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NP-Completeness Motivation Efficiency Complexity Classes COMS21103 Motivation I This course is mostly about efficient algorithms and data structures for NP-completeness solving computational problems. (or how to prove that problems are probably hard) I Today we take a break from this and look at whether we can prove that a problem has no efficient algorithm. Ashley Montanaro [email protected] I Why? Proving that a task is impossible can be helpful information, as Department of Computer Science, University of Bristol it stops us from trying to complete it. Bristol, UK 19 November 2013 I During this lecture we’ll take an informal approach to discussing this, and computational complexity in general – see the references at the end for more detail. Ashley Montanaro Ashley Montanaro [email protected] [email protected] COMS21103: Proving problems are hard Slide 1/31 COMS21103: Proving problems are hard Slide 2/31 Efficiency Complexity classes Our first task is to define what we mean by efficiency. I For the rest of this lecture we will restrict to decision problems, i.e. problems with a yes/no answer. I We think of an algorithm as being efficient if it runs in time polynomial in the input size. I When we say “problem”, we really mean a family of problems, rather than just one instance. I That is, if the input is n bits in size, the algorithm should run in time c O(n ) for some constant c which does not depend on the input. Examples of decision problems: I Examples of polynomial-time algorithms include Dijkstra’s algorithm, I CONNECTIVITY: decide whether a graph is connected; Kruskal’s algorithm, and in fact every algorithm you have seen in this I PRIMALITY: decide whether an integer is prime; course so far. I EDIT DISTANCE: given two strings and an integer k, decide whether their edit distance is at most k. I An example of an algorithm which is not polynomial-time: testing whether an integer N is prime by trying to divide it by all integers m The set of decision problems which have algorithms with runtime between 2 and √N. polynomial in the input size is known as P. I As N is specified by O(log N) bits, this algorithm runs in time I So we think of P as the class of decision problems which can be exponential in the input size. solved efficiently. Ashley Montanaro Ashley Montanaro [email protected] [email protected] COMS21103: Proving problems are hard Slide 3/31 COMS21103: Proving problems are hard Slide 4/31 Formalities Reductions One of the most fundamental questions in computer science is to Some notes about formalising this notion (which we’ll largely ignore for the determine which problems are in P. This can be done using the notion of rest of this lecture): reductions. I Imagine we want to solve a problem 1, but only know how to solve I A decision problem can be formally identified with a language, i.e. a L another problem 2. One way to solve 1 is simply to transform it into subset 0, 1 ∗, where 0, 1 ∗ is the set of bit-strings of arbitrary L L L ⊆ { } { } 2. length. L I That is, imagine we have a polynomial-time algorithm which, given an instance of 1, transforms it into an instance of 2 such that the I Each input bit-string x such that x corresponds to an input such L L ∈ L answer on the second instance is “yes” if and only if the answer on the that the answer should be “yes”; all strings x / correspond to inputs ∈ L first instance is “yes”. such that the answer should be “no”. I Then we can use our algorithm for to solve the second instance. L2 I The notion of “algorithm” should also be defined formally, in terms of I We say that reduces to if such a transformation exists. Turing machines. However, we omit the details for this lecture. L1 L2 I If P, and reduces to , then P. L2 ∈ L1 L2 L1 ∈ Ashley Montanaro Ashley Montanaro [email protected] [email protected] COMS21103: Proving problems are hard Slide 5/31 COMS21103: Proving problems are hard Slide 6/31 Verifying solutions to problems Verifying solutions to problems The class of decision problems for which, if the answer is “yes”, we can I There are many problems which we may not know how to solve verify this in time polynomial in the input size is known as NP. efficiently, but for which, if someone claims the answer is yes, we can verify the claim efficiently. I We can think of problems in this class as a game where we are given a proof that the answer is “yes” by an all-powerful wizard, and our job I For example, consider the FACTORISATION problem: given two is to check that the proof is correct. integers n and k, does n have a prime factor less than k? I If the answer is “yes”, we should accept a correct proof; but if the ACTORISATION = I An instance of F : n 820 580 620 832 258 609, answer is “no”, we should not be fooled by any incorrect proof. k = 364 797 008. Is the answer “yes”? I It is clear that P NP, as if we can solve a problem in polynomial ⊆ I If someone claims the answer is “yes”, they can give us the prime time, we can efficiently verify a claimed solution we are given: we just factors of n. We can then easily check whether they multiply together ignore it, and solve the problem ourselves. to give n, and whether one of them is less than k. I But whether or not P = NP (aka the P vs. NP question) is the biggest I Here the answer is indeed yes. unsolved problem in computer science! Ashley Montanaro Ashley Montanaro [email protected] [email protected] COMS21103: Proving problems are hard Slide 7/31 COMS21103: Proving problems are hard Slide 8/31 More on NP NP-hardness and NP-completeness I We say that a decision problem is NP-hard if, for every problem 0 I The initials “NP” stand for Nondeterministic Polynomial (for reasons L L ∈ NP, there is a polynomial-time reduction from 0 to . beyond the scope of this lecture. ), and not Non-Polynomial. L L I So, informally, NP-hard problems are at least as hard as the hardest I Indeed, the P vs. NP question precisely asks whether all problems in problems in NP. NP have polynomial-time algorithms. I If is NP-hard, and there exists a polynomial-time algorithm for , L L I Resolving P vs. NP would win you everlasting fame (as well as $1M then P = NP. So if we can prove that is NP-hard, this is evidence L from the Clay Mathematics Institute). that there is no polynomial-time algorithm that solves it. I Although we don’t know whether P = NP, most people consider this I We say that a problem is NP-complete if is NP-hard and NP. L L L ∈ very unlikely, as it would imply that whenever we have an efficient Informally, NP-complete problems are the hardest problems in NP. algorithm to verify a “yes” solution to a decision problem, we also have an efficient algorithm to solve the problem. I It is not obvious that any NP-complete problems should exist. Ashley Montanaro Ashley Montanaro [email protected] [email protected] COMS21103: Proving problems are hard Slide 9/31 COMS21103: Proving problems are hard Slide 10/31 P and NP in pictures An NP-complete problem The CIRCUIT SAT (short for “satisfiability”) problem is defined as follows. The picture if P=NP: The picture if P=NP: I The input to the problem is a circuit (i.e. a sequence of AND, OR and 6 NOT gates connected by wires in some order). I The circuit takes some bits as input and produces a single-bit output. NP-complete I The problem is to determine whether there exists an input such that the output is 1. For example: NP P=NP 1 1 AND OR P 0 NOT AND 1 CIRCUIT SAT is in NP: if the answer is “yes”, and we are given a claimed input such that the output is 1, we can simulate the circuit to check it. Ashley Montanaro Ashley Montanaro [email protected] [email protected] COMS21103: Proving problems are hard Slide 11/31 COMS21103: Proving problems are hard Slide 12/31 An NP-complete problem More NP-complete problems Claim I Now that we know that CIRCUIT SAT is NP-complete, we can use this CIRCUIT SAT is NP-hard. to prove that other problems are also NP-complete. Proof sketch I If we have a problem NP such that CIRCUIT SAT reduces to , L ∈ L then must be NP-complete. I We need to show that, for any NP, reduces to CIRCUIT SAT. L L ∈ L I If NP, then there is a polynomial-time algorithm which checks a I This can be seen as good evidence that there is no efficient algorithm L ∈ claimed solution to when the answer is “yes”. for . L L I We can write any such algorithm as a circuit with at most polynomially many gates by “compiling” it. I The first problem for which we will prove NP-completeness in this way I If there exists a proof that the answer should be “yes”, this is called 3-SAT.
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