08IntractabilityII

7/25/2019 08IntractabilityII

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Lecture slides by Kevin Wayne

Copyright 2005 Pearson-Addison Wesley

Copyright 2013 Kevin Wayne

http://www.cs.princeton.edu/~wayne/kleinberg-tardos

Last updated on Sep 8, 2013 6:41 AM

8. INTRACTABILITY II

! P vs. NP

! NP-complete

!

co-NP! NP-hard


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3-SATpoly-time reduces to all of

these problems (and many, many more)

2

Recap

3-SAT

DIR-HAM-CYCLEINDEPENDENT-SET

VERTEX-COVER

3-SAT

poly-

timer

educ

es

toIND

EPEN

DENT

-SET

GRAPH-3-COLOR

HAM-CYCLE

TSP

SUBSET-SUM

SCHEDULINGPLANAR-3-COLOR

SET-COVER


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SECTION 8.3


! P vs. NP

! NP-complete

!

co-NP! NP-hard


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3

Decision problems

Decision problem.

ProblemXis a set of strings.Instance s is one string.

AlgorithmAsolves problemX: A(s) =yesiff s!X.

Def. AlgorithmAruns in polynomial timeif for every string s,A(s)

terminates in at mostp(|s|)"steps", wherep(")is some polynomial.

Ex.

Problem PRIMES= { 2, 3, 5, 7, 11, 13, 17, 23, 29, 31, 37, . }.

Instance s= 592335744548702854681.

AKS algorithm PRIMES in O(|s|8) steps.

length of s


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P. Decision problems for which there is a poly-time algorithm.

Problem Description Algorithm yes no

MULTIPLE Isxa multiple ofy ? grade-school division 51, 17 51, 16

REL-PRIME Arexandyrelatively prime ? Euclid (300 BCE) 34, 39 34, 51

PRIMES Isxprime ? AKS (2002) 53 51

EDIT-DISTANCEIs the edit distance between

xandyless than 5 ?

dynamic

programming

niether

neither

acgggt

ttttta

L-SOLVEIs there a vector xthat

satisfiesAx = b ?

Gauss-Edmonds

elimination

ST-CONNIs there a path between s

and tin a graph G ?

depth-first search

(Theseus)

Definition of P

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NP

Certification algorithm intuition.

Certifier views things from "managerial" viewpoint.Certifier doesn't determine whether s!Xon its own;

rather, it checks a proposed proof tthat s!X.

Def. Algorithm C(s, t)is a certifierfor problemXif for every string s,

s!Xiff there exists a string tsuch that C(s, t) =yes.

Def. NPis the set of problems for which there exists a poly-time certifier.

C(s,t)is a poly-time algorithm.

Certificate t is of polynomial size: |t| !p(|s|)for some polynomialp(")

Remark. NPstands for nondeterministicpolynomial time.

6

"certificate" or "witness"


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COMPOSITES. Given an integer s, is scomposite?

Certificate. A nontrivial factor tof s. Such a certificate exists iff sis

composite. Moreover |t| !|s|.

Certifier. Check that 1< t < sand that sis a multiple of t.

Conclusion. COMPOSITES!NP.

Certifiers and certificates: composite

7

437,669 = 541 809

instance s 437669

certificate t 541 or 809


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3-SAT. Given a CNF formula ", is there a satisfying assignment?

Certificate. An assignment of truth values to the nboolean variables.

Certifier. Check that each clause in "has at least one true literal.

Conclusion. 3-SAT!NP.

Certifiers and certificates: 3-satisfiability

8

" = x

1 # x

2 # x

3( ) $ x1 # x2 # x3( ) $ x1 # x2 # x4( )instance s

certificate t x1= true, x2= true, x3= false, x4= false


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Certifiers and certificates: Hamilton path

HAM-PATH. Given an undirected graph G= (V,E), does there exist a simple

path Pthat visits every node?

Certificate. A permutation of the nnodes.

Certifier. Check that the permutation contains each node in Vexactly once,

and that there is an edge between each pair of adjacent nodes.

Conclusion. HAM-PATH!NP.

instance s certificate t

9


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NP. Decision problems for which there is a poly-time certifier.

Problem Description Algorithm yes no

L-SOLVEIs there a vectorxthat

satisfiesAx = b ?

Gauss-Edmondselimination

COMPOSITES Isxcomposite ? AKS (2002) 51 53

FACTORDoesxhave a nontrivial factor

less thany ?? (56159, 50) (55687, 50)

SATIs there a truth assignment that

satisfies the formula?? x1!x2x1!x2

x2x1!x2

x1!

x2

3-COLORCan the nodes of a graph Gbe

colored with 3 colors??

HAM-PATHIs there a simple path between

sand tthat visits every node??

Definition of NP

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Definition of NP

NP. Decision problems for which there is a poly-time certifier.

11

NP captures vast domains of computational, scientific, and mathematical

endeavors, and seems to roughly delimit what mathematicians and scientists

have been aspiring to compute feasibly. Christos Papadimitriou

In an ideal world it would be renamed P vs VP. Clyde Kruskal


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P, NP, and EXP

P. Decision problems for which there is a poly-time algorithm.

NP. Decision problems for which there is a poly-time certifier.EXP. Decision problems for which there is an exponential-time algorithm.

Claim. P # NP.

