I got nearly no knowledge of haskell and tried to solve some Project Euler Problems. While solving Number 5 I wrote this solution (for 1..10)
--Check if n can be divided by 1..max
canDivAll :: Integer -> Integer -> Bool
canDivAll max n = all (\x -> n `mod` x == 0) [1..max]
main = print $ head $ filter (canDivAll 10) [1..]
Now I found out, that all
is implemented like this:
all p = and . map p
Doesn't this mean, the condition is checked for every element? Wouldn't it be much faster to break upon the first 开发者_运维问答False-Result of the condition? This would make the execution of the code above faster.
Thanks
and
itself is short-circuited and since both map
and all
evaluation is lazy, you will only get as many elements as needed - not more.
You can verify that with a GHCi
session:
Prelude Debug.Trace> and [(trace "first" True), (trace "second" True)]
first
second
True
Prelude Debug.Trace> and [(trace "first" False), (trace "second" False)]
first
False
map
does not evaluate all its argument before and
executes. And and
is short-circuited.
Notice that in GHC all
isn't really defined like this.
-- | Applied to a predicate and a list, 'all' determines if all elements
-- of the list satisfy the predicate.
all :: (a -> Bool) -> [a] -> Bool
#ifdef USE_REPORT_PRELUDE
all p = and . map p
#else
all _ [] = True
all p (x:xs) = p x && all p xs
{-# RULES
"all/build" forall p (g::forall b.(a->b->b)->b->b) .
all p (build g) = g ((&&) . p) True
#-}
#endif
We see that all p (x:xs) = p x && all p xs
, so whenever p x
is false, the evaluation will stop.
Moreover, there is a simplification rule all/build
, which effectively transforms your all p [1..max]
into a simple fail-fast loop*, so I don't think you can improve much from modifying all
.
*. The simplified code should look like:
eftIntFB c n x0 y | x0 ># y = n
| otherwise = go x0
where
go x = I# x `c` if x ==# y then n else go (x +# 1#)
eftIntFB ((&&) . p) True 1# max#
This is a good program for the fusion optimization, as all your loops are expressed as fusible combinators. Thus you can write it using, e.g. Data.Vector, and get better performance than with lists.
From N=20, with lists as in your program:
- 52.484s
Also, use rem
instead of mod
.
- 15.712s
Where the list functions become vector operations:
import qualified Data.Vector.Unboxed as V
canDivAll :: Int -> Int -> Bool
canDivAll max n = V.all (\x -> n `rem` x == 0) (V.enumFromN 1 max)
main = print . V.head $ V.filter (canDivAll 20) $ V.unfoldr (\a -> Just (a, a+1)) 1
You're assuming that and
is not short-circuiting. and
will stop execution on the first false
result it sees, so it is "fast" as one might expect.
精彩评论