Approximation Algorithms Part Four: APTAS

Advanced Algorithms – COMS31900
Approximation algorithms part four
Asymptotic Polynomial Time Approximation Schemes
Benjamin Sach

Approximation Algorithms Recap
A polynomial time algorithm A is an α-approximation for problem P if,
it always outputs a solution s with
Opt
α s Opt (for a maximisation problem)
Opt s α · Opt (for a minimisation problem)
A poly-time approximation scheme (PTAS) is a family of algorithms:
For any constant > 0, A is a (1 + )-approximation for P
It is a (fully) FPTAS if A takes time polynomial in both n and 1/
A PTAS is allowed to have A which takes O(n1/ ) time
- which is polynomial in n (for any constant )
i.e. O((n/ )c)

Approximation Algorithms Recap
A polynomial time algorithm A is an α-approximation for problem P if,
it always outputs a solution s with
Opt
α s Opt (for a maximisation problem)
Opt s α · Opt (for a minimisation problem)
We have seen various c-approximations with constant c
A poly-time approximation scheme (PTAS) is a family of algorithms:
For any constant > 0, A is a (1 + )-approximation for P
Last lecture we saw an FPTAS for SUBSETSUM
It is a (fully) FPTAS if A takes time polynomial in both n and 1/
A PTAS is allowed to have A which takes O(n1/ ) time
- which is polynomial in n (for any constant )
i.e. O((n/ )c)

The SUBSETSUM problem
4 4
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t = 12

4 4
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t = 12
• Let S be a (multi) set of integers and t be a positive integer
here S = {4, 2, 4, 7, 2, 3} and t = 12

4 4
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t = 12
here S = {4, 2, 4, 7, 2, 3} and t = 12
Decision Problem Is there a subset, S ⊆ S with SIZE(S ) = t?

4 4
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t = 12
here S = {4, 2, 4, 7, 2, 3} and t = 12
where SIZE(S ) = a∈S a

t = 12
4
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2
7
32
here S = {4, 2, 4, 7, 2, 3} and t = 12

t = 12
7
3
2
4 42
here S = {4, 2, 4, 7, 2, 3} and t = 12

4 4
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322
t = 12
here S = {4, 2, 4, 7, 2, 3} and t = 12
(last lecture we discussed the NP-hard optimisation version)
The decision version is NP-complete

The PARTITION problem
4 4
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The PARTITION problem is a special case of SUBSETSUM

4 4
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The input S is a multi-set of integers

4 4
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but t is always half the sum of item sizes

4 4
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t = 11

4 4
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t = 11
t = SIZE(S)/2 = a∈S
a
2

4 4
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t = 11
Decision Problem Is there a subset, S ⊆ S with SIZE(S ) = SIZE(S)/2?
a
2

t = 11
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2
7
32
a
2

7
22
t = 11
a
2
4
4
3

4 4
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t = 11
a
2
Alternatively... Can we pack S into two bins of size SIZE(S)/2?

a
2
4 4
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t = 11

a
2
4
4
3
7
2
2
t = 11

a
2
4
4
3
7
2
2
t = 11
The PARTITION problem is also NP-complete

PARTITION and BINPACKING
Key Idea Solve the PARTITION problem by approximating BINPACKING

4 4
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t = 11

4 4
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t = 11
• Convert the PARTITION instance into a BINPACKING problem
by dividing all item and bin sizes by t = SIZE(S)/2

4
11
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7
11 3
112
11
2
11
1

• Now all items are have size at most 1 and the bins have size 1
4
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11 3
112
11
2
11
1

• The optimal number of bins Optb is 2 iff
the answer to the PARTITION instance is ‘yes’
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11 3
112
11
2
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1

• The optimal number of bins Optb is 2 iff
the answer to the PARTITION instance is ‘yes’
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11 3
112
11
2
11
1
• What does this tell us about approximating BINPACKING?

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112
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1
• Assume A is an α-approximation for BINPACKING with α < 3/2

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112
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2
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1
A outputs a solution using s bins with Optb s α · Optb

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112
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2
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1
• If Optb > 2 then s > 2

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1
• If Optb = 2 then 2 s α · Optb < (3/2) · Optb = 3

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11 3
112
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2
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1
so s = 2• If Optb = 2 then 2 s α · Optb < (3/2) · Optb = 3

4
11
4
11
7
11 3
112
11
2
11
1
so s = 2
• So A can solve the NP-complete PARTITION problem in polynomial time!

