1. Cook-Levin Theorem Proof and Illustration
WSAC2010 Talk
11st, January, 2010
Cook-Levin Theorem Proof and Illustration
Kwak, Nam-ju
Applied Algorithm Laboratory
KAIST
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Contents NP
NP-complete
SAT Problem
A Characteristic of NP-complete
Cook-Levin Theorem
Proof of Cook-Levin Theorem
Illustration
Corollary: 3SAT∈NPC
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3. Applied Algorithm Lab. KAISTNP
Definition
NP is the class of languages that have polynomial time verifiers.
Definition (informal)
A verifier V for a language A is an algorithm which accepts a given input w to A if a corresponding certificate or a proof c exists.
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4. Applied Algorithm Lab. KAISTNP
Theorem
A language is in NP iff it is decided by some nondeterministic polynomial time Turing machine.
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NP-complete
Definition
A language B is NP-complete if
B∈NP, and
∀A∈NP, A≤PB.
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NP-complete
Examples
SAT, 3SAT, CLIQUE, VERTEX-COVER, HAMPATH, SUBSET-SUM, and etc.
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SAT Problem
Definition
Boolean variables: variables that can take on the values TRUE and FALSE
Boolean operations: AND, OR, and NOT
Boolean formula: an expression involving Boolean variable and operations
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SAT Problem
Definition
satisfiable: if some assignment of 0s and 1s to the variables make the formula evaluate to 1
Example of a satisfiable Boolean formula
φ=(¬x∧y)∨(x∧¬z)
x=0, y=1, and z=0
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SAT Problem
Definition (SAT Problem)
SAT = {<φ>|φ is a satisfiable Boolean formula}.
Given a Boolean formula, is it satisfiable?
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10. A Characteristic of NP-complete
Theorem
If B∈NPC and B≤PC for C∈NP, then C∈NPC.
Is there a fundamental problem for the chain of polynomial time reducibility?
P1≤PP2≤P … ≤PPn≤P(?)
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11. Applied Algorithm Lab. KAIST
Cook-Levin Theorem
Cook-Levin Theorem
SAT∈NPC
P1≤PP2≤P … ≤PPn≤PSAT
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12. Proof of Cook-Levin Theorem
What to prove is …
SAT∈NP (very clear, not treated here)
∀A∈NP, A≤PSAT
We would show that for each language A∈NP with a given input w to it, it is possible to produce a Boolean formula which is satisfiable if w∈A.
Definition
A≤PB if ∃f such that w∈A⇔f(w)∈B and f is polynomial time computable.
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13. Proof of Cook-Levin Theorem
Input
Boolean formula φ NP
input w
Let’s construct this Boolean formula!!
Pick one. A
SAT
w∈A or not?
φ∈SAT or not?
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14. Proof of Cook-Levin Theorem
Proof idea
For each language A in NP, with a given input w for A, produce a Boolean formula φ that simulates the deciding NP Turing machine for A on input w.
Check if the Boolean formula is satisfiable.
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15. Proof of Cook-Levin Theorem
w∈A? A w
NP Turing machine
given
Focus on this step.
transformation
Boolean formula φ
SAT problem
If satisfiable, w∈A; otherwise, w∉A.
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16. Proof of Cook-Levin Theorem
Proof idea (cont.)
If w∈A, there exists a series of configurations that results in the accept state, given w as the input of the Turing machine.
We would construct a Boolean formula which is satisfiable if the Turing machine accepts with the input w.
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17. Proof of Cook-Levin Theorem
FYI: Turing machine notation example
δ(q0,0)={(q1,x,R)}, δ(q1,1)={(q1,y,R)}, δ(q1, ⊔)={(qaccept, ⊔,L)}
This Turing machine accepts {01+}
With 0111 as an input,
q00111⊔
xq1111⊔
xyq111⊔
xyyq11⊔
xyyyq1⊔
xyyyqa⊔
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18. Proof of Cook-Levin Theorem
w: input
A: language
N: NP Turing machine that decides A
Assume that N decides whether w∈A in nk steps, for some constant k.
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19. Proof of Cook-Levin Theorem
Proof (cont.)
“nk×nk-cell”tableau for N on input w # q0 w1 w2 … wn ⊔
start configuration
second configuration nk
nk th configuration nk
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20. Proof of Cook-Levin Theorem
Proof (cont.)
A variable could be represented as xi,j,s.
xi,j,s: true if cell[i,j] is s; otherwise, false.
cell[i,j]: the cell located on the ith row and the jth column.
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21. Proof of Cook-Levin Theorem# q0 w1 w2 … wn ⊔ x q1 #
q0q1 w1 w2 … wn ⊔ # q0 w1 w2 … wn ⊔ x q1 # x w1 q3 … wn ⊔
Proof of Cook-Levin Theorem
Proof (cont.)
The tableau, without any restriction, can contain many invalid series of configurations.
e.g. cells containing multiple symbols, not starting with the input w and start state q0, neighbor configurations not corresponding the transition rules, not resulting in the accept state, and etc.
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22. Proof of Cook-Levin Theorem
Proof (cont.)
Produce a Boolean formula which forces the tableau to be valid and result in the accept state.
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23. Proof of Cook-Levin Theorem
Proof (cont.)
One cell can contain exactly one symbol among a state, a tape alphabet, and a #. (φcell)
The first configuration should be corresponding to input w and the start state q0. (φstart)
A configuration is derivable from the immediately previous configuration according to the transition rule of the Turing machine. (φmove)
There should exist a cell containing the accept state. (φaccept)
φ=φcell∧φstart∧φmove∧φaccept
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24. Proof of Cook-Levin Theorem
Proof (cont.)
φ=φcell∧φstart∧φmove∧φaccept
Each cell contain at least one symbol.
