6

我正在尝试完成我的 Haskell 作业的最后一部分，但我被困住了，我的代码到目前为止：

data Entry = Entry (String, String)

class Lexico a where
    (<!), (=!), (>!) :: a -> a -> Bool

instance Lexico Entry where
    Entry (a,_) <! Entry (b,_) = a <  b
    Entry (a,_) =! Entry (b,_) = a == b
    Entry (a,_) >! Entry (b,_) = a >  b

entries :: [(String, String)]
entries =  [("saves", "en vaut"), ("time", "temps"), ("in", "<`a>"),
              ("{", "{"), ("A", "Un"), ("}", "}"), ("stitch", "point"),
              ("nine.", "cent."), ("Zazie", "Zazie")]

build :: (String, String) -> Entry
build (a, b) = Entry (a, b)

diction :: [Entry]
diction = quiksrt (map build entries)

size :: [a] -> Integer
size [] = 0
size (x:xs) = 1+ size xs

quiksrt :: Lexico a => [a] -> [a]
quiksrt [] = []
quiksrt (x:xs)
    |(size [y|y <- xs, y =! x]) > 0 = error "Duplicates not allowed."
    |otherwise = quiksrt [y|y <- xs, y <! x]++ [x] ++ quiksrt [y|y <- xs, y >! x] 


english :: String
english = "A stitch in time save nine."

show :: Entry -> String
show (Entry (a, b)) = "(" ++ Prelude.show a ++ ", " ++ Prelude.show b ++ ")"

showAll :: [Entry] -> String
showAll [] = []
showAll (x:xs) = Main.show x ++ "\n" ++ showAll xs

main :: IO ()
main = do putStr (showAll ( diction ))

问题问：

编写一个 Haskell 程序，获取英语句子“english”，使用二分搜索查找英法词典中的每个单词，执行逐字替换，组装法语翻译并将其打印出来。

函数“快速排序”拒绝重复条目（带有“错误”/中止），因此任何英语单词都有一个法语定义。使用原始 'raw_data' 和将 '("saves", "sauve")' 添加到 'raw_data' 后测试 'quicksort'。

这是二分搜索的冯诺依曼晚期停止版本。将字面音译为 Haskell。进入后，Haskell 版本必须立即验证递归“循环不变量”，如果无法保持，则以“错误”/中止终止。如果找不到英文单词，它也会以相同的方式终止。

function binsearch (x : integer) : integer
local j, k, h : integer
j,k := 1,n
do j+1 <> k --->
  h := (j+k) div 2
  {a[j] <= x < a[k]}        // loop invariant
  if x <  a[h] ---> k := h
   | x >= a[h] ---> j := h
  fi
od
{a[j] <= x < a[j+1]}        // termination assertion
found := x = a[j]
if found     ---> return j
 | not found ---> return 0
fi

在 Haskell 版本中

binsearch :: String -> Integer -> Integer -> Entry

因为类型为“[Entry]”的常量字典“a”是全局可见的。提示：输入“binsearch”后立即将您的字符串（英文单词）变成“Entry”。

高级数据类型“Entry”的编程价值在于，如果您可以在整数上设计这两个函数，那么将它们提升到对 Entry 进行操作是微不足道的。

有人知道我应该如何处理我的二进制搜索功能吗？

4

4 回答 4

4

讲师要求“直译”，因此请以相同的顺序使用相同的变量名。但请注意一些差异：

  • 给定的版本只需要1个参数，他给出的签名需要3个。嗯，
  • 给定的版本不是递归的，但他要求一个递归的版本。

另一个答案是转换为数组，但是对于这么小的练习（毕竟这是家庭作业），我觉得我们可以假装列表是直接访问的。我刚刚接受了你的 diction::[Entry] 并对其进行了索引。我确实必须在几个地方在 Int 和 Integer 之间进行转换。

小问题：您的英文值有误（bs 是我创建的 binSearch 的快捷方式）：

  *Main> map bs (words english)
[Entry ("A","Un"),Entry ("stitch","point"),Entry ("in","<`a>"),Entry ("time","te
mps"),*** Exception: Not found
*Main> map bs (words englishFixed)
[Entry ("A","Un"),Entry ("stitch","point"),Entry ("in","<`a>"),Entry ("time","te
mps"),Entry ("saves","en vaut"),Entry ("nine.","cent.")]
*Main>
于 2008-11-16T06:01:25.490 回答
3

二进制搜索需要随机访问，这在列表中是不可能的。因此，首先要做的可能是将列表转换为Array(with listArray)，然后对其进行搜索。

于 2008-11-16T00:28:53.900 回答
1

这是问题的英文部分的代码（我对其进行了测试，并且效果很好）：

module Main where

class Lex a where
    (<!), (=!), (>!) :: a -> a -> Bool

data Entry = Entry String String

instance Lex Entry where
    (Entry a _) <!  (Entry b _) = a <  b
    (Entry a _) =!  (Entry b _) = a == b
    (Entry a _) >!  (Entry b _) = a >  b
  -- at this point, three binary (infix) operators on values of type 'Entry'
  -- have been defined

type Raw = (String, String)

raw_data :: [Raw]
raw_data  =  [("than a", "qu'un"), ("saves", "en vaut"), ("time", "temps"),
                ("in", "<`a>"), ("worse", "pire"), ("{", "{"), ("A", "Un"),
                ("}", "}"), ("stitch", "point"), ("crime;", "crime,"),
                ("a", "une"), ("nine.", "cent."), ("It's", "C'est"),
                ("Zazie", "Zazie"), ("cat", "chat"), ("it's", "c'est"),
                ("raisin", "raisin sec"), ("mistake.", "faute."),
                ("blueberry", "myrtille"), ("luck", "chance"),
                ("bad", "mauvais")]

cook :: Raw -> Entry
cook (x, y) = Entry x y

a :: [Entry]
a = map cook raw_data

quicksort :: Lex a => [a] -> [a]
quicksort []     = []
quicksort (x:xs) = quicksort (filter (<! x) xs) ++ [x] ++ quicksort (filter (=! x) xs) ++ quicksort (filter (>! x) xs) 

getfirst :: Entry -> String
getfirst (Entry x y) = x

getsecond :: Entry -> String
getsecond (Entry x y) = y

binarysearch :: String -> [Entry] -> Int -> Int -> String
binarysearch s e low high 
    | low > high = " NOT fOUND "
    | getfirst ((e)!!(mid)) > s = binarysearch s (e) low (mid-1)
    | getfirst ((e)!!(mid)) < s = binarysearch s (e) (mid+1) high
    | otherwise = getsecond ((e)!!(mid))
        where mid = (div (low+high) 2)

translator :: [String] -> [Entry] -> [String]
translator [] y = []
translator (x:xs) y = (binarysearch x y 0 ((length y)-1):translator xs y)

english :: String
english = "A stitch in time saves nine."

compute :: String -> [Entry] -> String
compute x y = unwords(translator (words (x)) y)

main = do
    putStr (compute english (quicksort a))
于 2009-11-09T19:45:20.047 回答
0

一个重要的 Prelude 运算符是：

(!!) :: [a] -> Integer -> a
-- xs!!n returns the nth element of xs, starting at the left and
-- counting from 0.

因此，[14,7,3]!!1~~> 7。

于 2009-11-08T22:47:15.120 回答