Comments on: Transliterating python to haskell: Fibonacci in the state monad

By: TIK

TIK — Thu, 08 Dec 2011 09:25:20 +0000

Even simpler version:

fib = f n 0 1
f 0 a b = a
f n a b = f (n-1) b (a+b)

By: thomashartman1

thomashartman1 — Sun, 17 Jan 2010 18:42:34 +0000

I kind of doubt it, because if you try to to do any kind of recursive fib in python you run into “maximum recursion levels” exceeded error.

By: thomashartman1

thomashartman1 — Sun, 17 Jan 2010 18:41:19 +0000

My point is that what’s more idiomatic depends on what your context (or your primary programming language) is.

This post was aimed at pythonistas, and people that claim that haskell is a single paradigm language that is unfriendly to people that only have experience in imperative programming.

By: bryanstamour.com/blog » Blog Archive » Decking the halls with fibonaccis

Tue, 15 Dec 2009 03:33:25 +0000

[...] below method comes courtesy of Thomas Hartman from the Patch-Tag blog. I belive that it shows how elegant Haskell code can look, even while using the State Monad to hide [...]

By: Chaddaï

Chaddaï — Mon, 14 Dec 2009 17:53:48 +0000

I would probably write it :

> fib n = flip evalState (0,1) $ do
> replicateM n $ modify (\(a,b) -> (b,a+b))
> gets fst

By: Dave Leimbach

Dave Leimbach — Mon, 14 Dec 2009 15:29:25 +0000

Sorry, I forgot the part where I was wondering if what I just wrote could be written similarly in Python.

It’s good to see code expressed similarly between languages, but it makes me wonder if there’s another perhaps more expressive way to do the same in Python too!

By: Dave Leimbach

Dave Leimbach — Mon, 14 Dec 2009 15:26:39 +0000

Nice work!

How about

modify \x -> (snd x, (snd x) + (fst x))

instead of “get” then “put”

for your state update.

And you can do “return $ gets fst” rather than getting b again and not using it.

I actually wonder what the point-free version of that modify lambda expression would look like.

By: yitz

yitz — Mon, 14 Dec 2009 14:19:22 +0000

How about:

fib :: Integer -> Integer
fib n = flip evalState (0,1) $ do
replicateM n $ do
modify $ \(a,b) -> (b,a+b)
gets fst

By: JadeNB

JadeNB — Mon, 14 Dec 2009 05:56:45 +0000

Is this really more idiomatic than lifting the assignment to argument-passing in the usual way?

fib n =
let
fib 0 a b = a
fib n a b = fib $ n – 1 $ b $ a + b
in fib n 0 1

By: Brent

Brent — Sun, 13 Dec 2009 16:21:06 +0000

Instead of

(a,b) <- get
put (b, a+b)

you could also do

modify (snd &&& uncurry (+))

Although whether that is clearer is up for debate. =)