4
11
4
11
7
11 3
112
11
2
11
1
so s = 2
• So A can solve the NP-complete PARTITION problem in polynomial time!
which implies that P = NP

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11
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11 3
112
11
2
11
1
Lemma There is no α-approximation for BINPACKING with α < 3/2
unless P = NP

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11
4
11
7
11 3
112
11
2
11
1
unless P = NP
We saw that First Fit Decreasing (FFD) is a 3/2-approximation

4
11
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11 3
112
11
2
11
1
unless P = NP
so this is the best we can do?

4
11
4
11
7
11 3
112
11
2
11
1
unless P = NP
so this is the best we can do?
In fact, FFD gives a solution using s bins with
Optb s
11
9
· Optb + 1

Asymptotic Polynomial time approximation schemes
An asymptotic polynomial time approximation scheme (APTAS)
for problem P is a family of algorithms:
For any constant > 0, A runs in polynomial time (poly-time)
There is a constant c > 0 such that
and outputs a solution s with Opt s (1 + ) · Opt + c

(this is the minimisation version of the deﬁnition)

An APTAS does not have to have running time polynomial in 1/

An asymptotic fully PTAS (AFPTAS) is also polynomial in 1/

An asymptotic fully PTAS (AFPTAS) is also polynomial in 1/
We will see an APTAS for BINPACKING
which uses at most (1 + ) · Opt + 1 bins

A special case of BINPACKING
We will now see a special case of BINPACKING which we can solve optimally in polynomial time
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We have n items but

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We have n items but
◦ There are cs (a constant) number of different item sizes

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We have n items but
◦ At most cb (another constant) items ﬁt in each bin

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We have n items but
Here cs = 3 and cb = 2

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7
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We have n items but
Here cs = 3 and cb = 2
How many different ways are there to ﬁll a single bin?

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1 2 3 4 5 6 7
Type:
• We can describe any packing of items into a single bin by its type
cs diff. item sizes
cb items ﬁt in each bin

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1 2 3 4 5 6 7
Type:
• The type of a packed bin determines how many of each item is packed
cs diff. item sizes

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1 2 3 4 5 6 7
Type:
we ignore rearrangement of items
cs diff. item sizes

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1 2 3 4 5 6 7
Type:
3
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4
8
cs diff. item sizes

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1 2 3 4 5 6 7
Type:
• How many types can there be?
cs diff. item sizes

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8 3
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1 2 3 4 5 6 7
Type:
There are between 0 and cb items of any one size
cs diff. item sizes

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3
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3
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1 2 3 4 5 6 7
Type:
There are cs different sizes
cs diff. item sizes

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1 2 3 4 5 6 7
Type:
The number of types
cs diff. item sizes

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8 3
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1 2 3 4 5 6 7
Type:
(cb + 1) × (cb + 1) × . . . × (cb + 1)
cs
The number of types
cs diff. item sizes

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3
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3
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1 2 3 4 5 6 7
Type:
(cb + 1) × (cb + 1) × . . . × (cb + 1)
cs
= (cb + 1)csThe number of types
cs diff. item sizes

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3
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8 3
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3 4 5 6
• We can completely describe any packing of S into b n bins
by the number of bins of each type used
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cs diff. item sizes

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8 3
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3 4 5 6
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1× Type 3
2× Type 4
3× Type 5
1× Type 6
0× Type 1
0× Type 2
0× Type 7
cs diff. item sizes

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3
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8 3
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3 4 5 6
we ignore rearrangement of bins
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3
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1× Type 3
2× Type 4
3× Type 5
1× Type 6
0× Type 1
0× Type 2
0× Type 7
cs diff. item sizes

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8 3
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3 4 5 6
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1× Type 3
2× Type 4
3× Type 5
1× Type 6
0× Type 1
0× Type 2
0× Type 7
cs diff. item sizes

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8 3
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3 4 5 6
• How different packings can there be?
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1× Type 3
2× Type 4
3× Type 5
1× Type 6
0× Type 1
0× Type 2
0× Type 7
cs diff. item sizes

