Is there an algorithm to lead all possible combinations of given amount of three-valued logic values?
For example, F(2) should return this list:
t t
t u
t f
u t
u u
u f
f t
f u
f f
The function would look like this (in Haskell):
data Tril = FALSE | NULL | TRUE
all :: Int -> [[Tril]]
all amount = ???
all1 :: [Tril]
all1 = join (all 1)
all2 :: [(Tril, Tril)]
all2 = map (\[f, s] -> (f, s)) (all 2)
all3 :: [(Tril, Tril, Tril)]
all3 = map (\[f, s, t] -> (f, s, t)) (all 3)
You can do this very simply as a list comprehension:
all2 = [ (v1, v2) | v1 <- [FALSE, TRUE, NULL], v2 <- [FALSE, TRUE, NULL] ]
You can write it equivalently as a monadic do-block:
all2 = do
v1 <- [FALSE, TRUE, NULL]
v2 <- [FALSE, TRUE, NULL]
return (v1, v2)
And that gives us an idea for how we can write the variable-size one:
all 0 = [[]] -- Note: Empty list with one empty item.
all n = do
v <- [FALSE, TRUE, NULL]
vs <- all (n-1)
return (v:vs)
As it turns out — and this is slightly mind-bending — this is the net effect of the replicateM function. It takes a monadic action, does it N times, and gathers the results together.
all n = replicateM n [FALSE, TRUE, NULL]
replicateM does exactly that:
> import Control.Monad
> replicateM 2 [1,2,3]
[[1,1],[1,2],[1,3],[2,1],[2,2],[2,3],[3,1],[3,2],[3,3]]
Hence,
all :: Int -> [[Tril]]
all amount = replicateM amount [FALSE,NULL,TRUE]
I'd suggest to pick anouther name, since all is already taken by Prelude.all.
Related
My code aims to create a word search puzzle. There is a data called Orientation representing the direction of each word in the puzzle.
data Orientation =
Forward | Back | Up | Down | UpForward | UpBack | DownForward | DownBack
deriving (Eq, Ord, Show, Read)
Now given a input of strings which is [String], I want to randomly assign each string an orientation like [(Orientation, String)]
assignWordDir :: [String] -> [(Orientation, String)]
assignWordDir [] = []
assignWordDir (s:strs) = (ori, s) : assignWordDir
where ori = pickOri [Forward, Back, Up, Down, UpForward, UpBack, DownForward, DownBack]
pickOri :: [a] -> IO a
pickOri xs = do
i <- randomRIO (0, len)
pure $ xs !! i
where len = length xs - 1
I cannot compile because the output of pickOri is IO Orientation, is there any suggestions on how to modify my code? Thanks a lot
Couldn't match expected type ‘[(IO Orientation, String)]’
with actual type ‘[String] -> [(Orientation, String)]’
You might consider modifying the functions so that they stay pure by taking a RandomGen parameter. The pickOri function, for example, might be modified thusly:
pickOri :: RandomGen g => g -> [a] -> (a, g)
pickOri rnd xs =
let len = length xs - 1
(i, g) = randomR (0, len) rnd
in (xs !! i, g)
It's necessary to return the new RandomGen value g together with the selected list element, so that it'll generate another pseudo-random number the next time around.
Likewise, you can modify assignWordDir like this:
assignWordDir :: RandomGen g => g -> [b] -> [(Orientation, b)]
assignWordDir _ [] = []
assignWordDir rnd (s:strs) = (ori, s) : assignWordDir g strs
where (ori, g) =
pickOri rnd [Forward, Back, Up, Down, UpForward, UpBack, DownForward, DownBack]
Notice that when recursing into to assignWordDir, the recursive function call uses the g it receives from pickOri.
You can use mkStdGen or newStdGen to produce RandomGen values. Here's an example using newStdGen:
*Q65132918> rnd <- newStdGen
*Q65132918> assignWordDir rnd ["foo", "bar", "baz"]
[(UpBack,"foo"),(Up,"bar"),(UpBack,"baz")]
*Q65132918> assignWordDir rnd ["foo", "bar", "baz"]
[(UpBack,"foo"),(Up,"bar"),(UpBack,"baz")]
Notice that when you use the same RandomGen value, you get the same sequence. That's because assignWordDir is a pure function, so that's expected.
