import Data.Char

-- Exercitiul 1

-- a)

data Reteta =  Stop | R Ing Reteta
         deriving Show

data Ing = Ing String Int
         deriving Show


-- a)

-- Returneaza cantitatea maxima a ingredientului cu numele "nume", case insensitive.
-- Folosesc map toLower pentru a aplica functia "toLower" tuturor caracterelor dintr-un string.
r15pmMaxIng :: Reteta -> String -> Int
r15pmMaxIng Stop _ = 0
r15pmMaxIng (R (Ing nume_ing cantitate) r) nume =
    let maxRest = r15pmMaxIng r nume in
        if map toLower nume_ing == map toLower nume
            then max cantitate maxRest
        else maxRest

-- Verifica daca reteta din primul parametru contine ingredientul din al doilea parametru.
r15pmContine :: Reteta -> Ing -> Bool
r15pmContine Stop _ = False
r15pmContine (R ing r) toFind
    | ing == toFind = True
    | otherwise = r15pmContine r toFind

-- Face tot ce face si r15pmEx1 dar mai are si parametru reteta_initiala.
-- Aveam nevoie si de parametrul reteta_initiala pentru ca eu caut in reteta initiala cantitatea maxima a ingredientului curent.
-- Nu e clar din enunt ce sa fac daca am ingrediente duplicate, asa ca le-am eliminat, afisand mereu ultima aparitie.
-- Daca nu voiam sa le elimin, scoteam "&& not (r15pmContine r (Ing nume_ing cantitate))" din if.
r15pmEx1_initial :: Reteta -> Reteta -> Reteta
r15pmEx1_initial Stop _ = Stop
r15pmEx1_initial (R (Ing nume_ing cantitate) r) reteta_initiala =
    if cantitate == r15pmMaxIng reteta_initiala nume_ing && not (r15pmContine r (Ing nume_ing cantitate))
        then R (Ing nume_ing cantitate) (r15pmEx1_initial r reteta_initiala)
        else r15pmEx1_initial r reteta_initiala

r15pmEx1 :: Reteta -> Reteta
r15pmEx1 r = r15pmEx1_initial r r


-- Pentru testing:

ex :: Reteta
ex = R (Ing "faina" 500) (R  (Ing "Oua" 4) (R (Ing "faina" 300) Stop))

retetaGoala :: Reteta
retetaGoala = Stop

reteta1 :: Reteta
reteta1 = R (Ing "branza" 1) (R  (Ing "Oua" 5) (R (Ing "brAnza" 20) (R (Ing "OUA" 0) Stop)))

reteta2 :: Reteta
reteta2 = R (Ing "branza" 1) (R  (Ing "Oua" 5) (R (Ing "brAnza" 1) (R (Ing "OUA" 0) Stop)))

-- Teste:
-- 1. ex (dat in enunt): R (Ing "faina" 500) (R (Ing "Oua" 4) Stop)
-- 2. testReteta: Stop
-- 3. reteta1: R (Ing "Oua" 5) (R (Ing "brAnza" 20) Stop)
-- 4. reteta2: R (Ing "Oua" 5) (R (Ing "brAnza" 1) Stop)

-- b)

-- Verifica daca doua ingrediente sunt egale ca si nume (case insensitive) si cantitate.
instance Eq Ing where
    (Ing name1 quantity1) == (Ing name2 quantity2) = map toLower name1 == map toLower name2 && quantity1 == quantity2

-- Verifica daca reteta din primul parametru contine toate ingredientele din al doilea parametru.
r15pmContineTot :: Reteta -> Reteta -> Bool
r15pmContineTot _ Stop = True
r15pmContineTot Stop _ = False
r15pmContineTot r1 (R ing r)
    | r15pmContine r1 ing = r15pmContineTot r1 r
    | otherwise = False

-- Ca doua retete sa fie egale, am folosit dubla incluziune.
-- Adica prima reteta trebuie sa contina tot ce e in a doua si invers.
instance Eq Reteta where
    a == b =
        let x = r15pmEx1 a in
            let y = r15pmEx1 b in
                r15pmContineTot x y && r15pmContineTot y x


