Cook-Levin Theorem

• Circuit Satisfiability– A “first” NP-complete problem

• Reduction overview

• Example reduction

Circuit Satisfiability

• 5 types of gates– Constant (T/F)– And– Or– Not– Variable

• Input size is total number of gates

• One output• Question

– Is there an assignment of truth values to the variable gates that causes the output of the circuit to be true?

G 5

G 8

G6

G 10

G 9

G7

X1 X2 X3 T

Reduction Overview

• We need to develop a reduction that can be applied to ANY problem in NP

• This leads to two key issues– How do we represent the input for an arbitrary

problem in NP?– What do we know about an arbitrary problem

in NP?

Encoding Inputs

• We assume the input is encoded as a string of 0’s and 1’s

• Example: input is an undirected graph– First specify the number of nodes in binary – Then use an adjacency matrix to represent edges

1

2

3

01010101011

Arbitrary problem in NP

• If Π belongs to NP, then there exists a verification algorithm A along with a set of certificates C

• Given any input x to Π, we construct a circuit to simulate the computation of A on x with the certificate C represented as variable gates

RCircuit representing computation of A on x with certificate C

01010101011

Input x

Example

• We will show an example with the Hamiltonian Path Problem

• Key concept– Representation of computation of A on x as a

sequence of configurations– Configuration: With a configuration and the

code for A, you should be able to complete the computation of A on x

Verification algorithm, certificate for HP

bool HPV(graph G=(V,E), path P) {int i = 1;bool answer = true;bool used[V] = false; /* initialized to all false */1. if i = V goto 6;2. if (used[P[i]] = = true) answer = false; else used[i] = true;3. if (!E(P[i],P[i+1])) answer = false;4. i++;5. goto 1;6. if (used[P[i]] = = true) answer = false;7. return answer;

}Certificate: A path which is a permutation of the V nodes in the graph

Configurations of HPV computationbool HPV(graph G=(V,E), path P) {

int i = 1;bool answer = true;bool used[V] = false; /* initialized to all

false */1. if i = V goto 6;2. if (used[P[i]] = = true) answer =

false; else used[i] = true;3. if (!E(P[i],P[i+1])) answer = false;4. i++;5. goto 1;6. if (used[P[i]] = = true) answer =

false;7. return answer;

}Certificate: A path which is a permutation

of the V nodes in the graph1

2

3

01010101011

• What information should be recorded in a configuration of this program?

• Current instruction (PC)

• value of all variables• Input x• Certificate

PC G=(V,E) Path i answer used

Configuration Description

One computation of HPVbool HPV(graph G=(V,E), path P) {

int i = 1;bool answer = true;bool used[V] = false; /* initialized to all

false */1. if i = V goto 6;2. if (used[P[i]] = = true) answer =

false; else used[i] = true;3. if (!E(P[i],P[i+1])) answer = false;4. i++;5. goto 1;6. if (used[P[i]] = = true) answer =

false;7. return answer;

}Certificate: A path which is a permutation

of the V nodes in the graph1

2

3

01010101011

001 01010101001101101100011

PC G=(V,E) Path i answer used

Path P = 1, 2, 3

010 01010101001101101100011

011 01010101001101101110011

100 01010101001101101110011

101 01010101001101110110011

001 01010101001101110110011

011 01010101001101110111011

010 01010101001101110110011

One computation of HPV cont’dbool HPV(graph G=(V,E), path P) {

int i = 1;bool answer = true;bool used[V] = false; /* initialized to all

false */1. if i = V goto 6;2. if (used[P[i]] = = true) answer =

false; else used[i] = true;3. if (!E(P[i],P[i+1])) answer = false;4. i++;5. goto 1;6. if (used[P[i]] = = true) answer =

false;7. return answer;

}Certificate: A path which is a permutation

of the V nodes in the graph1

2

3

01010101011

PC G=(V,E) Path i answer used

Path P = 1, 2, 3

100 01010101001101110111011

101 01010101001101111111011

011 01010101001101110111011

001 01010101001101111111011

110 01010101001101111111011

111 01010101001101111111011

Second computation of HPVbool HPV(graph G=(V,E), path P) {

int i = 1;bool answer = true;bool used[V] = false; /* initialized to all

false */1. if i = V goto 6;2. if (used[P[i]] = = true) answer =

false; else used[i] = true;3. if (!E(P[i],P[i+1])) answer = false;4. i++;5. goto 1;6. if (used[P[i]] = = true) answer =

false;7. return answer;

}Certificate: A path which is a permutation

of the V nodes in the graph1

2

3

01010101011

001 01010101010110101100011

PC G=(V,E) Path i answer used

Path P = 2, 3, 1

010 01010101010110101100011

011 01010101010110101101011

100 01010101010110101101011

101 01010101010110110101011

001 01010101010110110101011

011 01010101010110110101111

010 01010101010110110101011

Second computation of HPV cont’dbool HPV(graph G=(V,E), path P) {

int i = 1;bool answer = true;bool used[V] = false; /* initialized to all

false */1. if i = V goto 6;2. if (used[P[i]] = = true) answer =

false; else used[i] = true;3. if (!E(P[i],P[i+1])) answer = false;4. i++;5. goto 1;6. if (used[P[i]] = = true) answer =

false;7. return answer;

