Haskell函数式编程基础习题(4)

缺乏、安全感 2022-08-06 06:27 161阅读 0赞

--Exercise 6.35
    
    --Exercise 6.34
    
    --Exercise 6.33
    
    --Exercise 6.32
    printImage :: Image -> IO()
    printImage i = print i
    --Exercise 6.31
    moveImage:: Image -> Int -> Int -> Image
    moveImage  (i,(x,y)) dx dy = (i,(x+dx,y+dy))
    
    --Exercise 6.30
    changePosition :: Image-> Position -> Image
    changePosition (i,p) p1 =  (i,p1)
    
    --Exercise 6.29
    makeImage :: String -> Position -> Image
    makeImage s p = (s,p)
    
    
    type Position = (Int,Int)
    type Image = (String,Position)
    --Exercise 6.28
    --Exercise 6.27
    --Exercise 6.26
    
    --Exercise 6.25
    
    --Exercise 6.24
    zipCompact::Eq a =>[(a,Int)]->[(a,Int)]->[(a,Int)]
    zipCompact a b 
    	| (length a) == 0 = b
    	| otherwise = zipCompact (tail a) (addzip (head a) b)
    
    addzip ::Eq a=> (a,Int)->[(a,Int)]->[(a,Int)]
    addzip x [] = [x]
    addzip (ax,ay) ((bx,by):xs)
    	| ax /= bx = (bx,by) : (addzip (ax,ay) xs)
    	| otherwise = (ax, ay+by) : xs
    	
    reverseCompact ::[(a,b)] -> [(b,a)]
    reverseCompact xs  =  [ (y,x)|(x,y)<-xs]
    
    --Exercise 6.23
    --Exercise 6.22
    --Exercise 6.21
    --Exercise 6.20
    --Exercise 6.19
    
    --Exercise 6.18
    
    --Exercise 6.17
    myReplicate :: Int -> a -> [a]
    myReplicate  n  x 
    	| n ==1  = [x]
    	| otherwise = x : (myReplicate (n-1) x)
    
    --Exercise 6.16
    --Exercise 6.15
    --Exercise 6.14
    --Exercise 6.13
    --Exercise 6.12
    --Exercise 6.11
    
    --Exercise 6.10
    
    --Exercise 6.9
    
    type Picture = [String]
    --Exercise 6.8
    rotate90 :: Picture -> Picture
    rotate90 (x:xs) 
    	| x == []  = [[]]
    	| xs==[[]] = [(leftCube (x:xs))]
    	| otherwise = (leftCube (x:xs)) :(rotate90 (rightCube (x:xs)))
    rightCube :: Picture -> Picture
    rightCube ((x1:x1s):xs) 
    	|  xs == []  = x1s : []
    	| otherwise = x1s : (rightCube xs)
    	
    leftCube :: Picture -> String
    leftCube (x:xs) 
    			| xs ==[]  = (head x) : []
    			| otherwise = (head x) : (leftCube xs)
    
    --Exercise 6.7
    printPicture :: Picture -> IO()
    printPicture (p:ps) 
    	|  ps== []  = print p
    	|  otherwise = do
    							print p
    							printPicture ps
    
    --Exercise 6.6
    --superimpose :: Picture -> Picture -> Picture
    
    --Exercise 6.5
    superimposeLine :: String -> String -> String
    superimposeLine (x:xs) (y:ys) 
    	| xs ==[] || ys == []  =  [(superimposeChar x y)]
    	| otherwise =  ( superimposeChar x y) : (superimposeLine xs ys)
    	
    --Exercise 6.4
    superimposeChar :: Char -> Char -> Char 
    superimposeChar '.' '.' = '.'
    superimposeChar _ _ = '#'
    
    --Exercise 6.3
    
    --Exercise 6.2
    
    --Exercise 6.1
    mySnd :: (Int,Int)-> Int 
    mySnd  (x,y) =y
    
    sing :: Int -> [Int]
    sing x = [x]
    
    --Exercise 5.34
    
    --Exercise 5.33
    
    --Exercise 5.32
    
    --Exercise 5.31
    
    --Exercise 5.30
    
    --Exercise 5.29
    returnLoan :: Database -> Person -> Book -> Database
    returnLoan database person book = [pair |pair<-database,pair /= (person,book)]
    
    --Exercise 5.28
    numBorrowed :: Database->Person -> Int 
    numBorrowed database p = length (borrowed database p)
    
    borrowers::Database -> Book ->[Person]
    borrowers database b = [person| (person,book)<- database,book==b]
    
    borrowed :: Database->Person -> [Book]
    borrowed database p = books database p
    
    --Exercise 5.27
    books :: Database -> Person -> [Book]
    books database p  = [book| (person,book)<- database,person==p]
    
    type Person  = String
    type Book = String
    type Database = [(Person,Book)]
    
    exampleBase :: Database
    exampleBase =[
    				("Alice","Tintin"),
    				("Anna","Little Women"),
    				("Alice","Asterix"),
    				("Rory","Tintin")]