You can, however, produce a new random sequence by creating or getting a new StdGen value:
*Q65132918> rnd <- newStdGen
*Q65132918> assignWordDir rnd ["foo", "bar", "baz"]
[(Up,"foo"),(Up,"bar"),(Forward,"baz")]
If you want to play with this in a compiled module, you can keep these functions as presented here, and then compose them with a newStdGen-generated StdGen in the main entry point.
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.
So I'm trying to define a function in Haskell that if given an integer and a list of integers will give a 'true' or 'false' whether the integer occurs only once or not.
So far I've got:
let once :: Eq a => a -> [a] -> Bool; once x l =
But I haven't finished writing the code yet. I'm very new to Haskell as you may be able to tell.
Start off by using pattern matching:
once x [] =
once x (y:ys) =
This won't give you a good program immediately, but it will lead you in the right direction.
Here's a solution that doesn't use pattern matching explicitly. Instead, it keeps track of a Bool which represents if a occurance has already been found.
As others have pointed out, this is probably a homework problem, so I've intentionally left the then and else branches blank. I encourage user3482534 to experiment with this code and fill them in themselves.
once :: Eq a => a -> [a] -> Bool
once a = foldr f False
where f x b = if x == a then ??? else ???
Edit: The naive implementation I was originally thinking of was:
once :: Eq a => a -> [a] -> Bool
once a = foldr f False
where f x b = if x == a then b /= True else b
but this is incorrect as,
λ. once 'x' "xxx"
True
which should, of course, be False as 'x' occurs more than exactly once.
However, to show that it is possible to write once using a fold, here's a revised version that uses a custom monoid to keep track of how many times the element has occured:
import Data.List
import Data.Foldable
import Data.Monoid
data Occur = Zero | Once | Many
deriving Eq
instance Monoid Occur where
mempty = Zero
Zero `mappend` x = x
x `mappend` Zero = x
_ `mappend` _ = Many
once :: Eq a => a -> [a] -> Bool
once a = (==) Once . foldMap f
where f x = if x == a then Once else Zero
main = do
let xss = inits "xxxxx"
print $ map (once 'x') xss
which prints
[False,True,False,False,False]
as expected.
The structure of once is similar, but not identical, to the original.
I'll answer this as if it were a homework question since it looks like one.
Read about pattern matching in function declarations, especially when they give an example of processing a list. You'll use tools from Data.List later, but probably your professor is teaching about pattern matching.
Think about a function that maps values to a 1 or 0 depending on whethere there is a match ...
match :: a -> [a] -> [Int]
match x xs = map -- fill in the thing here such that
-- match 3 [1,2,3,4,5] == [0,0,1,0,0]
Note that there is the sum function that takes a list of numbers and returns the sum of the numbers in the list. So to count the matches a function can take the match function and return the counts.
countN :: a -> [a] -> Int
countN x xs = ? $ match x xs
And finally a function that exploits the countN function to check for a count of only 1. (==1).
Hope you can figure out the rest ...
You can filter the list and then check the length of the resulting list. If length == 1, you have only one occurrence of the given Integer:
once :: Eq a => a -> [a] -> Bool
once x = (== 1) . length . filter (== x)
For counting generally, with import Data.List (foldl'), pointfree
count pred = foldl' (\ n x -> if pred x then n + 1 else n) 0
applicable like
count (< 10) [1 .. 10] == 9
count (== 'l') "Hello" == 2
gives
once pred xs = count pred xs == 1
Efficient O(n) short-circuit predicated form, testing whether the predicate is satisfied exactly once:
once :: (a -> Bool) -> [a] -> Bool
once pred list = one list 0
where
one [] 1 = True
one [] _ = False
one _ 2 = False
one (x : xs) n | pred x = one xs (n + 1)
| otherwise = one xs n
Or, using any:
none pred = not . any pred
once :: (a -> Bool) -> [a] -> Bool
once _ [] = False
once pred (x : xs) | pred x = none pred xs
| otherwise = one pred xs
gives
elemOnce y = once (== y)
which
elemOnce 47 [1,1,2] == False
elemOnce 2 [1,1,2] == True
elemOnce 81 [81,81,2] == False
I am trying to build a smallish haskell app that will translate a few key phrases from english to french.