}Certificate: A path which is a permutation

of the V nodes in the graph1

2

3

01010101011

PC G=(V,E) Path i answer used

Path P = 2, 3, 1

100 01010101001101110001111

101 01010101001101111001111

011 01010101010110110101111

001 01010101001101111001111

110 01010101001101111001111

111 01010101001101111001111

Computation represented as a Boolean Circuitbool HPV(graph G=(V,E), path P) {

int i = 1;bool answer = true;bool used[V] = false; /* initialized to all

false */1. if i = V goto 6;2. if (used[P[i]] = = true) answer =

false; else used[i] = true;3. if (!E(P[i],P[i+1])) answer = false;4. i++;5. goto 1;6. if (used[P[i]] = = true) answer =

false;7. return answer;

}Certificate: A path which is a permutation

of the V nodes in the graph1

2

3

01010101011

001 01010101010110101100011

Path P = 2, 3, 1

010 01010101010110101100011

011 01010101010110101101011

GATES

GATES

GATES

GATES

111 01010101001101111001111

Output of Circuitbool HPV(graph G=(V,E), path P) {

int i = 1;bool answer = true;bool used[V] = false; /* initialized to all

false */1. if i = V goto 6;2. if (used[P[i]] = = true) answer =

false; else used[i] = true;3. if (!E(P[i],P[i+1])) answer = false;4. i++;5. goto 1;6. if (used[P[i]] = = true) answer =

false;7. return answer;

}Certificate: A path which is a permutation

of the V nodes in the graph1

2

3

01010101011

001 01010101010110101100011

Path P = 2, 3, 1

010 01010101010110101100011

GATES

GATES

011 01010101010110101101011

GATES

GATES

111 01010101001101111001111

Constant Input Gatesbool HPV(graph G=(V,E), path P) {

int i = 1;bool answer = true;bool used[V] = false; /* initialized to all

false */1. if i = V goto 6;2. if (used[P[i]] = = true) answer =

false; else used[i] = true;3. if (!E(P[i],P[i+1])) answer = false;4. i++;5. goto 1;6. if (used[P[i]] = = true) answer =

false;7. return answer;

}Certificate: A path which is a permutation

of the V nodes in the graph1

2

3

01010101011

101101011000001 01010101011

Path P = 2, 3, 1

010 01010101010110101100011

GATES

GATES

011 01010101010110101101011

GATES

GATES

111 01010101001101111001111

Variable Input Gatesbool HPV(graph G=(V,E), path P) {

int i = 1;bool answer = true;bool used[V] = false; /* initialized to all

false */1. if i = V goto 6;2. if (used[P[i]] = = true) answer =

false; else used[i] = true;3. if (!E(P[i],P[i+1])) answer = false;4. i++;5. goto 1;6. if (used[P[i]] = = true) answer =

false;7. return answer;

}Certificate: A path which is a permutation

of the V nodes in the graph1

2

3

01010101011

? ? ? ? ? ?011000001 01010101011

010 010101010 ? ? ? ? ? ?01100011

GATES

GATES

011 010101010 ? ? ? ? ? ?01101011

GATES

GATES

111 010101010 ? ? ? ? ? ? 11001111

Answer-Preserving Nature

• The HP input below has an HP if and only if there exists an assignment of PATH variables in red such that the CIRCUIT will output 1.

? ? ? ? ? ?011000001 01010101011

GATES

GATES

111 010101010 ? ? ? ? ? ? 1111

1

2

3

01010101011

Polynomial Time• The number of “rows” is

polynomial in the input size since A is assumed to be a polynomial time verification algorithm

• The number of columns is polynomial in the input size since the certificate must have polynomial size and the variables used must be polynomial in number

• GATES is essentially constant-sized as GATES is essentially independent of the input; GATES depends mostly on the verification algorithm

? ? ? ? ? ?011000001 01010101011

GATES

GATES

111 010101010 ? ? ? ? ? ? 111101010101011

1

2

3

Different sized inputs

• There is a minor issue with size

• The number of columns does change depending on the size of the input

? ? ? ? ? ?011000001 010101010110101010101112

3

? ? ? ? ? ?011000001 011101110110111011101112

3

? ? ? ? ? ? 0 11 000001

010 101 0100012

34

1 0 1 00100

? ? 0 0010 101 010001 0 1 00100

General Case

• In the general case, the input represented as a string of bits is transformed in an answer-preserving fashion into a polynomial-sized Boolean circuit.

? ? ? ? ? ?001 01010101011

GATES

GATES

010101010 ? ? ? ? ? ?11

01010101011

PC Input Cert. Vars

Based on Verification Algorithm