Understanding NP-Complete Problems

1. NP-Complete Problems

2. Running Time v.s. Input Size • Concern with problems whose complexity may be described by exponential functions. • Tractable problems v.s. intractable problems

3. Decision Problems • Optimization problems can be formulated as decision problems. • A decision problem is a question that has two possible answers, yes and no. • A problem instance is the combination of the problem and a specific input. • The instance description part defines the information expected in the input. • The question part state the actual yes-or-no question; the question contains variables defined in the instance description. • A precise statement of the input is important • Instance: an undirected graph G = (V, E). Question: Does G contain a clique of k vertices? (A clique is a complete subgraph: every pair of vertices in the subgraph has an edge between them) • Instance: an undirected graph G = (V, E) and an integer k. Question: Does G contain a clique of k vertices?

4. Example Decision Problems • Bin packing Suppose we have an unlimted number of bins each of capacity one, and n objects with sizes s1, …, sn where 0 ≤ si ≤ 1. • Optimization problem: Determine the smallest number of bins into which the objects can be packed. • Decision problem: Given, in addition to the inputs described, an integer k, do the objects fit in k bins? • Knapsack Suppose we have a knapsack of capacity C (a positive integer) and n objects with sizes s1, …, sn and “profits” p1, …, pn. • Optimization problem: Find the largest total profit of any subset of the objects that fits in the knapsack. • Decision problem: Given k, is there a subset of the objects that fits in the knapsack and has total profit at least k?

5. Example Decision Problems • Subset sum The input is a positive integer C and n objects whose sizes are positive integers s1, …, sn • Optimization problem: Among subsets of the objects with sum at most C, what is the largest subset sum? • Decision problem: Is there a subset of the objects whose sizes add up to exactly C? • Satisfiability Given a Boolean expression S involving only Boolean variables and values in conjunctive normal form (CNF) • Decision problem: Can we find an assignment of Boolean values to the variables such that S is evaluated to the value of 1? Some concepts • A Boolean variable is a variable that may be assigned the value 1 or 0. • A literal is a Boolean variable or the negation of a Boolean variable. • A clause is a sequence of literals separated by the Boolean or operator (). • A Boolean formula is in conjunctive normal form (CNF) if it consists of a sequence of clauses separated by the Boolean and operator ().

6. The Class P • Polynomially bounded An algorithm is said to be polynomially bounded if its worst-case complexity is bounded by a polynomial function of the input size. • A problem is said to be polynomially bounded if there is a polynomially bounded algorithm for it. • The class P P is the class of decision problems that are polynomially bounded. • The problems in P may not be “tractable” in practice, but those not in P are intractable.

7. The Class NP • Nondeterministic algorithm A nondeterministic algorithm has two phases and output step: • The nondeterministic “guessing” phase: guess a “proposed solution” s (called certificate) • The deterministic “verifying” phase: A deterministic (i.e., ordinary) subroutine begins execution. In addition to the decision problems’ input, the subroutine may use s. Eventually it returns a value true or false  or it may get in an infinite loop and never halt. (Think of the verifying phase as checking s to see if it is a solution for the decision problem’s input, i.e., if it justifies a yes answer for the decision problem’s input.) • The output step. If the verifying phase returned true, the algorithm outputs yes. Otherwise, there is no output. • A nondeterministic algorithm is said to be polynomially bounded if there is a polynomial p such that for each input of size n for which the answer is yes, there is some execution of the algorithm that produces a yes output in at most p(n) time. • NP is the class of decision problems for which there is a polynomially bounded nondeterministic algorithm.

8. 1 4 2 5 3 The Class NP • Nondeterministic graph coloring Given an undirected graph G = (V, E) with n vertices and an integer k, is G colorable with k colors such that each edge connecting two vertices with different colors? • Guess a “proposed solution” s • The deterministic “verifying” phase • Check that each vertex is assigned a color • Totally there are at most k different colors • Scan the list of edges in the graph and check that the two vertices incident upon one edge have different colors. • The output step. If the verifying phase returned true, the algorithm outputs yes. Otherwise, there is no output.

9. The Class NP • Nondeterministic vertex covering Given an undirected graph G = (V, E) with n vertices and an integer k, is there a subset S of k vertices in G such that for each edge (u, v), either uS or vS? • Generate all subsets of k vertices of G; • Non-deterministically select a subset S of k vertices from G; • For every vertex v in G, check whether v is adjacent to someone in S. • P  NP • NP = P ? P  NP NP P

10. NP-Hard/NP-Complete • NP-complete problems are most “die-hard” problems in the class of NP. • Polynomial reductions Let T be a function from the input set for a decision problem P into the input set for a decision problem Q. T is a polynomial reduction from P to Q if all of the following hold: • T can be computed in polynomially bounded time. • For every instance x of P, the answer to x is yes if and only if the answer to instance T(x) of Q is yes. • Problem P is polynomially reducible to Q if there exists a polynomial reduction from P to Q. Polynomial reduction (encoding) P Q decoding

11. NP-Hard/NP-Complete encoding • Polynomial reductions are transitive. • NP-hard: A problem X is called an NP-hard problem if every problem in NP is polynomially reducible to X. encoding encoding C A B decoding decoding decoding X NP

12. NP-Hard/NP-Complete • NP-Complete: A problem X is NP-complete if the followings are true: • X is in NP (there is a non-deterministic polynomial time algorithm for X), and • X is NP-hard (for every problem in NP, there is a polynomial reduction to X). • Cook’s theorem (1971): The first NP-complete Problem The satisfiability problem (SAT) is NP-complete NP X

13. Proving NP-Completeness • Steps for proving a problem X is NP-complete. • Problem X is in NP • Find an NP-complete problem Y, and a polynomial reduction from Y to X • A good reduction R must do • Maps any instance of Y to some instances of X • The answer to Y is yes if and only if the answer to X is yes. • R should take polynomial time deterministically. R X Y

14. Examples • The clique problem Instance: an undirected graph G = (V, E) and an integer k. Question: Does G contain a clique of k vertices (A clique is a complete subgraph: every pair of vertices in the subgraph has an edge between them)? • Show the clique problem is NP-complete • Proof • Find a non-deterministic polynomial algorithm for the clique problem • Find a problem which is NP-complete (SAT), and a polynomial reduction. SAT: Instance: Given a CNF F = C1C2…Ck, where Ci = xi1xi2…xiki Question: Is F satisfiable? The reduction: Create a graph G = (V, E), V = {vij | xij  Ci} • The vertices from the same clause are not connected by edges • vij and vlk are not connected if • Connect everything else The reduction must make sense: There is a k-clique in G if and only if SAT is satisfiable.

15. Examples • Proof (cont.) The reduction must make sense: There is a k-clique in G if and only if SAT is satisfiable. “” If there is a k-clique in G, then F is satisfiable. For each vertex vij in the k-clique, let its corresponding literal xij or be 1 (true). And set all other boolean variable to be 0 (false). • No conflict assignment • For each clause, there are at most one literal in the k-clique • There are k clauses in F “” If F is satisfiable, then there is a k-clique. For each clause Ci = xi1xi2…xiki, if xij = 1, then vij is taken. • vij corresponding xijis taken, vlkcorresponding will not be taken, where