More NP-complete Problems
Fall 2006
Costas Busch - RPI
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Theorem: (proven in previous class)
If: Language A is NP-complete
Language B is in NP
A is polynomial time reducible to B
Then: B is NP-complete
Fall 2006
Costas Busch - RPI
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Using the previous theorem,
we will prove that 2 problems
are NP-complete:
Vertex-Cover
Hamiltonian-Path
Fall 2006
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Vertex Cover
Vertex cover of a graph
is a subset of nodes S such that every edge
in the graph touches one node in S
Example:
S = red nodes
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Size of vertex-cover
is the number of nodes in the cover
Example: |S|=4
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Corresponding language:
VERTEX-COVER = { G, k :
graph G contains a vertex cover
of size k }
Example:
G
G ,4  VERTEX - COVER
Fall 2006
Costas Busch - RPI
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Theorem: VERTEX-COVER is NP-complete
Proof:
1. VERTEX-COVER is in NP
Can be easily proven
2. We will reduce in polynomial time
3CNF-SAT to VERTEX-COVER
(NP-complete)
Fall 2006
Costas Busch - RPI
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Let  be a 3CNF formula
with m variables
and l clauses
Example:
  (x 1  x 2  x 3 )  ( x 1  x 2  x 4 )  ( x 1  x 3  x 4 )
Clause 1
Clause 2
Clause 3
m4
l 3
Fall 2006
Costas Busch - RPI
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Formula  can be converted
to a graph G such that:
 is satisfied
if and only if
G Contains a vertex cover
of size k  m  2l
Fall 2006
Costas Busch - RPI
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  (x 1  x 2  x 3 )  ( x 1  x 2  x 4 )  ( x 1  x 3  x 4 )
Clause 2
Clause 1
Clause 3
Variable Gadgets
x1
x1
x2
x3
x2
Clause Gadgets
x3
Clause 1
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nodes
x4
x3
x2
x4
3l nodes
x1
x1
x1
x2
2m
x4
Clause 2
Costas Busch - RPI
x3
x4
Clause 3
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  (x 1  x 2  x 3 )  ( x 1  x 2  x 4 )  ( x 1  x 3  x 4 )
Clause 2
Clause 1
x1
x1
x2
x3
Clause 1
Fall 2006
x4
x3
x2
x4
x1
x1
x1
x2
x3
x2
Clause 3
x4
Clause 2
Costas Busch - RPI
x3
x4
Clause 3
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First direction in proof:
If  is satisfied,
then G contains a vertex cover of size
k  m  2l
Fall 2006
Costas Busch - RPI
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Example:
  (x 1  x 2  x 3 )  ( x 1  x 2  x 4 )  ( x 1  x 3  x 4 )
Satisfying assignment
x1  1
x2  0
x3  0
x4  1
We will show that G contains
a vertex cover of size
k  m  2l  4  2  3  10
Fall 2006
Costas Busch - RPI
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  (x 1  x 2  x 3 )  ( x 1  x 2  x 4 )  ( x 1  x 3  x 4 )
x1  1
x1
x1
x2  0
x2
x3
x2
x4  1
x4
x3
x3
x2
x4
x1
x1
x1
x2
x3  0
x4
x3
x4
Put every satisfying literal in the cover
Fall 2006
Costas Busch - RPI
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  (x 1  x 2  x 3 )  ( x 1  x 2  x 4 )  ( x 1  x 3  x 4 )
x1  1
x1
x1
x2  0
x2
x3
x2
x4  1
x4
x3
x3
x2
x4
x1
x1
x1
x2
x3  0
x4
x3
x4
Select one satisfying literal in each clause gadget
in the cover
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2006 include the remaining
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This is a vertex cover since every edge is
adjacent to a chosen node
x1
x1
x2
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x4
x3
x3
x2
x4
x1
x1
x1
x2
x3
x2
x4
Costas Busch - RPI
x3
x4
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Explanation for general case:
x1
x1
x2
x2
x3
x3
x4
x4
Edges in variable gadgets
are incident to at least one node in cover
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Edges in clause gadgets
are incident to at least one node in cover,
since two nodes are chosen in a clause gadget
x2
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x1
x1
x1
x3
x2
x4
Costas Busch - RPI
x3
x4
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Every edge connecting variable gadgets
and clause gadgets is one of three types:
x1
x1
x2
Type 1
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x4
x3
x3
x2
x4
x1
x1
x1
x2
x3
x2
x4
x3
x4
Type 2
All adjacent to nodes in cover
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Type 3
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Second direction of proof:
If graph G contains a vertex-cover
of size k  m  2l
then formula  is satisfiable
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Example:
x1
x1
x2
Fall 2006
x4
x3
x3
x2
x4
x1
x1
x1
x2
x3
x2
x4
Costas Busch - RPI
x3
x4
21
To include “internal’’ edges to gadgets,
and satisfy k  m  2l
exactly one literal in each variable gadget is chosen
x1
x1
x2
x3
x2
x4
x3
x4
m chosen out of 2m
exactly two nodes in each clause gadget is chosen
x2
x1
x1
x1
x3
x2
x4
x3
x4
2l chosen out of 3l
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For the variable assignment choose the
literals in the cover from variable gadgets
x1
x1
x1  1
x2
x2
x2  0
x3
x3
x3  0
x4
x4
x4  1
  (x 1  x 2  x 3 )  ( x 1  x 2  x 4 )  ( x 1  x 3  x 4 )
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  (x 1  x 2  x 3 )  ( x 1  x 2  x 4 )  ( x 1  x 3  x 4 )
is satisfied with
x1  1
x1
x1
x2  0
x2
x3
x2
x4  1
x4
x3
x3
x2
x4
x1
x1
x1
x2
x3  0
x4
x3
x4
since the respective literals satisfy the clauses
Fall 2006
Costas Busch - RPI
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Theorem: HAMILTONIAN-PATH
is NP-complete
Proof:
1. HAMILTONIAN-PATH is in NP
Can be easily proven
2. We will reduce in polynomial time
3CNF-SAT to HAMILTONIAN-PATH
(NP-complete)
Fall 2006
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(x 1  x 2 )  (x 1  x 2 )
Gadget for
variable x 1
(x1  x2 )
(x 1  x 2 )
the directions
change
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(x 1  x 2 )  (x 1  x 2 )
Gadget for
variable x2
(x1  x2 )
(x 1  x 2 )
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(x 1  x 2 )  (x 1  x 2 )
x1
(x1  x2 )
(x 1  x 2 )
x2
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(x 1  x 2 )  (x 1  x 2 )  1
x1  1
(x1  x2 )
(x 1  x 2 )
x2  1
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(x 1  x 2 )  (x 1  x 2 )  1
x1  0
(x1  x2 )
(x 1  x 2 )
x2  1
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