NP-complete Languages
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Polynomial Time Reductions
Polynomial Computable function f :
There is a deterministic Turing machine M
such that for any string w computes f (w)
k
O
(|
w
|
)
in polynomial time:
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Observation:
the length of f (w) is bounded
| f (w ) | O (|w |k )
since, M cannot use
more than O (| w |k ) tape space
in time O (| w |k )
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Definition:
Language A
is polynomial time reducible to
language B
if there is a polynomial computable
function f such that:
w  A  f ( w)  B
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Theorem:
Suppose that A is polynomial reducible to
If B  P then A P .
B.
Proof:
Let
M
be the machine that decides B
in polynomial time
Machine M  to decide
On input string
w:
A
in polynomial time:
1. Compute f (w)
2. Run M on input f (w)
3. If f (w )  B acccept w
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Example of a polynomial-time reduction:
We will reduce the
3CNF-satisfiability problem
to the
CLIQUE problem
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3CNF formula:
variable or its
literal complement
( x1  x2  x3 )  ( x3  x5  x6 )  ( x3  x6  x4 )  ( x4  x5  x6 )
clause
Each clause has three literals
Language:
3CNF-SAT ={
w
:
w
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is a satisfiable
3CNF formula}
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A 5-clique in graph
G
Language:
CLIQUE = { G, k  : graph G
contains a
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k-clique}
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Theorem:
Proof:
3CNF-SAT is polynomial time
reducible to CLIQUE
give a polynomial time reduction
of one problem to the other
Transform formula to graph
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Transform formula to graph.
Example:
(x 1  x 2  x 4 )  (x 1  x 2  x 4 )  (x 1  x 2  x 3 )  (x 2  x 3  x 4 )
Clause 2
Create Nodes:
Clause 1
x1
x2
x4
Clause 3
x1
x1
x2
x2
x4
Clause 4
x2
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x3
x3
x4
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(x 1  x 2  x 4 )  (x 1  x 2  x 4 )  (x 1  x 2  x 3 )  (x 2  x 3  x 4 )
x1
x2
x4
x1
x1
x2
x2
x4
x3
x2
x3
x4
Add link from a literal  to a literal in every
other clause, except the complement 
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(x 1  x 2  x 4 )  (x 1  x 2  x 4 )  (x 1  x 2  x 3 )  (x 2  x 3  x 4 )
x1
x2
x4
x1
x1
x2
x2
x4
x3
x2
x3
x4
Resulting Graph
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(x 1  x 2  x 4 )  (x 1  x 2  x 4 )  (x 1  x 2  x 3 )  (x 2  x 3  x 4 )  1
x1  1
x2  0
x3  0
x4  1
x1
x2
x4
x1
x1
x2
x2
x4
x3
x2
x3
x4
The formula is satisfied if and only if
the Graph has a 4-clique
End of Proof
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NP-complete Languages
We define the class of NP-complete
languages
Decidable
NP
NP-complete
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A language L is NP-complete if:
• L is in NP, and
• Every language in NP
is reduced to L in polynomial time
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Observation:
If a NP-complete language
is proven to be in P then:
P  NP
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Decidable
NP
P
?
NP-complete
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An NP-complete Language
Cook-Levin Theorem:
Language SAT (satisfiability problem)
is NP-complete
Proof:
Part1: SAT is in NP
(we have proven this in previous class)
Part2: reduce all NP languages
to the SAT problem
in polynomial time
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Take an arbitrary language L  NP
We will give a polynomial reduction of L to SAT
Let M be the NonDeterministic
Turing Machine that decides L in polyn. time
For any string w we will construct
in polynomial time a Boolean expression (M,w )
such that: w  L   (M,w ) is satisfiabl e
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All computations
of M on string w
q0
qj
qi
depth
| w | n
…
…
…
…
reject
n
k
accept
accept
(deepest leaf)
reject
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Consider
an accepting
computation
q0
qj
qi
depth
…
…
…
…
reject
n
k
accept
accept
(deepest leaf)
reject
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Computation path
Sequence of
Configurations
q0
qi
qj
initial state
…
1:
2:
q0  1 2  n
  1 qi  2  n

qm
qa
nk  x :
  1  l qa  l1  n k
accept state
accept
w   1 2  n
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n
Machine M Tape
n
w
n
k
2n
k
k
Maximum working space area on tape
k
during n time steps
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1: #
2: #
Tableau of Configurations
   q0  1  2   n   
    1 qi  2  n   
#
#
……
x: # 
Accept configuration

