These are my slides from the third year Advanced Algorithms course I used to give at the University of Bristol, UK.

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

2. 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)

3. 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)

4. The SUBSETSUM problem
4 4
7
322
t = 12

5. The SUBSETSUM problem
4 4
7
322
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

6. The SUBSETSUM problem
4 4
7
322
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
Decision Problem Is there a subset, S ⊆ S with SIZE(S ) = t?

7. The SUBSETSUM problem
4 4
7
322
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
Decision Problem Is there a subset, S ⊆ S with SIZE(S ) = t?
where SIZE(S ) = a∈S a

8. The SUBSETSUM problem
t = 12
4
4
2
7
32
• 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
Decision Problem Is there a subset, S ⊆ S with SIZE(S ) = t?
where SIZE(S ) = a∈S a

9. The SUBSETSUM problem
t = 12
7
3
2
4 42
• 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
Decision Problem Is there a subset, S ⊆ S with SIZE(S ) = t?
where SIZE(S ) = a∈S a

10. The SUBSETSUM problem
4 4
7
322
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
Decision Problem Is there a subset, S ⊆ S with SIZE(S ) = t?
where SIZE(S ) = a∈S a
(last lecture we discussed the NP-hard optimisation version)
The decision version is NP-complete

11. The PARTITION problem
4 4
7
322
The PARTITION problem is a special case of SUBSETSUM

12. The PARTITION problem
4 4
7
322
The input S is a multi-set of integers
The PARTITION problem is a special case of SUBSETSUM

13. The PARTITION problem
4 4
7
322
The input S is a multi-set of integers
but t is always half the sum of item sizes
The PARTITION problem is a special case of SUBSETSUM

14. The PARTITION problem
4 4
7
322
t = 11
The input S is a multi-set of integers
but t is always half the sum of item sizes
The PARTITION problem is a special case of SUBSETSUM

15. The PARTITION problem
4 4
7
322
t = 11
The input S is a multi-set of integers
but t is always half the sum of item sizes
t = SIZE(S)/2 = a∈S
a
2
The PARTITION problem is a special case of SUBSETSUM

16. The PARTITION problem
4 4
7
322
t = 11
The input S is a multi-set of integers
but t is always half the sum of item sizes
Decision Problem Is there a subset, S ⊆ S with SIZE(S ) = SIZE(S)/2?
t = SIZE(S)/2 = a∈S
a
2
The PARTITION problem is a special case of SUBSETSUM

17. The PARTITION problem
t = 11
4
4
2
7
32
The input S is a multi-set of integers
but t is always half the sum of item sizes
Decision Problem Is there a subset, S ⊆ S with SIZE(S ) = SIZE(S)/2?
t = SIZE(S)/2 = a∈S
a
2
The PARTITION problem is a special case of SUBSETSUM

18. The PARTITION problem
7
22
t = 11
The input S is a multi-set of integers
but t is always half the sum of item sizes
Decision Problem Is there a subset, S ⊆ S with SIZE(S ) = SIZE(S)/2?
t = SIZE(S)/2 = a∈S
a
2
4
4
3
The PARTITION problem is a special case of SUBSETSUM

19. The PARTITION problem
4 4
7
322
t = 11
The input S is a multi-set of integers
but t is always half the sum of item sizes
Decision Problem Is there a subset, S ⊆ S with SIZE(S ) = SIZE(S)/2?
t = SIZE(S)/2 = a∈S
a
2
The PARTITION problem is a special case of SUBSETSUM

20. The PARTITION problem
4 4
7
322
t = 11
The input S is a multi-set of integers
but t is always half the sum of item sizes
Decision Problem Is there a subset, S ⊆ S with SIZE(S ) = SIZE(S)/2?
t = SIZE(S)/2 = a∈S
a
2
The PARTITION problem is a special case of SUBSETSUM

21. The PARTITION problem
4 4
7
322
t = 11
The input S is a multi-set of integers
but t is always half the sum of item sizes
Decision Problem Is there a subset, S ⊆ S with SIZE(S ) = SIZE(S)/2?
t = SIZE(S)/2 = a∈S
a
2
The PARTITION problem is a special case of SUBSETSUM
Alternatively... Can we pack S into two bins of size SIZE(S)/2?

