Related
I'm currently doing an assignment for college where we are implementing an polynomial calculator in Haskell.
The first part of the assignment is doing poly operations, and that is already done.
We get extra credit if we implement an parser for the polynomial, which I'm currently doing by turning a string to a tuple of [(factor, [(variable, exponent)])].
This means "-10y^4 - 5z^5" => "[(-10, [('y', 4)]), (-5, [('z', 5)].
The sub-problem I'm having trouble with is when I encounter polynomials like "5xy^2z^3" that should be stored as [(5, [('x',1), ('y', 2),('z',3)]], I don't know how to parse it.
Any suggestion on how I could approach this?
Thank you in advance for your help!
-- Slipts lists by chosen Char, only used with '+' in this project
split :: Char -> String -> [String]
split _ "" = []
split c s = firstWord : (split c rest)
where firstWord = takeWhile (/=c) s
rest = drop (length firstWord + 1) s
-- Remove all spaces from a string, for easier parsing
formatSpace :: String -> String
formatSpace = filter (not . isSpace)
-- Clever way to parse the polynomial, add an extra '+' before every '-'
-- so after we split the string by '+', it helps us keep the '-'
simplify_minus :: String -> String
simplify_minus [] = ""
simplify_minus (x:xs)
| x == '^' = x : head xs : simplify_minus (tail xs)
| x == '-' = "+-" ++ simplify_minus xs
| otherwise = x : simplify_minus xs
-- Splits an String by occurrences of '+' and creates a list of those sub-strings
remove_plus :: String -> [String]
remove_plus s = split '+' s
-- Removes multiplication on substrings
remove_mult :: [String] -> [[String]]
remove_mult [] = []
remove_mult (x:xs) = (remove_power (split '*' x)) : remove_mult xs
-- Function used to separate a variable that has an power. This translates ["y^2] to [["y", "2"]]
remove_power :: [String] -> [String]
remove_power [] = []
remove_power (x:xs) = (split '^' x) ++ remove_power xs
-- Wrapper function for all the functions necessary to the parser
parse_poly :: String -> [(Integer, String, Integer)]
parse_poly [] = []
parse_poly s = map (tuplify) (rem_m (remove_plus (simplify_minus (formatSpace s))))
rem_m :: [String] -> [String]
rem_m l = map (filter (not . (=='*'))) l
helper_int :: String -> Integer
helper_int s
| s == "" = 1
| s == "-" = -1
| otherwise = read s :: Integer
helper_char :: String -> String
helper_char s
| s == [] = " "
| otherwise = s
tuplify :: String -> (Integer, String, Integer)
tuplify l = (helper_int t1, helper_char t3, helper_int (drop 1 t4))
where (t1, t2) = (break (isAlpha) l)
(t3, t4) = (break (=='^') t2)
main :: IO()
main = do
putStr("\nRANDOM TESTING ON THE WAE\n")
putStr("--------------\n")
print(parse_poly "5*xyz^3 - 10*y^4 - 5*z^5 - x^2 - 5 - x")
-- [(5,"xyz",3),(-10,"y",4),(-5,"z",5),(-1,"x",2),(-5," ",1),(-1,"x",1)]
``
You have pretty much everything there already, but you do need to use break recursively to grab everything until the next variable. You probably should also use the similar span to first grab the coefficient.
parsePositiveMonomial :: String -> (Integer, [(Char, Integer)])
parsePositiveMonomial s = case span isDigit s of
([], varPows) -> (1, parseUnitMonomial varPows)
(coef, varPows) -> (read coef, parseUnitMonomial varPows)
where parseUnitMonomial [] = []
parseUnitMonomial (var:s') = case break isAlpha s' of
...
If I was given a string like skhfbvqa, how would I generate the next string? For this example, it would be skhfbvqb, and the next string of that would be skhfbvqc, and so on. The given string (and the answer) will always be N characters long (in this case, N=8).
