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# CSE 326: Data Structures: Graphs - PowerPoint PPT Presentation

CSE 326: Data Structures: Graphs. Lecture 24: Friday, March 7 th , 2003. Today. Finish NP complete problems Course evaluation forms. NP-Complete Problems. Recall: A Hamiltonean path = a paths that goes through each node exactly once How to find one ? Try out all paths.

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### CSE 326: Data Structures: Graphs

Lecture 24:Friday, March 7th, 2003

• Finish NP complete problems

• Course evaluation forms

Recall:

• A Hamiltonean path = a paths that goes through each node exactly once

• How to find one ? Try out all paths.

• Exponential time, and nobody knows better

• Why don’t we prove that there is no better algorithm ?

• Because we don’t know how to prove it

Recall:

• A problem is in P (or PTIME) if we can solve it in time O(nk), for some k > 0

• A problem is in NP if we can check a candidate solution in P

• Hamiltonean cycle (HC) is in NP

• In fact, P  NP

• But what about NP ⊈P or NP  P ? Nobody knows.

• SAT: Given a formula in Boolean logic, e.g.

determine if there is an assignment of values to the variables that makes the formula true (=1).

• Why is it in NP?

• Cook (1971) showed the following:

• Including Hamiltonean Cycle (HC)

• In some sense, SAT is the hardest problem in NP

• We say that “SAT is NP-Hard”

• A problem that is NP-Hard and in NP is called NP-complete

Theorem

Suppose that we can solve the SAT problem in polynomial time.

Then there is a way to solve ANY NP problem in polynomial time !!!

• Proof of Cook’s theorem:

• Suppose we can solve SAT in time O(m7), where m is the size of the formula

• Let some other problem in NP: we can check a candidate solution in, say, time O(n5), where n is the size of the problem’s input

• To solve that other problem, do the following

• We have a program A that checks some candidate solution in time O(n5)

• Construct a HUGE boolean formula that represents the execution of A: its variables are the candidate solution (which we don’t know) plus all memory bits

• Then check if this formula is satisfiable (i.e. there exists some candidate solution)

Boolean expressionsize = n5 n5

Boolean expressionsize = n5 n5

SAT is NP-Complete: Proof

Programcounter

Candidatesolution (unknown)

Memory(at most n5 memory words (why ?))

Input

Time = 0

Time = 1

Time = n5

Answer (0 or 1)

HUGE boolean formula of size O(n5 n5  n5)  check satisfiability in time O((n5 n5  n5)7)

• What is special about SAT ?

• Nothing ! There are hundreds of NP-complete problems:

• Directed Hamiltonean Path (DHP)

• Vertex Cover

• Clique

• etc, etc, ...

• Proof: by reducing SAT to DHP:

• Then use transitivity to argue than we can solve any NP problem in polynomial time

• I’ll show you how to prove the lemma...

Theorem Directed Hamiltonean Path (DHP) is NP Complete

Lemma If we can solve DHP in polynomial time, thenwe can solve SAT in polynomial time

• Suppose you are given a boolean formula in conjunctive normal form:

• Construct a directed graph G s.t. it admits a Hamiltonean cycle iff the formula is satisfiable

c1

c2

c3

c1

c2

c3

Step 1: construct this subgraph

c1

c2

c3

c1

c1

c1

c2

c2

c2

c3

c3

c3

Step 2: now replicate it once for each boolean variable

a

b

c

c1

c2

c3

c1

c1

c1

c2

c2

c2

c3

c3

c3

Step 3: now add a new node for each clause c1, c2, ...

c1

a

c2

c3

b

c

c1

c2

c3

c1

c1

c1

c2

c2

c2

c3

c3

c3

Step 4: now connect the variable graphs to the clause nodes in clever wayE.g. for c2:

Right-left fora

c1

a

c2

c3

b

Left-right for c

c

c1

c2

c3

Step 5: finally, the formula is satisfiable iff there exists a Hamiltonean path !

E.g a=1, b=0, c=1

c1 (true becauseof a)

a

c1

c2

c3

c2(true becauseof c)

b

c1

c2

c3

c3(true becauseof c)

c

c1

c2

c3

c1

c2

c3

E.g a=1, b=0, c=1

c1 (true becauseof a)

a

c1

c2

c3

c2(true becauseof c)

b

c1

c2

c3

c3(true becauseof c)

c

c1

c2

c3

• Computers and Intractability: A Guide to the Theory of NP-Completeness, by Michael S. Garey and David S. Johnson

• Nobody knows whether NP  P

• Proving or disproving this will bring you instant fame!

• It is generally believed that P  NP, i.e. there are problems in NP that are not in P

• But no one has been able to show even one such problem!

• Practically all of modern complexity theory is premised on the assumption that P  NP

• A very large number of useful problems are in NP

Alan Turing(1912-1954)

EXPTIME

• All currently known algorithms for NP-complete problems run in exponential worst case time

• Finding a polynomial time algorithm for any NPC problem would mean:

• Diagram depicts relationship between P, NP, and EXPTIME (class of problems that provably require exponential time to solve)

NPC

NP

P

It is believed that

P NP  EXPTIME

• Settle for algorithms that are fast on average: Worst case still takes exponential time, but doesn’t occur very often.

But some NP-Complete problems are also average-time NP-Complete!

• Settle for fast algorithms that give near-optimal solutions: But finding even approximate solutions to some NP-Complete problems is NP-Complete!

• Just get the exponent as low as possible! Much work on exponential algorithms for Boolean satisfiability: in practice can often solve problems with 1,000+ variables

But even 2n/100 will eventual hit the exponential curve!