1 CSE 326: Data Structures: Graphs Lecture 24: Friday, March 7 th, 2003

2 Today Finish NP complete problems Course evaluation forms

3 NP-Complete Problems 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

4 P and NP Recall: A problem is in P (or PTIME) if we can solve it in time O(n k ), 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.

5 Another NP Problem 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?

6 SAT is NP-Complete 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 !!! 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 !!!

7 SAT is NP-Complete Proof of Cook’s theorem: Suppose we can solve SAT in time O(m 7 ), where m is the size of the formula Let some other problem in NP: we can check a candidate solution in, say, time O(n 5 ), where n is the size of the problem’s input

8 SAT is NP-Complete: Proof To solve that other problem, do the following We have a program A that checks some candidate solution in time O(n 5 ) 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)

9 SAT is NP-Complete: Proof Time = 0 Memory (at most n 5 memory words (why ?)) Program counter Input Candidate solution (unknown) Time = 1 Time = n 5 Boolean expression size = n 5  n 5 Answer (0 or 1) HUGE boolean formula of size O( n 5  n 5  n 5 )  check satisfiability in time O(( n 5  n 5  n 5 ) 7 )

10 The Graph of NP-Completeness What is special about SAT ? Nothing ! There are hundreds of NP-complete problems: Directed Hamiltonean Path (DHP) Vertex Cover Clique etc, etc,...

11 Directed Hamiltonean Path is NP-Complete 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, then we can solve SAT in polynomial time

12 Directed Hamiltonean Path is NP-Complete 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

13 c1c1 c2c2 c3c3 c1c1 c2c2 c3c3 Step 1: construct this subgraph

14 c1c1 c2c2 c3c3 c1c1 c2c2 c3c3 Step 2: now replicate it once for each boolean variable c1c1 c2c2 c3c3 c1c1 c2c2 c3c3 a b c

15 c1c1 c2c2 c3c3 c1c1 c2c2 c3c3 Step 3: now add a new node for each clause c 1, c 2,... c1c1 c2c2 c3c3 c1c1 c2c2 c3c3 a b c c1c1 c2c2 c3c3

16 c1c1 c2c2 c3c3 c1c1 c2c2 c3c3 Step 4: now connect the variable graphs to the clause nodes in clever way E.g. for c 2 : c1c1 c2c2 c3c3 c1c1 c2c2 c3c3 a b c c1c1 c2c2 c3c3 Right-left for  a Left-right for c

17 c1c1 c2c2 c3c3 c1c1 c2c2 c3c3 Step 5: finally, the formula is satisfiable iff there exists a Hamiltonean path ! E.g a=1, b=0, c=1 c1c1 c2c2 c3c3 c1c1 c2c2 c3c3 aa bb cc c 1 (true because of a) c 2 (true because of c) c 3 (true because of c)

18 c1c1 c2c2 c3c3 c1c1 c2c2 c3c3 E.g a=1, b=0, c=1 c1c1 c2c2 c3c3 c1c1 c2c2 c3c3 aa bb cc c 1 (true because of a) c 2 (true because of c) c 3 (true because of c)

19 A Great Book You Should Own! Computers and Intractability: A Guide to the Theory of NP-Completeness, by Michael S. Garey and David S. Johnson

20 Your Chance to Win a Turing Award: P = NP? 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 ( )

21 P, NP, and Exponential Time Problems 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) It is believed that P  NP  EXPTIME EXPTIME NP P NPC

22 Coping with NP-Completeness 1.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! 2.Settle for fast algorithms that give near-optimal solutions: But finding even approximate solutions to some NP- Complete problems is NP-Complete! 3.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 2 n/100 will eventual hit the exponential curve!