NP Completeness and Polynomial Time Algorithms, Slides of Design and Analysis of Algorithms

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Lecture No. 44

NP Completeness

Polynomial Time Algorithms •^

On any inputs of size n, if worst-case running timeof algorithm is O(n

k^ ), for constant k, then it is called

polynomial time algorithm

-^

Every problem can not be solved in polynomial time

-^

There exists some problems which can not besolved by any computer, in any amount of time, e.g.Turing’s Halting Problem

-^

Such problems are called un-decidable

-^

Some problems can be solved but not in O(n

k^ )

NP Completeness

Decision & Optimization Problems

Decision problems •^

The problems which return yes or no for a given inputand a question regarding the same problem Optimization problems •^

Find a solution with best value, it can be maximum orminimum. There can be more than one solutions for it

-^

Optimization problems can be considered as decisionproblems which are easier to study.

-^

Finding a path between u and v using fewest edges inan un-weighted directed graph is O. P. but does a pathexist from u to v consisting of at most k edges is D. P.?

NP: Nondeterministic Problems

Class NP •^

Problems which are verifiable in polynomial time.

-^

Whether there exists or not any polynomial timealgorithm for solving such problems, we do not know.

-^

Can be solved by nondeterministic polynomial

-^

However if we are given a certificate of a solution, wecould verify that certificate is correct in polynomial time

-^

P = NP?

Example: Hamiltonian Cycle

-^

Given:

a directed graph G = (V, E), determine a

simple cycle that contains each vertex in V, whereeach vertex can only be visited once

-^

Certificate

:

• Sequence:



v^1

, v

, v 2

, …, v 3

n

• Generating certificates -^

Verification

:

1. (v

, vi

i+

)^



E for i = 1, …, n-

2. (v

, vn^

) 1 

E

It takes polynomial time

hamiltonian not hamiltonian

Reduction in Polynomial Time Algorithm

•^

Given two problems A, B, we say that A is reducible

to B in polynomial time (A

p^

B) if

1. There exists a function

f

that converts the input

of A to inputs of B in polynomial time

2. A(x) = YES

B(f(x)) = YES

where x is input for A and f(x) is input for B

NP Complete

-^

A problem A is

NP-complete

if

1. A



NP

2. B



p^

A for all B



NP

-^

If A satisfies only property No. 2 then B is

NP-hard

-^

No polynomial time algorithm has been discovered foran

NP-Complete

problem

-^

No one has ever proven that no polynomial timealgorithm can exist for any

NP-Complete

problem

Reduction and NP Completeness

•^

Let A and B are two problems, and also supposethat we are given–^

No polynomial time algorithm exists for problem A

If we have a polynomial reduction f from A to B

•^

Then no polynomial time algorithm exists for B

f^

Problem B

yes no

yes no

Problem A

Boolean Combinational Circuit•^

Boolean combinational elements wired together

-^

Each element takes a set of inputs and produces aset of outputs in constant number, assume binary

-^

Limit the number of outputs to 1

-^

Logic gates: NOT, AND, OR

-^

Satisfying assignment: a true assignment causingthe output to be 1.

-^

A circuit is satisfiable if it has a satisfyingassignment.

Problem Definitions: Circuit Satisfiability

Two Instances: Satisfiable and Un-satisfiable

Proof:•^

Suppose X is

any problem

in NP

•^

Construct polynomial time algorithm

F

that maps

every instance

x

in X to a circuit C =

f

( x

) such that

x

is YES

C

CIRCUIT-SAT (is satisfiable).

•^

Since X

NP, there is a polynomial time algorithm

A

which verifies X.

-^

Suppose the input length is

n

and Let

T

( n

) denote

the worst-case running time.

-^

Let

k

be the constant such that

T

( n

O

( n

k ) and the

length of the certificate is

O

( n

k ).

Lemma 2: CIRCUIT-SAT is NP Hard

•^

Represent computation of

A

as a sequence of

configurations,

c

c

c

, c i

i +

c

T (

n )

,^

each

c

can i^

be broken into various components

-^

c

is mapped to i^

c

i +

by the combinational circuit M

implementing the computer hardware.

-^

It is to be noted that A(x, y) = 1 or 0.

-^

Paste together all T(n) copies of the circuit M. Callthis as, F, the resultant algorithm

-^

Circuit Satisfiability Problem is NP-completePlease see the overall structure in the next slide

•^

Now it can be proved that:1.^

F correctly constructs reduction, i.e., C issatisfiable if and only if there exists a certificate y

, such that

A

( x

,^

y

F runs in polynomial time

(left as an assignment)

-^

Construction of C takes

O

( n

k ) steps, a step takes

polynomial time

-^

F takes polynomial time to construct C from

x

Circuit Satisfiability Problem is NP-complete

Lemma 3 •^

If X is problem such that P‘

p^

X for some P'

NPC,

then X is NP-hard. Moreover, X

NP

X

NPC.

Proof: •^

Since P’ is NPC hence for all P” in NP, we have

P”

p^

P’

•^

And P‘

p^

X

given

•^

By (1) and (2)

-^

P”

p^

P’

p^

X

P”

p^

X

hence X is NP-hard

•^

Now if X

NP

X

NPC

NP-completeness Proof Basis