What I tried:
I tried to generate the entire (infinite) list of possible combinations, and get the required (next) string of the given string, but unsurprisingly, it's so slow, that I don't even get the answer for N=6.
I used list comprehension:
allStrings = [ c : s | s <- "" : allStrings, c <- ['a'..'z'] ]
main = do
input <- readFile "k.in"
putStrLn . head . tail . dropWhile (not . (==) input) . map reverse $ allStrings
(Please excuse my incredibly bad Haskell-ing :) Still a noob)
So my question is, how can I do this? If there are multiple methods, a comparison between them is much appreciated. Thanks!
Here's a version with base conversion (this way you could add and subtract arbitrarily if you like):
encode x base = encode' x [] where
encode' x' z | x' == 0 = z
| otherwise = encode' (div x' base) ((mod x' base):z)
decode num base =
fst $ foldr (\a (b,i) -> (b + a * base^i,i + 1)) (0,0) num
Output:
*Main> map (\x -> toEnum (x + 97)::Char)
$ encode (decode (map (\x -> fromEnum x - 97) "skhfbvqa") 26 + 1) 26
"skhfbvqb"
I would go and make a helper function f :: Integer -> String and one g :: String -> Integer, where f 1 = "a", ... f 27 = "aa", f 28 = "ab" and so on and the inverse g.
Then incrementString = f . succ . g
Note: I omitted the implementation of f on purpose for learning
Update
for a different approach you could define a increment with carry function inc' :: Char -> (Char, Bool), and then
incString :: String -> String
incString = reverse . incString'
where incString' [] = []
incString' (x:xs) = case inc' x of (x',True) -> x': incString' xs
(x',False) -> x':xs
Note: this function is not tail recursive!
I found this to work. It just uses pattern matching to see if the string begins with a z and adds an additional a accordingly.
incrementString' :: String -> String
incrementString' [] = ['a']
incrementString' ('z':xs) = 'a' : incrementString' xs
incrementString' (x:xs) = succ x : xs
incrementString :: String -> String
incrementString = reverse . incrementString' . reverse
I have a problem making Backtracking on Haskell, I know how to do recursive functions but I get troubles when I try to get multiple solutions or the best one (backtracking).
There's a list with some strings, then I need to get the solutions to get from a string to another one changing one letter from the string, I will get the list, the first string and the last one. If there is solution return the count of steps that it did, if there is not solution it returns -1. here's an example:
wordF ["spice","stick","smice","stock","slice","slick","stock"] "spice" "stock"
Then I have my list and I need to start with "spice" and get to "stock"
and the best solution is ["spice","slice","slick","stick","stock"] with four steps to get from "spice" to "stock". then it return 4.
Another solution is ["spice","smice","slice","slick","stick","stock"] with five steps to get to "stock" then it return `5. But this is a wrong solution because there's another one that's better with lesser steps than this one.
I'm having troubles making a backtracking to get the best solution, because I don't know how to make that my code search another solutions and just not one..
Here's a code that i tried to make but i get some errors, btw i dont know if my way to "make" backtracking is good or if there are some mistakes that im not seeing..
