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Traveling Salesman
• Given n cities with a distance table, find a minimum total-distance tour to visit each city exactly once.
. distance
total tour witha find ,and cities those
between tabledistance a with cities Given
:TSP-Approx-
optr
n
n
r
Definition
Proof:
Given a graph G=(V,E), define a distance table on V as follows:
EvurV
Evuvud
),( if ,||
),( if ,1),(
hard.- is TSP-Approx- ,1any For NPrr
Theorem
solvable. time-polynomial being HC implies solvable
time-polynomial being TSP-Approx- r
Contradiction Argument
• Suppose r-approximation exists. Then we have a polynomial-time algorithm to solve Hamiltonian Cycle as follow:
r-approximation solution < r |V|
if and only if
G has a Hamiltonian cycle
Special Case
• Traveling around a minimum spanning tree is a 2-approximation.
solvable. time-polynomial
is TSP-Approx-2 ,inequality triangular
thesatisfies tabledistance n the Whe Theorem
• Minimum spanning tree + minimum-length perfect matching on odd vertices is 1.5-approximation
solvable. time-polynomial
is TSP-Approx-1.5 ,inequality triangular
thesatisfies tabledistance n the Whe Theorem
Minimum perfect matching on odd verticeshas weight at most 0.5 opt.
Knapsack
.any for Hence
.any for Assume
}.1 ,0{
t.s.
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solvable. time-polynomial
is Knapsack -Approx-2 Theorem
Proof.
solvable. time-polynomial
is Knapsack -Approx- ,1any For rr
Theorem
(PTAS). schemeion approximat
timepolynomial hasKnapsack s,other wordIn
• Classify: for i < m, ci < a= cG,
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MAX3SAT
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clauses. satisfied
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rr
This an important result proved using PCP system.
complete).-(APX
complete-SNP MAX is MAX3SAT
Theorem
Class MAX SNP (APX?)
solvable. time-polynomial
is PR-Approx-such that 1
constant a exists thereif SNP MAX tobelongs
PR problemon minimizatior on maximizatiA
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such that on of solutions feasible to
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instances to of instances from maps (L1)
such that 0, constants twoand , and
functions computable time-polynomial twoare thereif
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. degree with graph aconsider ,4For bGb
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Thus,
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PTASPTAS
MAX SNPMAX SNP
PTAS
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case
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case
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case
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problems.on maximizati are and Both :4
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case
x
MAX SNP-complete (APX-complete)
.
SNP, MAXany for and SNP MAX if
complete-SNP MAX is problemon optimizatiAn
pL
Theorem
PTAS. no has then , if i.e.,
hard,- is -Approx- ,1
, problem complete-SNP MAX
NPP
NPrr
MAX3SAT-3
clauses. satisfied of # themaximize
toassignmentan find times,most threeat
appears bleeach varia that 3CNF aGiven F
complete.-SNP MAX is 3-MAX3SAT
Theorem
3-MAX3SATMAX3SAT pL
VC-4 is MAX SNP-complete
graph. a is where
,)(construct , 3CNFeach For
4-VC of inputs3-SAT3MAX of inputs:
4-VC3-SAT3MAX
G
GFfF
f
pL
Proof.
1x
))(( 143231 xxxxxx
2x 3x 4x1x 2x 3x 4x
1c
13c
11c 12c2c
23c
22c21c
hard.-SNP MAX isCover -Set
Theorem
Proof. Cover-Set3-VC pL
. ofcover set a is }|{
ofcover - vertexa is
Then }.|{
set aconstruct ,each For
3.-VC of instancean be ),(Let
ECvsS
GVC
evEes
Vv
EVG
vC
v
Theorem
. unlessCover -Setfor
ionapproximat-) (ln time-polynomial no is There
PNP
no
).(
unlessCover -Setfor ion approximat- ln
time-polynomial no is there,1For
) (log nOnDTIMENP
n
Theorem
Proved using PCP system
).(
unless MCDSfor ion approximat- ln
time-polynomial no is there,1For
) (log nOnDTIMENP
n
Theorem
MCDS
y.cardinalit its minimize set to
dominating connected a find ,graph aGiven G
1x 2x 3x 4x 5x 6x
1S 2S 3S
}.,,{
},,,,{},,,{
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xxxxSxxxS
. unless CLIQUEfor
ionapproximat- time-polynomial no is there,1For
PNP
ns s
subgraph.) complete a is (A y.cardinalit its
maximize toclique a find,graph aGiven
clique
G
CLIQUE
Theorem
Proved with PCP system.
disjunct?- is
,1integer an and matrix binary aGiven
:-coin is problem following theProve
dM
dM
NP
matrix?binary -by-
disjunct- a thereis ,0,, integersGiven
:in is problem following theProve 2
nt
dtdn
p
1
2
Exercises
rows.) of # maximum thedeletingby submatrix
disjunct-2 a find ,matrix aGiven :DS-2-(Min
2. size has pool
every that case specialin hard- is DS-2-Min
:Prove
M
NP
3
hint
DS-2-MinVC pm
• Min-2-DS is MAX SNP-complete in the case that all given pools have size at most 2.
4 Prove that
5. Is TSP with triangular inequality MAX SNP-complete?
