## Fill order from smaller packages?

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The input is an integer that specifies the amount to be ordered. There are predefined package sizes that have to be used to create that order.

e.g.

Packs 3 for $5 5 for $9 9 for $16

for an input order 13 the output should be:

2x5 + 1x3

So far I've the following approach:

remaining_order = 13 package_numbers = [9,5,3] required_packages = [] while remaining_order > 0: found = False for pack_num in package_numbers: if pack_num <= remaining_order: required_packages.append(pack_num) remaining_order -= pack_num found = True break if not found: break

But this will lead to the wrong result:

1x9 + 1x3 remaining: 1

So, you need to fill the order with the packages such that the total price is maximal? This is known as Knapsack problem. In that Wikipedia article you'll find several solutions written in Python.

To be more precise, you need a solution for the unbounded knapsack problem, in contrast to popular 0/1 knapsack problem (where each item can be packed only once). Here is working code from Rosetta:

from itertools import product NAME, SIZE, VALUE = range(3) items = ( # NAME, SIZE, VALUE ('A', 3, 5), ('B', 5, 9), ('C', 9, 16)) capacity = 13 def knapsack_unbounded_enumeration(items, C): # find max of any one item max1 = [int(C / item[SIZE]) for item in items] itemsizes = [item[SIZE] for item in items] itemvalues = [item[VALUE] for item in items] # def totvalue(itemscount, =itemsizes, itemvalues=itemvalues, C=C): def totvalue(itemscount): # nonlocal itemsizes, itemvalues, C totsize = sum(n * size for n, size in zip(itemscount, itemsizes)) totval = sum(n * val for n, val in zip(itemscount, itemvalues)) return (totval, -totsize) if totsize <= C else (-1, 0) # Try all combinations of bounty items from 0 up to max1 bagged = max(product(*[range(n + 1) for n in max1]), key=totvalue) numbagged = sum(bagged) value, size = totvalue(bagged) size = -size # convert to (iten, count) pairs) in name order bagged = ['%dx%d' % (n, items[i][SIZE]) for i, n in enumerate(bagged) if n] return value, size, numbagged, bagged if __name__ == '__main__': value, size, numbagged, bagged = knapsack_unbounded_enumeration(items, capacity) print(value) print(bagged)

Output is:

23 ['1x3', '2x5']

Keep in mind that this is a NP-hard problem, so it will blow as you enter some large values :)

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You can use `itertools.product`

:

import itertools remaining_order = 13 package_numbers = [9,5,3] required_packages = [] a=min([x for i in range(1,remaining_order+1//min(package_numbers)) for x in itertools.product(package_numbers,repeat=i)],key=lambda x: abs(sum(x)-remaining_order)) remaining_order-=sum(a) print(a) print(remaining_order)

Output:

(5, 5, 3) 0

This simply does the below steps:

Get value closest to

`13`

, in the list with all the product values.Then simply make it modify the number of

`remaining_order`

.

If you want it output with `'x'`

:

import itertools from collections import Counter remaining_order = 13 package_numbers = [9,5,3] required_packages = [] a=min([x for i in range(1,remaining_order+1//min(package_numbers)) for x in itertools.product(package_numbers,repeat=i)],key=lambda x: abs(sum(x)-remaining_order)) remaining_order-=sum(a) print(' '.join(['{0}x{1}'.format(v,k) for k,v in Counter(a).items()])) print(remaining_order)

Output:

2x5 + 1x3 0

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For you problem, I tried two implementations depending on what you want, in both of the solutions I supposed you absolutely needed your remaining to be at 0. Otherwise the algorithm will return you `-1`

. If you need them, tell me I can adapt my algorithm.

As the algorithm is implemented via dynamic programming, it handles good inputs, at least more than 130 packages !

In the first solution, I admitted we fill with the biggest package each time. I n the second solution, I try to minimize the price, but the number of packages should always be 0.

remaining_order = 13 package_numbers = sorted([9,5,3], reverse=True) # To make sure the biggest package is the first element prices = {9: 16, 5: 9, 3: 5} required_packages = [] # First solution, using the biggest package each time, and making the total order remaining at 0 each time ans = [[] for _ in range(remaining_order + 1)] ans[0] = [0, 0, 0] for i in range(1, remaining_order + 1): for index, package_number in enumerate(package_numbers): if i-package_number > -1: tmp = ans[i-package_number] if tmp != -1: ans[i] = [tmp[x] if x != index else tmp[x] + 1 for x in range(len(tmp))] break else: # Using for else instead of a boolean value `found` ans[i] = -1 # -1 is the not found combinations print(ans[13]) # [0, 2, 1] print(ans[9]) # [1, 0, 0] # Second solution, minimizing the price with order at 0 def price(x): return 16*x[0]+9*x[1]+5*x[2] ans = [[] for _ in range(remaining_order + 1)] ans[0] = ([0, 0, 0],0) # combination + price for i in range(1, remaining_order + 1): # The not found packages will be (-1, float('inf')) minimal_price = float('inf') minimal_combinations = -1 for index, package_number in enumerate(package_numbers): if i-package_number > -1: tmp = ans[i-package_number] if tmp != (-1, float('inf')): tmp_price = price(tmp[0]) + prices[package_number] if tmp_price < minimal_price: minimal_price = tmp_price minimal_combinations = [tmp[0][x] if x != index else tmp[0][x] + 1 for x in range(len(tmp[0]))] ans[i] = (minimal_combinations, minimal_price) print(ans[13]) # ([0, 2, 1], 23) print(ans[9]) # ([0, 0, 3], 15) Because the price of three packages is lower than the price of a package of 9

