Doesn't iteration function save intermediate steps?

Question

I was just starting to study Haskell and how the exercise got into the Project Euler problem, where the Fibonacci numbers were. My current method is this function, which creates a new list with the following element:

fib :: (Integral a) => [a] -> [a] fib xs@ (x1:x2:_) = (x1+x2) : xs 

I found an iterate function that returns a function on the result. However, the result is a list of lists, [[2,1],[3,2,1],[5,3,2,1],..] . What is the iterate alternative when I'm not interested in intermediate results? I want to make takeWhile with a condition for the last generated number. Is it not at all to think about it?

(I saw better / shorter / faster ways to generate a Fibonacci sequence, so I am not looking for feedback on the fib function, but I would like to make it work, a suboptimal method or not)

Answer 1

Just use iterate ! Since Haskell is a pure language, all subscribers get shared access, and you pay almost no cost for creating all these mini-lists: [2, 1] actually 2, 1 in [3, 2, 1] , etc. .

You really don't want to takeWhile , because it will give you a lot of superfluous anthill, and you still need to get to the end with last . Use find instead.

Also note that if you plan to summarize the resulting list, you skipped 1 , so you will have one.

Answer 2

I'm not sure about using iterate , but see Haskell's Filtering the Fibonacci Sequence for a list of fibs.

Is it the same homework?

Answer 3

I would use take, as each subsequent approximation will be one whole more accurate than the last. Then you can do (reverse).

Note that “every” result is an intermediate result if the function you execute does not have a computable fixed point.

Answer 4

When you need a function in Haskell, just define it :)

The following is not the most elegant or idiomatic Haskell, but serves to show that you can always do something manually using a tail recursive loop if necessary

 apply_n fnx = if n = 0 then x else apply_n' f (n-1) (fx) n_th_fib n = apply_n fib n [1,1] 

I am sure there is an easier way to do this using folds (or a library function that I forgot about :))