Retrieve strings from Matrix - haskell

I'm stuck with my homework task, somebody help, please..
Here is the task:
Find all possible partitions of string into words of some dictionary
And here is how I'm trying to do it:
I use dynamical programming concept to fill matrix and then I'm stuck with how to retrieve data from it
-- Task5_2
retrieve :: [[Int]] -> [String] -> Int -> Int -> Int -> [[String]]
retrieve matrix dict i j size
| i >= size || j >= size = []
| index /= 0 = [(dict !! index)]:(retrieve matrix dict (i + sizeOfWord) (i + sizeOfWord) size) ++ retrieve matrix dict i (next matrix i j) size
where index = (matrix !! i !! j) - 1; sizeOfWord = length (dict !! index)
next matrix i j
| j >= (length matrix) = j
| matrix !! i !! j > 0 = j
| otherwise = next matrix i (j + 1)
getPartitionMatrix :: String -> [String] -> [[Int]]
getPartitionMatrix text dict = [[ indiceOfWord (getWord text i j) dict 1 | j <- [1..(length text)]] | i <- [1..(length text)]]
--------------------------
getWord :: String -> Int -> Int -> String
getWord text from to = map fst $ filter (\a -> (snd a) >= from && (snd a) <= to) $ zip text [1..]
indiceOfWord :: String -> [String] -> Int -> Int
indiceOfWord _ [] _ = 0
indiceOfWord word (x:xs) n
| word == x = n
| otherwise = indiceOfWord word xs (n + 1)
-- TESTS
dictionary = ["la", "a", "laa", "l"]
string = "laa"
matr = getPartitionMatrix string dictionary
test = retrieve matr dictionary 0 0 (length string)

Here is a code that do what you ask for. It doesn't work exactly like your solution but should work as fast if (and only if) both our dictionary lookup were improved to use tries as would be reasonable. As it is I think it may be a bit faster than your solution :
module Partitions (partitions) where
import Data.Array
import Data.List
data Branches a = Empty | B [([a],Branches a)] deriving (Show)
isEmpty Empty = True
isEmpty _ = False
flatten :: Branches a -> [ [ [a] ] ]
flatten Empty = []
flatten (B []) = [[]]
flatten (B ps) = concatMap (\(word, bs) -> ...) ps
type Dictionary a = [[a]]
partitions :: (Ord a) => Dictionary a -> [a] -> [ [ [a] ] ]
partitions dict xs = flatten (parts ! 0)
where
parts = listArray (0,length xs) $ zipWith (\i ys -> starting i ys) [0..] (tails xs)
starting _ [] = B []
starting i ys
| null words = ...
| otherwise = ...
where
words = filter (`isPrefixOf` ys) $ dict
go word = (word, parts ! (i + length word))
It works like this : At each position of the string, it search all possible words starting from there in the dictionary and evaluates to a Branches, that is either a dead-end (Empty) or a list of pairs of a word and all possible continuations after it, discarding those words that can't be continued.
Dynamic programming enter the picture to record every possibilities starting from a given index in a lazy array. Note that the knot is tied : we compute parts by using starting, which uses parts to lookup which continuations are possible from a given index. This only works because we only lookup indices after the one starting is computing and starting don't use parts for the last index.
To retrieve the list of partitions from this Branches datatype is analogous to the listing of all path in a tree.
EDIT : I removed some crucial parts of the solution in order to let the questioner search for himself. Though that shouldn't be too hard to complete with some thinking. I'll probably put them back with a somewhat cleaned up version later.

Related

Need help in subproblem of parser for polynomials (Haskell)

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
...

Implementing Backtracking on Haskell

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"]

counting frequencies in a collection of bins

I need to count values inbetween values in a list i.e. [135,136,138,140] would count all the numbers between 135-136,136-138,138-140. with the input list [135.2,135.3,137,139] would out put[2,1,1] using type [Float] [Float] [Int]. So far I have:
heightbetween :: Float -> Float -> [Float] -> Int
heightbetween _ _ [] = 0
heightbetween n s (x:xs)
| (n < x) && (s > x) = 1 + (heightbetween n s xs)
| otherwise = heightbetween n s xs
count :: [Float] -> [Float] -> [Int]
count [] [] = []
count [x,y] = [(x,y)]
count (x:y:ys) = (x,y):count (y:ys)
forEach fun lst = heightbetween op ([],lst)
where
op (start,[]) = Nothing
op (start,a:as) = Just (start++(fun a):as
,(start++[a],as))
forPairs fun lst lst2 = map (map fst)
$ forEach (\(a,b)->(fun a b,b))
$ zip lst lst2
Your count looks strange. It should be like this:
-- count -> ranges -> data -> [counts]
count :: [Float] -> [Float] -> [Int]
count [] _ = [] -- no ranges given -> empty list
count [_] _ = [] -- no ranges, but single number -> empty list
count _ [] = [] -- no data given -> empty list
count (x:y:xs) d =
(heightbetween x y d) : count (y:xs) d
heightbetween :: Float -> Float -> [Float] -> Int
heightbetween _ _ [] = 0
heightbetween n s (x:xs)
| (n < x) && (s > x) = 1 + (heightbetween n s xs)
| otherwise = heightbetween n s xs
The other lines are obsolete.
Then invoking
count [135,136,138,140] [135.2,135.3,137,139]
gives
[2,1,1]
First, make sure that your range list is in order....
rangePoints = [135,136,138,140]
orderedRangePoints = sort rangePoints
Next, you will find it much easier to work with actual ranges (which you can represent using a 2-tuple (low,high))
ranges = zip orderedRangePoints $ tail orderedRangePoints
You will need an inRange function (one already exists in Data.Ix, but unfortunately it includes the upperbound, so you can't use it)
inRange (low,high) val | val >= low && val < high = True
inRange _ _ = False
You will also want to order your input points
theData = sort [135.2,135.3,137,139]
With all of this out of the way, the binCount function is easy to write.
binCount'::[(Float, Float)]->[Float]->[Int]
binCount' [] [] = []
binCount' (range:rest) vals =
length valsInRange:binCount' rest valsAboveRange
where
(valsInRange, valsAboveRange) = span (`inRange` range) vals
Notice, that I defined a function called binCount', not binCount. I did this, because I consider this an unsafe function, because it only works on ordered ranges and values.... You should finalize this by writing a safer binCount function, which puts all of the stuff above in its where clause. You should probably add all the types and some error checking also (what happens if a value is outside of all ranges?).

