Complexity Classes: P and NP
CS 130 Theory of Computation
HMU Textbook: Chap 10
Turing machines and complexity
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Time and space complexity
The class P
Non-determinism
The class NP
Reduction and NP-Completeness
Time and space complexity
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Running time (or time complexity) for a TM
is f(n), where
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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
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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:
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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)
problems
solvable within
polynomial time
recursive
P
Extensions to the basic TM
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Multi-tape turing machine
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Nondeterminism:
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allow several possible transitions, given a state and symbol
Alternatives to TMs
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multiple tapes, input placed on first tape, other tapes filled
with blanks
multiple heads, moving independently
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
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An non-deterministic turing machine, or NDTM,
is a tuple M = (Q, , , , q0, B, F), where
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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
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Difference from a regular TM:
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Impact on the turing machine as a recognizer
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: Q    (Q    D)*
Multiple transitions, given a state and a symbol,
are now possible
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
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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
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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
solvable
(decidable)
problems
recursive
NP
solvable within
polynomial time
P
not yet proven
that this region
is empty,
but it likely isn’t
Some problems in NP
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Independent set
Hamiltonian cycle
Satisfiability
Vertex cover
“Student Reps”
Independent set
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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:
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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
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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
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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
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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
B,0,
B,1,
q1
B,0,
B,1,
q2
B,0,
B,1,
q3
q4
writes one of the following 3-bit strings on the tape:
000,001,010,011,100,101,110,111
Hamiltonian cycle
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Given a graph G = (V,E), is there a simple
cycle containing all vertices in V?
Easy brute-force method to solve this
problem:
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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
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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”:
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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
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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:
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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
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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:
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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
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Given:
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Question:
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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
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
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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
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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
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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
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Important condition: the reduction has to be carried
out in polynomial-time
How does one show that a problem P is
NP-complete?
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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
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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
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SAT is NP-complete by Cook’s theorem
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SAT reduces to Vertex Cover (VC)
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Proof is beyond the scope of this course
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)
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Which means IS and SR are NP-complete
SAT to VC
V = {a,b,c,d}
a
-
+
-
E = (c)(a+b)(b+c+d)
b
+
-
c
+
d
-
G
+
b
c
(c)
a
b
(a+b)
c
d
(b+c+d)
K=4+
0+1+2
=7
Reduction and NP-completeness
NP
all other NP
SAT
problems
VC
IS
All other NP
problems
HC
SR
Summary
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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
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Saying that a problem is NP-complete is a statement of
“hardness” of that problem
Proving NP-completeness: reduce from a known NPcomplete problem