1. Approximation Algorithms

2. Introduction
Approximation algorithms for optimization problems.
Approximation algorithm - an algorithm that is guaranteed to run quickly (in time polynomial in the input size) and to produce a solution for which the value of the objective function is quantifiably close to the optimal value.
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3. Bin Parking Problem: Given S={s1, s2, ...,sn}, 0<si1, i=1,2,...,n.
How to pack S into as few bins as possible, where which bin has capacity 1.
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4. Bin Parking capacity of 1 0.8 0.5 0.4 0.3 0.2 2019-02-19
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5. Bin Parking 4 bins used 0.5 0.8 0.4 0.3 0.2 2019-02-19 Dao Thanh Tinh

6. Bin Parking Nonincreasing order 0.8 0.5 0.4 0.3 0.2 2019-02-19
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7. Bin Parking Nonincreasing first fit 4 bins used 0.8 0.3 0.5 0.4 0.2
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8. Bin Parking
Problem: Given S={s1, s2, ...,sn}, 0< s i T, i = 1,2,...,n.
How to pack S into as few bins as possible, where which bin has capacity T.
Set yi = 1 if bin i is used, otherwise yi=0. Number of the used bins is
Set xij = 1 if item j is put into bin i, otherwise xij=0.
An each item can put exactly into only one bin!!!
Each bin have capacity T
minimize B
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9. Bin Parking Nonincreasing first fit (Niff)
Problem: Given S={s1, s2, ...,sn}, 0< s i T, i = 1,2,...,n.
How to pack S into as few bins as possible, where which bin has capacity T.
Nonincreasing first fit (Niff)
Input: s(1), s(2), ..., s(n), 0 < si ≤ 1, i=1,..,n.
Output: An array b(1),...,b(n), where b(k)=i if s(k) is placed into bin i.
sort s(1) ≥ s(2) ≥ ... ≥ s(n);
for (j=1; j≤n; j++) used(j)=0;
for (j=1 ; j≤n; j++)
i=1;
while (used(j)+s(i)>T) do i++;
used(i) = used(i) + s(j);
b(j) = i;
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10. Bin Parking Nonincreasing order 0.8 0.5 0.4 0.4 0.3 0.2 0.2 0.2
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11. Bin Parking The Optimal Solution 0.8 0.5 0.4 0.3 0.3 0.8 0.5 0.4 0.2
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12. Bin Parking The Optimal Solution Lemma 1.
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Lemma 1.
Let S={s1, s2, ..., sn} - input objects
op(S) be the minimum number of bins for S
All of the objects placed by Niff in the extra bins have size (weight) at most T/3 (1/3)
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13. Bin Parking Suppose that si>T/3 Niff: then: 1) s1, ...,si1>T/3
Lemma 1.
Let S={s1, s2, ..., sn} - input objects
op(S) be the minimum number of bins for S
All of the objects placed by Niff in the extra bins have size (weight) at most T/3 (1/3)
Proof.
Let i is index of the first objects placed by Niff in bin op(S)+1.
Suppose that si>T/3
then:
1) s1, ...,si1>T/3
2) bins bj, j=1,..,op(S), contain at
most 2 objects
Niff:
......
op(S) bins
Extra bins si
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14. Bin Parking bp bq st st” st’ si In an optimal solution:
op(S) bins
Extra bins bp bq
p < q st
st”
st’ si
if bin bp contains 2 objects
bins bp+1,..., bop(S) contain 2 objects
Suppose bq, p<q, contains st’.
Let 2 objects in bp: st, st”
Follow Niff:
st  st’ st”  si st  st”
T  st+st”  st’+si
 Niff would have put si into bq!
bins b1,.., bk contains s1,...,sk
bins bk+1,...bop(S) contains 2 objects
sk+1,...si-1
In an optimal solution:
sk+1,..., si1, si,... will be in bins bk,..,bop(S)
But it is imposible
therefore the assumtion that si>T/3
must be false. 
op(S) bins
Extra bins st
st” si
bop(S) bk
bk1 b1
...
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15. Bin Parking bp bq st st” st’ si Lemma 2. For S={s1,...,sn}
op(S) bins
Extra bins bp bq
p < q st
st”
st’ si
Lemma 2. For S={s1,...,sn}
Let F(S) is the number of objects which placed in extra bins bop(S)+1,... Then
F(S) < op(S)
Proof.
Suppose that F(S) = op(S)
Let: u1, ..., uop(S) be the objects which placed into extra bins
t1,..., top(S) be the final contents of bins b1,..., bop(S)
There are:
bj + uj > T, j =1,...,op(S)
which is impossible
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16. Bin Parking bp bq st st” st’ Remark. For S={s1,...,sn}
op(S) bins
Extra bins bp bq
p < q st
st”
st’
Remark. For S={s1,...,sn}
The number of extra bins is at most (op(S)-1)/3 si
Proof.
The number of objects placed by Niff in extra bins is at most op(S)-1
and size of each object at most T/3
Then, each extra bin contain at least 3 objects.
It show that the number of extra bins at most (op(S)-1)/3
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17. Stategies Remark. For S={s1,...,sn}
The number of extra bins is at most (op(S)-1)/3
Proof.
The number of objects placed by Niff in extra bins is at most op(S)-1
and size of each object at most T/3
Then, each extra bin contain at least 3 objects.
It show that the number of extra bins at most (op(S)-1)/3
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