22. The PARTITION problem
The input S is a multi-set of integers
but t is always half the sum of item sizes
Decision Problem Is there a subset, S ⊆ S with SIZE(S ) = SIZE(S)/2?
t = SIZE(S)/2 = a∈S
a
2
The PARTITION problem is a special case of SUBSETSUM
Alternatively... Can we pack S into two bins of size SIZE(S)/2?
4 4
7
322
t = 11

23. The PARTITION problem
The input S is a multi-set of integers
but t is always half the sum of item sizes
Decision Problem Is there a subset, S ⊆ S with SIZE(S ) = SIZE(S)/2?
t = SIZE(S)/2 = a∈S
a
2
The PARTITION problem is a special case of SUBSETSUM
Alternatively... Can we pack S into two bins of size SIZE(S)/2?
4
4
3
7
2
2
t = 11

24. The PARTITION problem
The input S is a multi-set of integers
but t is always half the sum of item sizes
Decision Problem Is there a subset, S ⊆ S with SIZE(S ) = SIZE(S)/2?
t = SIZE(S)/2 = a∈S
a
2
The PARTITION problem is a special case of SUBSETSUM
Alternatively... Can we pack S into two bins of size SIZE(S)/2?
4
4
3
7
2
2
t = 11
The PARTITION problem is also NP-complete

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

26. PARTITION and BINPACKING
Key Idea Solve the PARTITION problem by approximating BINPACKING
4 4
7
322
t = 11

27. PARTITION and BINPACKING
Key Idea Solve the PARTITION problem by approximating BINPACKING
4 4
7
322
t = 11
• Convert the PARTITION instance into a BINPACKING problem
by dividing all item and bin sizes by t = SIZE(S)/2

28. PARTITION and BINPACKING
Key Idea Solve the PARTITION problem by approximating BINPACKING
• Convert the PARTITION instance into a BINPACKING problem
by dividing all item and bin sizes by t = SIZE(S)/2
4
11
4
11
7
11 3
112
11
2
11
1

29. PARTITION and BINPACKING
Key Idea Solve the PARTITION problem by approximating BINPACKING
• Convert the PARTITION instance into a BINPACKING problem
by dividing all item and bin sizes by t = SIZE(S)/2
• Now all items are have size at most 1 and the bins have size 1
4
11
4
11
7
11 3
112
11
2
11
1

30. PARTITION and BINPACKING
Key Idea Solve the PARTITION problem by approximating BINPACKING
• Convert the PARTITION instance into a BINPACKING problem
by dividing all item and bin sizes by t = SIZE(S)/2
• Now all items are have size at most 1 and the bins have size 1
• The optimal number of bins Optb is 2 iff
the answer to the PARTITION instance is ‘yes’
4
11
4
11
7
11 3
112
11
2
11
1

31. PARTITION and BINPACKING
Key Idea Solve the PARTITION problem by approximating BINPACKING
• Convert the PARTITION instance into a BINPACKING problem
by dividing all item and bin sizes by t = SIZE(S)/2
• Now all items are have size at most 1 and the bins have size 1
• The optimal number of bins Optb is 2 iff
the answer to the PARTITION instance is ‘yes’
4
11
4
11
7
11 3
112
11
2
11
1
• What does this tell us about approximating BINPACKING?

32. PARTITION and BINPACKING
Key Idea Solve the PARTITION problem by approximating BINPACKING
4
11
4
11
7
11 3
112
11
2
11
1
• Assume A is an α-approximation for BINPACKING with α < 3/2

33. PARTITION and BINPACKING
Key Idea Solve the PARTITION problem by approximating BINPACKING
4
11
4
11
7
11 3
112
11
2
11
1
• Assume A is an α-approximation for BINPACKING with α < 3/2
A outputs a solution using s bins with Optb s α · Optb

34. PARTITION and BINPACKING
Key Idea Solve the PARTITION problem by approximating BINPACKING
4
11
4
11
7
11 3
112
11
2
11
1
• Assume A is an α-approximation for BINPACKING with α < 3/2
A outputs a solution using s bins with Optb s α · Optb
• If Optb > 2 then s > 2

35. PARTITION and BINPACKING
Key Idea Solve the PARTITION problem by approximating BINPACKING
4
11
4
11
7
11 3
112
11
2
11
1
• Assume A is an α-approximation for BINPACKING with α < 3/2
A outputs a solution using s bins with Optb s α · Optb
• If Optb > 2 then s > 2
• If Optb = 2 then 2 s α · Optb < (3/2) · Optb = 3