wordF :: [String] -> String -> String -> (String, String, Int)
wordF [] a b = (a, b, -1)
wordF list a b | (notElem a list || notElem b list) = (a, b, -1)
| otherwise = (a, b, (wordF2 list a b [a] 0 (length list)))
wordF2 :: [String] -> String -> String -> [String] -> Int -> Int -> Int
wordF2 list a b list_aux cont maxi | (cont==maxi) = 1000
| (a==b) = length list_aux
| (a/=b) && (cont<maxi) && notElemFound && (checkin /= "ThisWRONG") && (wording1<=wording2) = wording1
| (a/=b) && (cont<maxi) && notElemFound && (checkin /= "ThisWRONG") && (wording1>wording2) = wording2
| (a/=b) && (checkin == "ThisWRONG") = wordF2 list a b list_aux (cont+1) maxi
where
checkin = (check_word2 a (list!!cont) (list!!cont) 0)
wording1 = (wordF2 list checkin b (list_aux++[checkin]) 0 maxi)
wording2 = (wordF2 list checkin b (list_aux++[checkin]) 1 maxi)
notElemFound = ((any (==(list!!cont)) list_aux) == False)
check_word2 :: String -> String -> String -> Int -> String
check_word2 word1 word2 word3 dif | (dif > 1) = "ThisWRONG"
| ((length word1 == 1) && (length word2 == 1) && (head word1 == head word2)) = word3
| ((length word1 == 1) && (length word2 == 1) && (head word1 /= head word2) && (dif<1)) = word3
| ((head word1) == (head word2)) = check_word2 (tail word1) (tail word2) word3 dif
| otherwise = check_word2 (tail word1) (tail word2) word3 (dif+1)
My first function wordF2 get the list, the start, the end, an auxiliary list to get the current solution with the first element that always will be there ([a]), a counter with 0, and the max size of the counter (length list)..
and the second function check_word2 it checks if a word can pass to another word, like "spice" to "slice" if it cant like "spice" to "spoca" it returns "ThisWRONG".
This solution gets an error of pattern match failure
Program error: pattern match failure: wordF2 ["slice","slick"] "slice" "slick" ["slice"] 0 1
I was trying with little cases and nothing, and I'm restricting that i get a wrong position of the list with the count and the max.
Or may be I dont know how to implement backtracking on haskell to get multiple solutions, the best solution, etc..
UPDATE: I did a solution but its not backtracking
wordF :: [String] -> String -> String -> (String, String, Int)
wordF [] a b = (a, b, -1)
wordF list a b | (notElem a list || notElem b list) = (a, b, -1)
| otherwise = (a, b, (wordF1 list a b))
wordF1 :: [String] -> String -> String -> Int
wordF1 list a b | ((map length (wordF2 (subconjuntos2 (subconjuntos list) a b))) == []) = -1
| (calculo > 0) = calculo
| otherwise = -1
where
calculo = (minimum (map length (wordF2 (subconjuntos2 (subconjuntos list) a b))))-1
wordF2 :: [[String]] -> [[String]]
wordF2 [[]] = []
wordF2 (x:xs) | ((length xs == 1) && ((check_word x) == True) && ((check_word (head xs)) == True)) = x:xs
| ((length xs == 1) && ((check_word x) == False) && ((check_word (head xs)) == True)) = xs
| ((length xs == 1) && ((check_word x) == True) && ((check_word (head xs)) == False)) = [x]
| ((length xs == 1) && ((check_word x) == False) && ((check_word (head xs)) == False)) = []
| ((check_word x) == True) = x:wordF2 xs
| ((check_word x) == False ) = wordF2 xs
check_word :: [String] -> Bool
check_word [] = False
check_word (x:xs) | ((length xs == 1) && ((check_word2 x (head xs) 0) == True)) = True
| ((length xs >1) && ((check_word2 x (head xs) 0) == True)) = True && (check_word xs)
| otherwise = False
check_word2 :: String -> String -> Int -> Bool
check_word2 word1 word2 dif | (dif > 1) = False
| ((length word1 == 1) && (length word2 == 1) && (head word1 == head word2)) = True
| ((length word1 == 1) && (length word2 == 1) && (head word1 /= head word2) && (dif<1)) = True
| ((head word1) == (head word2)) = check_word2 (tail word1) (tail word2) dif
| otherwise = check_word2 (tail word1) (tail word2) (dif+1)
subconjuntos2 :: [[String]] -> String -> String -> [[String]]
subconjuntos2 [] a b = []
subconjuntos2 (x:xs) a b | (length x <= 1) = subconjuntos2 xs a b
| ((head x == a) && (last x == b)) = (x:subconjuntos2 xs a b)
| ((head x /= a) || (last x /= b)) = (subconjuntos2 xs a b)
subconjuntos :: [a] -> [[a]]
subconjuntos [] = [[]]
subconjuntos (x:xs) = [x:ys | ys <- sub] ++ sub
where sub = subconjuntos xs
Mmm may be its inefficient but at least it does the solution..
