Definir la función
suma :: Integer -> Integer |
tal que (suma n)
es la suma 1·1! + 2·2! + 3·3! + ... + n·n!
. Por ejemplo,
suma 1 == 1 suma 2 == 5 suma 3 == 23 suma 4 == 119 suma 5 == 719 take 9 (show (suma 70000)) == "823780458" |
Soluciones
import Test.QuickCheck (Positive (Positive), quickCheck) -- 1ª solución -- =========== suma1 :: Integer -> Integer suma1 n = sum [k * factorial k | k <- [1..n]] factorial :: Integer -> Integer factorial n = product [1..n] -- 2ª solución -- =========== suma2 :: Integer -> Integer suma2 n = sum (zipWith (*) [1..n] factoriales) factoriales :: [Integer] factoriales = scanl (*) 1 [2..] -- 3ª solución -- =========== -- Basada en los siguientes cálculos -- λ> [suma1 n | n <- [0..10]] -- [0,1,5,23,119,719,5039,40319,362879,3628799,39916799] -- λ> [factorial n | n <- [0..10]] -- [1,1,2,6,24,120,720,5040,40320,362880,3628800] -- λ> [factorial n | n <- [1..11]] -- [1,2,6,24,120,720,5040,40320,362880,3628800,39916800] -- λ> [factorial n - 1 | n <- [1..11]] -- [0,1,5,23,119,719,5039,40319,362879,3628799,39916799] suma3 :: Integer -> Integer suma3 n = factorial (n+1) - 1 -- Comprobación de equivalencia -- ============================ -- La propiedad es prop_suma :: Positive Integer -> Bool prop_suma (Positive n) = all (== suma1 n) [suma2 n, suma3 n] -- La comprobación es -- λ> quickCheck prop_suma -- +++ OK, passed 100 tests. -- Comparación de eficiencia -- ========================= -- La comparación es -- λ> take 5 (show (suma1 4000)) -- "73170" -- (5.04 secs, 16,225,195,448 bytes) -- λ> take 5 (show (suma2 4000)) -- "73170" -- (0.08 secs, 35,862,152 bytes) -- λ> take 5 (show (suma3 4000)) -- "73170" -- (0.01 secs, 12,896,968 bytes) -- -- -- λ> take 5 (show (suma2 40000)) -- "83669" -- (1.82 secs, 4,549,612,264 bytes) -- λ> take 5 (show (suma3 40000)) -- "83669" -- (0.24 secs, 1,620,976,984 bytes) |
El código se encuentra en GitHub.