IEOR 269, Spring 2010 Integer Programming and Combinatorial ...

IEOR269 notes, Prof. Hochbaum, 2010 37
The proof relies on the following two observations:
1. The algorithm A 1 can call at most O(p 1 (|I 1 |)) times the algorithm A 2 . This is true since each
call counts as one operation, and we know that A 1 performs O(p 1 (|I 1 |)) operations.
2. Each time the algorithm A 1 calls the algorithm A 2 , it gives it an instance of P 2 of size at
most O(p 1 (|I 1 |)). This is true since each bit of the created P 2 instance is either a bit of the
instance |I 1 |, or to create this bit we used at least one operation (and recall that A 1 performs
O(p 1 (|I 1 |) operations).
We conclude that the resulting algorithm for solving P 1 (and now counting all operations) performs
at most O(p 1 (|I 1 |) + p 1 (|I 1 |) ∗ p 2 ((p 1 (|I 1 |)))) operations. Since the multiplication and composition
of polynomials is still a polynomial, this is a polynomial time algorithm.
Corollary 18.8. If P 1 ∝ P 2 and P 1 ∉Pthen P 2 ∉P
18.5.2 NP-Completeness
Definition 18.9. A problem Q is said to be NP-hard if B ∝ Q ∀B ∈ NP. That is, if all problems
in NP are polynomially reducible to Q.
Definition 18.10.
A decision problem Q is said to be NP-Complete if:
1. Q ∈ NP, and
2. Q is NP-Hard.
It follows from Theorem 18.7 that if any NP-complete problem were to have a polynomial algorithm,
then all problems in NPwould. Also it follows from Corollary 18.8 that is we prove that any NPcomplete
problem has no polynomial time algorithm, then this would prove that no NP-complete
problem has a polynomial algorithm.
Note that when a decision problem is NP-Complete, it follows that its optimization problem is
NP-Hard.
Conjecture: P ≠ NP. So, we do not expect any polynomial algorithm to exist for an NP-complete
problem.
Now we will show specific examples of NP-complete problems.
SAT = { Given a boolean function in conjunctive normal 7 form (CNF), does it have a satisfying
assignment of variable? }
(X 1 ∨ X 3 ∨ ¯X 7 ) ∧ (X 15 ∨ ¯X 1 ∨ ¯X 3 ) ∧ ( ) . . . ∧ ( )
Theorem 18.11 (Cook-Levin (1970)). SAT is NP-Complete
Proof.
7 In boolean logic, a formula is in conjunctive normal form if it is a conjunction of clauses (i.e. clauses are “linked
by and”), and each clause is a disjunction of literals (i.e. literals are “linked by or”; a literal is a variable or the
negation of a variable).