 # Complexity Classes: P and NP

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Complexity Classes: P and NP
CS 130 Theory of Computation HMU Textbook: Chap 10

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

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

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)

The class P solvable (decidable) recursive problems P
solvable within polynomial time

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

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)

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

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)

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

Some problems in NP Independent set Hamiltonian cycle Satisfiability
Vertex cover “Student Reps”

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

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

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

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

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

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

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

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

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

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

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)

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

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)

Reduction and NP-completeness
all other NP problems All other NP problems SAT VC HC IS SR

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