Approximation
Algorithms
Introduction
2017-07-13

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.
Dao Thanh Tinh
2
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.
2017-07-13
Dao Thanh Tinh
3
Bin Parking
0.8
0.4
0.4
0.2
0.2
0.5
0.3
0.2
capacity of 1
2017-07-13
Dao Thanh Tinh
4
Bin Parking
0.2
0.5
0.2
0.2
0.8
0.4
0.4
0.3
4 bins used
2017-07-13
Dao Thanh Tinh
5
Bin Parking
0.8
0.5
0.4
0.4
0.3
0.2
0.2
0.2
Nonincreasing order
2017-07-13
Dao Thanh Tinh
6
Bin Parking
0.2
0.4
0.2
0.3
0.8
0.5
0.4
0.2
Nonincreasing first fit
4 bins used
2017-07-13
Dao Thanh Tinh
7
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
B
n
y
i 1
n
Set xij = 1 if item j is put into bin i, otherwise xij=0.
An each item can put exactly into only one bin!!!
x
j 1
ij
1
n
s
Each bin have capacity T
j 1
j
i
xij
 T
minimize B
2017-07-13
Dao Thanh Tinh
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.
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.
1) sort s(1) ≥ s(2) ≥ ... ≥ s(n);
2) for (j=1; j≤n; j++) used(j)=0;
3) 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;
2017-07-13
Dao Thanh Tinh
9
Bin Parking
Nonincreasing order
0.8
0.5
0.4
0.4
0.3
0.2
2017-07-13
Dao Thanh Tinh
0.2
0.2
10
Bin Parking
The Optimal Solution
0.8
0.5
0.4
0.4
0.3
0.2
0.2
0.2
0.2
0.3
0.4
0.5
0.4
0.2
0.2
0.8
2017-07-13
Dao Thanh Tinh
11
Bin Parking
The Optimal Solution
0.2
0.2
0.2
0.3
0.2
0.4
0.4
0.2
0.3
0.8
0.8
0.5
0.5
0.4
0.4
0.2
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)
2017-07-13
Dao Thanh Tinh
12
Bin Parking
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
Niff:
si>T/3
then:
1) s1, ...,si1>T/3
2) bins bj, j=1,..,op(S), contain at
most 2 objects
......
si
op(S) bins
2017-07-13
Extra bins
Dao Thanh Tinh
13
Bin Parking
op(S) bins
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!
op(S) bins
Extra bins
st”
st
st’
bp
si
bq
p<q
bins b1,.., bk
contains s1,...,sk
bins bk+1,...bop(S) contains 2 objects
sk+1,...si-1
Extra bins
In an optimal solution:
st”
sk+1,..., si1, si,... will be in bins
bk,..,bop(S)
...
But it is imposible
st
b1
bk1
2017-07-13
bk
therefore the assumtion that
si>T/3
must be false. 
si
bop(S)
Dao Thanh Tinh
14
Bin Parking
op(S) bins
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)
Extra bins
st”
st
st’
bp
si
bq
p<q
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)
T .op( S ) 
2017-07-13
n
opt ( S )
i 1
i 1
 si 


 t j  u j  T .op( S )
Dao Thanh Tinh
which is
impossible
15
Bin Parking
op(S) bins
Remark. For S={s1,...,sn}
The number of extra bins is at
most (op(S)-1)/3
Extra bins
st”
st
st’
bp
si
bq
p<q
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
2017-07-13
Dao Thanh Tinh
16
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
2017-07-13
Dao Thanh Tinh
17