1. Department of Computer Science & Engineering
NP-complete examples
CSCI3130 Tutorial 10
Chun-Ho Hung
Department of Computer Science & Engineering

2. Outline Review 2 problems P, NP, NPC Polynomial-time Reduction
Double-SAT
Dominating set

3. Review
P, NP, NPC
Polynomial-time Reduction

4. P P is the class of all languages that have poly-time algorithm
e.g., Shortest path on a directed graph, Sorting

5. NP NP is the class of all languages that have poly-time verifier
A verifier – a Turing Machine, V, s.t.
Given a potential x
x ∈ L  V accepts input <x, s> for some s
Solution s
V runs in polynomial time

6. NP (con’t)
While verifying a solution of a NP problem is easy (in poly-time), finding a solution could be more difficult
An 3SAT instance - Find a satisfying assignment for
Verifying
Given an assignment, just evaluate the truth value
Finding a solution?
No efficient algorithm has been discovered yet
f =
(x1∨x2 ) ∧ (x2 ∨x3 ∨x4) ∧ (x1)

7. P versus NP Every language, L in P, L is also in NP Therefore
Let Verifier = Poly-time TM that solves L
Therefore
P is contained in NP
Note: L in NP does not imply that efficient algorithm that decides L does not exist NP P

8. NPC A language C is NP-complete if: What is a reduction…? C is in NP
Every language L in NP, L poly-time reduces to C
What is a reduction…?

9. Reduction The direction of the reduction is very important
Saying “A is easier than B” and “B is easier than A” mean different things
“A (polynomially) reduces to B” means “B is not easier than A”

10. Reduction (Con’t) Consider 2 problems:
BFS on unweighted graph
Shortest path on weighted graph
Assume we have a TM, V, which solves 2)
We can reduce 1) to 2):
Given an instance of 1), convert it into an instance of 2):
Copy the graph, add weight=1 to every edge in 2)
Run this instance on V, output result
These two “yes” instances corresponds to each other

11. Poly-Time Reduction
How to show that a problem B is not easier than a problem A?
Informally, if B can be solved efficiently, we can solve A efficiently
Formally, we say A polynomially reduces to B if:
Given an instance a of problem, x
There is a polynomial time transformation to an instance of B, y = f(x)
x is a “yes” instance if and only if y is a “yes” instance

12. Poly-Time Reduction (Con’t)
Suppose A poly-time reduces to B
Then there exists a poly-time TM, R, s.t.,
Given an instance of A, x, transforms it to an instance of B, y = f(x), and
y is accepted  x is accepted

13. Poly-Time Reduction (Implication)
Suppose A reduces to B
If B is polynomial time solvable, then A is polynomial time solvable
If A is not polynomial time solvable, then B is not polynomial time solvable
Contrapositive
acc
Poly-time TM R
TM for L’ x y
rej

14. Poly-Time Reduction (Implication)
Suppose A reduces to B
Solving B cannot be easier than solving A
Suppose A is “difficult” while B is “easy”
However, by this reduction, you find a “easy” way to solve A
Consequently, if A is NPC, then B must be NPC
acc
Poly-time TM R
TM for L’ x y
rej

15. Poly-Time Reduction - P versus NP
To show P = NP, one could try to show that a NPC problem, C, can be solved in polynomial time. Why?
Every problem in NP poly-time reduces to C
If C can be solved in poly-time, so does each problem in NP
Then NP = P!!
But this is not that easy and it is counter-intuitive (to most people) too

16. P versus NP (Again)
Most believe that P ≠ NP, because intuitively searching for a solution is more difficult than verifying a solution
What does P = NP imply?
Know how to verify a solution in poly-time  Know how to find a solution in poly-time (?!)
Indeed we prefer P ≠ NP
Encryption algorithms heavily rely on the assumption that P ≠ NP
P = NP or P ≠ NP is still an open problem

17. Relations hard easy Is there any problem even harder than NP-C?
Yes! e.g. I-go P
easy

18. Methodology To show L is in NP, you can either
Show that solutions for L can be verified in polynomial-time, or
Describe a nondeterministic polynomial-time TM for L
(Come back to this if we have enough time)
To show L is NP-complete
Show that L is in NP
Poly-time reduce some NPC problem to L
i.e., design a polynomial-time reduction from some problem we know to be NP-complete

19. Example
Proving a problem being NPC

20. Double-SAT Problem: Goal:
Double-SAT = {<φ> | φ is a Boolean formula with at least two satisfying assignments}
Goal:
Show that Double-SAT is NP-Complete

21. Double-SAT (Proof Sketch)
Steps:
Show that Double-SAT ∈ NP
Show that Double-SAT is not easier than a certain NPC problem
For the NPC problem, we choose SAT
i.e., we want to poly-time reduce Double-SAT to SAT
Show the correspondence of “yes” instance between reduction

