Q. Predict the output of following code.
def express(x, n):
if n== 0:
return 1
elif n%2 == 0:
return express(x*x, n/2)
else:
return x * express (x, n-1)
express (2,5)
Output :-
32
>>>
Explanation :-
Now, let's analyze the expression express(2, 5) using the corrected code:
Since n is not 0, it moves to the else block.
It calls express(2, 4) because 5 is odd, reducing the exponent by 1.
In the next recursive call, express(2, 4) evaluates if 4 is even.
It calls express(2 * 2, 4 / 2) which is equivalent to express(4, 2).
The next recursive call, express(4, 2), evaluates if 2 is even.
It calls express(4 * 4, 2 / 2) which is equivalent to express(16, 1).
The next recursive call, express(16, 1), is not 0 and not even, so it moves to the else block.
It calls express(16, 0).
This time, n is 0, so it returns 1.
The previous recursive call express(16, 1) multiplies 16 by 1, resulting in 16.
The initial recursive call express(2, 4) receives the result of 16.
Since n is still not 0, it moves to the else block.
It calls express(2, 3) because 4 is odd, reducing the exponent by 1.
The process continues similarly until it reaches the base case.
The final result returned by the initial call express(2, 5) is 32.
How the output is 32 , please explain.
ReplyDeleteOk, I have explained it.
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