i search all posible solutions, i compare head == "slice" and last == "stock", then i filter the ones that are solution and print the shorter one,
thanks and if you guys have any suggest say it :)
Not thoroughly tested, but this hopefully will help:
import Data.Function (on)
import Data.List (minimumBy, delete)
import Control.Monad (guard)
type Word = String
type Path = [String]
wordF :: [Word] -> Word -> Word -> Path
wordF words start end =
start : minimumBy (compare `on` length) (generatePaths words start end)
-- Use the list monad to do the nondeterminism and backtracking.
-- Returns a list of all paths that lead from `start` to `end`
-- in steps that `differByOne`.
generatePaths :: [Word] -> Word -> Word -> [Path]
generatePaths words start end = do
-- Choose one of the words, nondeterministically
word <- words
-- If the word doesn't `differByOne` from `start`, reject the choice
-- and backtrack.
guard $ differsByOne word start
if word == end
then return [word]
else do
next <- generatePaths (delete word words) word end
return $ word : next
differsByOne :: Word -> Word -> Bool
differsByOne "" "" = False
differsByOne (a:as) (b:bs)
| a == b = differsByOne as bs
| otherwise = as == bs
Example run:
>>> wordF ["spice","stick","smice","stock","slice","slick","stock"] "spice" "stock"
["spice","slice","slick","stick","stock"]
The list monad in Haskell is commonly described as a form of nondeterministic, backtracking computation. What the code above is doing is allowing the list monad to take on the responsibility of generating alternatives, testing whether they satisfy criteria, and backtracking on failure to the most recent choice point. The bind of the list monad, e.g. word <- words, means "nondeterministically pick one of the words. guard means "if the choices so far don't satisfy this condition, backtrack and make a different choice. The result of a list monad computation is the list of all the results that stem from choices that did not violate any guards.
If this looks like list comprehensions, well, list comprehensions are the same thing as the list monad—I chose to express it with the monad instead of comprehensions.
There have been several articles published recently on classic brute-force search problems.
Mark Dominus published a simple example of using lists for a simple exhaustive search.
Justin Le followed up with a small modification to the previous article that simplified tracking the current state of the search.
I followed up with a further modification that allowed measuring the gains from early rejection of part of the search tree.
Note that the code in my article is quite slow because it's measuring the amount of work done as well as doing it. My article has good examples for how to quickly reject parts of the search tree, but it should be considered only an illustration - not production code.
A brute force approach using recursion:
import Data.List (filter, (\\), reverse, delete, sortBy)
import Data.Ord (comparing)
neighbour :: String -> String -> Bool
neighbour word = (1 ==) . length . (\\ word)
process :: String -> String -> [String] -> [(Int, [String])]
process start end dict =
let
loop :: String -> String -> [String] -> [String] -> [(Int,[String])] -> [(Int,[String])]
loop start end dict path results =
case next of
[] -> results
xs ->
if elem end xs
then (length solution, solution) : results
else results ++ branches xs
where
next = filter (neighbour start) dict'
dict' = delete start dict
path' = start : path
branches xs = [a | x <- xs, a <- loop x end dict' path' results]
solution = reverse (end : path')
in
loop start end dict [] []
shortestSolution :: Maybe Int
shortestSolution = shortest solutions
where
solutions = process start end dict
shortest s =
case s of
[] -> Nothing
xs -> Just $ fst $ head $ sortBy (comparing fst) xs
start = "spice"
end = "stock"
dict = ["spice","stick","smice","slice","slick","stock"]
Notes:
This code computes all possibles solutions (process) and select the shortest one (shortestSolution), as Carl said, you might want to prune parts of the search tree for better performance.
