Python Subset Sum

Question

I am trying to write a function that will not only determine whether the sum of a subset of a set adds to a desired target number, but also to print the subset that is the solution.

Here is my code for finding whether a subset exists:

def subsetsum(array,num):

    if num == 0 or num < 1:
        return False
    elif len(array) == 0:
        return False
    else:
        if array[0] == num:
            return True
        else:
            return subsetsum(array[1:],(num - array[0])) or subsetsum(array[1:],num)

How can I modify this to record the subset itself so that I can print it? Thanks in advance!

Answer 1

Based on your solution:

def subsetsum(array,num):

    if num == 0 or num < 1:
        return None
    elif len(array) == 0:
        return None
    else:
        if array[0] == num:
            return [array[0]]
        else:
            with_v = subsetsum(array[1:],(num - array[0])) 
            if with_v:
                return [array[0]] + with_v
            else:
                return subsetsum(array[1:],num)

Answer 2

Modification to also detect duplicates and further solutions when a match happened

def subset(array, num):
    result = []
    def find(arr, num, path=()):
        if not arr:
            return
        if arr[0] == num:
            result.append(path + (arr[0],))
        else:
            find(arr[1:], num - arr[0], path + (arr[0],))
        find(arr[1:], num, path)
    find(array, num)
    return result

Answer 3

You could change your approach to do that more easily, something like:

def subsetsum(array, num):
    if sum(array) == num:
        return array
    if len(array) > 1:
        for subset in (array[:-1], array[1:]):
            result = subsetsum(subset, num)
            if result is not None:
                return result

This will return either a valid subset or None.

Answer 4

Thought I'll throw another solution into the mix.

We can map each selection of a subset of the list to a (0-padded) binary number, where a 0 means not taking the member in the corresponsing position in the list, and 1 means taking it.

So masking [1, 2, 3, 4] with 0101 creates the sub-list [2, 4].

So, by generating all 0-padded binary numbers in the range between 0 and 2^LENGTH_OF_LIST, we can iterate all selections. If we use these sub-list selections as masks and sum the selection - we can know the answer.

This is how it's done:

#!/usr/bin/env python

# use a binary number (represented as string) as a mask
def mask(lst, m):
    # pad number to create a valid selection mask 
    # according to definition in the solution laid out 
    m = m.zfill(len(lst))
    return map(lambda x: x[0], filter(lambda x: x[1] != '0', zip(lst, m)))

def subset_sum(lst, target):
    # there are 2^n binary numbers with length of the original list
    for i in xrange(2**len(lst)):
        # create the pick corresponsing to current number
        pick = mask(lst, bin(i)[2:])
        if sum(pick) == target:
            return pick
    return False


print subset_sum([1,2,3,4,5], 7)

Output:

[3, 4]

To return all possibilities we can use a generator instead (the only changes are in subset_sum, using yield instead of return and removing return False guard):

#!/usr/bin/env python

# use a binary number (represented as string) as a mask
def mask(lst, m):
    # pad number to create a valid selection mask 
    # according to definition in the solution laid out 
    m = m.zfill(len(lst))
    return map(lambda x: x[0], filter(lambda x: x[1] != '0', zip(lst, m)))

def subset_sum(lst, target):
    # there are 2^n binary numbers with length of the original list
    for i in xrange(2**len(lst)):
        # create the pick corresponsing to current number
        pick = mask(lst, bin(i)[2:])
        if sum(pick) == target:
            yield pick

# use 'list' to unpack the generator
print list(subset_sum([1,2,3,4,5], 7))

Output:

[[3, 4], [2, 5], [1, 2, 4]]

---

Note: While not padding the mask with zeros may work as well, as it will simply select members of the original list in a reverse order - I haven't checked it and didn't use it.

I didn't use it since it's less obvious (to me) what's going on with such trenary-like mask (1, 0 or nothing) and I rather have everything well defined.

Answer 5

Slightly updated the below code to return all possible combinations for this problem. Snippet in the thread above will not print all possible combinations when the input is given as subset([4,3,1],4)

def subset(array, num):
    result = []
    def find(arr, num, path=()):
        if not arr:
            return
        if arr[0] == num:
            result.append(path + (arr[0],))
        else:
            find(arr[1:], num - arr[0], path + (arr[0],))
        find(arr[1:], num, path)
    find(array, num)
    return result

Answer 6

A bit different approach to print all subset through Recursion.

def subsetSumToK(arr,k):
    if len(arr)==0:
        if k == 0:
            return [[]]
        else:
            return []
    
    output=[]
    if arr[0]<=k: 
        temp2=subsetSumToK(arr[1:],k-arr[0])  #Including the current element 
        if len(temp2)>0:
            for i in range(len(temp2)):
                temp2[i].insert(0,arr[0])
                output.append(temp2[i])
    
    temp1=subsetSumToK(arr[1:],k)            #Excluding the current element
    if len(temp1)>0:
        for i in range(len(temp1)):
            output.append(temp1[i])
    return output

arr=[int(i) for i in input().split()]
k=int(input())
sub=subsetSumToK(arr,k)
for i in sub:
    for j in range(len(i)):
        if j==len(i)-1:
            print(i[j])
        else:
            print(i[j],end=" ")

Answer 7

Rather than using recursion, you could use the iterative approach.

def desiredSum(array, sum):

  numberOfItems = len(array)
  storage = [[0 for x in range(sum + 1)] for x in range(numberOfItems + 1)]

  for i in range(numberOfItems + 1):
    for j in range(sum + 1):

        value = array[i - 1]

        if i is 0: storage[i][j] = 0
        if j is 0: storage[i][j] = 1

        if value <= j:

            noTake = storage[i - 1][j]
            take = storage[i - 1][j - value]
            storage[i][j] = noTake + take

  return storage[numberOfItems][sum]