36. PARTITION and BINPACKING
Key Idea Solve the PARTITION problem by approximating BINPACKING
4
11
4
11
7
11 3
112
11
2
11
1
• Assume A is an α-approximation for BINPACKING with α < 3/2
A outputs a solution using s bins with Optb s α · Optb
• If Optb > 2 then s > 2
• If Optb = 2 then 2 s α · Optb < (3/2) · Optb = 3

37. PARTITION and BINPACKING
Key Idea Solve the PARTITION problem by approximating BINPACKING
4
11
4
11
7
11 3
112
11
2
11
1
• Assume A is an α-approximation for BINPACKING with α < 3/2
A outputs a solution using s bins with Optb s α · Optb
• If Optb > 2 then s > 2
so s = 2• If Optb = 2 then 2 s α · Optb < (3/2) · Optb = 3

38. PARTITION and BINPACKING
Key Idea Solve the PARTITION problem by approximating BINPACKING
4
11
4
11
7
11 3
112
11
2
11
1
• Assume A is an α-approximation for BINPACKING with α < 3/2
A outputs a solution using s bins with Optb s α · Optb
• If Optb > 2 then s > 2
so s = 2
• So A can solve the NP-complete PARTITION problem in polynomial time!
• If Optb = 2 then 2 s α · Optb < (3/2) · Optb = 3

39. PARTITION and BINPACKING
Key Idea Solve the PARTITION problem by approximating BINPACKING
4
11
4
11
7
11 3
112
11
2
11
1
• Assume A is an α-approximation for BINPACKING with α < 3/2
A outputs a solution using s bins with Optb s α · Optb
• If Optb > 2 then s > 2
so s = 2
• So A can solve the NP-complete PARTITION problem in polynomial time!
which implies that P = NP
• If Optb = 2 then 2 s α · Optb < (3/2) · Optb = 3

40. PARTITION and BINPACKING
Key Idea Solve the PARTITION problem by approximating BINPACKING
4
11
4
11
7
11 3
112
11
2
11
1
Lemma There is no α-approximation for BINPACKING with α < 3/2
unless P = NP

41. PARTITION and BINPACKING
Key Idea Solve the PARTITION problem by approximating BINPACKING
4
11
4
11
7
11 3
112
11
2
11
1
Lemma There is no α-approximation for BINPACKING with α < 3/2
unless P = NP
We saw that First Fit Decreasing (FFD) is a 3/2-approximation

42. PARTITION and BINPACKING
Key Idea Solve the PARTITION problem by approximating BINPACKING
4
11
4
11
7
11 3
112
11
2
11
1
Lemma There is no α-approximation for BINPACKING with α < 3/2
unless P = NP
We saw that First Fit Decreasing (FFD) is a 3/2-approximation
so this is the best we can do?

43. PARTITION and BINPACKING
Key Idea Solve the PARTITION problem by approximating BINPACKING
4
11
4
11
7
11 3
112
11
2
11
1
Lemma There is no α-approximation for BINPACKING with α < 3/2
unless P = NP
We saw that First Fit Decreasing (FFD) is a 3/2-approximation
so this is the best we can do?
In fact, FFD gives a solution using s bins with
Optb s
11
9
· Optb + 1

44. 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

45. 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)

46. 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)
An APTAS does not have to have running time polynomial in 1/
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)

47. 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)
An APTAS does not have to have running time polynomial in 1/
An asymptotic fully PTAS (AFPTAS) is also polynomial in 1/
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)

48. 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)
An APTAS does not have to have running time polynomial in 1/
An asymptotic fully PTAS (AFPTAS) is also polynomial in 1/
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)
We will see an APTAS for BINPACKING
which uses at most (1 + ) · Opt + 1 bins

49. A special case of BINPACKING
We will now see a special case of BINPACKING which we can solve optimally in polynomial time
4
8
7
8
3
8
1
4
8
4
8 7
8
7
8
4
8
3
8

50. A special case of BINPACKING
We will now see a special case of BINPACKING which we can solve optimally in polynomial time
4
8
7
8
3
8
1
4
8
4
8 7
8
7
8
4
8
3
8
We have n items but

51. A special case of BINPACKING
We will now see a special case of BINPACKING which we can solve optimally in polynomial time
4
8
7
8
3
8
1
4
8
4
8 7
8
7
8
4
8
3
8
We have n items but
◦ There are cs (a constant) number of different item sizes

