fib function in Haskell efficiently generates values from the Fibonacci series:
= 0:1:zipWith (+) fib (tail fib)fib
It works by zipping (i.e. combining elements from) the list
0:1:(0+1):(1+1):(1+2):... and the tail of the same list (i.e.
1:(0+1):(1+1):(1+2):...). The (almost) magic thing is that the sums in the list are sums of elements from the same list, just one and two elements behind, respectively.
Anyway… Today, for some reason, I was trying to write this function in Python on a whiteboard, and I’m quite sure I screwed up the python syntax (well, I know I did – I completely forgot the
yield keyword used in Python generators to name one thing).
So, to brush up on my Python skills, I had the Python interpreter yell at me until I got it right. The following generator generates an infinite list of Fibonacci numbers and
fib allows you to ask for a specific number.
from itertools import islice, chain, imap def fibgen(): def tail(iterable): return islice(iterable, 1, None) for n in chain([0,1], imap(lambda a,b: a+b, fibgen(), tail(fibgen()))): yield n def fib(n): return list(islice(fibgen(), n, n+1))
This actually turned out to be quite different from the Haskell version. Since we can only access the
next element of a generator, it doesn’t make sense to have two references to previous elements of the same list we’re generating (it’s not even a list, after all). So, let have the generator have two versions of itself, drop one element from one of them and then have the (outer) generator yield the sum of the next items from its two inner generators, ad infinitum.
However, these two inner generators will have their own generators, and so on, so this will be dog slow.
(Yes, yes, the way it should actually be done is to just keep track of the two previous values.)
def fibgen(): = 0, 1 n1, n2 while True: yield n1 = n2, n1 + n2 n1, n2