Other NP‐Complete Problems
10.4
Mohamed M. El Wakil
[email protected]
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Agenda
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Why?
How to describe?
Independent Set Problem
d
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Node Cover Problem
Directed Hamilton Circuit Problem
Undirected Hamilton Circuit Problem
Undirected Hamilton Circuit Problem
Traveling Salesman Problem
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Remember?
• A problem P is NP
A problem P is NP‐Complete
Complete If:
If:
– P is in NP
• For
For every problem L in NP, there is a polynomial time every problem L in NP there is a polynomial time
reduction from L to P.
• Or, If P1 is NP‐Complete, and there is polynomial time reduction from P1 to P, then P is NP‐Complete.
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Why?
• When
When a problem is proved to NP
a problem is proved to NP‐Complete
Complete, there is no need to attempt finding an optimal solution for it
solution for it.
• EEvery new NP‐Complete problem, re‐assures NP C
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that all other NP‐Complete problems require exponential time to solve.
i l i
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NP‐Complete
NP
Complete problems description problems description
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•
•
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Name, and abbreviation.
Name
and abbreviation
Input: how it is represented.
O
Output: when the output shall be “Yes”
h
h
h ll b “ ”
The other NP‐Complete problem used to prove the NP‐Completeness of the problem at hand.
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The Independent Set Problem
The Independent Set Problem
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Independent Set
Independent Set
• A
A subset I
subset I of the nodes of a graph G is called of the nodes of a graph G is called
an independent set, if no two nodes in I are connected by an edge in G
connected by an edge in G.
– The red
The red nodes are independent sets.
nodes are independent sets
Source: MathWorld
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The Independent Set Problem (IS)
The Independent Set Problem (IS)
• Given
Given a graph G=(V,E) and integer k (|V|>k>1), a graph G=(V E) and integer k (|V|>k>1)
does G has an independent set of k or more nodes?
• IS is
IS i NP‐Complete, as
NP C
l
1. It is NP.
2. There is polynomial‐time reduction from 3SAT to IS.
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1. IS is in NP
1. IS is
in NP
• An
An IS guess is verifiable in polynomial time IS guess is verifiable in polynomial time
using DTM:
– Guess a set Gu
Guess a set Gu of k nodes.
of k nodes
– Check that no two nodes make an edge in G.
• Hence, IS is in NP
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2. 3SAT is reducible to IS in polynomial time
• Input : Input :
– E=(e1)(e2)…(em). – E is 3‐CNF
E is 3 CNF
– i.e. E has 3m literals
• Output:
O t t
– A graph G with 3m nodes, and some edges
– k.
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Nodes
• Every node is given a name n[i,j]
– i is a clause number (m≥i≥1)
i
l
b ( ≥i≥1)
– j is a literal number (3≥j≥1)
• E=
E
– (x1+x2+x3)
– (~x1+x2+x4)
(~ 1 2 4)
– (~x2+x3+x5)
– (~x3+~x4+~x5)
• A column corresponds to a clause
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Edges 1/2
Edges 1/2
• Put
Put an edge between all pairs of nodes in a an edge between all pairs of nodes in a
column.
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Edges 2/2
Edges 2/2
• Put
Put an edge between every complementary an edge between every complementary
pair of nodes.
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k
• IS
IS’ss k is set to be 3SAT
k is set to be 3SAT’ss m
m
• That is all!
h i ll!
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Example
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The Vertex Cover Problem
The Vertex Cover Problem
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Node/Vertex Cover (NC/VC)
Node/Vertex Cover (NC/VC)
• A
A vertex cover of a graph G=(V,E), is a subset vertex cover of a graph G=(V E) is a subset
of V, such that for every edge in E=(u,v), either u or v is in V
u or v is in V.
• V’={1,3,5,6} is vertex cover
V’ {1 3 5 6} i
• V’’={2,4,5} is another vertex cover
Source: Wikipedia
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Vertex Cover Problem
Vertex Cover Problem
• Given
Given a graph G=(V,E), and integer k a graph G=(V E) and integer k
(|V|>k>0), does G have a vertex cover with k or fewer nodes?
or fewer nodes?
• VC is NP‐Complete, as
VC i NP C
l
1. It is NP.
2. There is polynomial‐time reduction from IS to VC
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1. VC is in NP
1. VC is in NP
• A
A VC guess is verifiable in polynomial time VC guess is verifiable in polynomial time
using DTM:
– Guess Gu
Guess Gu a set of k nodes.
a set of k nodes
– Check that each edge in G, has at least one end node in Gu
node in Gu.
• Hence, VC is in NP
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2. IS is reducible to VC in polynomial time
• VC
VC is the complement of IS!
is the complement of IS!
• If I is largest independent set of graph G=(V,E), then V I is a smallest vertex cover of G
then V‐I, is a smallest vertex cover of G.