52. A special case of BINPACKING
We will now see a special case of BINPACKING which we can solve optimally in polynomial time
4
8
7
8
3
8
1
4
8
4
8 7
8
7
8
4
8
3
8
We have n items but
◦ There are cs (a constant) number of different item sizes
◦ At most cb (another constant) items fit in each bin

53. A special case of BINPACKING
We will now see a special case of BINPACKING which we can solve optimally in polynomial time
4
8
7
8
3
8
1
4
8
4
8 7
8
7
8
4
8
3
8
We have n items but
◦ There are cs (a constant) number of different item sizes
◦ At most cb (another constant) items fit in each bin
Here cs = 3 and cb = 2

54. A special case of BINPACKING
We will now see a special case of BINPACKING which we can solve optimally in polynomial time
4
8
7
8
3
8
1
4
8
4
8 7
8
7
8
4
8
3
8
We have n items but
◦ There are cs (a constant) number of different item sizes
◦ At most cb (another constant) items fit in each bin
Here cs = 3 and cb = 2
How many different ways are there to fill a single bin?

55. A special case of BINPACKING
4
8
3
8
3
8
3
8
4
8
7
8 3
8
4
8
4
8
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

56. A special case of BINPACKING
4
8
3
8
3
8
3
8
4
8
7
8 3
8
4
8
4
8
1 2 3 4 5 6 7
Type:
• We can describe any packing of items into a single bin by its type
• The type of a packed bin determines how many of each item is packed
cs diff. item sizes
cb items fit in each bin

57. A special case of BINPACKING
4
8
3
8
3
8
3
8
4
8
7
8 3
8
4
8
4
8
1 2 3 4 5 6 7
Type:
• We can describe any packing of items into a single bin by its type
• The type of a packed bin determines how many of each item is packed
we ignore rearrangement of items
cs diff. item sizes
cb items fit in each bin

58. A special case of BINPACKING
4
8
3
8
3
8
3
8
4
8
7
8 3
8
4
8
4
8
1 2 3 4 5 6 7
Type:
• We can describe any packing of items into a single bin by its type
• The type of a packed bin determines how many of each item is packed
we ignore rearrangement of items
3
8
4
8
cs diff. item sizes
cb items fit in each bin

59. A special case of BINPACKING
4
8
3
8
3
8
3
8
4
8
7
8 3
8
4
8
4
8
1 2 3 4 5 6 7
Type:
• We can describe any packing of items into a single bin by its type
• The type of a packed bin determines how many of each item is packed
we ignore rearrangement of items
cs diff. item sizes
cb items fit in each bin

60. A special case of BINPACKING
4
8
3
8
3
8
3
8
4
8
7
8 3
8
4
8
4
8
1 2 3 4 5 6 7
Type:
• We can describe any packing of items into a single bin by its type
• The type of a packed bin determines how many of each item is packed
we ignore rearrangement of items
• How many types can there be?
cs diff. item sizes
cb items fit in each bin

61. A special case of BINPACKING
4
8
3
8
3
8
3
8
4
8
7
8 3
8
4
8
4
8
1 2 3 4 5 6 7
Type:
• We can describe any packing of items into a single bin by its type
• The type of a packed bin determines how many of each item is packed
we ignore rearrangement of items
• How many types can there be?
There are between 0 and cb items of any one size
cs diff. item sizes
cb items fit in each bin

62. A special case of BINPACKING
4
8
3
8
3
8
3
8
4
8
7
8 3
8
4
8
4
8
1 2 3 4 5 6 7
Type:
• We can describe any packing of items into a single bin by its type
• The type of a packed bin determines how many of each item is packed
we ignore rearrangement of items
• How many types can there be?
There are between 0 and cb items of any one size
There are cs different sizes
cs diff. item sizes
cb items fit in each bin

63. A special case of BINPACKING
4
8
3
8
3
8
3
8
4
8
7
8 3
8
4
8
4
8
1 2 3 4 5 6 7
Type:
• We can describe any packing of items into a single bin by its type
• The type of a packed bin determines how many of each item is packed
we ignore rearrangement of items
• How many types can there be?
There are between 0 and cb items of any one size
There are cs different sizes
The number of types
cs diff. item sizes
cb items fit in each bin