First, i have a list of ordered pairs of strings that represent and english word/phrase followed by the french translations:
icards = [("the", "le"),("savage", "violent"),("work", "travail"),
("wild", "sauvage"),("chance", "occasion"),("than a", "qu'un")...]
next i have a new data:
data Entry = Entry {wrd, def :: String, len :: Int, phr :: Bool}
deriving Show
then i use the icards to populate a list of Entrys:
entries :: [Entry]
entries = map (\(x, y) -> Entry x y (length x) (' ' `elem` x)) icards
for simplicity, i create a new type that will be [Entry] called Run.
Now, i want to create a hash table based on the number of characters in the english word. This will be used later to speed up searchings. So i want to create a function called runs:
runs :: [Run]
runs = --This will run through the entries and return a new [Entry] that has all of the
words of the same length grouped together.
I also have:
maxl = maximum [len e | e <- entries]
It just so happens that Hackage has a hashmap package! I'm going to create a small data type based on that HashMap, which I will call a MultiMap. This is a typical trick: it's just a hash map of linked lists. I'm not sure what the correct name for MultiMap actually is.
import qualified Data.HashMap as HM
import Data.Hashable
import Prelude hiding (lookup)
type MultiMap k v = HM.Map k [v]
insert :: (Hashable k, Ord k) => k -> a -> MultiMap k a -> MultiMap k a
insert k v = HM.insertWith (++) k [v]
lookup :: (Hashable k, Ord k) => k -> MultiMap k a -> [a]
lookup k m = case HM.lookup k m of
Nothing -> []
Just xs -> xs
empty :: MultiMap k a
empty = HM.empty
fromList :: (Hashable k, Ord k) => [(k,v)] -> MultiMap k v
fromList = foldr (uncurry insert) empty
I mimicked only the essentials of a Map: insert, lookup, empty, and fromList. Now it is quite easy to turn entries into a MutliMap:
data Entry = Entry {wrd, def :: String, len :: Int, phr :: Bool}
deriving (Show)
icards = [("the", "le"),("savage", "violent"),("work", "travail"),
("wild", "sauvage"),("chance", "occasion"),("than a", "qu'un")]
entries :: [Entry]
entries = map (\(x, y) -> Entry x y (length x) (' ' `elem` x)) icards
fromEntryList :: [Entry] -> MutiMap Int Entry
fromEntryList es = fromList $ map (\e -> (len e, e)) es
Loading that up into ghci, we can now lookup a list of entries with a given length:
ghci> let m = fromEntryList entries
ghci> lookup 3 m
[Entry {wrd = "the", def = "le", len = 3, phr = False}]
ghci> lookup 4 m
[Entry {wrd = "work", def = "travail", len = 4, phr = False},
Entry {wrd = "wild", def = "sauvage", len = 4, phr = False}]
(Note that this lookup is not the one defined in Prelude.) You could similarly use the English word as a key.
-- import Data.List (find) -- up with other imports
fromEntryList' :: [Entry] -> MultiMap String Entry
fromEntryList' es = fromList $ map (\e -> (wrd e, e)) es
eLookup :: String -> MultiMap String Entry -> Maybe Entry
eLookup str m = case lookup str m of
[] -> Nothing
xs -> find (\e -> wrd e == str) xs
Testing...
ghci> let m = fromEntryList' entries
ghci> eLookup "the" m
Just (Entry {wrd = "the", def = "le", len = 3, phr = False})
ghci> eLookup "foo" m
Nothing
Notice how in eLookup we first perform the Map lookup in order to determine if anything has been placed in that slot. Since we are using a hash set, we need to remember that two different Strings might have the same hash code. So in the event that the slot is not empty, we perform a find on the linked list there to see if any of the entries there actually match the correct English word. If you are interested in performance, you should consider using Data.Text instead of String.
groupBy and sortBy are both in Data.List.
import Data.List
import Data.Function -- for `on`
runs :: [Run]
runs = f 0 $ groupBy ((==) `on` len) $ sortBy (compare `on` len) entries
where f _ [] = []
f i (r # (Entry {len = l} : _) : rs) | i == l = r : f (i + 1) rs
f i rs = [] : f (i + 1) rs
Personally, I would use a Map instead
import qualified Data.Map as M
runs :: M.Map String Entry
runs = M.fromList $ map (\entry -> (wrd entry, entry)) entries
and lookup directly by English word instead of a two step length-of-English-word and then English-word process.
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.