Each cell contain no more than one different symbol.
Here, C=Q∪Γ∪{#} where Q and Γ are the state set and tape alphabet of the Turing machine N.
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25. Proof of Cook-Levin Theorem
Proof (cont.)
φ=φcell∧φstart∧φmove∧φaccept
Each cell of the first row has a symbol corresponding to the start configuration where the input is w.
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26. Proof of Cook-Levin Theorem
Proof (cont.)
φ=φcell∧φstart∧φmove∧φaccept
At least one cell is the accept state.
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27. Proof of Cook-Levin Theorem
Proof (cont.)
φ=φcell∧φstart∧φmove∧φaccept
φmove checks whether every 2×3 window is legal according to the transition rule of the Turing machine. # … # q0 x y z q1
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28. Proof of Cook-Levin Theorem
Proof (cont.)
φ=φcell∧φstart∧φmove∧φaccept
For example,
δ(q1,a)={(q1,b,R)}, δ(q1,b)={(q2,c,L), (q2,a,R)} a q1 b q2 c a q1 b # b a
some examples of legal 2×3 windows a b a q1 b b q1 q2
some examples of illegal 2×3 windows
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29. Proof of Cook-Levin Theorem
Proof (cont.)
φ=φcell∧φstart∧φmove∧φaccept
When Si,j is the set of legal (i, j)-windows and Ui,j is the set of all the (i, j)-windows,
Si,j⊆Ui,j, |Ui,j|=|C|6 → |Si,j|≤|C|6
|S|=|ΣSi,j|≤n2k|C|6
That is, the set of legal windows is finite.
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30. Proof of Cook-Levin Theorem
Proof (cont.)
φ=φcell∧φstart∧φmove∧φaccept
That is, for all legal (i, j)-windows, which is represented as … a1 a2 a3 a4 a5 a6
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31. Proof of Cook-Levin Theorem
Proof (cont.)
φ=φcell∧φstart∧φmove∧φaccept
Now, we get φ as we wished.
If φ is satisfiable, w∈A. That is, to see if w∈A, we just need to check whether φ is satisfiable or not.
It is just a SAT problem.
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32. Proof of Cook-Levin Theorem
What to prove is … (restated)
SAT∈NP (very clear, not treated here)
∀A∈NP, A≤PSAT
Definition (restated)
A≤PB if ∃f such that w∈A⇔f(w)∈B and f is polynomial time computable.
What we’ve shown is …
w, A, and w∈A
f(w)=φ, B=SAT, and φ=f(w)∈B=SAT
w∈A⇔φ∈SAT
∴A≤PSAT
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Illustration
N=(Q, Σ, Γ, δ, q0, qaccept)
Q={q0, q1, qaccept}
Σ={0, 1}, Γ={0, 1, ⊔}
δ(q0,0)={(q1,0,R)}
δ(q1,1)={(q1,1,R)}
δ(q1, ⊔)={(qaccept, ⊔,L)}
A=L(N)={01+}={01, 011, 0111, 01111, …}
w=0111
Then, w∈A?
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34. Applied Algorithm Lab. KAIST
Illustration
N can decide in n2 steps where n=|w|. Therefore n2 is 16.
C={q0, q1, qaccept, 0, 1, ⊔, #} # q0 1 ⊔ …
16×16 tableau
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Illustration
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36. Applied Algorithm Lab. KAIST
Illustration
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37. Applied Algorithm Lab. KAIST
Illustration
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38. Illustration For φmove, find out all the legal windows. # q0 q1 q0 q1q1 q0 q1 q0 1 q1 … q1 1 1 q1 q1 1 q1 1 … … 1 q1 ⊔ 1 q1 ⊔
qacc q1 ⊔
qacc ⊔
qacc … …
No more than n2k|C|6=42*246=410 legal windows exist. (finite)
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Illustration
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40. Applied Algorithm Lab. KAIST
Illustration
Now, we get φ=φcell∧φstart∧φmove∧φaccept. ∧
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41. Illustration ∧ Solve a SAT problem with this Boolean formula, φ.
If satisfiable, w=0111∈A={01+}. (Actually, it is right.)
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42. Applied Algorithm Lab. KAIST
Corollary: 3SAT∈NPC
CNFs can be converted into CNFs with three literals per clause (as explained soon).
At first, convert φcell, φstart, φmove, and φaccept into CNF.
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Corollary: 3SAT∈NPC
φcell, φstart, and φaccept are already in CNF.
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44. Applied Algorithm Lab. KAIST
Corollary: 3SAT∈NPC
φmove is converted into CNF using the distributive laws.
To CNFs have 3 literals per clause
For each clause with less than 3 literals, do literal replication.
For each clause with more than 3 literals, split it with new variables.
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Corollary: 3SAT∈NPC
Examples
(a1∨a2)≡(a1∨a2∨a2)
(a1∨a2∨a3∨a4)≡(a1∨a2∨z)∧(¬z∨a3∨a4)
(a1∨a2∨a3∨a4∨a5)≡ (a1∨a2∨z1)∧(¬z1∨a3∨z2)∧(¬z2∨a4∨a5 )
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Corollary: 3SAT∈NPC
Hence, a SAT problem is able to converted into a equivalent 3SAT problem (in polynomial time).
That is, SAT≤P3SAT.
Since SAT∈NPC, considering the definition of NPC, 3SAT∈NPC, too.
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47. Conclusion Theorem (restated)
SAT∈NPC
3SAT∈NPC
All the NP problems are polynomial time reducible into SAT.
<Important NP-complete problems>
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Q&A
Please, ask questions, if any.
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Thank you!
Contents covered are based on Introduction to the Theory of Computation, 2nd ed. written by Michael Sipser.
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