I want to display some Rational values in their decimal expansion. That is, instead of displaying 3 % 4, I would rather display 0.75. I'd like this function to be of type Int -> Rational -> String. The first Int is to specify the maximum number of decimal places, since Rational expansions may be non-terminating.
Hoogle and the haddocks for Data.Ratio didn't help me. Where can I find this function?
You can make it. Not elegant, but does the job:
import Numeric
import Data.Ratio
display :: Int -> Rational -> String
display n x = (showFFloat (Just n) $ fromRat x) ""
Here is an arbitrary precision solution that doesn't use floats:
import Data.Ratio
display :: Int -> Rational -> String
display len rat = (if num < 0 then "-" else "") ++ (shows d ("." ++ take len (go next)))
where
(d, next) = abs num `quotRem` den
num = numerator rat
den = denominator rat
go 0 = ""
go x = let (d, next) = (10 * x) `quotRem` den
in shows d (go next)
Arbitrary precision version that re-uses library code:
import Data.Number.CReal
display :: Int -> Rational -> String
display digits num = showCReal digits (fromRational num)
I know I've seen a function before that converts rationals into digits in a way that's easier to inspect (i.e. that makes it quite clear where the digits start repeating), but I can't seem to find it now. In any case, it's not hard to write, if that turns out to be a need; you just code up the usual long-division algorithm and watch for divisions you've already done.
Here's one that I wrote a few weeks ago. You can specify the number of decimals you want (correctly rounded), or just pass Nothing in which case it will print the full precision, including marking the repeated decimals.
module ShowRational where
import Data.List(findIndex, splitAt)
-- | Convert a 'Rational' to a 'String' using the given number of decimals.
-- If the number of decimals is not given the full precision is showed using (DDD) for repeating digits.
-- E.g., 13.7/3 is shown as \"4.5(6)\".
showRational :: Maybe Int -> Rational -> String
showRational (Just n) r =
let d = round (abs r * 10^n)
s = show (d :: Integer)
s' = replicate (n - length s + 1) '0' ++ s
(h, f) = splitAt (length s' - n) s'
in (if r < 0 then "-" else "") ++ h ++ "." ++ f
-- The length of the repeating digits is related to the totient function of the denominator.
-- This means that the complexity of computing them is at least as bad as factoring, i.e., it quickly becomes infeasible.
showRational Nothing r =
let (i, f) = properFraction (abs r) :: (Integer, Rational)
si = if r < 0 then "-" ++ show i else show i
decimals f = loop f [] ""
loop x fs ds =
if x == 0 then
ds
else
case findIndex (x ==) fs of
Just i -> let (l, r) = splitAt i ds in l ++ "(" ++ r ++ ")"
Nothing -> let (c, f) = properFraction (10 * x) :: (Integer, Rational) in loop f (fs ++ [x]) (ds ++ show c)
in if f == 0 then si else si ++ "." ++ decimals f
import Data.List as L
import Data.Ratio
display :: (Integral i, Show i) => Int -> Ratio i -> String
display len rat = (if num < 0 then "-" else "") ++ show ip ++ "." ++ L.take len (go (abs num - ip * den))
where
num = numerator rat
den = denominator rat
ip = abs num `quot` den
go 0 = ""
go x = shows d (go next)
where
(d, next) = (10 * x) `quotRem` den
Related
I am writing some code to work with arbitrary radix numbers in haskell. They will be stored as lists of integers representing the digits.
I almost managed to get it working, but I have run into the problem of converting a list of tuples [(a_1,b_1),...,(a_n,b_n)] into a single list which is defined as follows:
for all i, L(a_i) = b_i.
if there is no i such that a_i = k, a(k)=0
In other words, this is a list of (position,value) pairs for values in an array. If a position does not have a corresponding value, it should be set to zero.
I have read this (https://wiki.haskell.org/How_to_work_on_lists) but I don't think any of these methods are suitable for this task.
baseN :: Integer -> Integer -> [Integer]
baseN n b = convert_digits (baseN_digits n b)
chunk :: (Integer, Integer) -> [Integer]
chunk (e,m) = m : (take (fromIntegral e) (repeat 0))
-- This is broken because the exponents don't count for each other's zeroes
convert_digits :: [(Integer,Integer)] -> [Integer]
convert_digits ((e,m):rest) = m : (take (fromIntegral (e)) (repeat 0))
convert_digits [] = []
-- Converts n to base b array form, where a tuple represents (exponent,digit).