64. A special case of BINPACKING
4
8
3
8
3
8
3
8
4
8
7
8 3
8
4
8
4
8
1 2 3 4 5 6 7
Type:
• We can describe any packing of items into a single bin by its type
• The type of a packed bin determines how many of each item is packed
we ignore rearrangement of items
• How many types can there be?
There are between 0 and cb items of any one size
There are cs different sizes
(cb + 1) × (cb + 1) × . . . × (cb + 1)
cs
The number of types
cs diff. item sizes
cb items fit in each bin

65. A special case of BINPACKING
4
8
3
8
3
8
3
8
4
8
7
8 3
8
4
8
4
8
1 2 3 4 5 6 7
Type:
• We can describe any packing of items into a single bin by its type
• The type of a packed bin determines how many of each item is packed
we ignore rearrangement of items
• How many types can there be?
There are between 0 and cb items of any one size
There are cs different sizes
(cb + 1) × (cb + 1) × . . . × (cb + 1)
cs
= (cb + 1)csThe number of types
cs diff. item sizes
cb items fit in each bin

66. A special case of BINPACKING
4
8
3
8
3
87
8 3
8
4
8
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
7
8
4
3
8
3
8
5
3
8
3
8
5
cb items fit in each bin
cs diff. item sizes

67. A special case of BINPACKING
4
8
3
8
3
87
8 3
8
4
8
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
7
8
4
3
8
3
8
5
3
8
3
8
5
1× Type 3
2× Type 4
3× Type 5
1× Type 6
0× Type 1
0× Type 2
0× Type 7
cb items fit in each bin
cs diff. item sizes

68. A special case of BINPACKING
4
8
3
8
3
87
8 3
8
4
8
3 4 5 6
• We can completely describe any packing of S into b n bins
we ignore rearrangement of bins
by the number of bins of each type used
7
8
4
3
8
3
8
5
3
8
3
8
5
1× Type 3
2× Type 4
3× Type 5
1× Type 6
0× Type 1
0× Type 2
0× Type 7
cb items fit in each bin
cs diff. item sizes

69. A special case of BINPACKING
• We can completely describe any packing of S into b n bins
we ignore rearrangement of bins
by the number of bins of each type used
4
8
3
8
3
87
8 3
8
4
8
3 4 5 6
7
8
4
3
8
3
8
5
3
8
3
8
5
1× Type 3
2× Type 4
3× Type 5
1× Type 6
0× Type 1
0× Type 2
0× Type 7
cb items fit in each bin
cs diff. item sizes

70. A special case of BINPACKING
4
8
3
8
3
87
8 3
8
4
8
3 4 5 6
• We can completely describe any packing of S into b n bins
we ignore rearrangement of bins
by the number of bins of each type used
7
8
4
3
8
3
8
5
3
8
3
8
5
1× Type 3
2× Type 4
3× Type 5
1× Type 6
0× Type 1
0× Type 2
0× Type 7
cb items fit in each bin
cs diff. item sizes

71. A special case of BINPACKING
4
8
3
8
3
87
8 3
8
4
8
3 4 5 6
• We can completely describe any packing of S into b n bins
we ignore rearrangement of bins
• How different packings can there be?
by the number of bins of each type used
7
8
4
3
8
3
8
5
3
8
3
8
5
1× Type 3
2× Type 4
3× Type 5
1× Type 6
0× Type 1
0× Type 2
0× Type 7
cb items fit in each bin
cs diff. item sizes

72. A special case of BINPACKING
4
8
3
8
3
87
8 3
8
4
8
3 4 5 6
• We can completely describe any packing of S into b n bins
we ignore rearrangement of bins
• How different packings can there be?
There are between 0 and n bins of any type
by the number of bins of each type used
7
8
4
3
8
3
8
5
3
8
3
8
5
1× Type 3
2× Type 4
3× Type 5
1× Type 6
0× Type 1
0× Type 2
0× Type 7
cb items fit in each bin
cs diff. item sizes

73. A special case of BINPACKING
4
8
3
8
3
87
8 3
8
4
8
3 4 5 6
• We can completely describe any packing of S into b n bins
we ignore rearrangement of bins
• How different packings can there be?
There are between 0 and n bins of any type
There are at most (cb + 1)cs different types
by the number of bins of each type used
7
8
4
3
8
3
8
5
3
8
3
8
5
1× Type 3
2× Type 4
3× Type 5
1× Type 6
0× Type 1
0× Type 2
0× Type 7
cb items fit in each bin
cs diff. item sizes

