 # Complexity 11-1 Complexity Andrei Bulatov NP-Completeness.

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Complexity 11-1 Complexity Andrei Bulatov NP-Completeness

Complexity 11-2 NP-Completeness Proofs To prove that a language L is NP-complete we now just have to perform two steps: 1.Show that L belongs to NP 2.Find a known NP-complete problem (language) L and show L  L 100s of problems have now been shown to be NP-complete (for an earlier survey see Garey and Johnson) Note: If we can complete Step 2 but not Step 1, then we say that L is NP-hard

Complexity 11-3 NP-Completeness of Clique Instance: A graph G and a number k. Question: Does G contain a clique of size k ? Clique Theorem Clique is NP-complete Theorem Clique is NP-complete Step 1: The problem Clique is in NP : the list of vertices in the clique is the certificate

Complexity 11-4 Step 2: To show that Clique is NP-complete we shall reduce Satisfiability to Clique Given a formula  with clauses For each literal X in clause create a graph vertex Put an edge between every pair of vertices from distinct clauses, unless they are incompatible (i.e. X and  X ) This construction can be carried out in polynomial time The resulting graph has a clique of size k if and only if  is satisfiable (assign the value true to every variable occurring in the clique)

Complexity 11-5 Example Construction X Y Z U V YY  Z  X  U k  3

Complexity 11-6 NP-Completeness of Vertex Cover Instance: A graph G = (V,E), and a natural number k. Question: Is there a set M  N, with  M  k, such that for each edge (i,j)  E, {i,j}  M   ? Vertex Cover¹ ¹Sometimes called Node Cover Step 1: The problem Vertex Cover is in NP : the list of vertices in M is the certificate

Complexity 11-7 Given a formula  with clauses and n variables Step 2: To show that Vertex Cover is NP-complete we shall reduce Satisfiability to Vertex Cover This construction can be carried out in polynomial time For each variable X create two adjacent vertices and to represent the literals X and  X For each clause of size create a complete subgraph with vertices connected to corresponding literals Set

Complexity 11-8 Example Construction X Y Z U V

Complexity 11-9 At least one of each pair must be in the cover At least vertices from each complete graph must be in the cover If the formula is satisfiable, then choose the cover by choosing each literal assigned true plus all but one vertex in each (Omit a vertex which is connected to a satisfied literal.) Conversely, if a vertex cover exists, assign each Boolean variable according to whether ( true ) or ( false ) is in M. (By the choice of k, there must be one vertex in each clique which is not in M. This vertex must be adjacent to a vertex in M, hence the clause is satisfied.)

Complexity 11-10 Alternative Reductions We have shown that Clique and Vertex Cover are both NP-complete This means it must be possible to reduce Clique to Vertex Cover and vice versa!

Complexity 11-11 NP-Completeness of 3- SAT To show that 3- Satisfiability is NP-complete we reduce Satisfiability to 3- Satisfiability C is satisfiable if and only if C is, since at least one of the literals other than Y s must be true Replace every clause with k > 3 by

Complexity 11-12 NP-Completeness of SubsetSum Instance: A sequence of positive integers and a target integer t. Question: Is there a subset T  S such that ? SubsetSum Step 1: The problem SubsetSum is in NP : the set T is the certificate

Complexity 11-13 Step 2: To show that SubsetSum is NP-complete we shall reduce Satisfiability to SubsetSum Given a formula  with clauses Choose t so that T must contain exactly one of each pair and at least one from each clause This construction can be carried out in polynomial time For each clause of length create integers For each variable X create 2 integers and

Complexity 11-14 Example Construction

Complexity 11-15 NP-Completeness of HamCircuit Instance: A graph G. Question: Does G contain a Hamilton circuit? HamCircuit Theorem HamCircuit is NP-complete Theorem HamCircuit is NP-complete Step 1: The problem HamCircuit is in NP : the Hamilton circuit is the certificate

Complexity 11-16 Step 2: To show that HamCircuit is NP-complete we shall reduce 3- Satisfiability to HamCircuit Given a formula  with clauses and variables What is to be encoded? Boolean variables a choice between two values (for each variable) consistency: all occurrences of X must have the same truth value constraints on the possible values imposed by clauses

Complexity 11-17 the choice gadget We assume that all gadgets are connected with the rest of the graph only through their endpoints, shown as full dots; there are no edges connecting other vertices of the gadget to the rest of the graph This gadget will allow the Hamiltonian circuit, approaching from above, to pick either left or right edge, thus communicating to a truth value

the consistency gadget Complexity 11-18 - This graph can be traversed by the Hamiltonian circuit in one of the two ways

Complexity 11-19 the constraint gadget - If, using the choice and consistency devices, we have made sure that each side of the triangle is traversed by the Hamilton circuit if and only if the corresponding literal is false ; then at least one literal has to be true

Complexity 11-20 Properties of the Gadgets the choice gadget can be traversed in exactly two ways the internal vertices of the consistency gadget (“exclusive or” gadget) can be traversed in exactly two ways, so that exactly one pair of the external vertices is involved any Hamilton circuit traverses at most two of the edges of a constraint gadget

Complexity 11-21 X Y Z 1 2 true false all these vertices are connected

Complexity 11-22 NP-Completeness of TSP(D) Instance: A finite set of cities, a positive integer distance, between each pair and an integer B. Question: Is there a permutation  of such that Travelling Salesperson(D) Step 1: The problem TSP(D) is in NP : a route satisfying the inequality is the certificate

Complexity 11-23 Step 2: To show that TSP(D) is NP-complete we shall reduce HamCircuit to TSP(D) Given a graph G with vertex set V and edge set E For each vertex v create a city Set if (u,v)  E and otherwise Set B = |V| Then If G has a Hamilton circuit then there is a route of weight B (the Hamilton circuit) If there is a route of weight B, then in G this route goes through edges and therefore is a Hamilton circuit