Knapsack Problems#
This module implements a number of solutions to various knapsack problems, otherwise known as linear integer programming problems. Solutions to the following knapsack problems are implemented:
Solving the subset sum problem for super-increasing sequences.
General case using Linear Programming
AUTHORS:
Minh Van Nguyen (2009-04): initial version
Nathann Cohen (2009-08): Linear Programming version
Definition of Knapsack problems#
You have already had a knapsack problem, so you should know, but in case you do not, a knapsack problem is what happens when you have hundred of items to put into a bag which is too small, and you want to pack the most useful of them.
When you formally write it, here is your problem:
Your bag can contain a weight of at most \(W\).
Each item \(i\) has a weight \(w_i\).
Each item \(i\) has a usefulness \(u_i\).
You then want to maximize the total usefulness of the items you will store into your bag, while keeping sure the weight of the bag will not go over \(W\).
As a linear program, this problem can be represented this way (if you define \(b_i\) as the binary variable indicating whether the item \(i\) is to be included in your bag):
(For more information, see the Wikipedia article Knapsack_problem)
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)]]
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 \(L = (a_1, a_2, a_3, \dots, a_n)\) be a non-empty sequence of non-negative integers. Then \(L\) is said to be super-increasing if each \(a_i\) is strictly greater than the sum of all previous values. That is, for each \(a_i \in L\) the sequence \(L\) must satisfy the property
\[a_i > \sum_{k=1}^{i-1} a_k\]in order to be called a super-increasing sequence, where \(|L| \geq 2\). If \(L\) has only one element, it is also defined to be a super-increasing sequence.
If
seq
isNone
, 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 sage: seq = Superincreasing(); seq An empty sequence. sage: seq = Superincreasing([1, 3, 6]); seq Super-increasing sequence of length 3 sage: seq = Superincreasing([1, 2, 5, 21, 69, 189, 376, 919]); seq Super-increasing sequence of length 8
- is_superincreasing(seq=None)#
Determine whether or not
seq
is super-increasing.If
seq=None
then determine whether or notself
is super-increasing.Let \(L = (a_1, a_2, a_3, \dots, a_n)\) be a non-empty sequence of non-negative integers. Then \(L\) is said to be super-increasing if each \(a_i\) is strictly greater than the sum of all previous values. That is, for each \(a_i \in L\) the sequence \(L\) must satisfy the property
\[a_i > \sum_{k=1}^{i-1} a_k\]in order to be called a super-increasing sequence, where \(|L| \geq 2\). If \(L\) has exactly one element, then it is also defined to be a super-increasing sequence.
INPUT:
seq
– (default:None
) a sequence to test
OUTPUT:
If
seq
isNone
, then testself
to determine whether or not it is super-increasing. In that case, returnTrue
ifself
is super-increasing;False
otherwise.If
seq
is notNone
, then testseq
to determine whether or not it is super-increasing. ReturnTrue
ifseq
is super-increasing;False
otherwise.
EXAMPLES:
By definition, an empty sequence is not super-increasing:
sage: from sage.numerical.knapsack import Superincreasing sage: Superincreasing().is_superincreasing([]) False sage: Superincreasing().is_superincreasing() False sage: Superincreasing().is_superincreasing(tuple()) False sage: Superincreasing().is_superincreasing(()) False
But here is an example of a super-increasing sequence:
sage: L = [1, 2, 5, 21, 69, 189, 376, 919] sage: Superincreasing(L).is_superincreasing() True sage: L = (1, 2, 5, 21, 69, 189, 376, 919) sage: Superincreasing(L).is_superincreasing() True
A super-increasing sequence can have zero as one of its elements:
sage: L = [0, 1, 2, 4] sage: Superincreasing(L).is_superincreasing() True
A super-increasing sequence can be of length 1:
sage: Superincreasing([randint(0, 100)]).is_superincreasing() True
- largest_less_than(N)#
Return the largest integer in the sequence
self
that is less than or equal toN
.This function narrows down the candidate solution using a binary trim, similar to the way binary search halves the sequence at each iteration.
INPUT:
N
– integer; the target value to search for.
