Proof of Cook-Levin Theorem: Reducing Every NP Problem to SAT, Slides of Theory of Automata

Download Proof of Cook-Levin Theorem: Reducing Every NP Problem to SAT and more Slides Theory of Automata in PDF only on Docsity!

NP-complete problems

Polynomial-time reductions

• Language L polynomial-time reduces to L’ if

there exists a polynomial-time computable map R

that takes an instance x of L into instance y of L’

s.t.

x ∈ L if and only if y ∈ L’

L ( IS ) L’ ( CLIQUE )

x ( G , k ) y ( G’ , k’ )

x ∈ L y ∈ L ’ ( G has IS of size k ) ( G’ has clique of size k’ )

R

NP-hardness

• Language L is NP-hard if every L’ in NP reduces

to L

• Intuitively, NP-hard means “harder than all of NP”

• The Cook-Levin Theorem says

SAT is NP-hard

P

SAT NP

NP-complete

• L is NP-complete if L is NP-hard and L ∈ NP

• Intuitively, NP-complete means “hardest in

NP”

• Recall that SAT ∈ NP, so SAT is NP-complete

If SAT ∈ P, then P = NP

P

SAT NP

Proof of Cook-Levin Theorem

• All we know about L : It has a poly-time NTM M

• Let’s look at computation tableau of M on input w

M

w q0 w 1 w 2

qacc …

S- th configuration symbol at time T

S

T

Since M is nondeterministic, there may be many possible tableaus

Proof of Cook-Levin Theorem

q0 w 1 w 2

q (^) acc …

S

T

n = length of input w

height of tableau is p ( n ) for some polynomial p

width is at most p ( n )

k possible tableau symbols

x T , S , u =

true, if the ( T , S ) cell of tableau contains u false, if not

u

1 ≤ S ≤ p ( n ) 1 ≤ T ≤ p ( n ) 1 ≤ u ≤ k

Proof of Cook-Levin Theorem

• We want to construct (in time poly( n )) a formula f :

x T , S , u =

true, if the ( T , S ) cell of tableau contains u false, if not

1 ≤ S ≤ p ( n ) 1 ≤ T ≤ p ( n ) 1 ≤ u ≤ k

f ( x 1, 1, 1 , ..., xp ( n ), p ( n ), k ) =

true, if the x s represent a valid accepting tableau false, if not

Proof of Cook-Levin Theorem

q0 w 1 w 2

q (^) acc …

S

T

u

f = f cell ∧ f 0 ∧ f move ∧ f acc

f cell: (^) “Every cell contains

exactly one symbol”

“The first row is q 0 w 1 w 2 ... w (^) k ☐...☐”

f 0 :

f move : (^) “The moves between rows follow the transitions of M ”

f acc: (^) “qacc appears somewhere in the last row”

Proof of Cook-Levin Theorem

• Desired meaning

• Implementation:

f 0 : (^) “The first row is q 0 w 1 w 2 ... w (^) k ☐...☐” f acc: “qacc appears somewhere in the last row”

f 0 = x 1, 1, q0 ∧ x 1, 1, w 1 ∧ x 1, 1, w2 ∧ ... ∧ x 1, p ( n ),☐

f acc = x (^) p ( n ), 1, qacc ∨ x (^) p ( n ), 2, qacc ∨ ... ∨ x (^) p ( n ), p ( n ), qacc

Valid and invalid windows

… 6 c3a0t0 … … 0 c6a0p0 … 0

… 6 t 3 t 0 u 0 … … 0 t 6 t 0 u 0 … 0

valid window

invalid window

… 6 t 3 t 0 u 0 … … 0 t 6 t 0 q 3 … 0

valid window

… 6 t 3 q 3 u 0 … … 0 t 6 a 0 q 7 … 0

valid if δ(q 3 , u) = (q 7 , a, R)

… 6 q3t0u0 … … 0 k6t0q0 … 0

invalid window

… 6 c3a0t0 … … 0 b6a0t0 … 0

valid window

Other NP-complete problems

CLIQUE = {(G, k ): G is a graph with a clique of k vertices} IS = {(G, k ): G is a graph with an independent set of k vertices} VC = {(G, k ): G is a graph with a vertex cover of k vertices}

CLIQUE , IS and VC are NP-complete

SAT

NP

CLIQUE IS (^) VC

Proving NP-hardness

• To show L is NP-hard, it is enough to reduce

from some L’ we already know is NP-hard

• For now we can take L’ = SAT

• To show L is NP-complete,

we also need to argue that

L is in NP

• This is usually the easy part

roadmap:

SAT

3SAT

IS

CLIQUE VC

NP-hardness of 3SAT

• Theorem

• Proof: We describe a reduction R from SAT

Boolean formula f 3CNF formula f ’

f ’ is satisfiable

R

f is satisfiable

3SAT is NP-hard

Reducing SAT to 3SAT

• Example: f =( x 2 ∨( x 1 ∧ x 2 ))∧( x 1 ∧( x 1 ∨ x 2 ))

We give extra variables to every gate (“wire”)

x 3

x 6 x 7

x 8 x 9

x 4 x 5

x 10 AND

OR NOT

AND

NOT OR

AND

NOT

x 2 x 1 x 2 x 1 x 1 x 2

x 4 x 5 x 7 x 7 = x 4 ∧ x 5 T T T T T T F F T F T F T F F T F T T F F T F T F F T F F F F T

( x 4 ∨ x 5 ∨ x 7 ) ( x 4 ∨ x 5 ∨ x 7 ) ( x 4 ∨ x 5 ∨ x 7 )

( x 4 ∨ x 5 ∨ x 7 )

( x 4 ∨ x 5 ∨ x 7 ) ∧( x 4 ∨ x 5 ∨ x 7 ) ∧( x 4 ∨ x 5 ∨ x 7 ) ∧( x 4 ∨ x 5 ∨ x 7 )