– {1,3,5,6} is VCÎ {2,4} is an IS
– {2,4,5} is VC Î
{2 4 5} i VC Î {1,3,6} is an IS
{1 3 6} i
IS
Source: Wikipedia
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IS reduction to VC
IS reduction to VC
• IS Reduction to VC:
IS Reduction to VC:
– The VC instance G is a copy of the VC instance G.
– The VC instance k is |V|‐k
The VC instance k is |V| kis, where k
where kis is IS
is IS k.
k
• Obviously, this could be done in polynomial time.
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The Directed Hamilton The Directed Hamilton
Circuit Problem
Circuit Problem
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A Hamilton Circuit
A Hamilton Circuit
• A
A Hamiltonian circuit is a cycle in a graph Hamiltonian circuit is a cycle in a graph
which visits each vertex exactly once.
• Example:
– Graph: Blue
– HC: Black
Source: Wikipedia
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Directed Hamilton Circuit Problem (
(DHC)
)
• Given
Given a directed
a directed graph G, does it have a graph G, does it have a
Hamilton circuit?
• DHC is NP‐Complete, as
1. It
1
It is NP.
is NP
2. There is polynomial‐time reduction from 3SAT to DHCP
Source: Wikipedia
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1. DHC is in NP
1. DHC is in NP
• A
A DHC guess is verifiable in polynomial time DHC guess is verifiable in polynomial time
using DTM:
– Guess Gu
Guess Gu a cycle in G.
a cycle in G
– check if all needed edges are in G.
• Hence, DHC is in NP
H
DHC i i NP
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2. 3SAT is reducible to DHC in polynomial time
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• This is rather lengthy, so we will skip it!
This is rather lengthy so we will skip it!
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The Undirected Hamilton The Und
irected Hamilton
Circuit Problem
Circuit Problem
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Undirected Hamilton Circuit Problem ( )
(HC)
• Given
Given an undirected
an undirected graph G, does G have a graph G does G have a
Hamilton circuit?
• HC is NP‐Complete, as
– HC is in NP
– There is a polynomial time reduction from DHC to HC. 28
HC is in NP
HC is in NP
• A
A HC guess is verifiable in polynomial time HC guess is verifiable in polynomial time
using DTM:
– Guess Gu
Guess Gu a cycle in G.
a cycle in G
– check if all needed edges are in G.
• Hence, DHC is in NP
H
DHC i i NP
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2. DHC is reducible to HC in polynomial time
• Input:
– Directed graph Gd
• Output
– Undirected graph Gu
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Constructing Gu ‐ Nodes
Constructing Gu
• For
For every node v in Gd, there are three nodes every node v in Gd there are three nodes
(v0,v1,v2) in Gu
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Constructing Gu ‐ Edges
Constructing Gu
• Edges:
– For all nodes in Gd, there are two edges in Gu
(v0 v1) and (v1 v2)
(v0,v1), and (v1,v2).
– If there is an edge (vÆw) in Gd, then add edge (v2,w0) to Gu
(v2,w0) to Gu
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The Traveling Salesman
The Traveling Salesman
Problem
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Traveling Salesman
A salesman wants to visit all these cities, incurring the least
incurring the least possible cost.
In graph terminology, we want to have a Hamilton circuit with the minimum summation of edges weights.
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Traveling Salesman Problem (TSP)
Traveling Salesman Problem (TSP)
• Given
Given an undirected weighted graph G, and a an undirected weighted graph G, and a
number k, is there a Hamilton circuit in G, such that the sum of the weights on the edges of the HC is less than or equal k?
•
• TSP is NP‐Complete, as
– TSP is in NP.
– There is a polynomial time reduction from HC to TSP. 35
TSP is in NP
TSP is in NP
• A
A TSP guess is verifiable in polynomial time TSP guess is verifiable in polynomial time
using DTM:
– Guess an itinerary (i.e. cycle) in G
Guess an itinerary (i e cycle) in G
– Check if the sum of the weights of all involved edges is less than or equal k
edges is less than or equal k.
• Hence, TSP is in NP
Hence TSP is in NP
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2. HC is reducible to TSP in polynomial time
• Input:
– Undirected graph Ghc
• Output
– Undirected weighted graph Gtsp
– Number k
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Constructing TSP instance
Constructing TSP instance
• Gtsp nodes:
– A copy of Ghc nodes
• Gtsp edges:
d
– A copy of Ghc edges
– Set the weight of all edges to be 1
Set the weight of all edges to be 1
• k
– Set k to be the number of nodes of Ghc
• Evidently, this is polynomial time reduction
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Conclusion
• Proving NP‐Completeness i
is not easy!
!
– The problem must be proved to be NP
d b NP
– There should be a polynomial time reduction l
i l ti
d ti
from another NP‐Complete problem!
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That’ss all folks!
That
all folks!
Questions?
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