NPC
Millennium Prize Problems
http://en.wikipedia.org/wiki/Millennium_Prize_Problems
8 NP-complete problem
Hard problems: demo
NP-hard (Non-deterministic Polynomial-time hard), in computational complexity theory, is
a class of problems that are, informally, "at least ashard as the hardest problems in NP".
Informal Definition of NP

Solutions to the problem can be verified easily (in
polynomial time).

The idea of “guessing” the right answer or checking all
possible solutions in parallel to determine which is
correct is called non-determinism.

Any problem with an algorithm that runs on a nondeterministic machine in polynomial time is called a
problem in NP.

Is Towers of Hanoi problem in NP?
NPC
2-Coloring
2-coloring (2Col): Given a graph G, can each of its
vertices be labeled with one of 2 different “colors”,
such that no two adjacent vertices have the same label?




Is 2-coloring problem in NP ?
Yes
 Determine the graph is a
Is it in P?
bipartite
graph
or
not
Yes
3-Coloring
3-coloring (3Col): Given a graph G, can each of its
vertices be labeled with one of 3 different “colors”,
such that no two adjacent vertices have the same label?
3-Coloring
3-Coloring



Is 3-coloring problem in NP ?
Yes
Is it in P?
3-Coloring



Is 3-coloring problem in NP ?
Yes
It is in NPC
4-coloring

Four color conjecture – a planar map can be
colored properly using 4 colors.
4-coloring

Four color theorem – a planar map can be
colored properly using 4 colors.
The conjecture was first proposed in 1852
 1970s, was proved with help from computer
programs

Hamilton cycle problem

Hamilton cycle


Hamilton cycle problem:


a simple cycle contains all vertices
Does a given undirected graph have a Hamilton
cycle ?
Is Hamilton cycle problem in NP?

Yes
Traveling salesman problem

Given a list of cities and the distances between
each pair of cities, what is the shortest possible
route that visits each city exactly once and
returns to the origin city?
Vertex Cover


Given a graph G, a vertex cover for G is a subset C of
the vertices such that, for every edge (u,v) of G, uC or
vC. The optimization goal is to find as small a vertex
cover for G as possible.
Vertex cover problem: Given a graph G and an
integer k, determine whether there is a vertex cover for
G containing at most k vertices.
B
B
A
D
C
D
Vertex Cover


Is vertex cover problem in NP ?
Yes
2SAT

Often expressed as 2CNF (conjunctive normal form - a
conjunction of clauses, where a clause is a disjunction
of literals and there are n different literals)
Ex:
(xy)(yz)(xz)(zy)
The 2-satisfiability problem is to find a truth
assignment to these variables that makes a
formula of this type true. It’s a decision
problem.
exercise
(x  y) (xz)
(xy)(yz)(xz)(zy)
kSAT
Maximal number of
literals per clause
4
3
2
1
P
NP-hard
2SAT is in P
Theorem: 2SAT is polynomial-time decidable.
Proof: using path searches (DFS/BFS) in
graphs…
Graph Construction

Vertex for each variable and a negation of the
variable

Edge (,) iff there exists a clause equivalent to
()
Graph Construction: Example
(xy)(yz)(xz)(zy)
More edges?
Note: xy  yx
x
y
x
y
z
z
Observation
Claim: If the graph contains a path from  to , it
also contains a path from  to .
Proof: If there’s an edge (,), then there’s also an
edge (,).
Corollary
A 2-CNF formula  is unsatisfiable iff there
exists a variable x, such that:
 there is a path from x to x in the graph
AND
 there is a path from x to x in the graph
Graph Construction: Example
(xy)(yz)(xz)(zy)(zx)
x
y
x
y
z
z
IF


Suppose there are paths x..x and x..x for
some variable x, AND there’s also a satisfying
assignment .
If (x)=T (similarly for (x)=F):
() =?
x
T
() is false!
. . .
T

x
F
F
Only IF


Suppose there are no such paths.
Construct an assignment as follows:
1. pick an
unassigned literal ,
with no path from 
to , and assign it
T
2. assign T to all
reachable vertices
x
yy
xx
y
z
zz
3. assign F to their
negations
4. Repeat until all vertices are
assigned
Only IF (2)
Claim: The algorithm is well defined.
Proof: If there were a path from x to both y
and y, then there would have been a path
from x to y and from y to x which
indicate a path from x to x.
Corollary
Claim: a 2-CNF formula  is unsatisfiable iff
there exists a variable x, such that:
 there is a path from x to x in the graph
AND
 there is a path from x to x in the graph
2SAT is in P
We get the following efficient algorithm for 2SAT:
For each variable x find if there is a path from x to
x and vice-versa.
 Reject if any of these tests succeeded.
 Accept otherwise

 2SATP.
3SAT

Is E satisfiable?

3SAT is in NP – a decision problem that can be
solved in nondeterministic polynomial time

Final review