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
recursivesolvable(decidable)problems
Psolvable withinpolynomial 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
recursivesolvable(decidable)problems
solvable withinpolynomial time
NP
not yet proventhat this regionis empty,but it likely isn’t
P
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 Vsuch 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
B,1,B,0,
q3
B,1,B,0,
q4
B,1,B,0,
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 Vsuch 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:
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) andto 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)
+- +-+- +-a cb d
bac dc
b
(c) (a+b) (b+c+d)
K = 4 +0+1+2= 7
G
Reduction and NP-completeness
NP
SAT
VC HC
SRIS
all other NPproblems
All other NPproblems
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