-- This works, except it ignores digits which are zero. thus, I converted it to return (exponent, digit) pairs.
baseN_digits :: Integer -> Integer -> [(Integer,Integer)]
baseN_digits n b | n <= 0 = [] -- we're done.
| b <= 0 = [] -- garbage input.
| True = (e,m) : (baseN_digits (n-((b^e)*m)) b)
where e = (greedy n b 0) -- Exponent of highest digit
m = (get_coef n b e 1) -- the highest digit
-- Returns the exponent of the highest digit.
greedy :: Integer -> Integer -> Integer -> Integer
greedy n b e | n-(b^e) < 0 = (e-1) -- We have overshot so decrement.
| n-(b^e) == 0 = e -- We nailed it. No need to decrement.
| n-(b^e) > 0 = (greedy n b (e+1)) -- Not there yet.
-- Finds the multiplicity of the highest digit
get_coef :: Integer -> Integer -> Integer -> Integer -> Integer
get_coef n b e m | n - ((b^e)*m) < 0 = (m-1) -- We overshot so decrement.
| n - ((b^e)*m) == 0 = m -- Nailed it, no need to decrement.
| n - ((b^e)*m) > 0 = get_coef n b e (m+1) -- Not there yet.
You can call "baseN_digits n base" and it will give you the corresponding array of tuples which needs to be converted to the correct output
Here's something I threw together.
f = snd . foldr (\(e,n) (i,l') -> ( e , (n : replicate (e-i-1) 0) ++ l')) (-1,[])
f . map (fromIntegral *** fromIntegral) $ baseN_digits 50301020 10 = [5,0,3,0,1,0,2,0]
I think I understood your requirements (?)
EDIT:
Perhaps more naturally,
f xs = foldr (\(e,n) fl' i -> (replicate (i-e) 0) ++ (n : fl' (e-1))) (\i -> replicate (i+1) 0) xs 0
Im am making a function which compares two strings to see if one is a rearrangement of the other. for example "hhe" and "heh" would produce true but "hhe" and "hee" would be false. I thought I could do this by summing the elements of the string and seeing if they are the same. I am knew to haskell, so I dont know if I can sum chars like in C. Code so far:
comp :: String -> String-> Bool
comp x y = (sum x) == (sum y)
This produces an error when compiling.
You can first sort, then compare the strings
import Data.List
import Data.Function
comp = (==) `on` sort
which can then be used like this
"abcd" `comp` "dcba" --yields True
It doesn't make sense to "sum" two strings. Use permutations instead:
comp :: String -> String -> Bool
comp x = (`elem` permutations x)
Live demo
Though there are problems with your implementation, as suggested by others, the direct answer to your question is that you can first convert characters to Int (a type that supports arithmetic) with fromEnum.
> sum . map fromEnum $ "heh"
309
Taking your example code at face value, the problem with it is that Char doesn't implement Num, so sum :: Num a => [a] -> a is incompatible.
We can fix that, however, by using fromEnum to convert the Chars to Ints:
isPermutationOf :: String -> String-> Bool
isPermutationOf x y = hash x == hash y
where hash = sum . map fromEnum
And this will work on your example case:
λ isPermutationOf "hhe" "heh"
True
The downside is that it also has some false positives:
λ isPermutationOf "AAA" "ab"
True
We can try to reduce those somewhat by making sure that the lengths, maxes, and mins of the inputs are the same:
isPermutationOf :: String -> String-> Bool
isPermutationOf x y = hash x == hash y && maximum x == maximum y && minimum x == minimum y
where hash = sum . map fromEnum
But though that catches some cases
λ isPermutationOf "AAA" "ab"
False
It doesn't catch them all
λ isPermutationOf "abyz" "acxz"
True
To do that, we really need to make sure we've got the same number of each Char in both inputs. We could solve this by using a Data.Map.Map to store the counts of each Char or by using Data.List.sort to sort each of the inputs, but if we only want to use the Prelude, we'll need to roll our own solution.
There's any number of examples on how to write quicksort in haskell out there, so I'm not going to tell you how to do that. So here's a dumb isPermutationOf that uses math instead.
isPermutationOf xs ys = all (\k -> powsum k as == powsum k bs) [0..n]
where as = map fromEnum xs
bs = map fromEnum ys
n = length xs
powsum k zs = sum (map (^k) zs)
Basically, we can view an n-length string as a set of n unknowns. isPermutationOf checks the n+1 equations:
eq0: x00 + x10 + ... + xn-10 = y00 + y10 + ... + ym-10
eq1: x01 + x11 + ... + xn-11 = y01 + y11 + ... + ym-11
eq2: x02 + x12 + ... + xn-12 = y02 + y12 + ... + ym-12
...
eqn: x0n + x1n + ... + xn-1n = y0n + y1n + ... + ym-1n
eq0 is essentially a length check. Given xs, the other n equations work out to n equations for n unknowns, which will give us a solution for ys unique up to permutation.
