Haskell 在没有显式递归的情况下生成具有适当性的随机数

Question

Here I have a function to generate a stream of random numbers between 0 and 999.

randomHelp :: RandomGen g => g -> [Int]
randomHelp g = zipWith (mod) (map fst $ iterate (next . snd) $ next $ snd $ split g) $ repeat 1000

I would like to select all numbers from the stream defined above and each elem(i) and elem(i + 1) must respect a propriety. For example their gcd have to be one. All I can think is a fold function with because I can start with and accumulator which contains the number 1 (let's assume 1 will be the first element I want to show) then I check the propriety in fold's function and if it is respected i add the element to the accumulator, but the problem is the program blocks because of stackoverflow I think.

Here is the function:

randomFunc :: RandomGen g => g -> [Int]
randomFunc g = foldl (\acc x -> if (gcd x (last acc) == 1) then acc ++ [x] else acc) [1] (randomHelp g)

Note: I don't want to use explicit recursion.

Answer 1

A right fold would probably fit better, something like:

import System.Random (RandomGen, randomRs, mkStdGen)

randomFunc :: RandomGen g => g -> [Int]
randomFunc g = foldr go (const []) (randomRs (1, 20) g) 1
    where go x f lst = if gcd x lst == 1 then x: f x else f lst

then

\> take 20 . randomFunc $ mkStdGen 1
[16,7,6,19,8,15,16,1,9,2,15,17,14,3,11,17,15,8,1,5]

Doing so you may build the list using : instead of ++ which may cause quadratic performance cost, and you may bypass the call to last .