埃拉托色尼在Haskell的筛
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27-09-2019 - |
题
我解决哈斯克尔一些经典的问题,发展我的功能 技能,我有一个问题,以实现在此“实践编程能力”提出的优化网站:
我有三种解决这个问题,第三个是太慢 相比于第二溶液。有人建议可以给一些改进 我的代码?
我的实现方式是:
-- primeira implementação
primes n
| n < 2 = []
| n == 2 = [2]
| n `mod` 2 == 0 = primes'
| otherwise = if (find (\x -> n `mod` x == 0) primes') == Nothing then
n:primes'
else
primes'
where primes' = primes (n - 1)
-- segunda implementação
primes' :: Integer -> [Integer]
primes' n = sieve $ 2 : [3,5..n]
where sieve :: [Integer] -> [Integer]
sieve [] = []
sieve l@(x:xs)
| x*x >= n = l
| otherwise = x : sieve list'
where list' = filter (\y -> y `mod` x /= 0) xs
-- terceira implementação
primes'' :: Integer -> [Integer]
primes'' n = 2 : sieve 3 [3,5..n]
where sieve :: Integer -> [Integer] -> [Integer]
sieve _ [] = []
sieve m l@(x:xs)
| m*m >= n = l
| x < m*m = x : sieve m xs
| otherwise = sieve (m + 2) list'
where list'= filter (\y -> y `mod` m /= 0) l
解决方案
在我看来像你的第三次修改的问题是,你如何选择下一个元素进行筛选的。 你胡乱按2递增的问题是,你再上不必要的号码筛选。 例如,在这个版本中你最终要通过9米,你会做一个额外的递归滤波月9日,即使它甚至不是在列表中,这样的话你应该从来没有把它在首位(因为这将在3中的第一个过滤器已被移除)
即使第二版本不启动过去它进行筛选,对数的平方的滤波,它从未选择的不必要的筛选值。
在换句话说,我想你最终3和n之间过筛上的每个奇数。相反,你应该筛选对尚未由先前的通去除每个奇数。
我认为正确实现在当前筛值的平方开始筛的优化,则必须保留列表的前面,而在那里回来包含的元素后筛选> =该筛值的平方。我认为这将迫使你使用串连,我不那么肯定的优化是不够好,以抵消使用++引起的开销。
其他提示
首先,mod
是缓慢,在情况下使用rem
它并不重要(当你不使用底片处理,基本上)。其次,使用标准显示(自己)什么是速度更快,什么样的变化实际上是优化。我知道我不给一个完整的答案,你有这个问题,但它是一个好地方,你(和其他潜在的应答者)来启动,所以这里的一些代码:
import List
import Criterion.Main
main = do
str <- getLine
let run f = length . f
input = read str :: Integer
defaultMain [ bench "primes" (nf (run primes) input)
, bench "primes'" (nf (run primes') input)
, bench "primes''" (nf (run primes'') input)
, bench "primesTMD" (nf (run primesTMD) input)
, bench "primes'TMD" (nf (run primes'TMD) input)
, bench "primes''TMD" (nf (run primes''TMD) input)
]
putStrLn . show . length . primes'' $ (read str :: Integer)
-- primeira implementação
primes n
| n < 2 = []
| n == 2 = [2]
| n `mod` 2 == 0 = primes'
| otherwise = if (find (\x -> n `mod` x == 0) primes') == Nothing then
n:primes'
else
primes'
where primes' = primes (n - 1)
primesTMD n
| n < 2 = []
| n == 2 = [2]
| n `mod` 2 == 0 = primes'
| otherwise = if (find (\x -> n `rem` x == 0) primes') == Nothing then
n:primes'
else
primes'
where primes' = primesTMD (n - 1)
-- segunda implementação
primes' :: Integer -> [Integer]
primes' n = sieve $ 2 : [3,5..n]
where sieve :: [Integer] -> [Integer]
sieve [] = []
sieve l@(x:xs)
| x*x >= n = l
| otherwise = x : sieve list'
where list' = filter (\y -> y `mod` x /= 0) xs
primes'TMD :: Integer -> [Integer]
primes'TMD n = sieve $ 2 : [3,5..n]
where sieve :: [Integer] -> [Integer]
sieve [] = []
sieve l@(x:xs)
| x*x >= n = l
| otherwise = x : sieve list'
where list' = filter (\y -> y `rem` x /= 0) xs
-- terceira implementação
primes'' :: Integer -> [Integer]
primes'' n = 2 : sieve 3 [3,5..n]
where sieve :: Integer -> [Integer] -> [Integer]
sieve _ [] = []
sieve m l@(x:xs)
| m*m >= n = l
| x < m*m = x : sieve m xs
| otherwise = sieve (m + 2) list'
where list'= filter (\y -> y `mod` m /= 0) l
primes''TMD :: Integer -> [Integer]
primes''TMD n = 2 : sieve 3 [3,5..n]
where sieve :: Integer -> [Integer] -> [Integer]
sieve _ [] = []
sieve m l@(x:xs)
| m*m >= n = l
| x < m*m = x : sieve m xs
| otherwise = sieve (m + 2) list'
where list'= filter (\y -> y `rem` m /= 0) l
注意的改进的运行时使用的变体rem
:
$ ghc --make -O2 sieve.hs
$./sieve
5000
...
benchmarking primes
mean: 23.88546 ms, lb 23.84035 ms, ub 23.95000 ms
benchmarking primes'
mean: 775.9981 us, lb 775.4639 us, ub 776.7081 us
benchmarking primes''
mean: 837.7901 us, lb 836.7824 us, ub 839.0260 us
benchmarking primesTMD
mean: 16.15421 ms, lb 16.11955 ms, ub 16.19202 ms
benchmarking primes'TMD
mean: 568.9857 us, lb 568.5819 us, ub 569.4641 us
benchmarking primes''TMD
mean: 642.5665 us, lb 642.0495 us, ub 643.4105 us
当我看到你在做这个为自己的教育,其值得一提的素数对Haskell的相关链接.ORG 和上hackage快速素数包。
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