But really, you should use a (bucket) sort instead, because the above algorithm is O(n^2), which is slow for this kind of check.
if you do not want to use standard library(learning purpose) function, you can quickSort both string and check for equality of string (bonus: quickSort)
isEqual :: String -> String -> Bool
isEqual a b = sortString a == sortString b
where
sortString :: String -> String
sortString [] = []
sortString (x:xs) = sortString (filter (<x) xs) ++ [x] ++ sortString (filter (>=x) xs)
Below I have defined a function that converts a list of base-3 digits to the corresponding numerical value. For example:
f "201" = (2 * 3^2) + (0 * 3^1) + (1 * 3^0) = 19
f "12" = 5
f "1202" = 47
f "120221" = 430
Here is a definition using comprehension:
f :: String -> Int
f str = sum (listToFinal (stringToTuples str))
Helper functions:
-- 1) converts "201" to "102"
reverse "str"
-- 2) converts "102" to 102
stringToInt :: String -> Int
stringToInt str = read str :: Int
-- 3) converts 102 to ['1','0','2']
intToList :: Int -> [Int]
intToList 0 = []
intToList x = intToList (x `div` 10) ++ [x `mod` 10]
-- 4) converts "201" to [(1,0),(0,1),(2,2)] using reverse, stringToInt, intToList
stringToTuples :: String -> [(Int,Int)]
stringToTuples str = zip (intToList (stringToInt (reverse str))) [0..]
-- 5) converts [(1,0),(0,1),(2,2)] to [1*3^0, 0*3^1, 2*3^2]
listToFinal :: [(Int,Int)] -> [Int]
listToFinal list = [ x * (3^y) | (x,y) <- list ]
Now I'd like to do it with recursion only (well, using basic & library functions too, of course).
An idea: I was thinking of taking the head of each element in the list and simply multiplying it with 3^(length of string - 1). The only problem is, with each recursive call the power of three would have to decrease by 1, e.g. given:
recursive_version "201" = (2 * 3^2) + (0 * 3^1) + (1 * 3^0)
How to implement this?
Here is a much simpler approach; note that, through the use of foldl, it's only "implicitly" recursive, though. For information, digitToInt is exported by Data.Char.
import Data.Char
import Data.List ( foldl' )
--- horner x xs : the value of polynomial 'xs' at point 'x'
horner :: Int -> [Int] -> Int
horner x = foldl' (\c1 c0 -> c1 * x + c0) 0
-- f s : the integer whose representation in base 3 is string 's'
f :: String -> Int
f = horner 3 . map digitToInt
When you define it recursively, the natural way to decrement the length is trimming the array from the head. For example:
base3 x = base3' x 0 where
base3' (d:ds) v = base3' ds $ v + d * 3 ^ length ds
base3' [] v = v
Here is my first Haskell program. What parts would you write in a better way?
-- Multiplication table
-- Returns n*n multiplication table in base b
import Text.Printf
import Data.List
import Data.Char
-- Returns n*n multiplication table in base b
mulTable :: Int -> Int -> String
mulTable n b = foldl (++) (verticalHeader n b w) (map (line n b w) [0..n])
where
lo = 2* (logBase (fromIntegral b) (fromIntegral n))
w = 1+fromInteger (floor lo)
verticalHeader :: Int -> Int -> Int -> String
verticalHeader n b w = (foldl (++) tableHeader columnHeaders)
++ "\n"
++ minusSignLine
++ "\n"
where
tableHeader = replicate (w+2) ' '
columnHeaders = map (horizontalHeader b w) [0..n]
minusSignLine = concat ( replicate ((w+1)* (n+2)) "-" )
horizontalHeader :: Int -> Int -> Int -> String
horizontalHeader b w i = format i b w
line :: Int -> Int -> Int -> Int -> String
line n b w y = (foldl (++) ((format y b w) ++ "|" )
(map (element b w y) [0..n])) ++ "\n"
element :: Int -> Int -> Int -> Int -> String
element b w y x = format (y * x) b w
toBase :: Int -> Int -> [Int]
toBase b v = toBase' [] v where
toBase' a 0 = a
toBase' a v = toBase' (r:a) q where (q,r) = v `divMod` b
toAlphaDigits :: [Int] -> String
toAlphaDigits = map convert where
convert n | n < 10 = chr (n + ord '0')
| otherwise = chr (n + ord 'a' - 10)
format :: Int -> Int -> Int -> String
format v b w = concat spaces ++ digits ++ " "
where
digits = if v == 0
then "0"
else toAlphaDigits ( toBase b v )
l = length digits
spaceCount = if (l > w) then 0 else (w-l)
spaces = replicate spaceCount " "
Here are some suggestions:
To make the tabularity of the computation more obvious, I would pass the list [0..n] to the line function rather than passing n.
