Generating Fibonacci numbers in Haskell?

Here's a different and simpler function that calculates the n'th Fibonacci number:

fib :: Integer -> Integer
fib 0 = 0
fib 1 = 1
fib n = fib (n-1) + fib (n-2)

The implementation you are referring to relays on some observations about how values in Fibonacci relate to each other (and how Haskell can define data structures in terms of themselfs in effect creating infinite data structures)

The function in your question works like this:

Assume you already had an infinite list of the Fibonacci numbers:

   [ 1, 1, 2, 3, 5,  8, 13, .... ]

The tail of this list is

   [ 1, 2, 3, 5, 8, 13, 21, .... ]

zipWith combines two lists element by element using the given operator:

   [ 1, 1, 2, 3,  5,  8, 13, .... ]
+  [ 1, 2, 3, 5,  8, 13, 21, .... ]
=  [ 2, 3, 5, 8, 13, 21, 34, .... ]

So the infinite list of Fibonacci numbers can be calculated by prepending the elements 1 and 1 to the result of zipping the infinite list of Fibonacci numbers with the tail of the infinite list of Fibonacci numbers using the + operator.

Now, to get the n'th Fibonacci number, just get the n'th element of the infinite list of Fibonacci numbers:

fib n = fibs !! n

The beauty of Haskell is that it doesn't calculate any element of the list of Fibonacci numbers until its needed.

Did I make your head explode? :)

To expand on dtb's answer:

There is an important difference between the "simple" solution:

fib 0 = 1
fib 1 = 1
fib n = fib (n-1) + fib (n-2)

And the one you specified:

fibs = 1 : 1 : zipWith (+) fibs (tail fibs)

The simple solution takes O(1.618^NN) time to compute the Nth element, while the one you specified takes O(N²). That's because the one you specified takes into account that computing fib n and fib (n-1) (which is required to compute it) share the dependency of fib (n-2), and that it can be computed once for both to save time. O(N²) is for N additions of numbers of O(N) digits.

There are a number of different Haskell algorithms for the Fibonacci sequence here. The "naive" implementation looks like what you're after.

fibs :: [Integer]
fibs = 1 : 1 : zipWith (+) fibs (tail fibs)

at first, with fibs and tail fibs, we can get the 3rd element:

fibs                        : [1, 1, ?
tail fibs                   : [1, ?
zipWith (+) fibs (tail fibs): [2, ?

now, we know the 3rd is 2, we can get the 4th:

fibs                        : [1, 1, 2, ?
tail fibs                   : [1, 2, ?
zipWith (+) fibs (tail fibs): [2, 3, ?

now the 5th:

fibs                        : [1, 1, 2, 3, ?
tail fibs                   : [1, 2, 3, ?
zipWith (+) fibs (tail fibs): [2, 3, 5, ?

and so on ..

The definition of fibonaci(n) is:

fibonacci (n) = fibonacci (n-1) + fibonacci (n-2)

The naive implementation in Haskell

fibonacci :: Integer -> Integer
fibonacci 0 = 1
fibonacci 1 = 1
fibonacci x = fibonacci (x-1) + fibonacci (x-2)

All formulas can be traced back to this definition, some which run very quickly, some of which run very slowly. The implementation above has O(n) = 2^n

In the spirit of your question, let me remove the use of lists and give you something that runs in O(n) I.e. let's not hold all the fibonaccis from 0 to n in a list.

If we have a triple (a tuple with three members) that looks like:

(n, fibonacci[n-1], fibonacci[n])

Remembering the initial definition, we can calculate the next triple from the last triple:

(n+1, fibonacci[n], fibonacci[n-1] + fibonacci[n]) = (n+1, fibonacci[n], fibonacci[n+1])

And the next triple from the last triple: (n+2, fibonacci[n+1], fibonacci[n] + fibonacci[n+1]) = (n+1, fibonacci[n+1], fibonacci[n+2])

And so on...

n = 0 => (0,0,1) 
n = 1 => (1,1,1) - calculated from the previous triple
n = 2 => (2,1,2) - calculated from the previous triple
n = 3 => (3,2,3) - calculated from the previous triple
n = 4 => (4,3,5) - calculated from the previous triple
n = 5 => (5,5,8) - calculated from the previous triple

