Question

Here is the code to a subset problem with arguments n and k. n represents the total number of students and k represents the amount of the students I want to get out of n. The code attempts to give the number of possible combinations of pulling k number of students out of n number of students.

def subset(n, k): 
    if k == 0:
        return 1
    if n == k:
        return 1
    else:
        return subset(n-1, k-1) + subset(n-1, k)

I understand the first part of the recursive call, but I’m having trouble understanding the + subset(n-1, k) part. Can anyone explain this to me?

Answer 1

The recursion is based on a simple observation, for which I will give a combinatorial argument, as to why it is true, rather than a mathematical proof through formulae.

Whenever you choose k elements out of n, there are two cases:

1. You choose element #n
2. You don’t choose element #n

Since these events are mutually exclusive, the total amount of combinations is given by the amount of combinations when choosing #n, and those when you don’t choose #n.

Choosing element #n

Since we have already chosen one element, we need only choose another k-1 elements. Also, since we have decided upon one element – as to whether it is included or not – already, we only need to consider the remaining n-1 elements.

Thus, the amount of combinations for choosing element #n is given by

    subset(n - 1, k - 1)

Not choosing element #n

There are still k elements to choose, but since we have already made up our mind about element #n, there remain only n - 1 elements to choose from. Thus:

    subset(n - 1, k)

The base case

The recursion uses the fact, that we can usually differentiate between two situations, solutions where element n is part of that solution, and those where it is not.

However, such a distinction can not always be made:

• When choosing all elements (corresponding to case n == k in code below)
• or when choosing no elements at all (corresponding to case k == 0 in code below)

In these cases, there is only exactly one solution, hence

if k == 0:
    return 1
if n == k:
    return 1

Ensuring it works

To do that, we need to convince ourselves (or prove) that the base case is always hit at some point.

Let us assume, that n < k at some point. Since per our assumption, n was originally greater or equal to k, there must have been some point where n = k, because n and k decrease in unison or only n decreases by one, i.e. it follows

This implies, that there must have been a call to subset(n - 1, k) for it to happen, that n decreases below k. However, this is not possible since we have a base case on n = k where we return a constant 1.

We conclude that either n decreases at some point such that n = k, or decrease in unison exactly k times such that k = 0.

Thus, the base case works.