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**Complexity Classes: P and NP**

CS 130 Theory of Computation HMU Textbook: Chap 10

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**Turing machines and complexity**

Time and space complexity The class P Non-determinism The class NP Reduction and NP-Completeness

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**Time and space complexity**

Running time (or time complexity) for a TM is f(n), where n is the length of the input tape f(n) maximum number of steps/transitions the TM makes before halting Could be infinite if the TM does not halt on some inputs Space complexity is the maximum number of cells on the tape used/encountered by the TM during execution

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The class P The class P describes all languages (or problems) described by a Turing Machine decider, whose time complexity is bounded by a polynomial on n Examples: Divisibility of a number by another number Recognizing palindromes Matching symbols in a string Many other more complex problems (e.g., searching, shortest paths, min-cost spanning tree)

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**The class P solvable (decidable) recursive problems P**

solvable within polynomial time

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**Extensions to the basic TM**

Multi-tape turing machine multiple tapes, input placed on first tape, other tapes filled with blanks multiple heads, moving independently Nondeterminism: allow several possible transitions, given a state and symbol Alternatives to TMs Counter machines, stack machines, etc… None of these extensions extend the capability of TMs, but may impact on time/space complexity

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**Non-deterministic Turing Machine**

An non-deterministic turing machine, or NDTM, is a tuple M = (Q, , , , q0, B, F), where Q is a set of states is the input alphabet is the tape alphabet = {B} other tape symbols : Q (Q D)* is the state transition function mapping (state, symbol) to (state, symbol, direction) possibilities; D = {,}; may be empty/undefined for some pairs q0 is the start state of M B is the blank symbol (default symbol on input tape) F Q is the set of accepting states or final states of M (if applicable)

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**Non-deterministic Turing Machine**

Difference from a regular TM: : Q (Q D)* Multiple transitions, given a state and a symbol, are now possible Impact on the turing machine as a recognizer String is acceptable as long as at least one path of transitions leads to a final state Impact on the turing machine as a decider String is acceptable as long as at least one path of transitions leaves a YES on the tape; not acceptable if all paths leave a NO on the tape

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The class NP The class NP describes all languages (or problems) described by an NDTM decider, whose time complexity is bounded by a polynomial on n Clearly P NP, but it is not yet known or proven that P NP (though many believe this is true)

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**The classes P and NP not yet proven**

that this region is empty, but it likely isn’t solvable (decidable) problems recursive NP P solvable within polynomial time

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**Some problems in NP Independent set Hamiltonian cycle Satisfiability**

Vertex cover “Student Reps”

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Independent set Given a graph G = (V,E) and an integer K, is there a subset S of K vertices in V such that no two vertices in S are connected by an edge? There is an easy brute-force method to solve this problem: for each possible subset S of V (2n such subsets): check if S contains K vertices and then check if each pair in S is connected by an edge Answer yes if there is an S that satisfies the condition, answer no if all subsets do not

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**Independent set and TMs**

A vertex subset can be represented by an n-bit string (string of 0’s and 1’s: 1 means a vertex is part of the subset) Deterministic TM solution Loop that generates each subset on the tape and then checks if a subset satisfies the condition Exponential time complexity because there are 2n subsets NDTM solution Non-deterministically write a subset on the tape, then check if the subset satisfies the condition Polynomial-time complexity because there is no exponential loop involved

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**NDTM and possibilities**

q1 q2 q3 q4 writes one of the following 3-bit strings on the tape: 000,001,010,011,100,101,110,111

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Hamiltonian cycle Given a graph G = (V,E), is there a simple cycle containing all vertices in V? Easy brute-force method to solve this problem: for each possible permutation P of V (n! possibilities): check if the appropriate edges implied by the permutation exist, forming the cycle Answer yes if there is a P that forms a cycle, answer no if all permutations do not

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**Alternative characterization of NP**

A problem is in NP if a feasible solution to the problem can be verified in polynomial time A problem is in NP if it can be solved by the following “framework”: for each possibility P: check (in polynomial time) if the possibility P satisfies the condition stated in the problem Answer yes if there is a P that satisfies the condition, answer no if all possibilities do not

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Satisfiability Given a set V of variables, and a boolean expression E over V, consisting of a conjunction of clauses of disjunctions of literals (conjunctive normal form), is there a truth assignment for V that satisfies E (E evaluates to true under the assignment)? Example: V = {a,b,c}, E = (a+b)(b+c)(c) Assignment that satisfies E: A=true, B=true, c=false Easy brute-force method to solve this problem: for each possible truth assignment A (2n possibilities): evaluate E under A Answer yes if there is an A that satisfies E, answer no if all assignments do not

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Vertex cover Given a graph G = (V,E) and an integer K, is there a subset S of K vertices in V such that every edge in E has at least one endpoint in S? There is an easy brute-force method to solve this problem: for each possible subset S of V (2n such subsets): check if S contains K vertices and then check if edges in E have an incident vertex in S Answer yes if there is an S that satisfies the condition, answer no if all subsets do not

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**Student reps Given: Question:**

A set S of all students in a university A set O of student organizations, each having members that comprise a subset of S An integer K Question: Can I find K students from S such that all organizations are represented? Exercise: Formulate a brute-force solution to this problem following the framework mentioned, thereby showing that this problem is in NP

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NP-complete problems The problems we have identified so far are “hard” in the sense that there are no known polynomial-time solutions using a regular TM but there are “easy” exponential-time solutions (or, polynomial solutions in an NDTM) Some of these problems have been shown “complete” in the sense that all problems in NP reduce to these problems

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Reduction Reduction entails converting an instance of one problem into an equivalent instance of another If a problem A reduces to a problem B, then a solution to B can be used to solve A Means that B is at least as hard as A Remember HP and HPA? Cook’s Theorem: Satisfiability (SAT) is NP-complete; all problems in NP reduce to SAT What does this mean? If someone discovers a polynomial-time solution for SAT, all other problems are solved

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Reduction Important condition: the reduction has to be carried out in polynomial-time How does one show that a problem P is NP-complete? Use a proof similar to Cook’s theorem (too hard, and too much work!) Easier option: reduce a known NP-complete problem (such as SAT) to P, so that P is NP-complete by transitivity Thousands of problems in NP have already been shown NP-complete If any one of these problems turns out to be solvable in polynomial time, it is a proof that P=NP! ($1M prize)

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**Reduction and NP-completeness**

SAT is NP-complete by Cook’s theorem Proof is beyond the scope of this course SAT reduces to Vertex Cover (VC) Convert variables and clauses to a graph and an integer such that a truth assignment corresponds to a vertex cover in the converted graph With a successful polynomial-time reduction, this shows that VC is NP-complete VC reduces to Independent Set (IS) and to Student Reps (SR) Which means IS and SR are NP-complete

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**SAT to VC V = {a,b,c,d} E = (c)(a+b)(b+c+d) - + - + - + - +**

K = = 7 G b c a b c d (b+c+d) (c) (a+b)

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**Reduction and NP-completeness**

all other NP problems All other NP problems SAT VC HC IS SR

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**Summary Turing machines model computability**

The class P: problems (languages) that can be solved in polynomial time using a TM decider The class NP: problems that can be solved in polynomial time using a NDTM (they can be solved in exponential time using a regular TM) Not yet proven whether P NP There are problems in NP that are NP-complete; i.e., all other NP problems reduce to it Saying that a problem is NP-complete is a statement of “hardness” of that problem Proving NP-completeness: reduce from a known NP-complete problem

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