NP-Completeness

• A language B is NP-complete iff • B ∈ NP • ∀ A ∈ NP A ≤P B

This property means B is NP hard

1

3SAT is NP-complete

2

Result Idea:

• B is known to be NP complete • Use it to prove NP-Completeness of C

IF • C is NP-hard

• B is NP-complete and • B ≤P C and

• C ∈ NP THEN

• C is NP-complete3

Result• CLIQUE is NP-hard

• 3SAT is NP-complete and • 3SAT ≤P CLIQUE and

• CLIQUE ∈ NP

THEREFORE

• CLIQUE is NP-complete4

Result• VERTEX-COVER is NP-hard

• 3SAT is NP-complete and • 3SAT ≤P VERTEX-COVER and

• VERTEX-COVER ∈ NP

THEREFORE

• VERTEX-COVER is NP-complete5

Result

IF • B is NP-complete and • B ∈ P

THEN • P = NP

6

SAT ∈ P ⇔ P = NP3SAT ∈ P ⇔ P = NP

CLIQUE ∈ P ⇔ P = NPVERTEX-COVER ∈ P ⇔ P = NP

7

SAT = { Φ | Φ is a satisfiable Boolean formula}

SAT ∈ NP

SAT is NP-complete

Cook-Levin Theorem

In other words • SAT ∈ NP • ∀ A ∈ NP A ≤P SAT

9

To prove • ∀ A ∈ NP A ≤P SAT

define polynomial time computable function f: Σ* →Σ*

∀ w ∈ Σ* w ∈ A ⇔ f(w) ∈ SAT

∀ w ∈ Σ* w ∈ A ⇔ Φ ∈ SAT10

Consider any A ∈ NP ∃ NTM N that decides A in polytime nk For any input w ∈ Σ*

valid tableau of configurations

11

There is exactly one symbol in each cell The first row is the (‘legal’) start configuration Every subsequent row is generated ‘legally’

Properties of a Valid Tableau

12

For any input w ∈ A there is an accepting (valid) tableau of configurations

13

There is exactly one symbol in each cell The first row is the (‘legal’) start configuration Every subsequent row is generated ‘legally’ One of these rows is an accepting configuration

Properties of an Accepting Tableau

14

Given N and w construct a Boolean formula that is satisfiable exactly when N has an accepting tableau on input w

15

Proof Idea

Define Boolean formula with variables xijs for

• 1 ≤ i ≤ nk • 1 ≤ j ≤ nk • s ∈ State Set ∪ Tape Alphabet ∪ Delimiter

Want following semantics: xijs is T iff cell (i, j) contains symbol s

for some valid accepting tableau

16

Constructing the formula

xijs is T iff cell (i, j) contains symbol s x12q_0 is T whereas x12⊔ is F

17

There is exactly one symbol in each cellThe first row is the (‘legal’) start configuration Every subsequent row is generated ‘legally’ One of these rows is an accepting configuration

Properties of an Accepting Tableau

18

represent valid tableau with a Boolean formula with components

•Φcell • exactly one symbol per cell

for any pair (i,j)the cell contains at least one symbol

the cell contains at most one symbol19

There is exactly one symbol in each cell The first row is the (‘legal’) start configurationEvery subsequent row is generated ‘legally’ One of these rows is an accepting configuration

Properties of an Accepting Tableau

20

represent valid tableau with a Boolean formula with components

• Φcell • exactly one symbol per cell

• Φstart • legal starting configuration

21

There is exactly one symbol in each cell The first row is the (‘legal’) start configuration Every subsequent row is generated ‘legally’One of these rows is an accepting configuration

Properties of an Accepting Tableau

22

represent valid tableau with a Boolean formula with components

•Φcell • exactly one symbol per cell

•Φstart • legal starting configuration

•Φmove • legal moves

23

represent valid tableau with a Boolean formula with components

• Φmove • legal moves

# a b q1 b c ⊔ ⊔ ⊔ ⊔ ⊔ ⊔# a q2 b c c ⊔ ⊔ ⊔ ⊔ ⊔ ⊔

# a b q1 b c ⊔ ⊔ ⊔ ⊔ ⊔ ⊔# a b a q2 c ⊔ ⊔ ⊔ ⊔ ⊔ ⊔

q2

b → c, L

q1N

transition function

b → a, Ra → b, R

24

represent valid tableau with a Boolean formula with components

• Φmove • legal moves represented by legal windows

q2

b → c, L

q1N

transition function

b → a, Ra → b, R

25

represent valid tableau with a Boolean formula with components

• Φmove • legal moves represented by legal windows

26

There is exactly one symbol in each cell The first row is the (‘legal’) start configuration Every subsequent row is generated ‘legally’ One of these rows is an accepting configuration

Properties of an Accepting Tableau

27

represent valid accepting tableau with a Boolean formula with components

•Φcell • exactly one symbol per cell

•Φstart • legal starting configuration

•Φmove • legal moves

•Φaccept • legal accepting configuration

28

represent valid accepting tableau with Boolean formula Φcell ∧ Φstart ∧ Φmove ∧ Φaccept

29

∀ w ∈ Σ* w ∈ A ⇔

there is a valid accepting tableau ⇔

constructed formula is SATISFIABLE

30

SAT is NP-complete

Corollary 7.42: 3SAT is NP-complete

HALTTM = { M, w : M is a TM that halts on input w}

HALTTM is NP-hard

In other words ∀ A ∈ NP A ≤P HALTTM

31

define polynomial time computable function f: Σ* →Σ*

∀ w ∈ Σ* w ∈ 3SAT ⇔ f(w) ∈ HALTTM

3SAT ≤P HALTTM

32

Define TM M as follows

M: On input w:

If w is not a valid 3CNF encoding then loop

If w is a valid 3CNF encoding then

check if w evaluates to True for any possible assignment to variables in w If yes then accept else loop

3SAT ≤P HALTTM

∀ w ∈ Σ* w ∈ 3SAT ⇔ M, w ∈ HALTTM33

HALTTM = { M, w : M is a TM that halts on input w}

Is HALTTM NP-complete?

34

P

NP

All languages

NP-hard

NP-complete

35