Самостоятельная работа Haskell 12.04.2021

-- Задача №1

-- 1-ый способ:
allPred1 :: (a -> Bool) -> [a] -> Bool
allPred1 pr [] = True
allPred1 pr (x : xs) = if pr x == False then False else allPred1 pr xs

-- 2-ой способ:
allPred2 :: (a -> Bool) -> [a] -> Bool
allPred2 pr [] = True
allPred2 pr (x : xs)
    | pr x == False = False
    | otherwise = allPred2 pr xs

-- 3-ий способ:
allPred3 :: (a -> Bool) -> [a] -> Bool
allPred3 pr [] = True
allPred3 pr (x : xs) =
    case pr x of
         False -> False
         _ -> allPred3 pr xs

-- 4-ый способ:
allPred4 :: (a -> Bool) -> [a] -> Bool
allPred4 a = and . map a

-- 5-ый способ:
allPred5 :: (a -> Bool) -> [a] -> Bool
allPred5 pr [] = True
allPred5 pr a = foldr1 (&&) $ map pr a

-- Задача №2

-- 1-ый способ:
anyPred1 :: (a -> Bool) -> [a] -> Bool
anyPred1 pr [] = False
anyPred1 pr (x : xs) = if pr x == True then True else anyPred1 pr xs

-- 2-ой способ:
anyPred2 :: (a -> Bool) -> [a] -> Bool
anyPred2 pr [] = False
anyPred2 pr (x : xs)
    | pr x == True = True
    | otherwise = anyPred2 pr xs

-- 3-ий способ:
anyPred3 :: (a -> Bool) -> [a] -> Bool
anyPred3 pr [] = False
anyPred3 pr (x : xs) =
    case pr x of
         True -> True
         _ -> anyPred3 pr xs

-- 4-ый способ:
anyPred4 :: (a -> Bool) -> [a] -> Bool
anyPred4 a = or . map a

-- 5-ый способ:
anyPred5 :: (a -> Bool) -> [a] -> Bool
anyPred5 pr [] = False
anyPred5 pr a = foldr1 (||) $ map pr a