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Recursion In Python
Can someone please explain why the following code prints 300 instead of 203 def f(n): if n == 0: return 0 else: return f(n - 1) + 100 print(f(3))
5 Answers
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Hi, again, Abdulrahim Jamil !
But itâs recursion. Every value f(n) is built on previouse values f(n-1), in that way itâs showed below. So f(0) -> f(1), f(1) -> f(2), âŠ, f(n-1) -> f(n):
f(0) = 0
f(1) = f(0) + 100 = 100
f(2) = ff1) + 100 = 100 + 100 = 200
f(3) = f(2) + 100 = 200 + 100 = 300
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Hi, Abdulrahim Jamil !
Why should it be 203!
Itâs start with 0, and add 100 every time you ingrease n with 1
f(0) = 0
f(1) = 0 + 100
f(2) = 0 + 100 + 100
f(3) = 0 = 100 + 100 + 100
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Per Bratthammar thanks for clarifying that, now I remember that the function stack also works like a normal stack i.e it uses the LIFO technique, I now understand how it works. I had initially thought it would work like this:
f(0) = 0
f(1) = 0 + 100
f(2) = 1 + 100
f(3) = 2 + 100
I now see why it couldn't have worked that way and I think it's because it return statement wasn't written like this:
return n + f(n - 1) + 100
As you would have done when calculating the sum of numbers below a number n (without the 100 of course!).
Thanks again Per Bratthammar.
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Per Bratthammar it doesn't start with zero, it starts with a 3
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Mirielle I totally agree with you, as I mentioned in my reply to Per Bratthammar I approached it the same way I would a function calculating the sum of all numbers below a certain number n. Thanks