74. A special case of BINPACKING
4
8
3
8
3
87
8 3
8
4
8
3 4 5 6
• We can completely describe any packing of S into b n bins
we ignore rearrangement of bins
• How different packings can there be?
There are between 0 and n bins of any type
There are at most (cb + 1)cs different types
by the number of bins of each type used
7
8
4
3
8
3
8
5
3
8
3
8
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
cb items fit in each bin
cs diff. item sizes

75. A special case of BINPACKING
4
8
3
8
3
87
8 3
8
4
8
3 4 5 6
• We can completely describe any packing of S into b n bins
we ignore rearrangement of bins
• How different packings can there be?
There are between 0 and n bins of any type
There are at most (cb + 1)cs different types
(n + 1) × (n + 1) × . . . × (n + 1)
(cb + 1)cs
by the number of bins of each type used
7
8
4
3
8
3
8
5
3
8
3
8
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
cb items fit in each bin
cs diff. item sizes

76. A special case of BINPACKING
4
8
3
8
3
87
8 3
8
4
8
3 4 5 6
• We can completely describe any packing of S into b n bins
we ignore rearrangement of bins
• How different packings can there be?
There are between 0 and n bins of any type
There are at most (cb + 1)cs different types
(n + 1) × (n + 1) × . . . × (n + 1)
(cb + 1)cs
= (n+1)(cb+1)cs
by the number of bins of each type used
7
8
4
3
8
3
8
5
3
8
3
8
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
cb items fit in each bin
cs diff. item sizes

77. A special case of BINPACKING
4
8
3
8
3
87
8 3
8
4
8
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
7
8
4
3
8
3
8
5
3
8
3
8
5
1× Type 3
2× Type 4
3× Type 5
1× Type 6
0× Type 1
0× Type 2
0× Type 7
cb items fit in each bin
cs diff. item sizes

78. A special case of BINPACKING
4
8
3
8
3
87
8 3
8
4
8
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
7
8
4
3
8
3
8
5
3
8
3
8
5
• 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
cb items fit in each bin
cs diff. item sizes

79. A special case of BINPACKING
4
8
3
8
3
87
8 3
8
4
8
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
7
8
4
3
8
3
8
5
3
8
3
8
5
• However, not all these different packings are valid
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
cb items fit in each bin
cs diff. item sizes
(in particular the example packing uses too many items of size 3/8)

80. A special case of BINPACKING
4
8
3
8
3
87
8 3
8
4
8
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
7
8
4
3
8
3
8
5
3
8
3
8
5
• However, not all these different packings are valid
Many of them use too few or many items of a particular size
• 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
cb items fit in each bin
cs diff. item sizes
(in particular the example packing uses too many items of size 3/8)

81. A special case of BINPACKING
4
8
3
8
3
87
8 3
8
4
8
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
7
8
4
3
8
3
8
5
3
8
3
8
5
• However, not all these different packings are valid
Many of them use too few or many items of a particular size
• We check each of the different packings to see if it is valid
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
cb items fit in each bin
cs diff. item sizes
(in particular the example packing uses too many items of size 3/8)

82. A special case of BINPACKING
4
8
3
8
3
87
8 3
8
4
8
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
7
8
4
3
8
3
8
5
3
8
3
8
5
• However, not all these different packings are valid
Many of them use too few or many items of a particular size
• We check each of the different packings to see if it is valid
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)
cb items fit in each bin
cs diff. item sizes
(in particular the example packing uses too many items of size 3/8)

83. 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)

84. 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

85. 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)

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

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

88. Removing the small items
we pack the small items ( /2) on top of the large items greedily
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)
pack the small items anywhere you like - just but don’t use an extra bin unless you have to

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

90. Removing the small items
we pack the small items ( /2) on top of the large items greedily
Either: We don’t use any extra bins
or every bin (except possibly the last) is 1 − ( /2) full
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)

91. Removing the small items
we pack the small items ( /2) on top of the large items greedily
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:
After packing of large items (> /2)
b bins or (1 + )Opt + 1 bins
Either: We don’t use any extra bins
or every bin (except possibly the last) is 1 − ( /2) full

92. Removing the small items
we pack the small items ( /2) on top of the large items greedily
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:
After packing of large items (> /2)
Either: We don’t use any extra bins
or every bin (except possibly the last) is 1 − ( /2) full
b bins or (1 + )Opt + 1 bins