   1  2  3

qa l1

n #
qa l1

n #
k
indentical rows
nk : #
 
n
   1  2  3
n
k

k
k
2n  3
k
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Tableau Alphabet
C  { # }  {tape alphabet}  {set of states}
 { # }  {1 ,,  r }  {q1 ,, qt }
Finite size (constant)
| C | O (1)
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For every cell position
i, j
and
for every symbol in tableau alphabet
Define variable
s C
xi , j ,s
Such that if cell i , j contains symbol s
Then xi , j ,s  1
Else xi , j ,s  0
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Examples:
nk  3
1: #
2: #
   q0  1  2   n   
    1 qi  2  n   
x1,1,#  1
x2,n k 3,q  1
x1,1,  0
x2,n k 3,#  0
#
#
i
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(M,w )
is built from variables
xi , j ,s
 (M ,w )  cell   start   accept  move
When the formula is satisfied,
it describes an accepting computation
in the tableau
of machine M on input w
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cell
cell
makes sure that every
cell in the tableau contains
exactly one symbol




   xi , j ,s   xi , j ,s  xi , j ,t
all i , j  s C
 s ,t C
 s t



Every cell contains
at least one symbol






Every cell contains
at most one symbol
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Size of
cell
:


    xi , j ,s    xi , j ,s  xi , j ,t
all i , j  s C
 s ,t C
 s t


cell



(2n  3)n  ( |C |
k
k
 O (n
2k





|C |2 )
)
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start
makes sure that the tableau
starts with the initial configuration
start  x1,1,#  x1,2,    x1,n
k
1, 
 x1,n k 2,q  x1,n k 3,    x1,n k n 2,
0
1
n
 x1,n k n 3,  x1,2n k 2,    x1,2n k 2,#
Describes the initial configuration
in row 1 of tableau
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Size of
start :
start  x1,1,#  x1,2,  
 x1,n k 1,  x1,n k 2,q  x1,n k 3,  
0
 x1,2n k 2,  x1,2n k 3,#
1
2n  3  O (n )
k
k
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
makes sure that the computation
accept leads to acceptance
 accept   xi , j ,q
all i , j
all q F
Accepting states
An accept state should appear somewhere
in the tableau
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Size of
 accept :
 accept   xi , j ,q
all i , j
all q F
(2n  3)n  O(n )
k
k
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2k
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move
move
makes sure that the tableau
gives a valid sequence
of configurations
is expressed in terms of
legal windows
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Tableau
Window
a q1 b
q2 a c
2x6 area of cells
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Possible Legal windows
a  b, R
q1
b  c, L
b  a,R
q2
a q1 b
q2 a c
a q1 b
a a q2
a a q1
a a b
a b a
a b q2
Legal windows obey the transitions
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Possible illegal windows
a b a
a a a
a  b, R
q1
b  c, L
b
q2
q2
b  a,R
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a q1 b
q1 a a
q1 b
b q2
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move   (window (i, j) is legal)
all i,j
window (i,j) is legal:
j
j
j
i a q1 b i a q1 b i a a q1
q2 a c
a a q2
a a b
((is legal)  (is legal)  (is legal))

all possible legal windows
in position (i , j )
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j
i a q1 b
q2 a c
(is legal)
Formula:
xi , j ,a  xi , j 1,q1  xi , j  2,b
 xi 1, j ,q2  xi 1, j 1,a  xi 1, j  2,c
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move
Size of
:
Size of formula for a legal window
in a cell i,j: 6
Number of possible legal windows
in a cell i,j: at most |C |6
k
k
(
2
n

3
)
n
Number of possible cells:

| C | (2n  3)n  O(n )
6
k
k
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2k
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Size of (M,w ) :
 (M ,w )  cell   start   accept  move
O (n 2k )
 O (n

2k
O (n k )

O (n 2k )

O (n 2k )
)
it can also be constructed in time O (n 2k )
polynomial in
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 (M ,w )  cell   start   accept  move
we have that:
w  L  (M,w ) is satisfiabl e
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Since,
w  L  (M,w ) is satisfiabl e
and
(M,w ) is constructed
in polynomial time
L
is polynomial-time reducible to SAT
END OF PROOF
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Observation 1:
The (M,w ) formula can be converted
to CNF (conjunctive normal form) formula
in polynomial time
 (M ,w )  cell   start   accept  move
Already CNF
NOT CNF
But can be converted to CNF
using distributive laws
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Distributive Laws:
P  (Q  R )  (P  Q )  (P  R )
P  (Q  R )  (P  Q )  (P  R )
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Observation 2:
The (M,w ) formula can also
be converted to a 3CNF formula
in polynomial time
a1  a2    al 
convert
a1  a2  z1   (z1  a3  z 2 )  (z 2  a4  z 3 )    (zl 3  al 1  zl )
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From Observations 1 and 2:
CNF-SAT and
3CNF-SAT are
NP-complete languages
(they are known NP languages)
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Theorem:
If: a. Language A is NP-complete
b. Language B is in NP
c. A is polynomial time reducible to B
Then: B is NP-complete
Proof:
Any language L in NP
is polynomial time reducible to A.
Thus, L is polynomial time reducible to B
(sum of two polynomial reductions,
gives a polynomial reduction)
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Corollary: CLIQUE is NP-complete
Proof:
a. 3CNF-SAT is NP-complete
b. CLIQUE is in NP (shown in last class)
c. 3CNF-SAT is polynomial reducible to CLIQUE
(shown earlier)
Apply previous theorem with
A=3CNF-SAT
and
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B=CLIQUE
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