l 30 nn pnp complete

8/12/2019 l 30 Nn Pnp Complete

1/58

P, NP Classes


2/58

Deliverables

Unsolvability

Satisfiability

Problem ClassificationReductions

NP-Complete and NP-HardCopyright @ gdeepak.com6/15/2012


3/58

Easy Vs. Hard

Euler Circuit vs. Hamiltonian CircuitShortest Path vs. Longest Path

Minimum Spanning Tree vs. Dominating Set

Edge Cover vs. Vertex Cover

Min Cut vs. Max Cut

2-Dim Matching vs. 3-Dim Matching

2-Colorability vs. Colourability

2-Satisfiability vs. Satisfiability

Copyright @ gdeepak.com6/15/2012


4/58

We can categorize the problem spaceinto two parts

Solvable Problems

Unsolvable problems

Solvable and Unsolvable



5/58

Halting Problem

Given a description of a program and a finite input, decide wheththe program finishes running in a finite time or will run forever,given that input

The halting problem is famous because it was one of the firstproblems proved undecidable, which means there is no computeprogram capable of correctly answering the question for all possiinputs.

Given an polynomial , Does it has an integer solution



6/58

Diagonalization

In a small town of Sundargarh All those who dont know how tdo make-up go to the only beauty parlor of Sunaina in the towand those who know how to do make-up do it themselves.

Set of all Finite State Machines that dont accept itself will go t

Finite state machine that runs them

It is similar to the powerful idea of diagonalization

6/15/2012 Copyright @ gdeepak.com


7/58


8/58


9/58

All those algorithms that can be solved polynomial time with worst case running time O(nk). So these problems are regarded as tractabK can not be a variable, but has to be fixed.

A problem that can be solved polynomially in omodel can also be solved polynomially in othmodel.

P Class of Problems



10/58

P

Copyright @ gdeepak.com

Minimum Spanning TreeShortest PathSortingMin-MaxMax FlowConvex Hull

6/15/2012


11/58

Internal Structure of P

O(nk) K is fixed

O(n3)

O(n2)

O(n)

O(logn)

O(1)



12/58

Big Picture

Unsolvable

EXPSPACE

EXPTIME

PSPACE

NP

P



13/58

NP


Vertex CoverGraph ColoringTravelling Salesman

Max-Cut in a graphSteiner TreeSubgraph isomorphismDominating Set

6/15/2012


14/58

How many

There are thousands of known NP-complete probleKnown till now. There are definitely thousands of othwhich have not been explored yet. More and more coming into the net as the researchers prove the reductof a problem into already known NP-Complete Problem.



15/58


16/58


17/58

A Non deterministic algorithm for searching an elemex in a given set of elements A[1..n]

1. j= choice(1,n)

2. if A[j]=x then write (j); Success

3. else write(0); Failure

Non determinism



18/58

Algorithm Nsort For i=1 to n do B[i]=0;

for i=1to n do

{ j= choice(1,n);

if b[j]0 then failure(); B[j] = a[i] }

for i= 1 to n-1 do

if B[i] > B[i+1] then failure;

success();

Non determinsitic sorting



19/58

It is the set of all problems that can be solvedpolynomial time with non-deterministic concept.

However this concept only exists in theory. There areimplementations what so ever.

Non-deterministic Polynomial (NP)



20/58

A decision problem is that problem which can have otwo options as its solution space i.e. yes or no.

Whether x is a prime number is also a decision problthat will have answer yes or no.

For a Travelling salesperson problem the question twhether a tour of cost < k exists is a decision problem.

Decision Problems



21/58

So for many problems it will be like guess & vesituation.

We will guess a particular solution and then verify twhether this is the solution (yes) or not ( no)

If this guessing and verifying can be done in polynomtime with the help of a non deterministic concept thwe say that the problem is in NP

Guess & verify



22/58

Non Determinism

Non deterministic machine checks all of them at once Can be done hypothetically. Means we need exponen

number of parallel processors to do the same task ideterministic way in polynomial time.

So Non deterministic machine will take 3n If you have a good guess, then it can be done in

polynomial time, otherwise total number of guesses exponential.