93. Removing the small items
we pack the small items ( /2) on top of the large items greedily
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:
After packing of large items (> /2)
Either: We don’t use any extra bins
or every bin (except possibly the last) is 1 − ( /2) full
b bins or (1 + )Opt + 1 bins

94. Removing the small items
we pack the small items ( /2) on top of the large items greedily
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:
After packing of large items (> /2)
Either: We don’t use any extra bins
or every bin (except possibly the last) is 1 − ( /2) full
b bins or (1 + )Opt + 1 bins
small items don’t fit in here (so they are very well packed)
1 − 2

95. Removing the small items
we pack the small items ( /2) on top of the large items greedily
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:
After packing of large items (> /2)
Either: We don’t use any extra bins
or every bin (except possibly the last) is 1 − ( /2) full
b bins or (1 + )Opt + 1 bins
small items don’t fit in here (so they are very well packed)
1 − 2
this is the +1

96. 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
We can now ignore all the small items
and focus on finding a good packing of the large items

97. 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
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?

98. 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
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. . .

99. 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
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/

100. 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
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/ which is a constant :)

101. 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

102. 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
• We define a new set of items S where each item is rounded up
so each item in a group has the same size

103. 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
• We define a new set of items S where each item is rounded up
so each item in a group has the same size

104. 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
• We define a new set of items S where each item is rounded up
so each item in a group has the same size
and the largest group is removed

105. 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
• We define a new set of items S where each item is rounded up
so each item in a group has the same size
• Notice that S contains only n/k distinct item sizes
and the largest group is removed

106. 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
• We define a new set of items S where each item is rounded up
so each item in a group has the same size
• Notice that S contains only n/k distinct item sizes
and the largest group is removed
(k will be big enough so that n/k 4/ 2 - which is a constant)

107. Reducing the number of item sizes
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

108. Reducing the number of item sizes
Lemma Let S be S after linear grouping (with groups of size k).
Opt(S ) Opt(S) Opt(S ) + k .
.
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
using at most k extra bins

109. Reducing the number of item sizes
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

110. Reducing the number of item sizes
Lemma Let S be S after linear grouping (with groups of size k).
Opt(S ) Opt(S) Opt(S ) + k .
S
S

111. Reducing the number of item sizes
Proof
Lemma Let S be S after linear grouping (with groups of size k).
Opt(S ) Opt(S) Opt(S ) + k .
S
S

112. Reducing the number of item sizes
Proof
Lemma Let S be S after linear grouping (with groups of size k).
Opt(S ) Opt(S) Opt(S ) + k .
S
S
If you can pack S into b bins,
you can pack S into b bins

113. Reducing the number of item sizes
Proof
Lemma Let S be S after linear grouping (with groups of size k).
Opt(S ) Opt(S) Opt(S ) + k .
S
S
If you can pack S into b bins,
you can pack S into b bins
Take the packing of S
and replace each item as shown
unpack
these

114. Reducing the number of item sizes
Proof
Lemma Let S be S after linear grouping (with groups of size k).
Opt(S ) Opt(S) Opt(S ) + k .
S
S
If you can pack S into b bins,
you can pack S into b bins
Take the packing of S
and replace each item as shown
Each item from S is replaced
with one no larger from S
unpack
these

115. Reducing the number of item sizes
Proof
Lemma Let S be S after linear grouping (with groups of size k).
Opt(S ) Opt(S) Opt(S ) + k .
S
S
If you can pack S into b bins,
you can pack S into b bins
Take the packing of S
and replace each item as shown
Each item from S is replaced
with one no larger from S
So the packing is valid and hence
Opt(S ) Opt(S)
unpack
these

116. Reducing the number of item sizes
Proof
Lemma Let S be S after linear grouping (with groups of size k).
Opt(S ) Opt(S) Opt(S ) + k .
S
S
If you can pack S into b bins,
you can pack S into b + k bins

117. Reducing the number of item sizes
Proof
Lemma Let S be S after linear grouping (with groups of size k).
Opt(S ) Opt(S) Opt(S ) + k .
S
S
If you can pack S into b bins,
you can pack S into b + k bins
Take the packing of S
and replace each item as shown

118. Reducing the number of item sizes
Proof
Lemma Let S be S after linear grouping (with groups of size k).
Opt(S ) Opt(S) Opt(S ) + k .
S
S
If you can pack S into b bins,
you can pack S into b + k bins
Take the packing of S
and replace each item as shown

