from each clause gives a set V’ of k vertices. V’ is a clique ( Why? ) If G has a clique V’ of size k, it must contain one vertex in each triple (clause) ( Why? ) We can assign 1 to each literal corresponding with a vertex in V’, without fear of contradiction
• 18.
  Clique  Vertex Cover A vertex cover for a graph G is a set of vertices incident to every edge in G The vertex cover problem : what is the minimum size vertex cover in G? Restated as a decision problem: does a vertex cover of size k exist in G? Thm 36.12: vertex cover is NP-Complete
• 19.
  Clique  Vertex Cover First, show vertex cover in NP ( How? ) Next, reduce k -clique to vertex cover The complement G C of a graph G contains exactly those edges not in G Compute G C in polynomial time G has a clique of size k iff G C has a vertex cover of size |V| - k
• 20.
  Clique  Vertex Cover Claim: If G has a clique of size k , G C has a vertex cover of size |V| - k Let V’ be the k -clique Then V - V’ is a vertex cover in G C Let ( u,v ) be any edge in G C Then u and v cannot both be in V’ ( Why? ) Thus at least one of u or v is in V-V’ ( why? ), so edge ( u , v ) is covered by V-V’ Since true for any edge in G C , V-V’ is a vertex cover
• 21.
  Clique  Vertex Cover Claim: If G C has a vertex cover V’  V, with |V’| = |V| - k , then G has a clique of size k For all u , v  V, if ( u , v )  G C then u  V’ or v  V’ or both ( Why? ) Contrapositive: if u  V’ and v  V’, then ( u , v )  E In other words, all vertices in V-V’ are connected by an edge, thus V-V’ is a clique Since |V| - |V’| = k , the size of the clique is k
• 22.
  General Comments Literallyhundreds of problems have been shown to be NP-Complete Some reductions are profound, some are comparatively easy, many are easy once the key insight is given You can expect a simple NP-Completeness proof on the final
• 23.
  Other NP-Complete ProblemsSubset-sum : Given a set of integers, does there exist a subset that adds up to some target T ? 0-1 knapsack : when weights not just integers Hamiltonian path : Obvious Graph coloring : can a given graph be colored with k colors such that no adjacent vertices are the same color? Etc…
• 24.
  The End