PTAS for Bin-Packing

1
PTAS for Bin-Packing
2
Special Cases of Bin Packing

• 1. All item sizes smaller than
• Claim 1
• Proof If then
• So assume
• Therefore

3

• But for
• since
• and
• Therefore

4
2. Assume that there are only k different sizes
(k fixed). Claim 2 Optimum number of bins
can be found in time Proof For
every subset of items compute the optimal
number of bins it can be packed in.
Total number of subsets is Why? Sort
items by size
size
size
size
5
Denote a subset by a k-tuple of the number of
elements of each size in the subset. Number of
different such tuples Compute optimal number of
bins for each subset using dynamic
programming. Find subsets that can be packed in 1
bin Find subsets that can be packed in 2
bins Find subsets that can be packed in OPT
bins (Note OPT can not exceed n)
6
Number of values per iteration Number of
iterations n IMPLEMENTATION 1.
Sort subsets by sums of sizes. Those with sum not
exceeding 1 are packed in a single bin. Call them
1-bin subsets. 2.For all pairs of 1-bin subsets,
consider their union. Every union that is not a
1-bin subset is a 2-bin subset.

If entire set is packed done.
o/w continue to next step.
. . .
7
i1. For all pairs consisting of a 1-bin subset
and an i-bin subset, consider their union.
If it is not an i-bin subset or
less , then it is a (i1)-bin subset.

If entire set is packed done.
o/w continue to next step.
. . . Time We
consider all pairs at every step, thus
time per step. At most
OPT steps. OPT Conclude
8
3. Conclude Given and
Assume that all items have size
then (By claim 1). 4. Assume that all
items have size Claim 2 There exists a packing
algorithm PA for which Proof
Consider the following approximation
algorithm.
9
Fix k (we will later decide the value of
k) 1.Sort I in non-decreasing order. Split into
groups of k elements each. 2. Pack
group G1 in at most k bins. 3. Change all
numbers in to largest
number in group. Call new item set I.
10
EXAMPLE G1 G2
G3 G4 I 10 9 8
7 7 6 6 6 5 3 2 I
7 7 7 6 6 6
3 3 4. Find OPT(I) (by Claim 2).
11
Lemma OPT(I) OPT(I) Proof Consider I
constructed as Discard
For change every
element in group to smallest. EXAMPLE
G1 G2 G3
G4 I 10 9 8 7 7 6
6 6 5 3 2
I 8 8 8 6 6 6 5 5
5 Clearly OPT(I) OPT(I) (since I has
less elements and they are smaller)
12

• OPT(I) OPT(I)
• Why?
• I may have less elements. ( ,
  and all other sets are equal sized in both)
• I has smaller total sum.
• (
  )
• Conclude
• OPT(I) OPT(I) OPT(I)

13
Algorithm Running Time Approximation Factor
PA(I)
OPT(I)k NOW CHOOSE Therefore Approximation
Factor (1e) OPT(I)1
14
What about the time? But for
all i, so Thus Algorithm Running
Time Since e is fixed, this is a polynomial.
15

• Algorithm for the General Case
• Split items into two sets
  
  small (e/2) and large (gte/2).
• 2. Handle large items as in Claim 4.
• 3. Use FF to pack small items into remaining
  spaces of large item bins. When all such space is
  used, open new bins using FF.
• Time Clearly polynomial.
• Approximation Factor
• Case 1 All large item bins are filled, and new
  ones opened.
• Case 2 Not all large item bins are filled.

16

• Case 1 The situation is
• This is the case of claim 1, so approximation is
• (1e) OPT(I)1
• Case 2 Let IL be the set of large items. The
  number of bins produced by our algorithm is
• OPT(I)k
• (1e) OPT(I)1