🔒 very simple recursion

[knotty aspen]

'''
Write a recursive function power(x, n), where n is 0 or a postive integer. For example,
power(2, 10) will return 1024. Write a suitable base case, and for the general case use the
idea that xn = x * xn-1 .
'''
def power(x,n):
    if n < 0:
        return 1
    else:
        return x * power(x, n-1)
    
print(power(2,10))

someone please help me fix my code, without telling me the answer

[fleet jacinth]

@knotty aspen

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[onyx linden]

What's the problem?

[knotty aspen]

I get 2048 as the output

[timid magnet]

your base case is wrong

where n is 0 or a postive integer
yet your condition is n < 0

which is only satisfied when n is, well, less than 0, ie it is negative.

if n is an integer and is reduced by 1 each time the function recurses, it will stop when n = -1, not n = 0. because 0 is not < 0.

so essentially you are always recursing one extra time

[knotty aspen]

oh ok

got it

if n == 0: would be better

[timid magnet]

yes

[fleet jacinth]

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🔒 very simple recursion