Using a Maybe instead of returning -1 when a function can fail to return results is preferred.
Another way using a tree with breadth-first search:
import Data.Tree
import Data.List( filter, (\\), delete )
import Data.Maybe
node :: String -> [String] -> Tree String
node label dict = Node{ rootLabel = label, subForest = branches label (delete label dict) }
branches :: String -> [String] -> [Tree String]
branches start dict = map (flip node dict) (filter (neighbour start) dict)
neighbour :: String -> String -> Bool
neighbour word = (1 ==) . length . (\\ word)
-- breadth first traversal
shortestBF tree end = find [tree] end 0
where
find ts end depth
| null ts = Nothing
| elem end (map rootLabel ts) = Just depth
| otherwise = find (concat (map subForest ts)) end (depth+1)
result = shortestBF tree end
tree :: Tree String
tree = node start dict
start = "spice"
end = "stock"
dict = ["spice","stick","smice","slice","slick","stock"]
So the problem I'm working on matching a pattern to a list, such like this:
match "abba" "redbluebluered" -> True or
match "abba" "redblueblue" -> False, etc. I wrote up an algorithm that works, and I think it's reasonable understandable, but I'm not sure if there's a better way to do this without explicit recursion.
import Data.HashMap.Strict as M
match :: (Eq a, Eq k, Hashable k) => [k] -> [a] -> HashMap k [a] -> Bool
match [] [] _ = True
match [] _ _ = False
match _ [] _ = False
match (p:ps) s m =
case M.lookup p m of
Just v ->
case stripPrefix v s of
Just post -> match ps post m
Nothing -> False
Nothing -> any f . tail . splits $ s
where f (pre, post) = match ps post $ M.insert p pre m
splits xs = zip (inits xs) (tails xs)
I would call this like match "abba" "redbluebluered" empty. The actual algorithm is simple. The map contains the patterns already matched. At the end it is [a - > "red", b -> "blue"]. If the next pattern is one we've seen before, just try matching it and recurse down if we can. Otherwise fail and return false.
If the next pattern is new, just try mapping the new pattern to every single prefix in the string and recursing down.
This is very similar to a parsing problem, so let's take a hint from the parser monad:
match should return a list of all of the possible continuations of the parse
if matching fails it should return the empty list
the current set of assignments will be state that has to carried through the computation
To see where we are headed, let's suppose we have this magic monad. Attempting to match "abba" against a string will look like:
matchAbba = do
var 'a'
var 'b'
var 'b'
var 'a'
return () -- or whatever you want to return
test = runMatch matchAbba "redbluebluered"
It turns out this monad is the State monad over the List monad. The List monad provides for backtracking and the State monad carries the current assignments and input around.
