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Consider a list of integers, ranging from 0 to 50. I would like to generate and print all possible permutations of this list that sum to a given number. An example is shown below:

If the sum is 41:
One output is [4,5,7,9,16]
...
mullena
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Anonymousss
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  • Do the combinations need to be a specific length? And in your example, would something like `[16, 9, 7, 5, 4]` (reversed) also be an answer? – Ben Soyka Dec 29 '20 at 14:47
  • Do you have to follow a specific method or can you brute force it? – Guimoute Dec 29 '20 at 14:48
  • standard leetcode problem https://leetcode.com/problems/combination-sum/discuss/16510/Python-dfs-solution SO dup: https://stackoverflow.com/questions/34517540/find-all-combinations-of-a-list-of-numbers-with-a-given-sum/34519260 – Equinox Dec 29 '20 at 14:49
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    Does this answer your question? [Find all combinations of a list of numbers with a given sum](https://stackoverflow.com/questions/34517540/find-all-combinations-of-a-list-of-numbers-with-a-given-sum) – Equinox Dec 29 '20 at 14:49
  • @BenSoyka Any combination would work and there is no specific length. – Anonymousss Dec 29 '20 at 14:51
  • Does the posts help you out? – Daniel Hao Dec 29 '20 at 15:33

4 Answers4

0

You can use combinations(...) function from itertools

import itertools

nums = range(51)
for n in range(1, len(nums) + 1):
    for p in itertools.combinations(nums, n):
        if sum(p) == 41:
            print(*p)
AWhiteFox
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0

The following code repeatedly generates random sequences of numbers and adds them to a tell until the list has the desired mean. In the example below, the code will output a list of six numbers, ranging from 9 to 17, with an average of 14.5. These can obviously be changed as desired.

```python
def stats():
    x =[0,0,0,0,0,0]
    while mean(x) != 14.5:
        x =[0,0,0,0,0,0]
        for i in range(6):
             a = random.randint(9,17)
             x[i] = a
return x
```

The mean function very simply accepts the list and returns the mean as desired. You will need to change the divisor (6 in the example below) for different lengths of lists.

```python
def mean(x):
    y = sum(x)/6
return y
```

At the start of the code, you need to import the random library to allow the code to work properly. The one caveat of this code, is that it only generates one permutation at a time. However a for loop can be used to generate more:

stats()
>>> [11, 12, 16, 17, 16, 15]
mullena
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0

You could write a recursive generator to efficiently produce only the combinations of positive integers that sum up to the target number:

def sumSets(numbers,target):
    if not target  : yield []
    if not numbers or target <= 0: return
    yield from sumSets(numbers[1:],target)
    yield from (numbers[:1]+ss for ss in sumSets(numbers[1:],target-numbers[0]))

nums = list(range(1,50,3))
for ss in sumSets(nums,41): print(ss)
      
[19, 22]
[16, 25]
[13, 28]
[10, 31]
[7, 34]
[4, 37]
[1, 40]
[1, 4, 7, 13, 16]
[1, 4, 7, 10, 19]

Note that, if you're looking for all combinations of numbers from 1 to 50 that sum up to 41, you're going to get a lot of them:

nums = list(range(1,51))
print(sum(1 for _ in sumSets(nums,41))) # 1260
Alain T.
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-1
import itertools

def find_permutations_with_sum(nums, target_sum):
    # Generate all permutations of the input list
    all_permutations = list(itertools.permutations(nums))

    # Filter permutations that sum to the target_sum
    valid_permutations = []
    for perm in all_permutations:
        if sum(perm) == target_sum:
            valid_permutations.append(perm)

    return valid_permutations

# Example usage
nums_range = range(0, 51)  # List of integers from 0 to 50
target = 10  # The desired sum
valid_permutations = find_permutations_with_sum(nums_range, target)

# Print the valid permutations
for perm in valid_permutations:
    print(perm)
Arya Tavana
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