-- Pentru testing:
r1 :: Reteta
r1 =  R (Ing "faina" 500) (R (Ing "oua" 4) (R  (Ing "zahar" 500) (R (Ing "faina" 300) Stop)))
r2 :: Reteta
r2 =  R (Ing "fAIna" 500) (R (Ing "zahar" 500) (R (Ing "Oua" 4) Stop ))
r3 :: Reteta
r3 =  R (Ing "fAIna" 500) (R (Ing "zahar" 500) (R  (Ing "Oua" 55) Stop))
r4 :: Reteta
r4 =  R (Ing "fAIna" 499) (R (Ing "zahar" 500) (R  (Ing "Oua" 55) Stop))
r5 :: Reteta
r5 = Stop
r6 :: Reteta
r6 = R (Ing "fAIna" 499) (R (Ing "zahar" 500) Stop) -- este continuta de r4

-- Teste:
-- 1. r1 == r2 (dat in enunt): True
-- 2. r2 == r3 (dat in enunt): False
-- 3. r3 == r3: False
-- 4. r1 == r5: False
-- 5. r5 == r5: True
-- 6. r1 == r1: True
-- 7. r4 == r6: False
-- 8. r6 == r4: False


-- c)
-- Combina 2 retete (duce b-ul la finalul lui a)
r15pmCombina :: Reteta -> Reteta -> Reteta
r15pmCombina a Stop = a
r15pmCombina Stop a = a
r15pmCombina (R ing Stop) b = R ing b
r15pmCombina (R _ r) b = r15pmCombina r b

data Arb = Leaf Int String | Node Arb Int String Arb
         deriving  Show

r15pmArbToReteta :: Arb -> Reteta
r15pmArbToReteta (Leaf x nume) = R (Ing nume x) Stop
r15pmArbToReteta (Node leftArb x nume rightArb) =
    let leftRightReteta = r15pmCombina (r15pmArbToReteta leftArb) (r15pmArbToReteta rightArb) in
        R (Ing nume x) leftRightReteta



-- Exercitiul 2
data E x = A | M x Bool (E x)

-- a)
instance Foldable E where
    foldMap _ A = mempty
    foldMap f (M x _ e) = foldMap f e `mappend` f x

-- Pentru testing:
fTest0 :: Bool
fTest0 = maximum (M 1 True (M 5 False (M 3 True (M 2 True A)))) == 3
fTest1 :: Bool
fTest1 = maximum (M 1 True (M 5 False (M 8 True (M 2 True A)))) == 8
fTest2 :: Bool
fTest2 = sum (M 1 True (M 5 True (M 8 True (M 2 True A)))) == 16
fTest3 :: Bool
fTest3 = foldr (*) 1 (M 1 True (M 5 True (M 8 True A))) == 40


-- Teste:
-- toate True

-- b)
class C e where
  cFilter :: (a -> Bool) -> e a -> e a
  toList :: e a -> [a]

instance C E where
    cFilter _ A = A
    cFilter f (M x b e)
        | not (f x) = M x False (cFilter f e)
        | otherwise = M x b (cFilter f e)
    toList A = []
    toList (M x b e)
        | b = x : toList e
        | otherwise = toList e


-- Pentru testing:
cTest0 :: Bool
cTest0 = toList  (cFilter (>2) (M 1 True (M 5 False (M 3 True (M 2 True A))))) == [3]
cTest1 :: Bool
cTest1 = toList  (cFilter (>2) (M 1 True (M 5 True (M 3 True (M 2 True A))))) == [5, 3]
cTest2 :: Bool
cTest2 = toList  (cFilter (> 100) (M 1 True (M 5 True (M 3 True (M 2 True A))))) == []
cTest3 :: Bool
cTest3 = toList  (cFilter (== 5) (M 1 True (M 5 False (M 3 True (M 2 True A))))) == []
cTest4 :: Bool
cTest4 = toList  (cFilter (== 5) (M 1 True (M 5 False (M 5 True A)))) == [5]

-- Teste:
-- toate True