proof of cook levin theorem (presentation by xiechuan, song and shuo)

Proof of Cook-

Levin TheoremTeam: 921S

Member: Xiechuan Liu

Song Song

Shuo Su

1CS 6212/Arora/Fall 2015Xiechuan/Song/Shuo

The Content of Theorem

• Simply: SAT is NP-complete problem

• Actually, there is an explicit CNF formula f

of length O(𝑓(𝑛)3) which is satisfiable if

and only if N accepts w. In particular, when

f(n) is a polynomial, f has polynomial

length in terms of n.

CNF f=True

(length= O(𝑓(𝑛)3) )NTM N accepts

language w

iff


The importance of Cook-Levin

Theorem• Find the first NPC problem: SAT.

• The theorem shows that every problem in

NP-complete can be reduced to a SAT.

• A breakthrough that if SAT = P, Then NP =

P.


The proof process

• 1. SAT is NP

Easily proved because any assignment can be verified

whether is satisfiable in polynomial time.


The proof process

• 2. SAT is NP-complete

Proof idea: construct a NTM N that accepts w, and reduce

this accepting process into a CNF formula.

CNF f=True

(length= O(𝑓(𝑛)3) )NTM N accepts

language w

iff


Definition of NTM N

NTM N (Q, , , , 𝑞0,𝑞𝑎𝑐𝑐𝑒𝑝𝑡,𝑞𝑟𝑒𝑗𝑒𝑐𝑡)

1. Q is the set of states.

2. is the input alphabet not containing the special blank symbol ˽ ,

3. is the tape alphabet, where ˽∈ and ⊆ ,

4. : Q × → Q × × {L,R} is the transition function,

5. 𝑞0 ∈ Q is the start state,

6. 𝑞𝑎𝑐𝑐𝑒𝑝𝑡 ∈ Q is the accept state, and

7. 𝑞𝑟𝑒𝑗𝑒𝑐𝑡 ∈ Q is the reject state, where 𝑞𝑟𝑒𝑗𝑒𝑐𝑡 ≠ 𝑞𝑎𝑐𝑐𝑒𝑝𝑡

Essence: Use the Boolean formula as the constraints to assure the

whole accepting process is a legal computation history that NTM

accepts w.


1→Y,R

A

REJECT

CB

ACCEPT

Any

wrong

symbol

0→X,R

˽ → ˽ ,R

0→X,R

L=01*0

w=01110

1 2 3 4 5 6 7 8 9 10

# state w1 w2 w3 w4 w5 w6 w7 #

1 # A 0 1 1 1 0 ˽ ˽ #

2 # B X 1 1 1 0 ˽ ˽ #

3 # B X Y 1 1 0 ˽ ˽ #

4 # B X Y Y 1 0 ˽ ˽ #

5 # B X Y Y Y 0 ˽ ˽ #

6 # C X Y Y Y X ˽ ˽ #

7 #ACC

EPTX Y Y Y X ˽ ˽ #

Constraints: cell ∧ start ∧ accept ∧ move𝑋𝑖,𝑗,𝑠 (for all 1≤ i, j ≤ 𝑁𝑘 and s ∈ Q ∪ ∪ {#}) = True iff the cell[i][j]

contains “s”

For example, 𝑋3,5,“1” = True iff the cell[3][5] contains “1”

7Xiechuan/Song/Shuo CS 6212/Arora/Fall 2015

cell

1. Every cell must contains one symbol

s ∈ Q ∪ ∪ {#}

𝑋𝑖,𝑗,𝑠

2. Every cell can only contains one symbol.

𝑠≠𝑡

(¬𝑋𝑖,𝑗,𝑠 ∨ ¬𝑋𝑖,𝑗,𝑡)

means (for example):

𝑋1,1,“#” ∨ 𝑋1,2,“0” ∨ 𝑋1,3,“1” ∨……

means (for example):