I would further split out the computation of the horizontal and vertical axes so that they are passed as arguments to mulTable rather than computed there.
Haskell is higher-order, and almost none of the computation has to do with multiplication. So I would change the name of mulTable to binopTable and pass the actual multiplication in as a parameter.
Finally, the formatting of individual numbers is repetitious. Why not pass \x -> format x b w as a parameter, eliminating the need for b and w?
The net effect of the changes I am suggesting is that you build a general higher-order function for creating tables for binary operators. Its type becomes something like
binopTable :: (i -> String) -> (i -> i -> i) -> [i] -> [i] -> String
and you wind up with a much more reusable function—for example, Boolean truth tables should be a piece of cake.
Higher-order and reusable is the Haskell Way.
You don't use anything from import Text.Printf.
Stylistically, you use more parentheses than necessary. Haskellers tend to find code more readable when it's cleaned of extraneous stuff like that. Instead of something like h x = f (g x), write h = f . g.
Nothing here really requires Int; (Integral a) => a ought to do.
foldl (++) x xs == concat $ x : xs: I trust the built-in concat to work better than your implementation.
Also, you should prefer foldr when the function is lazy in its second argument, as (++) is – because Haskell is lazy, this reduces stack space (and also works on infinite lists).
Also, unwords and unlines are shortcuts for intercalate " " and concat . map (++ "\n") respectively, i.e. "join with spaces" and "join with newlines (plus trailing newline)"; you can replace a couple things by those.
Unless you use big numbers, w = length $ takeWhile (<= n) $ iterate (* b) 1 is probably faster. Or, in the case of a lazy programmer, let w = length $ toBase b n.
concat ( (replicate ((w+1)* (n+2)) "-" ) == replicate ((w+1) * (n+2)) '-' – not sure how you missed this one, you got it right just a couple lines up.
You do the same thing with concat spaces, too. However, wouldn't it be easier to actually use the Text.Printf import and write printf "%*s " w digits?
Norman Ramsey gave excellent high-level (design) suggestions; Below are some low-level ones:
First, consult with HLint. HLint is a friendly program that gives you rudimentary advice on how to improve your Haskell code!
In your case HLint gives 7 suggestions. (mostly about redundant brackets)
Modify your code according to HLint's suggestions until it likes what you feed it.
More HLint-like stuff:
concat (replicate i "-"). Why not replicate i '-'?
Consult with Hoogle whenever there is reason to believe that a function you need is already available in Haskell's libraries. Haskell comes with tons of useful functions so Hoogle should come in handy quite often.
Need to concatenate strings? Search for [String] -> String, and voila you found concat. Now go replace all those folds.
The previous search also suggested unlines. Actually, this even better suits your needs. It's magic!
Optional: pause and thank in your heart to Neil M for making Hoogle and HLint, and thank others for making other good stuff like Haskell, bridges, tennis balls, and sanitation.
Now, for every function that takes several arguments of the same type, make it clear which means what, by giving them descriptive names. This is better than comments, but you can still use both.
So
-- Returns n*n multiplication table in base b
mulTable :: Int -> Int -> String
mulTable n b =
becomes
mulTable :: Int -> Int -> String
mulTable size base =
To soften the extra characters blow of the previous suggestion: When a function is only used once, and is not very useful by itself, put it inside its caller's scope in its where clause, where it could use the callers' variables, saving you the need to pass everything to it.
So
line :: Int -> Int -> Int -> Int -> String
line n b w y =
concat
$ format y b w
: "|"
: map (element b w y) [0 .. n]
element :: Int -> Int -> Int -> Int -> String
element b w y x = format (y * x) b w
becomes
line :: Int -> Int -> Int -> Int -> String
line n b w y =
concat
$ format y b w
: "|"
: map element [0 .. n]
where
element x = format (y * x) b w
You can even move line into mulTable's where clause; imho, you should.
