Ytukyg

going by the definition, every item of the fibonacci series is the sum of the previous two terms. putting this definition in to lazy haskell gives u this!

fibo a b = a:fibo b (a+b)

now just take n items from fibo starting with 0,1

take 10 (fibo 0 1)

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.

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

This implementation calculates 100,000th Fibonacci number almost instantly:

fib = fastFib 1 1



fastFib _ _ 0 = 0

fastFib _ _ 1 = 1

fastFib _ _ 2 = 1

fastFib a b 3 = a + b

fastFib a b c = fastFib (a + b) a (c - 1)

A lazy way of generating infinite Fibonacci series can easily be achieved by unfoldr as follows;

fibs :: [Integer]

fibs = unfoldr ((f,s) -> Just (f,(s,f+s))) (0,1)

LOL, I love Haskell pattern matching but it is rendered useless in standard Fibonacci functions. The standard list is constructed from the right. To use pattern matching and cons, the list must be constructed from the left. Well, one consolation, at least, is this is really fast. ~O(n), it should be. A helper function is needed to reverse the infinite list (things you can only do in Haskell, joy) and this function outputs each subsequent list of the run so 'last' is also used in the helper function pipeline.

f (x:y:xs) = (x+y):(x:(y:xs))

The helper

fib n = reverse . last . take n $ iterate f [1,0]

This is a list version and, I think, it explicates how the list is constructed which is the purpose. I want to do a tuple version.

Edit 3/15/2018

First off, Will Ness enlightened me with the knowledge that an entire list being generated at each iteration was unnecessary and that only the last two values used were needed and that the values for the result list were the first values of each list or pair generated. It was so funny. After Will told me the values for the list were the first values of the lists, I ran it and saw the values 0,1,1,2,3,5,8,13 as each head of each list, I said WTF, did Will change my code on my PC? The values were there but how!? After a while, I realized they were there all along but I just didn't see them. ugh.
Will's version of the function and helper function are:

f = ((x:y:xs) -> (x+y):x:xs) -- notice, no y: put back only x+y & x

and his helper function rewrite

fib n = map head . take n $iterate f [0,1]

I think, too, that they now can be combined:

fib n = take n . map head $ iterate ((x:y:xs) -> (x+y):x:xs) [0,1]

As an irrelevant aside, the function can be with tuples, too

fib n = take n . map fst $ iterate ((a,b) -> (b,a+b)) (0,1)

Another form, a list comprehension form, can also be written for all:

fib n = take n [ fst t | t <- iterate ((a,b) -> (b,a+b)) (0,1)]

These are all iterative and robust. The fastest is the map with lists at 12.23 seconds for fib 5000. The tuple comprehension was second fastest for fib 5000 at 13.58 seconds.

using iterate

fibonaci = map fst (iterate f (0,1)) where f (x,y) = (y,x+y)

using

take 10 fibonaci



[0,1,1,2,3,5,8,13,21,34,55,89,144,233,377]