r"""
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):
.. MATH::
\mbox{Maximize: }\sum_i b_i u_i \\
\mbox{Such that: }
\sum_i b_i w_i \leq W \\
\forall i, b_i \mbox{ binary variable} \\
(For more information, see the :wikipedia:`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]
"""
from sage.misc.latex import latex
from sage.rings.integer import Integer
from sage.structure.sage_object import SageObject
class Superincreasing(SageObject):
r"""
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
.. MATH::
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`` 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
sage: seq = Superincreasing(); seq
An empty sequence.
sage: seq = Superincreasing([1, 3, 6]); seq
Super-increasing sequence of length 3
sage: seq = Superincreasing(list([1, 2, 5, 21, 69, 189, 376, 919])); seq
Super-increasing sequence of length 8
"""
def __init__(self, seq=None):
r"""
Constructing a super-increasing sequence object from ``seq``.
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
"""
if seq is None:
self._seq = None
else:
if self.is_superincreasing(seq):
self._seq = seq
else:
raise ValueError, "seq must be a super-increasing sequence"
def __cmp__(self, other):
r"""
Comparing ``self`` to ``other``.
TESTS::
sage: from sage.numerical.knapsack import Superincreasing
sage: L = [1, 2, 5, 21, 69, 189, 376, 919]
sage: seq = Superincreasing(L)
sage: seq == loads(dumps(seq))
True
"""
return cmp(self._seq, other._seq)
def __repr__(self):
r"""
Return a string representation of this super-increasing
sequence object.
EXAMPLES::
sage: from sage.numerical.knapsack import Superincreasing
sage: seq = Superincreasing(); seq
An empty sequence.
sage: seq = Superincreasing([1, 3, 6]); seq
Super-increasing sequence of length 3
sage: seq = Superincreasing(list([1, 2, 5, 21, 69, 189, 376, 919])); seq
Super-increasing sequence of length 8
"""
if self._seq is None:
return "An empty sequence."
else:
return "Super-increasing sequence of length %s" % len(self._seq)
def largest_less_than(self, N):
r"""
Return the largest integer in the sequence ``self`` that is less than
or equal to ``N``.
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 to ``N``. If
no solution exists, then return ``None``.
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) == None
True
TESTS:
The target ``N`` must be an integer::
sage: from sage.numerical.knapsack import Superincreasing
sage: L = [2, 3, 7, 25, 67, 179, 356, 819]
sage: Superincreasing(L).largest_less_than(2.30)
Traceback (most recent call last):
...
TypeError: N (= 2.30000000000000) must be an integer.
The sequence that ``self`` represents must also be non-empty::
sage: Superincreasing([]).largest_less_than(2)
Traceback (most recent call last):
...
ValueError: seq must be a super-increasing sequence
sage: Superincreasing(list()).largest_less_than(2)
Traceback (most recent call last):
...
ValueError: seq must be a super-increasing sequence
"""
from sage.functions.other import Function_floor
floor = Function_floor()
if len(self._seq) == 0:
raise TypeError, "self must be a non-empty list of integers."
if (not isinstance(N, Integer)) and (not isinstance(N, int)):
raise TypeError, "N (= %s) must be an integer." % N
low = 0
high = len(self._seq) - 1
while low <= high:
mid = floor((low + high) / 2)
if N == self._seq[mid]:
return self._seq[mid]
if N < self._seq[mid]:
high = mid - 1
else:
low = mid + 1
if N >= self._seq[high]:
return self._seq[high]
else:
return None
def _latex_(self):
r"""
Return LaTeX representation of ``self``.
EXAMPLES::
sage: from sage.numerical.knapsack import Superincreasing
sage: latex(Superincreasing())
\left[\right]
sage: seq = Superincreasing([1, 2, 5, 21, 69, 189, 376, 919])
sage: latex(seq)
<BLANKLINE>
\left[1,
2,
5,
21,
69,
189,
376,
919\right]
"""
if self._seq is None:
return latex([])
else:
return latex(self._seq)
def is_superincreasing(self, seq=None):
r"""
Determine whether or not ``seq`` is super-increasing.
If ``seq=None`` then determine whether or not ``self`` 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
.. MATH::
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`` is ``None``, then test ``self`` to determine whether or
not it is super-increasing. In that case, return ``True`` if
``self`` is super-increasing; ``False`` otherwise.
- If ``seq`` is not ``None``, then test ``seq`` to determine whether
or not it is super-increasing. Return ``True`` if ``seq`` 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
TESTS:
The sequence must contain only integers::
sage: from sage.numerical.knapsack import Superincreasing
sage: L = [1.0, 2.1, pi, 21, 69, 189, 376, 919]
sage: Superincreasing(L).is_superincreasing()
Traceback (most recent call last):
...
TypeError: Element e (= 1.00000000000000) of seq must be a non-negative integer.
sage: L = [1, 2.1, pi, 21, 69, 189, 376, 919]
sage: Superincreasing(L).is_superincreasing()
Traceback (most recent call last):
...
TypeError: Element e (= 2.10000000000000) of seq must be a non-negative integer.
