1. NP-completeness The Chinese University of Hong Kong Fall 2008
CSC 3130: Automata theory and formal languages
NP-completeness
Andrej Bogdanov

2. P and NP P = languages that can be decided on a TM
with polynomial running time
(problems that admit efficient algorithms)
NP =
languages decidable on a nondeterministic
TM with polynomial running time
(solutions can be checked efficiently)
decidable
We believe that NP is bigger than P,
but we are not 100% sure NP P

3. Evidence that NP is bigger than P
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}
SAT = {f: f is a satisfiable Boolean formula}
These (and many others) are in NP
Their solutions, once found, are easy to verify
But no efficient algorithms are known for any of them
The fastest known algorithms take time ≈2n

4. Equivalence of certain NP languages
We strongly suspect that problems like CLIQUE, SAT, etc. require time ≈2n to solve
We do not know how to prove this, but what we can prove is that
If any one of them is tractable (by a polynomial-
time algorithm), then all of them are tractable

5. Equivalence of certain NP languages
All these problems are as hard as one another
Moreover, they are at the “frontier” of NP
They are at least as hard as any problem in NP NP
clique
independent set
vertex-cover P
satisfiability

6. Polynomial-time reductions
What do we mean when we say, for example,
We mean that
Or, we can convert an imaginary poly-time algorithm for IS into one for CLIQUE
“CLIQUE is at least as hard as IS”
If CLIQUE has a polynomial-time algorithm,
so does IS

7. Polynomial-time reductions
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}
Theorem 1 2
If CLIQUE has a polynomial-time
algorithm, so does IS 3 4
{1, 4}, {2, 3, 4}, {1} are cliques
{1, 2}, {1, 3}, {4} are independent sets

8. Polynomial-time reductions
If CLIQUE has a polynomial-time algorithm, so does IS
Proof: Suppose CLIQUE has an algorithm A
We want to use it to solve IS
G, k
reject
if not
accept
if G has IS of size k
reject
if not
accept
if G’ has clique of size k A
G’, k’

9. Reducing IS to CLIQUE
We look for a polynomial-time transformation s.t.:
If G’ has a clique of size k’,
then G has an IS of size k R
G, k
G’, k’
If G’ does not have a clique of size k’,
then G has no IS of size k 1 2 3 4 1 2 3 4
flip all edges G G’
set k’ = k
ISs of size k
cliques of size k’

10. ✓ ✓ Reducing IS to CLIQUE R On input (G, k)
Construct G’ by flipping all edges of G
Set k’ = k
Output (G’, k’) R
G, k
G’, k’
independent sets in G
cliques in G’
If G’ has a clique of size k’,
then G has an IS of size k ✓
If G’ does not have a clique of size k’,
then G has no IS of size k ✓

11. Reduction recap
We showed that by converting an imaginary algorithm for CLIQUE into one for IS
To do this, we came up with a reduction that transforms instances of IS into ones for CLIQUE
If CLIQUE has a polynomial-time algorithm,
so does IS

12. Polynomial-time reductions
Language L polynomial-time reduces to L’ if there exists a polynomial-time computable map R that takes an instance x of L into instance y of L’ s.t.
x ∈ L if and only if y ∈ L’ L
(IS) L’
(CLIQUE) x
(G, k) R y
(G’, k’)
x ∈ L
y ∈ L’
(G has IS of size k)
(G’ has clique of size k’)

13. Meaning of poly-time reductions
Saying L reduces to L’ means L is easier than L’
In other words, if we can solve L’, then we can also solve L
Theorem
Proof
If L reduces to L’ and L’ ∈ P, then L ∈ P
acc R
algorithm for L’ x y
rej
x ∈ L
y ∈ L’
algorithm accepts

14. Notes about reductions
The direction of the reduction is very important
This makes sense, because “A is easier than B” and “B is easier than A” mean different things
However, it is possible that L reduces to L’ and L’ reduces to L
This means that L and L’ are as hard as one another
For example, IS and CLIQUE reduce to one another

15. The Cook-Levin Theorem
Every L ∈ NP reduces to SAT
SAT = {f: f is a satisfiable Boolean formula}
So every problem in NP is easier than SAT
But SAT itself is in NP, so SAT must be the “hardest problem” in NP:
If SAT ∈ P, then P = NP

16. Interpretation of Cook-Levin Theorem
Optimistic view:
Pessimistic view:
If we manage to solve SAT, then we can also solve
CLIQUE, scheduling, and almost anything
Since we do not believe P = NP, it is unlikely that
we will ever have a fast algorithm for SAT