Np complete reductions

1. © 2020 Wael Badawy
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More NP-complete Problems
1
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3. © 2020 Wael Badawy

4. © 2020 Wael Badawy
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Theorem:
If: Language is NP-complete
Language is in NP
is polynomial time reducible to
A
A B
B
Then: is NP-completeB
(proven in previous class)

5. © 2020 Wael Badawy
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Using the previous theorem,
we will prove that 2 problems
are NP-complete:
Vertex-Cover
Hamiltonian-Path

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Vertex Cover
S
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Vertex cover of a graph
is a subset of nodes such that every edge
in the graph touches one node in S
S = red nodes
Example:

7. © 2020 Wael Badawy
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|S|=4Example:
Size of vertex-cover
is the number of nodes in the cover

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graph contains a vertex cover
of size }
VERTEX-COVER = { :kG,
G
k
Corresponding language:
Example:
G 
COVER-VERTEX4, G

9. © 2020 Wael Badawy
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Theorem:
1. VERTEX-COVER is in NP
2. We will reduce in polynomial time
3CNF-SAT to VERTEX-COVER
Can be easily proven
VERTEX-COVER is NP-complete
Proof:
(NP-complete)

10. © 2020 Wael Badawy
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Let be a 3CNF formula
with variables
and clauses

m
l
Example:
)()()( 431421321 xxxxxxxxx 
4m
3l
Clause 1 Clause 2 Clause 3

11. © 2020 Wael Badawy
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Formula can be converted
to a graph such that:

G
 is satisfied
if and only if
G Contains a vertex cover
of size lmk 2

12. © 2020 Wael Badawy
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)()()( 431421321 xxxxxxxxx 
1x 1x 2x 2x 3x 3x 4x 4x
1x
2x 3x
1x
2x 4x
1x
3x 4x
Clause 1 Clause 2 Clause 3
Clause 1 Clause 2 Clause 3
Variable Gadgets
Clause Gadgets
m2 nodes
l3 nodes

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)()()( 431421321 xxxxxxxxx 
1x 1x 2x 2x 3x 3x 4x 4x
1x
2x 3x
1x
2x 4x
1x
3x 4x
Clause 1 Clause 2 Clause 3
Clause 1 Clause 2 Clause 3

14. © 2020 Wael Badawy
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If is satisfied,
then contains a vertex cover of size

G
lmk 2
First direction in proof:

15. © 2020 Wael Badawy
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)()()( 431421321 xxxxxxxxx 
Satisfying assignment
1001 4321  xxxx
Example:
We will show that contains
a vertex cover of size
103242  lmk
G

16. © 2020 Wael Badawy
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)()()( 431421321 xxxxxxxxx 
1x 1x 2x 2x 3x 3x 4x 4x
1x
2x 3x
1x
2x 4x
1x
3x 4x
1001 4321  xxxx
Put every satisfying literal in the cover

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)()()( 431421321 xxxxxxxxx 
1x 1x 2x 2x 3x 3x 4x 4x
1x
2x 3x
1x
2x 4x
1x
3x 4x
1001 4321  xxxx
Select one satisfying literal in each clause gadget
and include the remaining literals in the cover

18. © 2020 Wael Badawy
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1x 1x 2x 2x 3x 3x 4x 4x
1x
2x 3x
1x
2x 4x
1x
3x 4x
This is a vertex cover since every edge is
adjacent to a chosen node

19. © 2020 Wael Badawy
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Explanation for general case:
1x 1x 2x 2x 3x 3x 4x 4x
Edges in variable gadgets
are incident to at least one node in cover

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1x
2x 3x
1x
2x 4x
1x
3x 4x
Edges in clause gadgets
are incident to at least one node in cover,
since two nodes are chosen in a clause gadget

21. © 2020 Wael Badawy
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1x 1x 2x 2x 3x 3x 4x 4x
1x
2x 3x
1x
2x 4x
1x
3x 4x
Every edge connecting variable gadgets
and clause gadgets is one of three types:
Type 1 Type 2 Type 3
All adjacent to nodes in cover

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If graph contains a vertex-cover
of size
then formula is satisfiable
G

lmk 2
Second direction of proof:

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Example:
1x 1x 2x 2x 3x 3x 4x 4x
1x
2x 3x
1x
2x 4x
1x
3x 4x

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1x 1x 2x 2x 3x 3x 4x 4x
1x
2x 3x
1x
2x 4x
1x
3x 4x
exactly one literal in each variable gadget is chosen
exactly two nodes in each clause gadget is chosen
To include “internal’’ edges to gadgets,
and satisfy lmk 2
m chosen out of m2
l2 l3chosen out of

