Cecil Westerhof <ce...@decebal.nl> writes:
> To show why it is often easy, but wrong to use recursive functions I
> wrote the following two Fibonacci functions:
> def fib_ite(n):
> if not type(n) is int:
> raise TypeError(f'Need an integer ({n})')
> if n < 0:
> raise ValueError(f'Should not be negative ({n})')
>
> if n in [0, 1]:
> return n
>
> # a is previous fibonacy (starts with fib(0))
> # b is current fibonaccy (starts with fib(1))
> a, b = 0, 1
> # range goes to n - 1, so after loop b contains fib(n)
> for i in range(1, n):
> a, b = b, a + b
> return b
>
>
> def fib_rec(n):
> if not type(n) is int:
> raise TypeError(f'Need an integer ({n})')
> if n < 0:
> raise ValueError(f'Should not be negative ({n})')
>
> if n in [0, 1]:
> return n
>
> return fib_rec(n - 2) + fib_rec(n - 1)
>
> The first eight lines are the same. And I did change the description
> of the errors, which had to be done in both functions. What would be
> the best way to circumvent this?
> Two options are:
> - Call an init function.
> - Call the 'master' function with a lambda.
>
> What is the preferable way, or is there a better way?
I have chosen this implementation with inner functions:
def fibonacci(n, implementation = 'iterative'):
def ite(n):
# a is previous fibonacy (starts with fib(0))
# b is current fibonaccy (starts with fib(1))
a, b = 0, 1
# range goes to n - 1, so after loop b contains fib(n)
for i in range(1, n):
a, b = b, a + b
return b
def rec(n):
if n in [0, 1]:
return n
return rec(n - 2) + rec(n - 1)
if not type(n) is int:
raise TypeError(f'Need an integer ({n})')
if n < 0:
raise ValueError(f'Should not be negative ({n})')
if n in [0, 1]:
return n
if implementation == 'iterative':
return ite(n)
elif implementation == 'recursive':
return rec(n)
raise ValueError(f'Got a wrong function implementation type: {type}')
--
Cecil Westerhof
Senior Software Engineer
LinkedIn: http://www.linkedin.com/in/cecilwesterhof
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