Let's implement this in Haskell and use self explanatory variable names:

nextTripleIfCurrentNIsLessThanN :: (Int, Integer, Integer) -> Int -> (Int, Integer, Integer)
nextTripleIfCurrentNIsLessThanN (currentN, x, y) n = if currentN < n
then nextTripleIfCurrentNIsLessThanN (currentN + 1, y, x + y) n
else (currentN, x, y)

thirdElementOfTriple :: (x,y,z) -> z
thirdElementOfTriple (x,y,z) = z

fibonacci :: Int -> Integer
fibonacci n = thirdElementOfTriple (nextTripleIfCurrentNIsLessThanN (0,0,1) n)

This will work in O(n) [It is mildly quadratic which shows up in large numbers. The reason for that is that adding big numbers is more costly than adding small ones. But that's a separate discussion about model of computation.]

fibonacci 0
1
fibonacci 1
1
fibonacci 2
2
fibonacci 3
3
fibonacci 4
5
fibonacci 5
8
fibonacci 5000
6276302800488957086035253108349684055478528702736457439025824448927937256811663264475883711527806250329984690249846819800648580083040107584710332687596562185073640422286799239932615797105974710857095487342820351307477141875012176874307156016229965832589137779724973854362777629878229505500260477136108363709090010421536915488632339240756987974122598603591920306874926755600361865354330444681915154695741851960071089944015319300128574107662757054790648152751366475529121877212785489665101733755898580317984402963873738187000120737824193162011399200547424034440836239726275765901190914513013217132050988064832024783370583789324109052449717186857327239783000020791777804503930439875068662687670678802914269784817022567088069496231111407908953313902398529655056082228598715882365779469902465675715699187225655878240668599547496218159297881601061923195562143932693324644219266564617042934227893371179832389642895285401263875342640468017378925921483580111278055044254198382265567395946431803304304326865077742925818757370691726168228648841319231470626

Put in code, your definition is

fib :: Int -> Integer
fib 0 = 1
fib 1 = 1
fib n = fib (n-1) + fib (n-2)
  -- i.e.
  -- fib (n+2) = fib (n+1) + fib n

Int -> a ~= [a] because

from f = map f [0..]     -- from :: (Int -> a) -> [a]
to = (!!)                -- to :: [a] -> (Int -> a)

Thus

fibs :: [Integer]
fibs = from fib 

fibs !! 0 = 1
fibs !! 1 = 1
fibs !! (n+2)    = fibs !! (n+1)     +  fibs !! n
-- or,
drop 2 fibs !! n = drop 1 fibs !! n  +  fibs !! n
                 = zipWith (+) (tail fibs) fibs !! n
-- i.e.
take 2 fibs = [1,1]
drop 2 fibs = zipWith (+) (tail fibs) fibs
-- hence, 
fibs = take 2 fibs ++ drop 2 fibs
     = 1 : 1 : zipWith (+) (tail fibs) fibs

Or, as a, b = (0,1) : (b, a+b):

fibs :: [Integer]
fibs = a
  where
  (a,b) = unzip $ (0,1) : zip b (zipWith (+) a b)

I was doing the homework6 of CIS194 and find that you could write this way. Computing the first n elements requires only O(n) addition operations.

fibs2 :: [Integer]
fibs2 = [0, 1] ++ [fibs2 !! (n-1) + fibs2 !! (n-2) | n <- [2..]]

I tried to reimplement this in python3. The goal was to get a similar algorithm in python which is obviously the same, but not to mimic all aspects of Haskell.

I came up with the following code.

fibs.py:

# python version of Haskell's code
#    fibs = 1 : 1 : zipWith (+) fibs (tail fibs)

from operator import add

fibsList = [1, 1]  # growing

def fibs(n):
  if n >= len(fibsList):                # lazy evaluation
    x=zipWith(n-2,add,fibs,tail(fibs))  # or: ...,fibs,tailfibs)
    fibsList.append(x)
  return fibsList[n]

def zipWith(n,op,list1,list2):
  return op(list1(n),list2(n))

def tail(list):                         # or: def tailfibs(n):
  return lambda n : list(n + 1)         #       return fibs(n+1)

# test
print (fibs(10))
print (*fibsList)

Running it will output

$ python fibs.py
89
1 1 2 3 5 8 13 21 34 55 89

This will do the same as the Haskell code, but it is a step by step version where you can add some logging