1. More on NP-completeness
The Chinese University of Hong Kong
Fall 2011
CSCI 3130: Formal languages and automata theory
More on NP-completeness
Andrej Bogdanov

2. NP problems clique independent set vertex cover1 2 1 2 1 2 3 4 3 4 3 4
clique
independent set
vertex cover
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}

3. SAT and 3SAT SAT = {〈f〉: f is a satisfiable Boolean formula}
((x1∨x2 ) ∧ (x1∨x2)) ∨ ((x1∨(x2 ∧ x3)) ∧ x3)
3SAT = {〈f〉: f is a satisfiable 3CNF formula}
(x1∨x2∨x2 ) ∧ (x2∨x3∨x4) ∧ (x2∨x3∨x5) ∧ (x1∨x5∨x5)

4. P, NP, and reductions reduction from CLIQUE to IS (G, k) (G’, k’)
SAT IS
CLIQUE VC
3SAT
reduces to
(G, k)
(G’, k’)
G has a
k-clique
G’ has a
k’-ind set NP P
(easy to verify solution)
(easy to solve)

5. The plan for today Why? Because everything in NP reduces to SATIS
reduces to VC
CLIQUE
Because everything
in NP reduces to SAT
3SAT
SAT
NP-complete
(the Cook-Levin theorem) NP P
… so all these are
NP-complete
(easy to verify solution)
(easy to solve)

6. Reducing IS to VC Proof: We describe a reduction from IS to VC Example
(G, k)
(G’, k’)
G has an IS of size k
G’ has a VC of size k’
vertex covers
independent sets 1 2 3 4
{2, 4}, {3, 4},
{1, 2, 3}, {1, 2, 4},
{1, 3, 4}, {2, 3, 4},
{1, 2, 3, 4}
∅, {1}, {2}, {3}, {4},
{1, 2}, {1, 3}

7. Reducing IS to VC Claim S is an independent set of G if and Proof1 2 3 4
S is an independent set of G if and
only if S is a vertex cover of G ∅
{1}
{2}
{3}
{4}
{1, 2}
{1, 3}
{2, 4}
{3, 4}
{1, 2, 3}
{1, 2, 4}
{1, 3, 4}
{2, 3, 4}
{1, 2, 3, 4} IS VC
S is an independent set of G
no edge has both endpoints in S
every edge has an endpoint in S
S is a vertex cover of G

8. Reducing IS to VC ✓ R (G, k) (G’, k’) ✓ On input (G, k), R:
SAT
3SAT IS
CLIQUE VC ✓ R
(G, k)
(G’, k’) ✓
On input (G, k), R:
Output (G, n – k).
G has a VC of size n – k
G has an IS of size k

9. Reducing 3SAT to CLIQUE Proof: We give a reduction from 3SAT to CLIQUE
3SAT = {f: f is a satisfiable Boolean formula in 3CNF}
CLIQUE = {(G, k): G is a graph with a clique of k vertices}
3CNF formula f R
(G, k)
G has a clique of size k
f is satisfiable

10. Reducing 3SAT to CLIQUE Example: f =
(x1∨x1∨x2 ) ∧ (x1∨x2∨x2) ∧ (x1∨x2∨x3) x1 x2 x1 x2 x1 x2 x3
Put a vertex for every literal
Put an edge for every consistent pair

11. Reducing 3SAT to CLIQUE 3CNF formula f R (G, k) R:
On input f, where f is a 3CNF formula with m clauses
Construct the following graph G:
G has 3m vertices, divided into m groups,
one for each literal in f
If a and b are in different groups and a ≠ b,
put an edge (a, b)
Output (G, m)

12. Reducing 3SAT to CLIQUE 3CNF formula f R (G, m)
G has a clique of size m
f is satisfiable x1 x1 x1 x1 x2 x2 x2 x2 x3
f = (x1∨x1∨x2 ) ∧ (x1∨x2∨x2) ∧ (x1∨x2∨x3) T T F F F T F F T

13. Reducing 3SAT to CLIQUE 3CNF formula f R (G, m)
G has a clique of size m
f is satisfiable x1 x1 x1 x1 x2 x2 x2 x2 x3
f = (x1∨x1∨x2 ) ∧ (x1∨x2∨x2) ∧ (x1∨x2∨x3) F F T T F F T T T

14. Reducing 3SAT to CLIQUE
3CNF formula f R
(G, m)
f is satisfiable
G has a clique of size m
SAT
3SAT IS
CLIQUE VC ✓
Every satisfying assignment of f gives a clique of size m in G
Conversely, every clique of size m in G gives a consistent satisfying assignment of f. ✓

15. SAT and 3SAT SAT = {〈f〉: f is a satisfiable Boolean formula}
((x1∨x2 ) ∧ (x1∨x2)) ∨ ((x1∨(x2 ∧ x3)) ∧ x3)
3SAT = {〈f〉: f is a satisfiable 3CNF formula}
(x1∨x2∨x2 ) ∧ (x2∨x3∨x4) ∧ (x2∨x3∨x5) ∧ (x1∨x5∨x5)

16. Reducing SAT to 3SAT Example: f = (x2∨(x1∧x2 ))∧(x1∧(x1∨x2 )) x4 x5 x7
AND OR
NOT x2 x1
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
(x4∨x5∨x7)
(x4∨x5∨x7)
(x4∨x5∨x7)
∧(x4∨x5∨x7)
We add extra variables for every “wire”

17. Turning gates into 3CNFsz z z
Gj:
AND OR
NOT x y x y x x y z
z = y ∧ x x y z
z = y ∨ x x z
z = x
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
T T T T T T F F T F T T T F F F F T T T F T F F F F T F F F F T
T T F T F T F T T F F F
(x∨z)∧(x∨z)
(x∨y∨z)∧(x∨y∨z)
∧(x∨y∨z)∧(x∨y∨z)
(x∨y∨z)∧(x∨y∨z)
∧(x∨y∨z)∧(x∨y∨z)
fj:
(x∨x∨z)∧(x∨x∨z)

18. Reducing SAT to 3SAT Boolean formula f R 3CNF formula f’ R:=
On input f, where f is a boolean formula
Construct and output the following 3CNF formula f’:
f’ has extra variables xn+1, ..., xn+t
one for each gate Gj in f
For each gate Gj, construct the formula fj
Output
f’ = fn+1 ∧ fn+2 ∧ ... ∧ ft ∧ (xn+t ∨xn+t ∨xn+t )
requires that output of f is true

19. Reducing SAT to 3SAT
Boolean formula f R
3CNF formula f’ ✔
f’ is satisfiable
f is satisfiable
Every satisfying assignment of f extends uniquely to a satisfying assignment of f’
In the other direction, in every satisfying assignment of f’ the part x1, ..., xn is satisfies f