Ris - 1 year ago 113

Python Question

I am interested in a iterative algorithm for Fibonacci numbers, so I found the formula on wiki...it looks straight forward so I tried it in Python...it doesn't have a problem compiling and formula looks right...not sure why its giving the wrong output...did I not implement it right ?

`def fib (n):`

if( n == 0):

return 0

else:

x = 0

y = 1

for i in range(1,n):

z = (x + y)

x = y

y = z

return y

for i in range(10):

print (fib(i))

output

`0`

None

1

1

1

1

1

1

1

1

1

1

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Answer Source

The problem is that your `return y`

is within the loop of your function. So after the first iteration, it will already stop and return the first value: 1. Except when `n`

is 0, in which case the function is made to return `0`

itself, and in case `n`

is 1, when the for loop will not iterate even once, and no `return`

is being execute (hence the `None`

return value).

To fix this, just move the `return y`

outside of the loop.

Following KebertX’s example, here is a solution I would personally make in Python. Of course, if you were to process many Fibonacci values, you might even want to combine those two solutions and create a cache for the numbers.

```
def f(n):
a, b = 0, 1
for i in range(0, n):
a, b = b, a + b
return a
```

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