Approximation Algorithm
Prepared by:
Lamiya El_Saedi
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Introduction:
 There are many hard combinatorial
optimization problems that can’t
be solved efficiently using
backtracking or randomization.
The alternative way for talking
some of these problem is to devise
an approximation algorithm.
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 The approximation is depend on the
reasonable solution that
approximations as optimal solution
 There is a performance bound that
guarantees that the solution to a
given instance will not be far away
from the neighborhood of the exact
solution.
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 A marking characteristic of approximation
algorithms is that they are fast, as they
are mostly greedy heuristics.
 The proof of correctness of greedy
algorithm may be complex.
 In general, the better the performance
bound the harder it becomes to prove the
correctness of an approximation
algorithms.
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Basic Definition:
Combinatorial
optimization
problem
For each solution
A set
DII
of instances
For each
There is
I in DII
SII(I) of
Candidate solution
σ
In SII(I) there is
A value
fII(σ)
Called the solution
value of σ
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Note:
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Cont.
 In simple word:
assume that:
DII={I1,…,In}
SII(Ii)={σ1,…, σn}
fII(σi)={v1,…,vn}
fII(σ)=A(I)
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Subset-sum problem:
 Is a special case of the Knapsack problem in which
the item values are identical to their sizes.
 Ex:
I= {I1,I2,I3,I4}
S= {1,2,3,4}
V= {1,2,3,4}
C (Knapsack capacity)= 5
 The objective is to find a subset of the items
that maximizes the total sum of their sizes
without exceeding the Knapsack capacity.
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Subset-sum algorithm:
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Cont.
 Time complexity of algorithm is
exactly the size of the table Θ(nC) as
filling each entry requires Θ(1) time.
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Cont.
 When I apply the example by using subsetsum algorithm the results appear like this:
0
1
2
3
4
5
0
0
0
0
0
0
0
s1
0
1
1
1
1
1
s2
0
1
2
3
3
3
s3
0
1
2
3
4
5
s4
0
1
2
3
4
5
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Cont.
 So, from the table:
OPT(4)={1} <4
OPT(3)={1,2} <3
OPT(2)={0} <2 does not exist in DII
OPT(1)= {0} <1 does not exist in DII
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Now:
 We develop an approximation
algorithm
for some positive integer k.
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