23/58

Satisfiability is the problem of determining if the variables given Boolean formula can be assigned in such a way asmake the formula evaluate to TRUE. Equally important idetermine whether no such assignments exist, which woimply that the function expressed by the formula is identic

FALSE for all possible variable assignments. In this latter case, we would say that the function

unsatisfiable; otherwise it is satisfiable.

To emphasize the binary nature of this problem, itfrequently referred to as Boolean or propositional satisfiabil

Satisfiability Problem



24/58

if f(x1,x2,xn) = True xi= 0 or 1 we have to find x1,x2,x3xn such that f(b1,b2,b3bn)=1 where b1= 0 or 1,b2= 0 or 1

bn = 0 or 1 2n different permutations, someone out of that can g

answer as 1 (true) guess

?f(1,0,1,0,0,1 ..,1,0) = 1 verify ( evaluate the function) ability to guess and verify in polynomial time

Satisfiability Problem



25/58

The complexity class P is contained in NP but the vversa has not been proved and proving whether p= remains the biggest research question.

Cook Formulated a question that is there any sinproblem in NP such that if we showed it to be in P, ththat would imply P=NP

COOK theorem



26/58

Given an instance x of a problem A, use a polynomtime reduction algorithm to transform it to instance yproblem B.

Run the polynomial time algorithm for B on instance y

use the answer for y as the answer for x.

reducibility



27/58

NP

These can do things in parallel as long as the resultanswers can be connected with OR and any one of this the answer.

Or in other words it can choose one of the possichoices at any point of time at the same time.



28/58

3-SAT

(x++ z) (x+ y + )

(p++)

.

. .

Mth clause


N different variables being used in m set of clauses

Whether there is a way to assign truth values todifferent variables so that whole formula is true

Total number of such possible assignments for nvariables 2n

So Total time will be 3m.2n

This was the first problem to be proved NP-Comple

6/15/2012


29/58

NP version

For every possible set of assignment to n-variables it will check th

truth value of the assignment. Either this or this or or that

The solutions can be checked with a combination of Ors and one them may be true if it has an assignment otherwise all will be fals

Theoretically can be done with 2n Parallel processors in polynom

deterministic time. Non deterministic machine will take 3m.

Well, if you are a good boy or GOD-boy and you can guess the thimagically, then you can guess in the first go and truth value can bdetermined in polynomial time. But, we are yet to find such a BO



30/58

SAT

(x) (x++ z)

(x+ )

(p+++t)

. .

.

Mth clause


N different variables being used in m set of clausesA clause can contain any number of variable whichwas fixed to 3 in 3-SATWhether there is a way to assign truth values to

different variables so that whole formula is trueTotal number of such possible assignments for nvariables 2n

6/15/2012


31/58

General SAT

SAT

3-SAT Claim is that if you can solve 3-SAT you can solve any

version.

Any three variable clause will remain as it is

(x++ z) (x++ z) (x+) (x++c) (x++ )

(p+++t) (p++) (++t)

(p+++t + u+v) (p++1) (1 ++L2)(2 +t+L3)(3 +u+v)



32/58

Solving 2-SAT

(x1 +x2) (x2+x3) (x1+x2) (x3 +x4) (x3+x5) (x4+x5) +x4).

Number of variables n = 5 and number of clauses m = 7. general, m n.)

Start by makingx1 TRUE. This makes the clause (x1 +x2) TR

accordingly, this clause may be removed. Under this assignmx2must be TRUE (i.e.,x2 must be FALSE) in order to satisfyclause (x1+x2). Makingx2 FALSE satisfies the clause (x2+as well, leaving only the shorter formula and so on

(x3 +x4) (x3+x5) (x4+x5) (x3+x4)6/15/2012 Copyright @ gdeepak.com


33/58

3-colorability SAT

Can you color a Graph in with 3 colors or Not. Let us say th3 colors to be used are Red, Yellow and Green.

For every node the logical relation has to be

(AR+ AG+ AY)

Similar Relation will be formed for every node with 3 clause


A B

C D

6/15/2012


34/58

3-colorability

6 logical relations has to be created using above rule

AR

AGG

AY

BR

BGG BY

Total 3n+6e clauses will be there in the corresponding 3-SArelation where n is the number of nodes and e is the numbeof edges



35/58

Reductions

SAT

3-SAT

Means SAT is at least as hard as 3-SAT or if I can solve 3-SAthen I can solve SAT also.