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3
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8 3
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3 4 5 6
There are between 0 and n bins of any type
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1× Type 3
2× Type 4
3× Type 5
1× Type 6
0× Type 1
0× Type 2
0× Type 7
cs diff. item sizes

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8 3
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3 4 5 6
There are at most (cb + 1)cs different types
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1× Type 3
2× Type 4
3× Type 5
1× Type 6
0× Type 1
0× Type 2
0× Type 7
cs diff. item sizes

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3
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3
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8 3
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4
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3 4 5 6
7
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3
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3
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number of packings
1× Type 3
2× Type 4
3× Type 5
1× Type 6
0× Type 1
0× Type 2
0× Type 7
cs diff. item sizes

4
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3
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3
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8 3
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3 4 5 6
(n + 1) × (n + 1) × . . . × (n + 1)
(cb + 1)cs
7
8
4
3
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3
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5
3
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3
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5
number of packings
1× Type 3
2× Type 4
3× Type 5
1× Type 6
0× Type 1
0× Type 2
0× Type 7
cs diff. item sizes

4
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3
8
3
87
8 3
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3 4 5 6
(n + 1) × (n + 1) × . . . × (n + 1)
(cb + 1)cs
= (n+1)(cb+1)cs
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3
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3
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5
3
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number of packings
1× Type 3
2× Type 4
3× Type 5
1× Type 6
0× Type 1
0× Type 2
0× Type 7
cs diff. item sizes

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8 3
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3 4 5 6
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• However, not all these different packings are valid
1× Type 3
2× Type 4
3× Type 5
1× Type 6
0× Type 1
0× Type 2
0× Type 7
cs diff. item sizes

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Many of them use too few or many items of a particular size
1× Type 3
2× Type 4
3× Type 5
1× Type 6
0× Type 1
0× Type 2
0× Type 7
cs diff. item sizes
(in particular the example packing uses too many items of size 3/8)

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• We check each of the different packings to see if it is valid
1× Type 3
2× Type 4
3× Type 5
1× Type 6
0× Type 1
0× Type 2
0× Type 7
cs diff. item sizes

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8 3
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3 4 5 6
7
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5
and output the valid one which uses the fewest bins
1× Type 3
2× Type 4
3× Type 5
1× Type 6
0× Type 1
0× Type 2
0× Type 7
cs diff. item sizes

4
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8 3
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3 4 5 6
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5
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5
and output the valid one which uses the fewest bins
• This takes O(n · (n + 1)(cb+1)cs
) time and exactly solves BINPACKING
1× Type 3
2× Type 4
3× Type 5
1× Type 6
0× Type 1
0× Type 2
0× Type 7
(i.e. it outputs an optimal packing)
cs diff. item sizes

Towards an APTAS
The APTAS for BINPACKING will use a four step process:
Step 1 Remove all the small items
Step 2 Divide the items into a constant number of groups
Only cb of the remaining large items will ﬁt into a single bin
Sizes of items in each group are then rounded up
to match the size of the largest member
This will leave only cs different item sizes
Step 3 Use the poly-time algorithm to optimally pack the remaining special case
the constants cb and cs will depend on
Step 4 Reverse the grouping from Step 2 and then
Remember that the goal is to use b bins where Opt b (1 + ) · Opt + c (for some c)
greedily pack all the small items removed in Step 1
(and the largest group is removed)

Removing the small items
Lemma Let 0 < < 1. Given a packing of the items a ∈ S with size a > /2 into b bins,
in polynomial time we can ﬁnd a packing of all items in S which either uses:
b bins or (1 + )Opt + 1 bins

After packing of large items (> /2)

we pack the small items ( /2) on top of the large items greedily

pack the small items anywhere you like - just but don’t use an extra bin unless you have to

Either: We don’t use any extra bins
or every bin (except possibly the last) is 1 − ( /2) full

small items don’t ﬁt in here (so they are very well packed)
1 − 2

small items don’t ﬁt in here (so they are very well packed)
1 − 2
this is the +1

We can now ignore all the small items
and focus on ﬁnding a good packing of the large items

How many large items ﬁt in a single bin?