OUTPUT:
The largest integer in
self
that is less than or equal toN
. If no solution exists, then returnNone
.EXAMPLES:
When a solution is found, return it:
sage: from sage.numerical.knapsack import Superincreasing sage: L = [2, 3, 7, 25, 67, 179, 356, 819] sage: Superincreasing(L).largest_less_than(207) 179 sage: L = (2, 3, 7, 25, 67, 179, 356, 819) sage: Superincreasing(L).largest_less_than(2) 2
But if no solution exists, return
None
:sage: L = [2, 3, 7, 25, 67, 179, 356, 819] sage: Superincreasing(L).largest_less_than(-1) is None True
- subset_sum(N)#
Solving the subset sum problem for a super-increasing sequence.
Let \(S = (s_1, s_2, s_3, \dots, s_n)\) be a non-empty sequence of non-negative integers, and let \(N \in \ZZ\) be non-negative. The subset sum problem asks for a subset \(A \subseteq S\) all of whose elements sum to \(N\). This method specializes the subset sum problem to the case of super-increasing sequences. If a solution exists, then it is also a super-increasing sequence.
Note
This method only solves the subset sum problem for super-increasing sequences. In general, solving the subset sum problem for an arbitrary sequence is known to be computationally hard.
INPUT:
N
– a non-negative integer.
OUTPUT:
A non-empty subset of
self
whose elements sum toN
. This subset is also a super-increasing sequence. If no such subset exists, then return the empty list.
ALGORITHMS:
The algorithm used is adapted from page 355 of [HPS2008].
EXAMPLES:
Solving the subset sum problem for a super-increasing sequence and target sum:
sage: from sage.numerical.knapsack import Superincreasing sage: L = [1, 2, 5, 21, 69, 189, 376, 919] sage: Superincreasing(L).subset_sum(98) [69, 21, 5, 2, 1]
- sage.numerical.knapsack.knapsack(seq, binary, max=True, value_only=1, solver=False, verbose=None, integrality_tolerance=0)#
Solves the knapsack problem
For more information on the knapsack problem, see the documentation of the
knapsack module
or the Wikipedia article Knapsack_problem.INPUT:
seq
– Two different possible types:A sequence of tuples
(weight, value, something1, something2, ...)
. Note that only the first two coordinates (weight
andvalues
) will be taken into account. The rest (if any) will be ignored. This can be useful if you need to attach some information to the items.A sequence of reals (a value of 1 is assumed).
binary
– When set toTrue
, an item can be taken 0 or 1 time. When set toFalse
, an item can be taken any amount of times (while staying integer and positive).max
– Maximum admissible weight.value_only
– When set toTrue
, only the maximum useful value is returned. When set toFalse
, both the maximum useful value and an assignment are returned.solver
– (default:None
) Specify a Mixed Integer Linear Programming (MILP) solver to be used. If set toNone
, the default one is used. For more information on MILP solvers and which default solver is used, see the methodsolve
of the classMixedIntegerLinearProgram
.verbose
– integer (default:0
). Sets the level of verbosity. Set to 0 by default, which means quiet.integrality_tolerance
– parameter for use with MILP solvers over an inexact base ring; seeMixedIntegerLinearProgram.get_values()
.
OUTPUT:
If
value_only
is set toTrue
, only the maximum useful value is returned. Else (the default), the function returns a pair[value,list]
, wherelist
can be of two types according to the type ofseq
:The list of tuples \((w_i, u_i, ...)\) occurring in the solution.
A list of reals where each real is repeated the number of times it is taken into the solution.
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)]] sage: knapsack( [(1,2), (1.5,1), (0.5,3)], max=2, value_only=True) 5.0
Besides weight and value, you may attach any data to the items:
sage: from sage.numerical.knapsack import knapsack sage: knapsack( [(1, 2, 'spam'), (0.5, 3, 'a', 'lot')]) [3.0, [(0.500000000000000, 3, 'a', 'lot')]]
In the case where all the values (usefulness) of the items are equal to one, you do not need embarrass yourself with the second values, and you can just type for items \((1,1), (1.5,1), (0.5,1)\) the command:
sage: from sage.numerical.knapsack import knapsack sage: knapsack([1,1.5,0.5], max=2, value_only=True) 2.0