22. Double-SAT - (1) NP It is trivial to see that Double-SAT ∈ NP
Given 2 assignments for φ, and verify whether both of them satisfy φ
We can just evaluate the truth value in poly-time

23. Double-SAT - (2) Reduction
On input φ(x1, , xn):
1. Introduce a new variable w
2. Output formula
φ’(x1, , xn, y) = φ(x1, , xn) ∧ ( w ∨ w ).
acc R
TM for L’ x y
rej
SAT
Double-SAT
x ∈ L
y ∈ L’
TM accepts

24. Double-SAT - (3) Correspondence
x ∈ L  y ∈ L’
 : Suppose there is an satisfying assignment, X, for φ(x1, , xn), we can find two satisfying assignments for φ’(x1, , xn, w):
Assignment 1 = {X, w=True}
Assignment 2 = {X, w=False}
φ’(x1, , xn, w) = φ(x1, , xn) ∧ ( w ∨ w )
For {xi}, assign X,
then this part = True
No matter what w is,
this part = True

25. Double-SAT - (3) Correspondence
x ∈ L  y ∈ L’
: We use contrapositive
i.e., to show x ∉ L ⇒ y ∉ L’
Indeed, if x ∉ L, φ(x1, , xn)=False
Then, no matter what the value of y is
φ’(x1, , xn, y)=False

26. Dominating Set Problem: Goal:
Dominating-set = {<G, K> | A dominating set of size K for G exists}
Goal:
Show that Dominating-set is NP-Complete

27. Dominating Set (Definition)
Problem:
Dominating-set = {<G, K> | A dominating set of size (at most) K for G exists}
Let G=(V,E) be an undirected graph
A dominating set D is a set of vertices that covers all vertices
i.e., every vertex of G is either in D or is adjacent to at least one vertex from D

28. Dominating Set (Example)
Size-2 example : {Yellow vertices} e

29. Dominating Set (Proof Sketch)
Steps:
Show that Dominating-set ∈ NP.
Show that Dominating-set is not easier than a NPC problem
We choose this NPC problem to be Vertex cover
Reduction from Vertex-cover to Dominating-set
Show the correspondence of “yes” instances between the reduction

30. Dominating Set - (1) NP It is trivial to see that Dominating-set ∈ NP
Given a vertex set D of size K, we check whether (V-D) are adjacent to D
i.e., for each vertex, v, in (V-D), whether v is adjacent to some vertex u in D

31. Dominating Set - (2) Reduction
Reduction - Graph transformation
Construct a new graph G' by adding new vertices and edges to the graph G as follows: T G G’
<G,k> ∈ L
Vertex-cover
Dominating-set
<G’, k> ∈ L’

32. Dominating Set - (2) Reduction
Reduction - Graph transformation (Con’t)
For each edge (v, w) of G, add a vertex vw and the edges (v, vw) and (w, vw) to G'
Furthermore, remove all vertices with no incident edges; such vertices would always have to go in a dominating set but are not needed in a vertex cover of G
We skip the discussion of this subtle part in the followings T G G’
<G,k> ∈ L
Vertex-cover
Dominating-set
<G’, k> ∈ L’

33. [Recap] Vertex cover
A vertex cover, C, is a set of vertices that covers all edges
i.e., each edge is at least adjacent to some node in C 1 2 3 4
{2, 4}, {3, 4}, {1, 2, 3}
are vertex covers

34. Dominating Set – Graph Transformation Examplew v z u vz wu vw zu vu G' v w z u G

35. Dominating Set - (3) Correspondence
A dominating set of size K in G’  A vertex cover of size K in G
 Let D be a dominating set of size K in G’
Case 1): D contains only vertices from G
Then, all new vertices have an edge to a vertex in D
D covers all edges
D is a valid vertex cover of G

36. Dominating Set - (3) Correspondence
A dominating set of size K in G’  A vertex cover of size K in G
 Let D be a dominating set of size K in G’
Case 2): D contains some new vertices (vertex in the form of uv)
(We show how to construct a vertex cover using only old vertices, otherwise we cannot obtain a vertex cover for G)
For each new vertex uv, replace it by u (or v)
If u ∈ D, this node is not needed
Then the edge u-v in G will be covered
After new edges are removed, it is a valid vertex cover of G (of size at most K)

37. Dominating Set - (3) Correspondence
A dominating set of size K in G’  A vertex cover of size K in G
 Let C be a vertex cover of size K in G
For an old vertex, v ∈ G’ :
By the definition of VC, all edges incident to v are covered
v is also covered
For a new vertex, uv ∈ G’ :
Edge u-v must be covered, either u or v ∈ C
This node will cover uv in G’
Thus, C is a valid dominating for G’ (of size at most K)

38. Dominating Set - (3) Correspondencevw v w v w vz wu vu z u z u zu
Vertex-cover in G
Dominating-set in G'

39. End
Any questions? (There should be some)