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For the variable assignment choose the
literals in the cover from variable gadgets
1x 1x 2x 2x 3x 3x 4x 4x
1001 4321  xxxx
)()()( 431421321 xxxxxxxxx 

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1x 1x 2x 2x 3x 3x 4x 4x
1x
2x 3x
1x
2x 4x
1x
3x 4x
1001 4321  xxxx
)()()( 431421321 xxxxxxxxx 
since the respective literals satisfy the clauses
is satisfied with

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1x 1x
1x
2x 3x
It is impossible to have this scenario
because one edge wouldn’t be covered
3x 3x2x 2x

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End of proof
The proof can be generalized for arbitrary 
we have reduced in polynomial time
3CNF-SAT to VERTEX-COVER
Therefore:
The graph is constructed in polynomial
time with respect to the size of
G


29. © 2020 Wael Badawy
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Theorem:
1. HAMILTONIAN-PATH is in NP
2. We will reduce in polynomial time
3CNF-SAT to HAMILTONIAN-PATH
Can be easily proven
HAMILTONIAN-PATH
is NP-complete
Proof:
(NP-complete)

30. © 2020 Wael Badawy
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)()( 2121 xxxx 
Consider an arbitrary CNF formula
Whatever we will do applies to
3CNF formulas as well

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Gadget for
variable 1x
33 lNumber of nodes in row:
for clausesl
)()( 2121 xxxx 

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)()( 2121 xxxx 
)( 21 xx 
)( 21 xx 
Gadget for
variable 1x
clause node
clause node
A pair of nodes in gadget for each clause node;
Pairs separated by a node
pair 1 pair 2

33. © 2020 Wael Badawy
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)()( 2121 xxxx 
Gadget for
variable
)( 21 xx 
clause node
If variable appears as is in clause:
first edge from gadget outgoing (left to right)
second edge from gadget incoming
pair 1
1x

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)()( 2121 xxxx 
)( 21 xx 
Gadget for
variable 1x
the directions change
clause node
pair 2
If variable appears inverted in clause:
first edge from gadget incoming (left to right)
second edge from gadget outgoing

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)()( 2121 xxxx 
)( 21 xx 
)( 21 xx 
Gadget for
variable 1x
clause node
clause node
pair 1 pair 2

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)()( 2121 xxxx 
)( 21 xx 
)( 21 xx 
Gadget for
variable
clause node
clause node
pair 1 pair 2
2x

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)( 21 xx 
)( 21 xx 
1x
2x
)()( 2121 xxxx 
Complete Graph

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1)()( 2121  xxxx
Satisfying assignment: 11 x 12 x
Satisfying assignment: 01 x 12 x

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)( 21 xx 
)( 21 xx 
11 x
1)()( 2121  xxxx
Hamiltonian
path
If variable is set to 1 traverse its gadget
from left to right
Visit clauses satisfied from the variable assignment
Gadget for 1x

40. © 2020 Wael Badawy
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)( 21 xx 
)( 21 xx 
01 x
1)()( 2121  xxxx
Hamiltonian
path
If variable is set to 0 traverse its gadget
from right to left
Visit clauses satisfied from the variable assignment
Gadget for 1x

41. © 2020 Wael Badawy
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)( 21 xx 
)( 21 xx 
11 x
12 x
1)()( 2121  xxxx
A satisfying
assignment
Hamiltonian
path

42. © 2020 Wael Badawy
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)( 21 xx 
)( 21 xx 
01 x
12 x
1)()( 2121  xxxx
Another satisfying
assignment
Hamiltonian
path

43. © 2020 Wael Badawy
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For each clause pick one variable that
satisfies it and visit respective node once
from that variable gadget
)( 21 xx 
)( 21 xx 
11 x
12 x
1)()( 2121  xxxx

44. © 2020 Wael Badawy
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if the graph has a Hamiltonian path,
then the formula has a satisfying assignment
Each clause node is visited exactly once
from some variable gadget, and that variable
satisfies the clause
Symmetrically, it can also be shown:
Basic idea:

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)( 21 xx 
)( 21 xx 
It is impossible a clause node to be traversed
from different variable gadgets
Node not visited
Node repeated

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Therefore:
The formula is satisfied if and only if
the respective graph has a Hamiltonian path

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End of proof
The proof can apply to any 3CNF formula
we have reduced in polynomial time
3CNF-SAT to HAMILTONIAN-PATH
Therefore:
The graph is constructed in polynomial
time with respect to the size of the formula