1. CSC 4170
Theory of Computation
The class NP
Section 7.3

2. and target node are fixed.
The HAMPATH problem
7.3.a
A Hamiltonian path in a directed graph G is a directed path that goes through each
node exactly once. We consider a special case of this problem where the start node
and target node are fixed.
HAMPATH = {<G,s,t> | G is a directed graph with a Hamiltonian path from s to t} 1 5 3
 PATH? s t
 HAMPATH? 4 2 6 1 5 3
 HAMPATH? s t 4 2 6

3. Polynomial verifiability
Does this graph have a
Hamiltonian path? s t

4. The COMPOSITES problem
COMPOSITES = {x | x=pq for some integers p,q>1}
15  COMPOSITES ?
17  COMPOSITES ?
77  COMPOSITES ?
10403  COMPOSITES ?
997,111,111,911,111,119,131,119,871  COMPOSITES ?
COMPOSITES is very easily seen to be verifiable in polynomial time.
But only a few years ago it was proven that it is also decidable in
polynomial time. The same question, however, remains open for
HAMPATH --- one of the seven $1,000,000 questions!

5. NP and NP-completeness
Definition NP is the class of languages decided by some nondeterministic polynomial time Turing machine.
The class of so called NP-complete languages is a subset of NP
consisting of the “hardest” languages from NP.
It is known that if any of the NP-complete problems has a polynomial
time deterministic algorithm, then so do all problems from NP and
thus P=NP.
In fact, a polynomial time deterministic algorithm for any NP-complete
problem can be easily/automatically “translated” into a polynomial time
deterministic algorithm for any other problem from NP.
HAMPATH is NP-complete. So are CLIQUE, SUBSET-SUM, SAT and many
other natural languages (but not COMPOSITES).

6. A k-clique in an undirected graph is a subgraph
The CLIQUE problem
7.3.e
A k-clique in an undirected graph is a subgraph
with k nodes, wherein every two nodes are
connected by an edge. On the right we see a
5-clique
CLIQUE = {<G,k> | G is an undirected graph with a k-clique}
Theorem CLIQUENP.
Proof. Here is a polynomial time NTM N deciding CLIQUE:
N = “On input <G,k>:
1. Nondeterministically select a subset c of k nodes of G.
2. Test whether G contains all edges connecting nodes in c.
3. If yes, accept; otherwise reject.”

7. The SUBSET-SUM problem
7.3.f
SUBSET-SUM = {<S,t> | S is a multiset of integers and, for some
RS, the sum of all elements of R equals t}
Theorem SUBSET-SUMNP.
Proof. Here is a polynomial time NTM N deciding SUBSET-SUM:
N = “On input <S,t>:
1. Nondeterministically select a subset c of S.
2. Test whether the elements of c sum up to t.
3. If yes, accept; otherwise reject.”

8. (1  (0  1))  (0  1) (1  (1  0))  (0  0)
Boolean formulas
7.3.g
Boolean variables x,y,… take one of the two values 0 (false) or 1 (true).
Boolean operations:  (NOT),  (AND),  (OR). We write A for A.
Boolean formulas are constructed from variables and operations in the
standard way. Once a truth assignment for variables is given, the
value of a compound formula is calculated as follows:
0 =  0 =  0 = 0
1 =  1 =  1 = 1
1  0 =  0 = 1
1  1 =  1 = 1
If x=0 and y=1, what is the value
of the following formula?
(y  (x  y))  (x  y)
(1  (0  1))  (0  1)
(1  (1  0))  (0  0)
(1  )  0
 0 1

9. of 0s and 1s to its variables that makes the formula evaluate to 1.
The SAT problem
7.3.h
We say that a Boolean formula is satisfiable iff there is an assignment
of 0s and 1s to its variables that makes the formula evaluate to 1.
Are the following formulas satisfiable?
x(xy)
x(xy)
Yes. An (in fact, the) satisfying assignment: x=0, y=1 No
SAT = {<> |  is a satisfiable Boolean formula}
SAT  P?
SAT  NP?
Nobody knows!
Yes. A satisying assignment can serve as a
membership certificate

10. P vs. NP vs. coNP vs. EXPTIME
Definitions
coNP = {L | L is the complement of some language in NP}
EXPTIME = TIME(2n1)  TIME(2n2)  TIME(2n3)  …
This is what we know:
P NP
coNP EXPTIME
NP  coNP ≠ EXPTIME
P: membership can be decided quickly
NP: membership can be verified quickly
coNP: membership can be refuted quickly
EXPTIME: none of the above
This is what we do not know:
P = NP?
NP = coNP?
P = NP  coNP?