Each large item is larger than /2. . .

so at most 2/

so at most 2/ which is a constant :)

Reducing the number of item sizes
• We divide the items by size, into groups of size k
k k k
1
the smallest group might contain fewer than k items

k k k
1
• We deﬁne a new set of items S where each item is rounded up
so each item in a group has the same size

k k k
1
and the largest group is removed

k k k
1
• Notice that S contains only n/k distinct item sizes

k k k
1
• Notice that S contains only n/k distinct item sizes
(k will be big enough so that n/k 4/ 2 - which is a constant)

Lemma Let S be S after linear grouping (with groups of size k).
Opt(S ) Opt(S) Opt(S ) + k .
. Further, any packing of S can be converted into a packing of S in polynomial time
using at most k extra bins

.
Here Opt(S) is the optimal number of bins required to pack S
Similarly Opt(S ) is the optimal number of bins required to pack S
Further, any packing of S can be converted into a packing of S in polynomial time

S
S

Proof
S
S

Proof
S
S
If you can pack S into b bins,
you can pack S into b bins

Proof
S
S
Take the packing of S
and replace each item as shown
unpack
these

Proof
S
S
Each item from S is replaced
with one no larger from S
unpack
these

Proof
S
S
So the packing is valid and hence
Opt(S ) Opt(S)
unpack
these

Proof
S
S
you can pack S into b + k bins

Proof
S
S

Proof
S
S
The k largest items are
given their own extra bins
give these k items a bin each

Proof
S
S
This gives a valid packing of S

Proof
S
S
Opt(S) Opt(S ) + k
Hence,

Proof
S
S
Opt(S) Opt(S ) + k
Hence,
Note that both transformations
take polynomial time

.
We set k = n · ( 2/2) and let S be the set of large items which implies that. . .
n = |S|

.
k · Opt(S)
n = |S|

.
k · Opt(S) (because each of the n items in S has size at least /2)
n = |S|

.
If we can ﬁnd the optimal packing of S , which uses Opt(S ) bins
we can convert it into a packing of S which uses
Opt(S ) + k (1 + )Opt(S) bins
n = |S|

.
S contains 4/ 2 distinct item sizes
and only 2/ items ﬁt in each bin. . .
n = |S|

.
I.e. we can optimally pack S in polynomial time. . .
S contains 4/ 2 distinct item sizes
and only 2/ items ﬁt in each bin. . .
n = |S|

The overall APTAS
Step 1 Remove all the small items
Step 2 Divide the items into k = n · ( 2/2) different groups
Only cb = 2/ of the remaining large items will ﬁt into a single bin
Sizes of items in each group are then rounded up
to match the size of the largest member
This will leave only cs = 4/ 2 different item sizes
Step 3 Use the poly-time algorithm to optimally pack the remaining special case
Step 4 Reverse the grouping from Step 2 and then
greedily pack all the small items removed in Step 1
(and the largest group is removed)
This takes O n · (n + 1)(4/ 2
+1)
2/
time
Theorem For any 0 < < 1, the algorithm presented runs in polynomial time and returns a packing
of any set of items using at most (1 + )Opt + 1 bins
3 4 5 64 5 5

Conclusions
• There is no α-approximation for BINPACKING with α < 3/2
• We saw an APTAS for BINPACKING. (which uses at most (1 + ) · Opt + 1 bins)
• There is a poly-time algorithm which outputs a solution using at most,
unless P = NP
• This in turn implies that there is no PTAS for BINPACKING
unless P = NP
• The First Fit Decreasing algorithm uses at most,
11
9
· Opt + 1 bins
there is also an AFPTAS for bin packing
Opt+O(log2 Opt) = 1 +
O(log2 Opt)
Opt
·Opt bins

Conclusions
• There is no α-approximation for BINPACKING with α < 3/2
• We saw an APTAS for BINPACKING. (which uses at most (1 + ) · Opt + 1 bins)
• There is a poly-time algorithm which outputs a solution using at most,
unless P = NP
• This in turn implies that there is no PTAS for BINPACKING
unless P = NP
• The First Fit Decreasing algorithm uses at most,
11
9
· Opt + 1 bins
there is also an AFPTAS for bin packing
Opt+O(log2 Opt) = 1 +
O(log2 Opt)
Opt
·Opt bins
Whether there is a poly-time algorithm which uses at most Opt + 1 bins is an open problem

Approximation Algorithms Part Four: APTAS

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Approximation Algorithms Part Four: APTAS