Reputation: 181
I'd like to modify the subset_sum()
python function from Finding all possible combinations of numbers to reach a given sum so that:
I've successfully accomplished #2, but I need assistance with #1:
def subset_sum(numbers, target, length, partial=[]):
s = sum(partial)
# check if the partial sum is equals to target
if s == target and len(partial) == length:
print(f"sum({partial})={target}")
if s >= target:
return # if we reach the number why bother to continue
for i in range(len(numbers)):
n = numbers[i]
remaining = numbers[i+1:]
subset_sum(remaining, target, length, partial + [n])
The desired output should be:
>>> subset_sum([3,9,8,4,5,7,10],target=15,length=3)
sum([3, 8, 4])=15
sum([3, 4, 8])=15
sum([4, 3, 8])=15
sum([4, 8, 3])=15
sum([8, 3, 4])=15
sum([8, 4, 3])=15
sum([3, 5, 7])=15
sum([3, 7, 5])=15
sum([5, 3, 7])=15
sum([5, 7, 3])=15
sum([7, 3, 5])=15
sum([7, 5, 3])=15
Upvotes: 2
Views: 2541
Reputation: 181
This does it:
import itertools
numbers = [3,9,8,4,5,7,10]
length = 3
target = 15
iterable = itertools.permutations(numbers,length)
predicate = lambda x: (sum(x) == target)
vals = filter(predicate,iterable)
list(vals)
Or a one-liner:
vals = [x for x in itertools.permutations(numbers,length) if sum(x) == target]
Results:
[(3, 8, 4),
(3, 4, 8),
(3, 5, 7),
(3, 7, 5),
(8, 3, 4),
(8, 4, 3),
(4, 3, 8),
(4, 8, 3),
(5, 3, 7),
(5, 7, 3),
(7, 3, 5),
(7, 5, 3)]
Upvotes: 2
Reputation: 77910
Since you've solved the problem of identifying one solution in each equivalence group, my advice is: do not alter that algorithm. Instead, harness itertools.permutations
to generate those items:
return list(itertools.permutations(numbers))
Upvotes: 2