(¬𝑋1,1,“#” ∨ ¬𝑋1,1,“0” ∨ ¬𝑋1,1,“1” ) ∧

(¬𝑋1,2,“#” ∨ ¬𝑋1,2,“0” ∨ ¬𝑋1,2,“1” ) ∧

(¬𝑋1,3,“#” ∨ ¬𝑋1,3,“0” ∨ ¬𝑋1,3,“1” ) ∧….


start

1. The first row describes the initial configuration.

start = 𝑋1,1,"#" ∧ 𝑋1,2,"𝐴" ∧ 𝑋1,2,"0" ∧…

… ∧ 𝑋1,𝑛𝑘−2,"˽" ∧ 𝑋1,𝑛𝑘−1,"˽" ∧

𝑋1,𝑛𝑘−1,"#"

accept

1. The ACCEPT state is reached in the

configuration history.

accept =

1≤ i, j ≤ 𝑁𝑘

𝑋𝑖,𝑗,"𝑞𝐴𝐶𝐶𝐸𝑃𝑇"


• move

Every configuration can legally follow the

previous configuration according to the details

of the Non-deterministic Turing Machine’s

transition function.

𝑞7 𝑞8b → a,R

a q7 b

a a q8

b q7 b

b a q8

c q7 b

c a q8

move = 𝑋𝑖,𝑗−1,"𝑎" ∧ 𝑋𝑖,𝑗,"𝑞7" ∧ 𝑋𝑖,𝑗+1,"𝑏" ∧

𝑋𝑖+1,𝑗−1,"𝑎" ∧ 𝑋𝑖+1,𝑗,"𝑎" ∧ 𝑋𝑖+1,𝑗+1,"𝑞8"


cell =

1≤ i, j ≤ 𝑁𝑘

[(

s ∈ Q ∪ ∪ {#}

𝑋𝑖,𝑗,𝑠) ∧ (

𝑠≠𝑡

(¬𝑋𝑖,𝑗,𝑠 ∨ ¬𝑋𝑖,𝑗,𝑡))]

start = 𝑋1,1,s1 ∧ 𝑋1,2,s2 ∧ 𝑋1,3,𝑠3 ∧…

… ∧ 𝑋1,𝑛𝑘−2,"˽" ∧ 𝑋1,𝑛𝑘−1,"˽" ∧ 𝑋1,𝑛𝑘−1,"#"

accept =

1≤ i, j ≤ 𝑁𝑘

𝑋𝑖,𝑗,"𝑞𝐴𝐶𝐶𝐸𝑃𝑇"

move =

1≤ i, j ≤ 𝑁𝑘

(

𝑎𝑙𝑙 𝑡ℎ𝑒 𝑙𝑒𝑔𝑎𝑙𝑙𝑦 𝑤𝑖𝑛𝑑𝑜𝑤

(𝑋 ∧ 𝑋 ∧ 𝑋 ∧ 𝑋 ∧ 𝑋 ∧ 𝑋))

= cell ∧ start ∧ accept ∧ move


Summary

1. We build a CNF formula

= cell ∧ start ∧ accept ∧ move

which only contains 𝑋𝑖,𝑗,𝑠(true or false),

ANDs and ORs.

2. The whole process of NTM accepting w is

described by , if and only if the process is

correct, = True


The last step-Time Analysis

• The time we consume is the running time

of NTM, set i ≤ f(n)

• cell = O(𝑓(𝑛)2)

• start = O(f(n))

• accept = O(1)

• move = O(𝑓(𝑛)2)

So = cell ∧ start ∧ accept ∧ move

and = O(𝑓(𝑛)2)


Time Analysis

So clauses = O(𝑓(𝑛)2), there are O(𝑓(𝑛)2) of 𝑋𝑖,𝑗,𝑠( i × j), each encodeable in space

O(log f(n)), the size of is

O(log(f(n)) 𝑓(𝑛)2), which is polynomial.


proof of cook levin theorem (presentation by xiechuan, song and shuo)

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