Pf. Consider any problemX!P.

By definition, there exists a poly-time algorithm A(s)that solvesX.

Certificate t= #, certifier C(s, t) =A(s). !

Claim. NP # EXP.

Pf. Consider any problemX!NP.

By definition, there exists a poly-time certifier C(s, t)forX.

To solve input s, run C(s, t)on all strings twith |t| !p(|s|).

Returnyesif C(s, t)returnsyesfor any of these potential certificates. !

Remark. Time-hierarchy theorem implies P ! EXP.

12


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The main question: P vs. NP

Q. How to solve an instance of 3-SATwith nvariables?

A. Exhaustive search: try all 2ntruth assignments.

Q. Can we do anything substantially more clever?

Conjecture. No poly-time algorithm for 3-SAT.

13

"intractable"


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The main question: P vs. NP

Does P = NP? [Cook 1971, Edmonds, Levin, Yablonski, Gdel]

Is the decision problem as easy as the certification problem?

If yes. Efficient algorithms for 3-SAT, TSP, 3-COLOR, FACTOR,

If no. No efficient algorithms possible for 3-SAT, TSP, 3-COLOR,

Consensus opinion. Probably no.

EXP NP

P

If P !NP If P = NP

EXP

P = NP

14


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Possible outcomes

P "NP.

15

I conjecture that there is no good algorithm for the traveling salesman

problem. My reasons are the same as for any mathematical conjecture:

(i) It is a legitimate mathematical possibility and (ii) I do not know.

Jack Edmonds 1966


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Possible outcomes

P "NP.

16

In my view, there is no way to even make intelligent guesses about the

answer to any of these questions. If I had to bet now, I would bet that

P is not equal to NP. I estimate the half-life of this problem at 2550

more years, but I wouldnt bet on it being solved before 2100.

Bob Tarjan

We seem to be missing even the most basic understanding of the

nature of its difficulty. All approaches tried so far probably (in

some cases, provably) have failed. In this sense P =NP is different

from many other major mathematical problems on which a gradual

progress was being constantly done (sometimes for centuries)

whereupon they yielded, either completely or partially.

Alexander Razborov


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Possible outcomes

P = NP.

17

P = NP. In my opinion this shouldnt really be a hard problem; its just

that we came late to this theory, and havent yet developed any

techniques for proving computations to be hard. Eventually, it will

just be a footnote in the books. John Conway


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Other possible outcomes

P= NP, but only !(n100)algorithm for 3-SAT.

P"NP, but with O(nlog*n)algorithm for 3-SAT.

P= NPis independent (of ZFC axiomatic set theory).

18

It will be solved by either 2048 or 4096. I am currently somewhat

pessimistic. The outcome will be the truly worst case scenario:

namely that someone will prove P = NP because there are only

finitely many obstructions to the opposite hypothesis; hence there

will exists a polynomial time solution to SAT but we will never

know its complexity! Donald Knuth


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Millennium prize

Millennium prize. $1 million for resolution of P = NPproblem.

19


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Looking for a job?

Some writers for the Simpsons and Futurama.

J. Steward Burns. M.S. in mathematics (Berkeley '93).

David X. Cohen. M.S. in computer science (Berkeley '92).

Al Jean. B.S. in mathematics. (Harvard '81).

Ken Keeler. Ph.D. in applied mathematics (Harvard '90).

Jeff Westbrook. Ph.D. in computer science (Princeton '89).

Copyright 2000, Twentieth Century FoxCopyright 1990, Matt Groening20


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21

Princeton CS Building, West Wall, Circa 2001


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Princeton CS Building, West Wall, Circa 2001

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SECTION 8.4


! P vs. NP

! NP-complete

! co-NP

! NP-hard


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Polynomial transformation

Def. ProblemXpolynomial (Cook) reducesto problem Yif arbitrary

instances of problemXcan be solved using:Polynomial number of standard computational steps, plus

Polynomial number of calls to oracle that solves problem Y.

Def. ProblemXpolynomial (Karp) transformsto problem Yif given any

inputxtoX, we can construct an inputysuch thatxis ayesinstance ofXiffyis ayesinstance of Y.

Note. Polynomial transformation is polynomial reduction with just one call

to oracle for Y, exactly at the end of the algorithm forX. Almost all previousreductions were of this form.

Open question. Are these two concepts the same with respect to NP?

we require |y| to be of size polynomial in |x|

we abuse notation #pand blur distinction24


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NP-complete

NP-complete. A problem Y!NPwith the property that for every

problemX!NP,X!pY.

Theorem. Suppose Y!NP-complete. Then Y!Piff P = NP.

Pf. $ If P = NP, then Y!Pbecause Y!NP.

Pf. % Suppose Y!P.

Consider any problemX!NP. SinceX!pY, we haveX!P.This implies NP# P.

We already know P# NP. Thus P = NP. !

Fundamental question. Do there exist "natural" NP-complete problems?

25


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CIRCUIT-SAT. Given a combinational circuit built from AND, OR, and NOTgates,

is there a way to set the circuit inputs so that the output is 1?

!

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1 0 ? ? ?

output

variable inputshard-coded inputs

yes: 1 0 1

Circuit satisfiability

26


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The "first" NP-complete problem

Theorem. CIRCUIT-SAT!NP-complete. [Cook 1971, Levin 1973]

27

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