119. Reducing the number of item sizes
Proof
Lemma Let S be S after linear grouping (with groups of size k).
Opt(S ) Opt(S) Opt(S ) + k .
S
S
If you can pack S into b bins,
you can pack S into b + k bins
Take the packing of S
and replace each item as shown
Each item from S is replaced
with one no larger from S

120. Reducing the number of item sizes
Proof
Lemma Let S be S after linear grouping (with groups of size k).
Opt(S ) Opt(S) Opt(S ) + k .
S
S
If you can pack S into b bins,
you can pack S into b + k bins
Take the packing of S
and replace each item as shown
Each item from S is replaced
with one no larger from S
The k largest items are
given their own extra bins
give these k items a bin each

121. Reducing the number of item sizes
Proof
Lemma Let S be S after linear grouping (with groups of size k).
Opt(S ) Opt(S) Opt(S ) + k .
S
S
If you can pack S into b bins,
you can pack S into b + k bins

122. Reducing the number of item sizes
Proof
Lemma Let S be S after linear grouping (with groups of size k).
Opt(S ) Opt(S) Opt(S ) + k .
S
S
If you can pack S into b bins,
you can pack S into b + k bins
This gives a valid packing of S

123. Reducing the number of item sizes
Proof
Lemma Let S be S after linear grouping (with groups of size k).
Opt(S ) Opt(S) Opt(S ) + k .
S
S
If you can pack S into b bins,
you can pack S into b + k bins
Opt(S) Opt(S ) + k
This gives a valid packing of S
Hence,

124. Reducing the number of item sizes
Proof
Lemma Let S be S after linear grouping (with groups of size k).
Opt(S ) Opt(S) Opt(S ) + k .
S
S
If you can pack S into b bins,
you can pack S into b + k bins
Opt(S) Opt(S ) + k
This gives a valid packing of S
Hence,
Note that both transformations
take polynomial time

125. Reducing the number of item sizes
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

126. Reducing the number of item sizes
Lemma Let S be S after linear grouping (with groups of size k).
Opt(S ) Opt(S) Opt(S ) + k .
.
We set k = n · ( 2/2) and let S be the set of large items which implies that. . .
n = |S|
Further, any packing of S can be converted into a packing of S in polynomial time
using at most k extra bins

127. Reducing the number of item sizes
Lemma Let S be S after linear grouping (with groups of size k).
Opt(S ) Opt(S) Opt(S ) + k .
.
We set k = n · ( 2/2) and let S be the set of large items which implies that. . .
k · Opt(S)
n = |S|
Further, any packing of S can be converted into a packing of S in polynomial time
using at most k extra bins

128. Reducing the number of item sizes
Lemma Let S be S after linear grouping (with groups of size k).
Opt(S ) Opt(S) Opt(S ) + k .
.
We set k = n · ( 2/2) and let S be the set of large items which implies that. . .
k · Opt(S) (because each of the n items in S has size at least /2)
n = |S|
Further, any packing of S can be converted into a packing of S in polynomial time
using at most k extra bins

129. Reducing the number of item sizes
Lemma Let S be S after linear grouping (with groups of size k).
Opt(S ) Opt(S) Opt(S ) + k .
.
We set k = n · ( 2/2) and let S be the set of large items which implies that. . .
k · Opt(S) (because each of the n items in S has size at least /2)
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|
Further, any packing of S can be converted into a packing of S in polynomial time
using at most k extra bins

130. Reducing the number of item sizes
Lemma Let S be S after linear grouping (with groups of size k).
Opt(S ) Opt(S) Opt(S ) + k .
.
We set k = n · ( 2/2) and let S be the set of large items which implies that. . .
k · Opt(S) (because each of the n items in S has size at least /2)
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
S contains 4/ 2 distinct item sizes
and only 2/ items fit in each bin. . .
n = |S|
Further, any packing of S can be converted into a packing of S in polynomial time
using at most k extra bins

131. Reducing the number of item sizes
Lemma Let S be S after linear grouping (with groups of size k).
Opt(S ) Opt(S) Opt(S ) + k .
.
We set k = n · ( 2/2) and let S be the set of large items which implies that. . .
k · Opt(S) (because each of the n items in S has size at least /2)
I.e. we can optimally pack S in polynomial time. . .
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
S contains 4/ 2 distinct item sizes
and only 2/ items fit in each bin. . .
n = |S|
Further, any packing of S can be converted into a packing of S in polynomial time
using at most k extra bins

132. 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

133. 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

134. 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