Chun-Ho Hung chhung@cse.cuhk.edu.hk
CSCI3130 Tutorial 10
1

 Review
 P, NP, NPC
 Polynomial-time Reduction
 2 problems
 Double-SAT
 Dominating set
 http://en.wikipedia.org/wiki/Dominating_set_problem
2

P, NP, NPC
Polynomial-time Reduction
3

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

 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
5

 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 f = (x
1
∨ x
2
)
∧
(x
2
∨ x
3
∨ x
4
)
∧
(x
1
)
 Verifying
 Given an assignment, just evaluate the truth value
 Finding a solution?
 No efficient algorithm has been discovered yet
6

 Every language, L in P, L is also in NP
 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
7

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

 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”
9

 Consider 2 problems:
1) BFS on unweighted graph
2) 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
10




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:
1.
Given an instance a of problem, x
2.
3.
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
11

 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
12




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 x R
Poly-time TM y
TM for L’ acc rej
13

 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 x R
Poly-time TM y
TM for L’ acc rej
14



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
15

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


(i)
(ii)
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)



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 NPcomplete
18

Proving a problem being NPC
19

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

 Steps:
1) Show that Double-SAT ∈ NP
2)


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
3) Show the correspondence of “yes” instance between reduction
21

 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
22

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

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

 x ∈ L  y ∈ L
  : We use contrapositive
 i.e., to show x ∉ L ⇒ y ∉ L’


Indeed, if x ∉ L, φ(x
1
, . . . , x n
)= False
Then, no matter what the value of y is
 φ’(x
1
, . . . , x n
, y)= False
25

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


 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
27

 Size-2 example : {Yellow vertices} e
28

 Steps:
1) Show that Dominating-set ∈ NP.
2)


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
3) Show the correspondence of “yes” instances between the reduction
29

 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
30

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

 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
G
<G,k> ∈ L
Vertex-cover
T G’
<G’, k> ∈ L ’
Dominating-set
32

 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
33

vw v w v w vz vu z u z u
G zu
G' wu
34

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
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 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)
36

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)
37

Dominating Set - (3) Correspondence vw v w v w z u
in G vz wu vu z u zu
G'
38

Any questions? (There should be some)
39