If you find a where clause nested inside another where clause troubling, then I suggest to change your indentation habits. My recommendation is to use consistent indentation of always 2 or always 4 spaces. Then you can easily see, everywhere, where the where in the other where is at. ok
Below's what it looks like (with a few other changes in style):
import Data.List
import Data.Char
mulTable :: Int -> Int -> String
mulTable size base =
unlines $
[ vertHeaders
, minusSignsLine
] ++ map line [0 .. size]
where
vertHeaders =
concat
$ replicate (cellWidth + 2) ' '
: map horizontalHeader [0 .. size]
horizontalHeader i = format i base cellWidth
minusSignsLine = replicate ((cellWidth + 1) * (size + 2)) '-'
cellWidth = length $ toBase base (size * size)
line y =
concat
$ format y base cellWidth
: "|"
: map element [0 .. size]
where
element x = format (y * x) base cellWidth
toBase :: Integral i => i -> i -> [i]
toBase base
= reverse
. map (`mod` base)
. takeWhile (> 0)
. iterate (`div` base)
toAlphaDigit :: Int -> Char
toAlphaDigit n
| n < 10 = chr (n + ord '0')
| otherwise = chr (n + ord 'a' - 10)
format :: Int -> Int -> Int -> String
format v b w =
spaces ++ digits ++ " "
where
digits
| v == 0 = "0"
| otherwise = map toAlphaDigit (toBase b v)
spaces = replicate (w - length digits) ' '
0) add a main function :-) at least rudimentary
import System.Environment (getArgs)
import Control.Monad (liftM)
main :: IO ()
main = do
args <- liftM (map read) $ getArgs
case args of
(n:b:_) -> putStrLn $ mulTable n b
_ -> putStrLn "usage: nntable n base"
1) run ghc or runhaskell with -Wall; run through hlint.
While hlint doesn't suggest anything special here (only some redundant brackets), ghc will tell you that you don't actually need Text.Printf here...
2) try running it with base = 1 or base = 0 or base = -1
If you want multiline comments use:
{- A multiline
comment -}
Also, never use foldl, use foldl' instead, in cases where you are dealing with large lists which must be folded. It is more memory efficient.
A brief comments saying what each function does, its arguments and return value, is always good. I had to read the code pretty carefully to fully make sense of it.
Some would say if you do that, explicit type signatures may not be required. That's an aesthetic question, I don't have a strong opinion on it.
One minor caveat: if you do remove the type signatures, you'll automatically get the polymorphic Integral support ephemient mentioned, but you will still need one around toAlphaDigits because of the infamous "monomorphism restriction."
I'm trying to complete the last part of my Haskell homework and I'm stuck, my code so far:
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 ))
The question asks:
Write a Haskell programs that takes
the English sentence 'english', looks
up each word in the English-French
dictionary using binary search,
performs word-for-word substitution,
assembles the French translation, and
prints it out.
The function 'quicksort' rejects
duplicate entries (with 'error'/abort)
so that there is precisely one French
definition for any English word. Test
'quicksort' with both the original
'raw_data' and after having added
'("saves", "sauve")' to 'raw_data'.
Here is a von Neumann late-stopping
version of binary search. Make a
literal transliteration into Haskell.
Immediately upon entry, the Haskell
version must verify the recursive
"loop invariant", terminating with
'error'/abort if it fails to hold. It
also terminates in the same fashion if
the English word is not found.
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
In the Haskell version
binsearch :: String -> Integer -> Integer -> Entry
as the constant dictionary 'a' of type
'[Entry]' is globally visible. Hint:
Make your string (English word) into
an 'Entry' immediately upon entering
'binsearch'.
The programming value of the
high-level data type 'Entry' is that,
if you can design these two functions
over the integers, it is trivial to
lift them to to operate over Entry's.
Anybody know how I'm supposed to go about my binarysearch function?
The instructor asks for a "literal transliteration", so use the same variable names, in the same order. But note some differences:
the given version takes only 1
parameter, the signature he gives
requires 3. Hmmm,
the given version is not recursive, but he asks for a
recursive version.
Another answer says to convert to an Array, but for such a small exercise (this is homework after all), I felt we could pretend that lists are direct access. I just took your diction::[Entry] and indexed into that. I did have to convert between Int and Integer in a few places.
Minor nit: You've got a typo in your english value (bs is a shortcut to binSearch I made):
*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>
A binary search needs random access, which is not possible on a list. So, the first thing to do would probably be to convert the list to an Array (with listArray), and do the search on it.
here's my code for just the English part of the question (I tested it and it works perfectly) :
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))
An important Prelude operator is:
(!!) :: [a] -> Integer -> a
-- xs!!n returns the nth element of xs, starting at the left and
-- counting from 0.
Thus, [14,7,3]!!1 ~~> 7.