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In case you need a solution for a small number of possible

package_numbers

but a possibly very big

remaining_order,

in which case all the other solutions would fail, you can use this to reduce remaining_order:

import numpy as np remaining_order = 13 package_numbers = [9,5,3] required_packages = [] sub_max=np.sum([(np.product(package_numbers)/i-1)*i for i in package_numbers]) while remaining_order > sub_max: remaining_order -= np.product(package_numbers) required_packages.append([max(package_numbers)]*np.product(package_numbers)/max(package_numbers))

Because if any package is in required_packages more often than (np.product(package_numbers)/i-1)*i it's sum is equal to np.product(package_numbers). In case the package max(package_numbers) isn't the one with the samllest price per unit, take the one with the smallest price per unit instead.

Example:

remaining_order = 100 package_numbers = [5,3]

Any part of remaining_order bigger than 5*2 plus 3*4 = 22 can be sorted out by adding 5 three times to the solution and taking remaining_order - 5*3. So remaining order that actually needs to be calculated is 10. Which can then be solved to beeing 2 times 5. The rest is filled with 6 times 15 which is 18 times 5.

In case the number of possible package_numbers is bigger than just a handful, I recommend building a lookup table (with one of the others answers' code) for all numbers below sub_max which will make this immensely fast for any input.

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Since no declaration about the object function is found, I assume your goal is to maximize the package value within the pack's capability.

Explanation: time complexity is fixed. Optimal solution may not be filling the highest valued item as many as possible, you have to search all possible combinations. However, you can reuse the possible optimal solutions you have searched to save space. For example, `[5,5,3]`

is derived from adding `3`

to a previous `[5,5]`

try so the intermediate result can be "cached". You may either use an array or you may use a set to store possible solutions. The code below runs the same performance as the rosetta code but I think it's clearer.

To further optimize, use a priority set for `opts`

.

costs = [3,5,9] value = [5,9,16] volume = 130 # solutions opts = set() opts.add(tuple([0])) # calc total value cost_val = dict(zip(costs, value)) def total_value(opt): return sum([cost_val.get(cost,0) for cost in opt]) def possible_solutions(): solutions = set() for opt in opts: for cost in costs: if cost + sum(opt) > volume: continue cnt = (volume - sum(opt)) // cost for _ in range(1, cnt + 1): sol = tuple(list(opt) + [cost] * _) solutions.add(sol) return solutions def optimize_max_return(opts): if not opts: return tuple([]) cur = list(opts)[0] for sol in opts: if total_value(sol) > total_value(cur): cur = sol return cur while sum(optimize_max_return(opts)) <= volume - min(costs): opts = opts.union(possible_solutions()) print(optimize_max_return(opts))

If your requirement is "just fill the pack" it'll be even simpler using the volume for each item instead.

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##### Comments

- As an aside, in Python, a
`for`

can have an`else`

. - Write down how
**you**would solve this problem yourself (how would you decide to use 2x5 + 1x3 instead of 1x9 + 1x3), then you'll have the correct algorithm.... - What should be the ouput for remaining_order = 9 ? 3x3 or 1x9? Just to be sure before I start writting a solution
- Can you specify boundaries? Will it be 3 package sizes max or may there be 20000? Biggest input?
- @BlueSheepToken lowest price I think, so 9. I'd even think that combination of biggest inputs should suffice, prices don't normally favor smaller package_sizes, but op should answer that.
- Am... link-only answers will be deleted in the low quality posts review page.
- @U9-Forward there's a meta discussion around somewhere, this is, barely, but still, an answer because it can be reproduced even if the link goes dead -> knapsack problem, wikipedia, done.
- I entered 130000 as remaining_order and had to kill it with fire ;), I then entered 130 and it's still not finished
- @DonQuiKong Well... any code doing that number would have to be killed and you would actually have your window being not responding, so restart shell...
- 130 -> MemoryError
- @DonQuiKong That's still kinda big... can ya think of other examples, anyway, bye, i have to go.
- Try integer factorisation with
`itertools`

next :) - Mmmhh... this is right, I will remove the side note for the moment, will put it back later if I find the mathematical proof :p
- It should be 9*(5*3-1) + 5*(3*9-1) + 3*(9*5-1) because for example for 5 and 3 5*3 would be 15, but two times 5 and two times 3 is 16 which is a possible number but can't be solved with mod 15.
- However, if any package is in there for self_value*product_of_rest, the mod takes it out. So any combination of a single package bigger than each package times product_of_all_other_package_values-1 can be ignored.