Sage Reference Manual: Numerical Optimization - Mirrors

Sage Reference Manual: Numerical Optimization, Release 6.1.1
1.2 Examples
If your knapsack problem is composed of three items (weight, value) defined by (1,2), (1.5,1), (0.5,3), and a bag of
maximum weight 2, you can easily solve it this way:
sage: from sage.numerical.knapsack import knapsack
sage: knapsack( [(1,2), (1.5,1), (0.5,3)], max=2)
[5.0, [(1, 2), (0.500000000000000, 3)]]
1.3 Super-increasing sequences
We can test for whether or not a sequence is super-increasing:
sage: from sage.numerical.knapsack import Superincreasing
sage: L = [1, 2, 5, 21, 69, 189, 376, 919]
sage: seq = Superincreasing(L)
sage: seq
Super-increasing sequence of length 8
sage: seq.is_superincreasing()
True
sage: Superincreasing().is_superincreasing([1,3,5,7])
False
Solving the subset sum problem for a super-increasing sequence and target sum:
sage: L = [1, 2, 5, 21, 69, 189, 376, 919]
sage: Superincreasing(L).subset_sum(98)
[69, 21, 5, 2, 1]
class sage.numerical.knapsack.Superincreasing(seq=None)
Bases: sage.structure.sage_object.SageObject
A class for super-increasing sequences.
Let L = (a 1 , a 2 , a 3 , . . . , a n ) be a non-empty sequence of non-negative integers. Then L is said to be superincreasing
if each a i is strictly greater than the sum of all previous values. That is, for each a i ∈ L the sequence
L must satisfy the property
∑i−1
a i >
k=1
in order to be called a super-increasing sequence, where |L| ≥ 2. If L has only one element, it is also defined to
be a super-increasing sequence.
If seq is None, then construct an empty sequence. By definition, this empty sequence is not super-increasing.
INPUT:
•seq – (default: None) a non-empty sequence.
EXAMPLES:
sage: from sage.numerical.knapsack import Superincreasing
sage: L = [1, 2, 5, 21, 69, 189, 376, 919]
sage: Superincreasing(L).is_superincreasing()
True
sage: Superincreasing().is_superincreasing([1,3,5,7])
False
a k
2 Chapter 1. Knapsack Problems