lec20, Thu 12/05

Recursion + Exam Review

Code from lecture

def sumOdd( L ):
    """
return a sum of odd integers in L
recursive solution
If the list is empty or contains odd values
return the sentinel value of None
    """
    if len(L) == 1: # base case
        if (L[0] % 2 == 1): # odd
            return L[0]
        elif (L[0] % 2 == 0): # even
            return None
    else: # recursive case
        last_el = L.pop()
        if (last_el % 2 == 1):
            return last_el + sumOdd(L) # input the smaller list
        else:
            return sumOdd(L)

These were the cases (input and output) we identified that we can test our algorithm on:

[] --> None
[1] --> 1
[2] --> None
[1,1] --> 2
[2,2] --> None
[1,2] --> 1

Notice that the above solution does not handle the empty list. How would you modify the code to correctly return None if the list is empty?

This is just one possible implementation of this function. Alternatively, you can be processing the list starting from the first element instead.

A great question/discussion during the lecture pointed out that None as a sentinel value allows us to distinguish it from 0, which is a valid sum if our list is not empty and contains, for example, [-1, 1].