Haskell Creating list of numbers

12,832

Solution 1

Actually, I am not sure if I get your idea. But Is this what you want?

generator list = list ++ generator next
    where
    next = (map (\n -> 2 * n + 1) list) ++ (map (\n -> 3 * n + 1) list)

Oh, you can use generator [1] to fire up. like this:

take 100 $ generator [1]

Solution 2

merge xs [] = xs
merge [] ys = ys
merge (x:xs) (y:ys) | x == y = x : merge xs ys
                    | x < y = x : merge xs (y:ys)
                    | otherwise = y : merge (x:xs) ys 

print $ take 10 $ merge [1,3..] [1,4..]
--[1,3,4,5,7,9,10,11,13,15]

Solution 3

As luqui said, we could use info such as do duplicates matter and does order matter. If the answers are no and no then a simple concatMap works fine:

myList = 1 : concatMap (\n -> 2*n+1 : 3*n+1 : []) myList

Results in:

> take 20 myList
[1,3,4,7,10,9,13,15,22,21,31,19,28,27,40,31,46,45,67,43]

If the answers are yes and yes then I imagine it could be cleaner, but this is sufficient:

myList = abs
  where
  abs = merge as bs
  as = 1 : map (\n -> 2*n+1) abs
  bs = 1 : map (\n -> 3*n+1) abs
  merge (x:xs) (y:ys)
        | x == y = x : merge xs ys
        | x < y  = x : merge xs (y:ys)
        | otherwise = y : merge (x:xs) ys

Results in:

> take 20 myList
[1,3,4,7,9,10,13,15,19,21,22,27,28,31,39,40,43,45,46,55]
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Updated on June 04, 2022

Comments

  • Admin
    Admin almost 2 years

    Hi Im new to Haskell and wish to write a simple code. I want to write a function which creates a list of numbers. Where it starts of with 1 and increase with 2n+1 and 3n+1 so for example output should be like take 6 myList = [1,3,4,7,9,10]

    I think i need to use recursion but not sure how to do it in list format.

    Any help will be appreciated. Thanks