Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
964 views
in Technique[技术] by (71.8m points)

haskell - How to have multiple infinite ranges in list comprehensions?

In haskell I have a list comprehension like this:

sq = [(x,y,z) | x <- v, y <- v, z <- v, x*x + y*y == z*z, x < y, y < z]
    where v = [1..]

However when I try take 10 sq, it just freezes... Is there a way to handle multiple infinite ranges?

Thanks

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Answer

0 votes
by (71.8m points)

In addition to the other answers explaining the problem, here is an alternative solution, generalized to work with level-monad and stream-monad that lend themselves for searches over infinite search spaces (It is also compatible with the list monad and logict, but those won't play nicely with infinite search spaces, as you already found out):

{-# LANGUAGE MonadComprehensions #-}

module Triples where

import Control.Monad

sq :: MonadPlus m => m (Int, Int, Int)
sq = [(x, y, z) | x <- v, y <- v, z <- v, x*x + y*y == z*z, x < y, y < z]
    where v = return 0 `mplus` v >>= (return . (1+))

Now, for a fast breadth first search:

*Triples> :m +Control.Monad.Stream
*Triples Control.Monad.Stream> take 10 $ runStream sq
[(3,4,5),(6,8,10),(5,12,13),(9,12,15),(8,15,17),(12,16,20),(7,24,25),
(15,20,25),(10,24,26),(20,21,29)]

Alternatively:

*Triples> :m +Control.Monad.Levels
*Triples Control.Monad.Levels> take 5 $ bfs sq   -- larger memory requirements
[(3,4,5),(6,8,10),(5,12,13),(9,12,15),(8,15,17)]
*Triples Control.Monad.Levels> take 5 $ idfs sq  -- constant space, slower, lazy
[(3,4,5),(5,12,13),(6,8,10),(7,24,25),(8,15,17)]

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...