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permutations solution in Clear category for Most Efficient Cutting by Sim0000
from itertools import permutations
def most_efficient_cutting(bom):
def f(bom):
total = 0
remain = 6000
for item in bom:
if item <= remain:
remain -= item
else:
total += remain
remain = 6000 - item
total += remain
return total
return min(f(x) for x in permutations(bom))
#These "asserts" using only for self-checking and not necessary for auto-testing
if __name__ == '__main__':
assert(most_efficient_cutting([3000, 2200, 2000, 1800, 1600, 1300]) == 100)
assert (most_efficient_cutting([4000, 4000, 4000]) == 6000), "wasted: 3x2000"
assert(most_efficient_cutting([1]) == 5999), "5999"
assert(most_efficient_cutting([3001, 3001]) == 5998), "2x2999"
assert(most_efficient_cutting([3000, 2200, 1900, 1800, 1600, 1300]) == 200), "2x5900"
assert(most_efficient_cutting([3000]) == 3000)
assert(most_efficient_cutting([3000, 2200, 2000, 1800, 1600, 1400]) == 0)
print('"Run" is good. How is "Check"?')
Nov. 19, 2017
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