我正在尝试完成我的Haskell作业的最后一部分而且我被卡住了,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 ))
问题是:
编写一个需要的Haskell程序 英文句子'english',看起来 英语 - 法语中的每个单词 字典使用二进制搜索, 进行逐字替换, 组装法语翻译,和 打印出来。
“快速排序”功能拒绝 重复的条目('错误'/中止) 所以只有一个法国人 任何英文单词的定义。测试 两个原始的'quicksort' 'raw_data'并在添加之后 '(“save”,“sauve”)'''raw_data'。
这是一个冯诺伊曼的迟到 二进制搜索的版本。做一个 字面音译到Haskell。 一进入,Haskell 版本必须验证递归 “循环不变”,终止于 '错误'/中止,如果它没有成功。它 如果,也会以相同的方式终止 找不到英文单词。
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
作为常量字典'a'的类型 '[Entry]'在全球范围内可见。暗示: 把你的字符串(英文单词)变成 进入后立即进入“进入” 'BINSEARCH'。
的编程值 高级数据类型'Entry'是, 如果你可以设计这两个功能 在整数上,它是微不足道的 将它们抬起来操作Entry's。
有人知道我应该如何处理我的二元搜索功能吗?
答案 0 :(得分:4)
教师要求“文字音译”,因此请使用相同的变量名称,顺序相同。但请注意一些差异:
另一个答案是转换为数组,但对于这么小的练习(毕竟这是功课),我觉得我们可以假装列表是直接访问。我刚刚接受了你的用词:: [Entry]并将其编入索引。我确实需要在几个地方之间转换Int和Integer。
Minor nit:你的英语价值有误(bs是我做的binSearch的捷径):
*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>
答案 1 :(得分:3)
二进制搜索需要随机访问,这在列表中是不可能的。因此,要做的第一件事可能是将列表转换为Array
(使用listArray
),并对其进行搜索。
答案 2 :(得分:1)
这是我的代码,只是问题的英文部分(我测试了它,它完美地运行):
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))
答案 3 :(得分: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
~~&gt; 7。