CODEWITHSUNDEEP
pythonrecursiontuplespascals-triangle
aijnij
aijnij

Reputation: 91

Pascal Triangles with Tuples in Recursion in Python

I am asked to write a recursion for Pascal's triangle, with tuples. This is my code:

def pascal(n):
    if n == 1:
        return (1,)
    if n == 2:
        return ((1,),(1,1))
    else:
        new_row = ()
        for i in range(-1, n-1):
            if i == -1:
                new_row = new_row + (1,)
            elif i == n - 1:
                new_row = new_row + (1,)
            else:
                a = pascal(n-1)[n-2][i] + pascal(n-2)[n-1][i+1]
                new_row = new_row + (a,)
        return pascal(n-1), new_row

I am getting an error that says tuple index out of range if I call pascal(3) or higher.

This is the trace output:

Traceback (most recent call last):
  File "tup.py", line 19, in <module>
    print pascal(3)
  File "tup.py", line 15, in pascal
    a = pascal(n-1)[n-2][i] + pascal(n-2)[n-1][i+1]
IndexError: tuple index out of range

How can I fix this error?

Upvotes: 1

Views: 1350

Answers (1)

user2390182
user2390182

Reputation: 73470

The following will work. Loop from 0 to n-2 (exclusive) on the previous row:

def pascal(n):
    if n == 1:  # one base case is enough
        return ((1,),)  # return tuple of tuples to be consistent
    prev = pascal(n-1)
    new_row = [1] + [prev[-1][i]+prev[-1][i+1] for i in range(n-2)] + [1]
    return prev + tuple(new_row)

>>> pascal(2)
((1,), (1, 1))
>>> pascal(3)
((1,), (1, 1), (1, 2, 1))
>>> pascal(4)
((1,), (1, 1), (1, 2, 1), (1, 3, 3, 1))

Explanation: The n-th row has n elements, 2 of which are 1. Thus n-2 of the elements are formed by appropriate sums in the loop, hence the range.

Upvotes: 4

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