Lecture 21 NP

complete problems
Why do we care about NP

complete problems?
Because if we wish to solve the P=NP problem, we need to
deal with the hardest problems in NP.
Why do we want to solve the P=NP problem?
Because it will solve other 3000 NPC problems.
It is one of the fundamental mathematical problems:
Millennium Prize (CMI)

$1M for each problem:
P versus NP
The Hodge conjecture
The Poincaré conjecture

solved, by
Grigori Perelman
The Riemann hypothesis
Yang

Mills existence and mass gap
Navier

Stokes existence and smoothness
The Birch and Swinnerton

Dyer conjecture
Theorem
. If there is a polynomial

time algorithm for any
one NP

complete problem, then there is a polynomial

time algorithm for every problem in NP.
(Proved in previous lecture.)
Corollary
. If no polynomial

time algorithm exists for some
problem in NP, then there is no polynomial

time
algorithm for any NP

complete problem.
NP

completeness
Easy vs Hard
Here are some examples of easy (P) and hard
problems (NP

hard).
Easy Hard
2SAT 3SAT
Minimum spanning tree Traveling salesman
Shortest path Longest path
linear programming integer linear programming
Eulerian cycle Hamiltonian cycle
“First Natural” NP

hard problem
Circuit Satisfiability problem. An instance of the
problem is a Boolean circuit (using AND, OR, and
NOT gates connected by wires) that has n Boolean
inputs and a single Boolean output. The circuit has
no cycles. The
size
of the circuit is defined to be the
total number of gates and wires.
CIRCUIT

SAT = the decision problem, given a
Boolean circuit C, is it satisfiable?
CIRCUIT

SAT can be solved by simply trying all
possible 2
n
assignments, not polynomial time.
By S.A. Cook, 1971 STOC
Circuit SAT
AND
OR
AND
AND
OR
OR
AND
NOT
OR
NOT
x
1
x
2
x
3
x
4
x
5
AND
AND
NOT
NOT
AND
OR
NOT
AND
OR
AND
Find a satisfying assignment
Thm. CIRCUIT

SAT is NP

complete
Proof by hand

waving. CIRCUIT

SAT is clearly in NP. The "certificate” is a
satisfying assignment. Given this, we can easily verify in polynomial time
that the circuit outputs a 1.
We now give a polynomial

time transformation from
every
problem L in NP
to CIRCUIT

SAT. If L is in NP, then there is a verifier A(x,y) that runs in
time T(x) = O(x
k
) for some k. We now construct a single boolean circuit
M that maps one "configuration" of a machine that carries out the
computation of A(x,y) (recording such things as the memory state, program
counter, etc.) to the next "configuration".
We now hook together T(x) of these circuits together, making the inputs to
the circuit at the top the value of y, and the output the single bit that reflects
the value of A(x,y). This big circuit C(x) is satisfiable (by a value of y) if and
only if x was a "yes" instance of L.
The size of this circuit is polynomial in x, and the transformation can be
done in polynomial time. QED
4 step routine for proving NPC
Four

step routine
for proving NP

completeness of a
decision problem A:
1. Prove A is in NP by describing the polynomial

time
verifier V that verifies "yes" instances of A. What is the
certificate? How is it verified?
2. Select a problem B that you already know to be NP

complete.
3. Design a function f that maps "yes" instances of B to
"yes" instances of A, and “no” instances of B to “no”
instances of A, and justify that.
4. Show that f can be computed in polynomial time.
Proving other NPC problems
The following is the scheme of reductions we will use
to prove some problems to be NP

complete (there
are over 3000 of them
–
actually many more now):
CIRCUIT

SAT

SAT

3

CNF

SAT
/
\
CLIQUE SUBSET

SUM
/
VERTEX

COVER
SAT problem.
SAT. A
boolean formula
consists of boolean (1/0)
variables joined by connectives: AND, OR, IMPLIES,
IFF, NOT. Also, parentheses may be used.
An example,
F = ((x
1
IMPLIES x
2
) OR NOT ((NOT x
1
IFF x
3
) OR
x
4
)) AND NOT x
2
.
An
assignment
is a specification of truth values of the
various variables. For example,
(x
1
,x
2
,x
3
,x
4
)=(0,0,0,1)
is an assignment that makes F true (equal to 1). Such
an assignment is called a
satisfying
assignment. If a
formula F has a satisfying assignment, it is called
satisfiable
.
Theorem
. SAT is NP

complete
(
Given a boolean formula F, is it satisfiable?)
Proof.
First, we need to show that SAT is in NP. Clearly there is a
polynomial

time algorithm A(x,y) to verify that x is a "yes"
instance of SAT, using y as the purported satisfying assignment
for the boolean formula x.
Second, we choose CIRCUIT

SAT (we do not have other
choices at this moment) to reduce it to SAT.
Third, we design a function that maps circuits to Boolean
formulas.
For each gate, label each wire coming out of a gate with a new
variable name. Thus, for example, if an AND gate has inputs x
6
, x
7
,
and x
8
, and output x
9
, introduce a clause
x
9
IFF (x
6
AND x
7
AND x
8
)
Add a clause that just consists of x
output
, the label for output wire.
Take the AND of all these clauses.
NPC of SAT, proof continues
Clearly, C is a satisfiable circuit if and only if the formula f(C) is a
satisfiable boolean formula.
If C is satisfiable, then there exists some assignment to the input
wires that results in the output being 1. Therefore, the exists some
assignment to the variables input wires, plus the additional
variables representing the output wires, that results in the formula
f(C) evaluating to 1.
Similarly, if the formula f(C) is satisfiable, then it corresponds to the
circuit being satisfiable, and each output wire being correctly
computed in terms of the gate type.
Step 4. Clearly, f can be computed in polynomial time. We can
simply do a breadth

first search on the graph representation of
the circuit, outputting a formula for each gate.
This completes our proof that SAT is NP

complete. QED
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