2-colorability

SAT (Useless Reduction)

Theoretically it means that if I can solve SAT, I can also solv2-colorability but that is already solved and is very easy usithe bipartite graph. So we never reduce an easy problem to hard problem.

So we try to solve a problem for many days, if it cannot besolved in a polynomial time then we try to reduce it to someNP-complete problem and the iterating like this between th

two stages.



36/58


37/58

Strongly Connected Components

Each strongly connected components can be collapsed ione. If all the complements are in different conneccomponents then 2-SAT has a true assignment.

M is NP-Complete if NP M and if M Q then Q is NComplete.

However if Q M that may not give us anything.



38/58

h


39/58

XY then YX

Independent Set problem: largest set of nodes which are not connected to each ot

Segment intersection: To find the number of lines that dont intersect with any otline.

We claim that maximum number of lines that dont intersect with each other is sas maximum independent set.


ed

c

b

a

e

c

d

a

b

6/15/2012

ll d i


40/58

All reductions are not easy

If we want to reduce independent set to segmentintersection, it will be an exponential reduction.


H d d i


41/58

How to do reduction

Experience

Feature extraction

Familiarity with both the problems

Boolean Algebra



42/58


43/58

SAT V t C


44/58

3-SAT Vertex Cover

Dotted is the vertex cover.


x y z

6/15/2012


45/58

3 SAT V t C


46/58

3-SAT Vertex Cover

It does not satisfies our equation.


x y z

6/15/2012

NP Complete


47/58

NP- Complete

It needs to be in NP

This problem is at least as hard as every other problemNP


Hamiltonian Circuit Problem


48/58

Hamiltonian Circuit Problem

ABCDEFG are the nodes

Guess: BDEFGCAB

I can check whether these are connected in this mannein polynomial time, if all the edges exist I say , yes andany edge is missing, I will say No and then make anothguess.

There are n! guesses

No body knows any better algorithm


Clique Problem


49/58

Clique Problem

Given a graph and a number k. In this graph can you fk or more nodes which are all connected to each other

Or find the max clique of a graph

If n is 3 or 4 then it is a polynomial algorithm, otherwik can go to n.


3D matching


50/58

3D matching

M F Martian

100 100 100

We can have n3 combinations of preferences

Guess 100 triples

Take your guess, for ever triple check if it is there in thpreferences then n*n3

How many guesses : exponential


Reduction


51/58

ReductionSorting Convex Hull

Sorting reduces to convex Hull

8 3 17

Reduction

8,64 3,9 17,

Solve convex hull

Solution to convex hull after again converting back will be solution to sorting

Reduction should be easy, not the bottom neck of your algorithm andweakest link of your chain

Reductions in the case of NP-complete problems should be polynomial reductions



52/58

NP-Hard


53/58

NP-Hard

Y is NP-Complete

1) Y belongs to NP2) For any X belonging to NP

X can be reduced in polynomial time to Y

If we remove the first condition then it is NP-Hard

Optimization version of the TSP where you say that find the optischedule instead of saying whether a Tour of less than K exists.

All problems in Complexity classes above NP-Complete are sometalso called NP-hard because they are more harder then any hproblem in NP.


Conclusion


54/58

So we are able to correlate one problem which is NP hard w

the another problem which is also NP hard and so we canthat if one of them can be solved in polynomial time tanother can also be solved in polynomial time. Similarly it be proved for all problems in NP.

But the problem is that we donthave any polynomial solut

for any of the problems in NP included both discussed. Aalso there is almost no hope. Only thing we can say here is tit has not been proved that P=NP and neither it has bproved that P NP so that can give us a little hope to workand on.

Conclusion


Questions Comments and Suggestion


55/58

Questions, Comments and Suggestion



56/58

Question 2


57/58

Question 2

What would prove that X is NP-hard?

(A) Showing a polynomial-time reduction from 3-Sat to X

(B) Showing a polynomial-time reduction from X to 3-Sa

(C) Either of the above

(D) None of the above


Question 3


58/58

Question 3

3-SAT is

A) P

B) NP

C) NP Complete

D) NP Hard


l 30 nn pnp complete

Documents