Las matrices se pueden representar como listas de listas de la misma longitud, donde cada uno de sus elementos representa una fila de la matriz.
Definir la función
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diagonal :: [[a]] -> [a] |
tal que (diagonal xss) es la diagonal de la matriz xss. Por ejemplo,
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diagonal [[3,5,2],[4,7,1],[6,9,0]] == [3,7,0] diagonal [[3,5],[4,7],[6,9]] == [3,7] diagonal [[3,5,2],[4,7,1]] == [3,7] sum (diagonal (replicate 20000 [1..20000])) == 200010000 |
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import Data.Array ((!), listArray) import Data.Matrix (fromLists, getDiag) import Data.Vector (toList) -- 1ª definición (por recursión): diagonal1 :: [[a]] -> [a] diagonal1 ((x:_):xss) = x : diagonal1 [tail xs | xs <- xss] diagonal1 _ = [] -- 2ª definición (por comprensión): diagonal2 :: [[a]] -> [a] diagonal2 xss = [xs!!k | (xs,k) <- zip xss [0..n]] where n = length (head xss) - 1 -- 3ª definición (con Data.Matrix) diagonal3 :: [[a]] -> [a] diagonal3 = toList . getDiag . fromLists -- 4ª definición (con Data.Array) diagonal4 :: [[a]] -> [a] diagonal4 xss = [p!(i,i) | i <- [1..k]] where m = length xss n = length (head xss) k = min m n p = listArray ((1,1),(m,n)) (concat xss) -- Comparación de eficiencia -- λ> let n = 3000 in sum (diagonal1 (replicate n [1..n])) -- 4501500 -- (2.08 secs, 754,089,992 bytes) -- λ> let n = 3000 in sum (diagonal2 (replicate n [1..n])) -- 4501500 -- (0.06 secs, 0 bytes) -- λ> let n = 3000 in sum (diagonal3 (replicate n [1..n])) -- 4501500 -- (1.22 secs, 1,040,787,360 bytes) -- λ> let n = 3000 in sum (diagonal4 (replicate n [1..n])) -- 4501500 -- (0.96 secs, 624,463,632 bytes) |
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diagonal (xss) := block ([A], A : apply (matrix, xss), [m,n] : matrix_size (A), k : min (m,n), makelist (A[i,i],i,k))$ |