Using recursion to solve the subset sum problem

Question

I want to calculate the number of subsets of the array A = [2, 1, 1, 4] that sum to 2. There are 2 ways: (A[0]) and (A[1], A[2]). My code for computing this is:

 def W(number, index):
     A = [2, 1, 1, 4]
     if number < 0 or index < 0:
          return 0
     elif number==0:
          return 1
     else: 
          return W(number, index-1) + W(number - A[index], index)

Right now, when I call the function with W(2,3), I get 4 instead of 2. My problem is that my code also calculates the possibility (A[1], A[1]) and (A[2], A[2]). Is there any way to fix it while still using recursion?

See that when you use a number from an array (decrease your number by it) you don't want to reuse it again :) — E. Shcherbo
– E. Shcherbo, Commented Feb 22, 2022 at 21:16

BrokenBenchmark · Accepted Answer · 2022-02-23 02:38:47Z

3

The call to W(number - A[index], index) should be W(number - A[index], index - 1); otherwise, you allow for the possibility of double-counting an element in your subset sum.

Here is a code snippet that fixes this issue. For each element, we decide whether or not to add it to our sum. If the element allows our target to reach 0, we add 1 to our total count of possibilities, then recurse to see if there are any other ways to reach the target without adding the element we're currently examining:

A = [2, 1, 1, 4]

def W(number, index):
     if number < 0 or index < 0 :
          return 0
     elif number - A[index] == 0:
         return 1 + W(number, index - 1)
     else: 
          return W(number, index - 1) + W(number - A[index], index - 1)
          
print(W(1, 3)) # Prints 2

edited Feb 23, 2022 at 2:38

answered Feb 22, 2022 at 21:18

BrokenBenchmark

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8 Comments

Cubix48 Over a year ago

Shouldn't that be 2 for W(4, 3), and 2 for W(5, 3)?

E. Shcherbo Over a year ago

I think the order of the conditions can be changed. The first check is if number == 0: return 1 and then elif number < 0 or index < 0: return 0. This also works, but not sure about other test cases

D Malan Over a year ago

W(1, 3) prints 1, but I think it should be 2 because A[1] and A[2] works.

BrokenBenchmark Over a year ago

Ah, yeah, you'll need to store the original target in that case. I'll fix shortly, I have to hop off for a bit.

BrokenBenchmark Over a year ago

Alright, edited.

|

quamrana · Accepted Answer · 2022-02-22 21:35:39Z

2

This code seems to work for cases finding sums of 1..8:

A = [2, 1, 1, 4]

def W(number, index):
     if number == 0:
          return 1
     elif number < 0 or index < 0:
          return 0
     else:
         return W(number, index-1) + W(number - A[index], index - 1)

Testcases:

print(W(1, 3))
print(W(2, 3))
print(W(3, 3))
print(W(4, 3))
print(W(5, 3))
print(W(6, 3))
print(W(7, 3))
print(W(8, 3))

Output

answered Feb 22, 2022 at 21:35

quamrana

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