Download or view knapsack.frink in plain text format
/** This file contains solvers for various knapsack problems.
See:
Knapsack Problems book:
http://www.or.deis.unibo.it/kp.html
http://hjemmesider.diku.dk/~pisinger/codes.html
https://www.algorist.com/problems/Knapsack_Problem.html
https://www.geeksforgeeks.org/0-1-knapsack-problem-dp-10/
https://en.m.wikipedia.org/wiki/Knapsack_problem
THINK ABOUT: What to return when no solution is found.
*/
/** This is a solver for the 0/1 knapsack problem.
It takes an array of integer values, array of integer weights,
and an integer capacity. It tries to optimize so that the sum of values
is the largest for which the sum of the weights is less than or equal to
capacity.
it returns:
[maxvalue, indices, values, weights]
where
maxvalue: The sum of the values in the best-fitting packing
indices : An array of the indices selected
values : An array of the values selected
weights : An array of the weights selected
This algorithm is O(capacity * numElements) in space and time
which is inefficient if the capacity is large.
Enumerating all the subsets would be O(2^numElements). For example, with
a capacity of 6404180 and 24 elements, this algorithm evaluates
153,700,320 cases, where evaluating the 2^24 subsets would be only
16,777,216 cases. Another algorithm should be chosen when the capacity
is large.
This algorithm is adapted from:
https://www.es.ele.tue.nl/education/5MC10/Solutions/knapsack.pdf
*/
knapsack[values, weights, capacity] :=
{
n = length[values]
if length[weights] != n
{
println["knapsack: length of values does not equal length of weights."]
return undef
}
V = new array[[n+1, capacity+1], 0]
keep = new array[[n+1, capacity+1], false]
for i = 1 to n
{
for w = 0 to capacity
{
if weights@(i-1) <= w and (values@(i-1) + V@(i-1)@(w - weights@(i-1)) > V@(i-1)@w)
{
V@i@w = values@(i-1) + V@(i-1)@(w-weights@(i-1))
keep@i@w = true
} else
{
V@i@w = V@(i-1)@w
keep@i@w = false
}
}
}
included = new array
K = capacity
for i = n to 1 step -1
if keep@i@K == true
{
included.push[ [i-1, values@(i-1), weights@(i-1)] ]
K = K - weights@(i-1)
}
return concat[V@n@capacity, included.transpose[]]
}
/** Brute-force algorithm for evaluating the knapsack problem. The time
complexity is O(2^numElements) so this is unfeasible if there is a large
number of elements. However, as noted above, this is can be significantly
more efficient than the above algorithm if the capacity is large.
This also works with floating-point numbers or units of measure.
It takes an array of values, array of weights,
and a capacity. It tries to optimize so that the sum of values
is the largest for which the sum of the weights is less than or equal to
capacity.
it returns:
[maxvalue, indices, values, weights]
where
maxvalue: The sum of the values in the best-fitting packing
indices : An array of the indices selected
values : An array of the values selected
weights : An array of the weights selected
*/
knapsackBrute[values, weights, capacity] :=
{
n = length[values]
if length[weights] != n
{
println["knapsack: length of values does not equal length of weights."]
return undef
}
// Make a bunch of loops that alternate between false and true.
loops = makeArray[[n], [false, true]]
// This is to make this work with units of measure.
zeroWeight = 0 weights@0
zeroValue = 0 values@0
bestValue = zeroValue
bestIndices = undef
LOOP:
multifor v = loops
{
weight = zeroWeight
value = zeroValue
for i = 0 to n-1
if v@i
{
weight = weight + weights@i
if weight > capacity
next LOOP i
value = value + values@i
}
if value > bestValue
{
bestValue = value
bestIndices = v.shallowCopy[]
}
}
// Build the result
indices = new array
outValues = new array
outWeights = new array
for i = 0 to n-1
if bestIndices@i
{
indices.push[i]
outValues.push[values@i]
outWeights.push[weights@i]
}
return [bestValue, indices, outValues, outWeights]
}
/** Brute-force algorithm for evaluating the multiple-choice knapsack problem
(MCKP). Each weight or value is an array of choices from which exactly
one item is selected.