"""
if seq is None:
if (self._seq is None) or len(self._seq) == 0:
return False
if (not isinstance(self._seq[0], Integer)) and (not isinstance(self._seq[0], int)):
raise TypeError, "Element e (= %s) of self must be a non-negative integer." % self._seq[0]
if self._seq[0] < 0:
raise TypeError, "Element e (= %s) of self must be a non-negative integer." % self._seq[0]
cumSum = self._seq[0]
for e in self._seq[1:]:
if (not isinstance(e, Integer)) and (not isinstance(e, int)):
raise TypeError, "Element e (= %s) of self must be a non-negative integer." % e
if e < 0:
raise TypeError, "Element e (= %s) of self must be a non-negative integer." % e
if e <= cumSum:
return False
cumSum += e
return True
else:
if len(seq) == 0:
return False
if (not isinstance(seq[0], Integer)) and (not isinstance(seq[0], int)):
raise TypeError, "Element e (= %s) of seq must be a non-negative integer." % seq[0]
if seq[0] < 0:
raise TypeError, "Element e (= %s) of seq must be a non-negative integer." % seq[0]
cumSum = seq[0]
for e in seq[1:]:
if (not isinstance(e, Integer)) and (not isinstance(e, int)):
raise TypeError, "Element e (= %s) of seq must be a non-negative integer." % e
if e < 0:
raise TypeError, "Element e (= %s) of seq must be a non-negative integer." % e
if e <= cumSum:
return False
cumSum += e
return True
def subset_sum(self, N):
r"""
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 to ``N``. 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 [HPS08]_.
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]
TESTS:
The target ``N`` must be a non-negative integer::
sage: from sage.numerical.knapsack import Superincreasing
sage: L = [0, 1, 2, 4]
sage: Superincreasing(L).subset_sum(-6)
Traceback (most recent call last):
...
TypeError: N (= -6) must be a non-negative integer.
sage: Superincreasing(L).subset_sum(-6.2)
Traceback (most recent call last):
...
TypeError: N (= -6.20000000000000) must be a non-negative integer.
The sequence that ``self`` represents must only contain non-negative
integers::
sage: L = [-10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1]
sage: Superincreasing(L).subset_sum(1)
Traceback (most recent call last):
...
TypeError: Element e (= -10) of seq must be a non-negative integer.
REFERENCES:
.. [HPS08] J. Hoffstein, J. Pipher, and J.H. Silverman. *An
Introduction to Mathematical Cryptography*. Springer, 2008.
"""
if not self.is_superincreasing():
raise TypeError, "self is not super-increasing. Only super-increasing sequences are currently supported."
if (not isinstance(N, Integer)) and (not isinstance(N, int)):
raise TypeError, "N (= %s) must be a non-negative integer." % N
if N < 0:
raise TypeError, "N (= %s) must be a non-negative integer." % N
candidates = []
a = self.largest_less_than(N)
while a is not None:
candidates.append(a)
a = self.largest_less_than(N - sum(candidates))
lst = list(set(candidates))
if len(lst) != len(candidates):
return []
if sum(candidates) == N:
return candidates
else:
return []
def knapsack(seq, binary=True, max=1, value_only=False, solver=None, verbose=0):
r"""
Solves the knapsack problem
For more information on the knapsack problem, see the documentation of the
:mod:`knapsack module <sage.numerical.knapsack>` or the
:wikipedia:`Knapsack_problem`.
INPUT:
- ``seq`` -- Two different possible types:
- A sequence of tuples ``(weight, value, something1, something2,
...)``. Note that only the first two coordinates (``weight`` and
``values``) 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 to ``True``, an item can be taken 0 or 1 time.
When set to ``False``, an item can be taken any amount of times (while
staying integer and positive).
- ``max`` -- Maximum admissible weight.
- ``value_only`` -- When set to ``True``, only the maximum useful value is
returned. When set to ``False``, both the maximum useful value and an
assignment are returned.
- ``solver`` -- (default: ``None``) Specify a Linear Program (LP) solver to
be used. If set to ``None``, the default one is used. For more information
on LP solvers and which default solver is used, see the documentation of
class :class:`MixedIntegerLinearProgram
<sage.numerical.mip.MixedIntegerLinearProgram>`.
- ``verbose`` -- integer (default: ``0``). Sets the level of verbosity. Set
to 0 by default, which means quiet.
OUTPUT:
If ``value_only`` is set to ``True``, only the maximum useful value is
returned. Else (the default), the function returns a pair ``[value,list]``,
where ``list`` can be of two types according to the type of ``seq``:
- 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
"""
reals = not isinstance(seq[0], tuple)
if reals:
seq = [(x,1) for x in seq]
from sage.numerical.mip import MixedIntegerLinearProgram
p = MixedIntegerLinearProgram(maximization=True, solver=solver)
present = p.new_variable()
p.set_objective(p.sum([present[i] * seq[i][1] for i in range(len(seq))]))
p.add_constraint(p.sum([present[i] * seq[i][0] for i in range(len(seq))]), max=max)
if binary:
p.set_binary(present)
else:
p.set_integer(present)
if value_only:
return p.solve(objective_only=True, log=verbose)
else:
objective = p.solve(log=verbose)
present = p.get_values(present)
val = []
if reals:
[val.extend([seq[i][0]] * int(present[i])) for i in range(len(seq))]
else:
[val.extend([seq[i]] * int(present[i])) for i in range(len(seq))]
return [objective,val]