Here's the code:
import Data.List
import Control.Monad
import Control.Monad.State
import Control.Monad.Trans
import Data.Maybe
import qualified Data.Map as M
import Data.Monoid
type Assigns = M.Map Char String
splits xs = tail $ zip (inits xs) (tails xs)
var p = do
(assigns,input) <- get
guard $ (not . null) input
case M.lookup p assigns of
Nothing -> do (a,b) <- lift $ splits input
let assigns' = M.insert p a assigns
put (assigns', b)
return a
Just t -> do guard $ isPrefixOf t input
let inp' = drop (length t) input
put (assigns, inp')
return t
matchAbba :: StateT (Assigns, String) [] Assigns
matchAbba = do
var 'a'
var 'b'
var 'b'
var 'a'
(assigns,_) <- get
return assigns
test1 = evalStateT matchAbba (M.empty, "xyyx")
test2 = evalStateT matchAbba (M.empty, "xyy")
test3 = evalStateT matchAbba (M.empty, "redbluebluered")
matches :: String -> String -> [Assigns]
matches pattern input = evalStateT monad (M.empty,input)
where monad :: StateT (Assigns, String) [] Assigns
monad = do sequence $ map var pattern
(assigns,_) <- get
return assigns
Try, for instance:
matches "ab" "xyz"
-- [fromList [('a',"x"),('b',"y")],fromList [('a',"x"),('b',"yz")],fromList [('a',"xy"),('b',"z")]]
Another thing to point out is that code which transforms a string like "abba" to the monadic value do var'a'; var'b'; var 'b'; var 'a' is simply:
sequence $ map var "abba"
Update: As #Sassa NF points out, to match the end of input you'll want to define:
matchEnd :: StateT (Assigns,String) [] ()
matchEnd = do
(assigns,input) <- get
guard $ null input
and then insert it into the monad:
monad = do sequence $ map var pattern
matchEnd
(assigns,_) <- get
return assigns
I would like to modify your signature and return more than Bool. Your solution then becomes:
match :: (Eq a, Ord k) => [k] -> [a] -> Maybe (M.Map k [a])
match = m M.empty where
m kvs (k:ks) vs#(v:_) = let splits xs = zip (inits xs) (tails xs)
f (pre, post) t =
case m (M.insert k pre kvs) ks post of
Nothing -> t
x -> x
in case M.lookup k kvs of
Nothing -> foldr f Nothing . tail . splits $ vs
Just p -> stripPrefix p vs >>= m kvs ks
m kvs [] [] = Just kvs
m _ _ _ = Nothing
Using the known trick of folding to produce a function we can obtain:
match ks vs = foldr f end ks M.empty vs where
end m [] = Just m
end _ _ = Nothing
splits xs = zip (inits xs) (tails xs)
f k g kvs vs = let h (pre, post) = (g (M.insert k pre kvs) post <|>)
in case M.lookup k kvs of
Nothing -> foldr h Nothing $ tail $ splits vs
Just p -> stripPrefix p vs >>= g kvs
Here match is the function folding all keys to produce a function taking a Map and a string of a, which returns a Map of matches of the keys to substrings. The condition for matching the string of a in its entirety is tracked by the last function applied by foldr - end. If end is supplied with a map and an empty string of a, then the match is successful.
The list of keys is folded using function f, which is given four arguments: the current key, the function g matching the remainder of the list of keys (i.e. either f folded, or end), the map of keys already matched, and the remainder of the string of a. If the key is already found in the map, then just strip the prefix and feed the map and the remainder to g. Otherwise, try to feed the modified map and remainder of as for different split combinations. The combinations are tried lazily as long as g produces Nothing in h.
Here is another solution, more readable, I think, and as inefficient as other solutions:
import Data.Either
import Data.List
import Data.Maybe
import Data.Functor
splits xs = zip (inits xs) (tails xs)
subst :: Char -> String -> Either Char String -> Either Char String
subst p xs (Left q) | p == q = Right xs
subst p xs q = q
match' :: [Either Char String] -> String -> Bool
match' [] [] = True
match' (Left p : ps) xs = or [ match' (map (subst p ixs) ps) txs
| (ixs, txs) <- tail $ splits xs]
match' (Right s : ps) xs = fromMaybe False $ match' ps <$> stripPrefix s xs
match' _ _ = False
match = match' . map Left
main = mapM_ (print . uncurry match)
[ ("abba" , "redbluebluered" ) -- True
, ("abba" , "redblueblue" ) -- False
, ("abb" , "redblueblue" ) -- True
, ("aab" , "redblueblue" ) -- False
, ("cbccadbd", "greenredgreengreenwhiteblueredblue") -- True
]
The idea is simple: instead of having a Map, store both patterns and matched substrings in a list. So when we encounter a pattern (Left p), then we substitute all occurrences of this pattern with a substring and call match' recursively with this substring being striped, and repeat this for each substring, that belongs to inits of a processed string. If we encounter already matched substring (Right s), then we just try to strip this substring, and call match' recursively on a successive attempt or return False otherwise.
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.