See section 2.12 in http://www.or.deis.unibo.it/kp.html
The time complexity is O(choices^numElements) so this is unfeasible if
there is a large number of elements.
This also works with floating-point numbers or units of measure.
It takes an array of array of values, array of array of weights,
and a capacity. It tries to optimize so that the sum of values
is the largest for which the sum of the weights is less than or equal to
capacity.
it returns:
[maxvalue, indices, values, weights]
where
maxvalue: The sum of the values in the best-fitting packing
indices : An array of the indices selected
values : An array of the values selected
weights : An array of the weights selected
*/
knapsackMultiBrute[values, weights, capacity] :=
{
n = length[values]
if length[weights] != n
{
println["knapsack: length of values does not equal length of weights."]
return undef
}
// Make a bunch of loops that alternate between the weights
loops = new array
for i=0 to n-1
loops.push[rangeOf[weights@i]]
// This is to make this work with units of measure.
zeroWeight = 0 weights@0@0
zeroValue = 0 values@0@0
bestValue = zeroValue
bestIndices = undef
LOOP:
multifor v = loops
{
weight = zeroWeight
value = zeroValue
for i = 0 to n-1
{
vi = v@i
weight = weight + weights@i@vi
if weight > capacity
next LOOP i
value = value + values@i@vi
}
if value > bestValue
{
bestValue = value
bestIndices = v.shallowCopy[]
}
}
// Build the result
indices = new array
outValues = new array
outWeights = new array
for i = 0 to n-1
{
bi = bestIndices@i
indices.push[bi]
outValues.push[values@i@bi]
outWeights.push[weights@i@bi]
}
return [bestValue, indices, outValues, outWeights]
}
/** Returns all the solutions for the subset-sum problem, that is, finds all
the subsets of the input that sum to a desired number. */
subsetSum[vals, target] :=
{
select[toSet[vals].subsets[], {|x, t| sum[x] == t}, target]
}
/** Returns all the solutions for the subset-product problem, that is, finds all
the subsets of the input that multiply to a desired number. */
subsetProduct[vals, target] :=
{
select[toSet[vals].subsets[], {|x, t| product[x] == t}, target]
}
/** This uses dynamic programming to find a maximum value for the subset-sum
problem for integers. That is, it finds the largest subset of weights
which sum to target *or less.*
arguments: [weights, target]
weights: an array of integers
target: an integer that we want to sum to.
returns [max, weights]
Where
max: an integer containing the sum
weights: an array of values that sum to target or less
TODO: Adopt the code from knapsack and simplify it to not pass in weights
twice.
*/
maxSubsetSum[weights, target] :=
{
[maxvalue, indices, values] = knapsack[weights, weights, target]
return [maxvalue, values]
}
/** This is a solver for the "partition problem": given a set of numbers W,
it is desired to separate these numbers into two subsets W1 and W2 so that
the sums of the numbers in each subset are equal.
The partition problem is a special case of the subset sum problem, in
which the target sum is one half the sum of the entire set W.
args:
weights: An array of numeric values
returns:
an array of pairs [W1, W2] where W1 and W2 are the subsets.
*/
equalPartitions[weights] :=
{
result = new array
sum = sum[weights] / 2
if ! isInteger[sum]
return result
for sol = subsetSum[weights, sum]
{
s = deepCopy[weights]
for elem = sol
s.removeValue[elem]
result.push[ [sol, s] ]
}
return result
}
Download or view knapsack.frink in plain text format
This is a program written in the programming language Frink.
For more information, view the Frink
Documentation or see More Sample Frink Programs.
Alan Eliasen was born 19970 days, 11 hours, 31 minutes ago.