Comparing 3 output lists in haskell

I am doing another Project Euler problem and I need to find when the result of these 3 lists is equal (we are given 40755 as the first time they are equal, I need to find the next:
hexag n = [ n*(2*n-1) | n <- [40755..]]
penta n = [ n*(3*n-1)/2 | n <- [40755..]]
trian n = [ n*(n+1)/2 | n <- [40755..]]
I tried adding in the other lists as predicates of the first list, but that didn't work:
hexag n = [ n*(2*n-1) | n <- [40755..], penta n == n, trian n == n]
I am stuck as to where to to go from here.
I tried graphing the function and even calculus but to no avail, so I must resort to a Haskell solution.
Your functions are weird. They get n and then ignore it?
You also have a confusion between function's inputs and outputs. The 40755th hexagonal number is 3321899295, not 40755.
If you really want a spoiler to the problem (but doesn't that miss the point?):
binarySearch :: Integral a => (a -> Bool) -> a -> a -> a
binarySearch func low high
| low == high = low
| func mid = search low mid
| otherwise = search (mid + 1) high
where
search = binarySearch func
mid = (low+high) `div` 2
infiniteBinarySearch :: Integral a => (a -> Bool) -> a
infiniteBinarySearch func =
binarySearch func ((lim+1) `div` 2) lim
where
lim = head . filter func . lims $ 0
lims x = x:lims (2*x+1)
inIncreasingSerie :: (Ord a, Integral i) => (i -> a) -> a -> Bool
inIncreasingSerie func val =
val == func (infiniteBinarySearch ((>= val) . func))
figureNum :: Integer -> Integer -> Integer
figureNum shape index = (index*((shape-2)*index+4-shape)) `div` 2
main :: IO ()
main =
print . head . filter r $ map (figureNum 6) [144..]
where
r x = inIncreasingSerie (figureNum 5) x && inIncreasingSerie (figureNum 3) x
Here's a simple, direct answer to exactly the question you gave:
*Main> take 1 $ filter (\(x,y,z) -> (x == y) && (y == z)) $ zip3 [1,2,3] [4,2,6] [8,2,9]
[(2,2,2)]
Of course, yairchu's answer might be more useful in actually solving the Euler question :)
There's at least a couple ways you can do this.
You could look at the first item, and compare the rest of the items to it:
Prelude> (\x -> all (== (head x)) $ tail x) [ [1,2,3], [1,2,3], [4,5,6] ]
False
Prelude> (\x -> all (== (head x)) $ tail x) [ [1,2,3], [1,2,3], [1,2,3] ]
True
Or you could make an explicitly recursive function similar to the previous:
-- test.hs
f [] = True
f (x:xs) = f' x xs where
f' orig (y:ys) = if orig == y then f' orig ys else False
f' _ [] = True
Prelude> :l test.hs
[1 of 1] Compiling Main ( test.hs, interpreted )
Ok, modules loaded: Main.
*Main> f [ [1,2,3], [1,2,3], [1,2,3] ]
True
*Main> f [ [1,2,3], [1,2,3], [4,5,6] ]
False
You could also do a takeWhile and compare the length of the returned list, but that would be neither efficient nor typically Haskell.
Oops, just saw that didn't answer your question at all. Marking this as CW in case anyone stumbles upon your question via Google.
The easiest way is to respecify your problem slightly
Rather than deal with three lists (note the removal of the superfluous n argument):
hexag = [ n*(2*n-1) | n <- [40755..]]
penta = [ n*(3*n-1)/2 | n <- [40755..]]
trian = [ n*(n+1)/2 | n <- [40755..]]
You could, for instance generate one list:
matches :: [Int]
matches = matches' 40755
matches' :: Int -> [Int]
matches' n
| hex == pen && pen == tri = n : matches (n + 1)
| otherwise = matches (n + 1) where
hex = n*(2*n-1)
pen = n*(3*n-1)/2
tri = n*(n+1)/2
Now, you could then try to optimize this for performance by noticing recurrences. For instance when computing the next match at (n + 1):
(n+1)*(n+2)/2 - n*(n+1)/2 = n + 1
so you could just add (n + 1) to the previous tri to obtain the new tri value.
Similar algebraic simplifications can be applied to the other two functions, and you can carry all of them in accumulating parameters to the function matches'.
That said, there are more efficient ways to tackle this problem.

Doing a binary search on some elements in Haskell

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.

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