maximum

Mínimo número de cambios para igualar una lista

PorJosé A. Alonso 27-abril-20164-mayo-2016

Definir la función

   nMinimoCambios :: Ord a => [a] -> Int

1	nMinimoCambios :: Ord a => [a] -> Int

tal que (nMinimoCambios xs) es el menor número de elementos de xs que hay que cambiar para que todos sean iguales. Por ejemplo,

   nMinimoCambios [3,5,3,7,9,6]      ==  4
   nMinimoCambios [3,5,3,7,3,3]      ==  2
   nMinimoCambios "Salamanca"        ==  5
   nMinimoCambios (4 : [1..500000])  ==  499999

nMinimoCambios [3,5,3,7,9,6] == 4

nMinimoCambios [3,5,3,7,3,3] == 2

nMinimoCambios "Salamanca" == 5

nMinimoCambios (4 : [1..500000]) == 499999

En el primer ejemplo, los elementos que hay que cambiar son 5, 7, 9 y 6.

Soluciones

import Data.List (group, nub, sort)

-- 1ª definición
-- =============

nMinimoCambios1 :: Ord a => [a] -> Int
nMinimoCambios1 xs = 
    length xs - fst (last (sort (frecuencias xs)))

-- (frecuencias xs) es la lista de los pares de los elementos de xs y el
-- número de veces que ocurren en xs. Por ejemplo,
--    frecuencias [3,5,3,7,9,6]  ==  [(2,3),(1,5),(1,7),(1,9),(1,6)]
--    frecuencias [3,5,3,7,5,5]  ==  [(2,3),(3,5),(1,7)]
frecuencias :: Ord a => [a] -> [(Int,a)]
frecuencias xs = [(cuenta x xs,x) | x <- nub xs]

-- (cuenta x ys) es el número de veces que ocurre x en ys. Por ejemplo,
--    cuenta 3 [3,5,3,7,9,6]  ==  2
cuenta :: Ord a => a -> [a] -> Int
cuenta x ys = length [y | y <- ys, y == x]

-- 2ª definición
-- =============

nMinimoCambios2 :: Ord a => [a] -> Int
nMinimoCambios2 xs = 
    length xs - fst (last (sort (frecuencias2 xs)))

-- (frecuencias2 xs) es la lista de los pares de los elementos de xs y el
-- número de veces que ocurren en xs. Por ejemplo,
--    frecuencias2 [3,5,3,7,9,6]  ==  [(2,3),(1,5),(1,7),(1,9),(1,6)]
--    frecuencias2 [3,5,3,7,5,5]  ==  [(2,3),(3,5),(1,7)]
frecuencias2 :: Ord a  => [a] -> [(Int,a)]
frecuencias2 xs = 
    [(1 + length ys, y) | (y:ys) <- group (sort xs)] 

-- 3ª definición
-- =============

nMinimoCambios3 :: Ord a => [a] -> Int
nMinimoCambios3 xs = sum ys - maximum ys
    where ys = [length ys | ys <- group (sort xs)]

import Data.List (group, nub, sort)

-- 1ª definición

-- =============

nMinimoCambios1 :: Ord a => [a] -> Int

nMinimoCambios1 xs =

length xs - fst (last (sort (frecuencias xs)))

-- (frecuencias xs) es la lista de los pares de los elementos de xs y el

-- número de veces que ocurren en xs. Por ejemplo,

-- frecuencias [3,5,3,7,9,6] == [(2,3),(1,5),(1,7),(1,9),(1,6)]

-- frecuencias [3,5,3,7,5,5] == [(2,3),(3,5),(1,7)]

frecuencias :: Ord a => [a] -> [(Int,a)]

frecuencias xs = [(cuenta x xs,x) | x <- nub xs]

-- (cuenta x ys) es el número de veces que ocurre x en ys. Por ejemplo,

-- cuenta 3 [3,5,3,7,9,6] == 2

cuenta :: Ord a => a -> [a] -> Int

cuenta x ys = length [y | y <- ys, y == x]

-- 2ª definición

-- =============

nMinimoCambios2 :: Ord a => [a] -> Int

nMinimoCambios2 xs =

length xs - fst (last (sort (frecuencias2 xs)))

-- (frecuencias2 xs) es la lista de los pares de los elementos de xs y el

-- número de veces que ocurren en xs. Por ejemplo,

-- frecuencias2 [3,5,3,7,9,6] == [(2,3),(1,5),(1,7),(1,9),(1,6)]

-- frecuencias2 [3,5,3,7,5,5] == [(2,3),(3,5),(1,7)]

frecuencias2 :: Ord a => [a] -> [(Int,a)]

frecuencias2 xs =

[(1 + length ys, y) | (y:ys) <- group (sort xs)]

-- 3ª definición

-- =============

nMinimoCambios3 :: Ord a => [a] -> Int

nMinimoCambios3 xs = sum ys - maximum ys

where ys = [length ys | ys <- group (sort xs)]

Máxima suma de elementos consecutivos

PorJosé A. Alonso 17-marzo-20161-mayo-2016

Definir la función

   sumaMaxima :: [Integer] -> Integer

1	sumaMaxima :: [Integer] -> Integer

tal que (sumaMaxima xs) es el valor máximo de la suma de elementos consecutivos de la lista xs. Por ejemplo,

   sumaMaxima []             ==  0 
   sumaMaxima [2,-2,3,-3,4]  ==  4
   sumaMaxima [-1,-2,-3]     ==  0
   sumaMaxima [2,-1,3,-2,3]  ==  5
   sumaMaxima [1,-1,3,-2,4]  ==  5
   sumaMaxima [2,-1,3,-2,4]  ==  6
   sumaMaxima [1..10^6]      ==  500000500000

sumaMaxima [] == 0

sumaMaxima [2,-2,3,-3,4] == 4

sumaMaxima [-1,-2,-3] == 0

sumaMaxima [2,-1,3,-2,3] == 5

sumaMaxima [1,-1,3,-2,4] == 5

sumaMaxima [2,-1,3,-2,4] == 6

sumaMaxima [1..10^6] == 500000500000

Comprobar con QuickCheck que

   sumaMaxima xs == sumaMaxima (reverse xs)

1	sumaMaxima xs == sumaMaxima (reverse xs)

Soluciones

import Data.List (inits, tails)
import Test.QuickCheck

-- 1ª definición
-- =============

sumaMaxima1 :: [Integer] -> Integer
sumaMaxima1 [] = 0
sumaMaxima1 xs =
    maximum (0 : map sum [sublista xs i j | i <- [0..length xs - 1],
                                            j <- [i..length xs - 1]])

sublista :: [Integer] -> Int -> Int -> [Integer]
sublista xs i j =
    [xs!!k | k <- [i..j]]

-- 2ª definición
-- =============

sumaMaxima2 :: [Integer] -> Integer
sumaMaxima2 [] = 0
sumaMaxima2 xs = sumaMaximaAux 0 0 xs
    where m = maximum xs

sumaMaximaAux :: Integer -> Integer -> [Integer] -> Integer
sumaMaximaAux m v [] = max m v
sumaMaximaAux m v (x:xs)
    | x >= 0    = sumaMaximaAux m (v+x) xs
    | v+x > 0   = sumaMaximaAux (max m v) (v+x) xs
    | otherwise = sumaMaximaAux (max m v) 0 xs

-- 3ª definición
-- =============

sumaMaxima3 :: [Integer] -> Integer
sumaMaxima3 [] = 0
sumaMaxima3 xs = maximum (map sum (segmentos xs))

-- (segmentos xs) es la lista de los segmentos de xs. Por ejemplo 
--    segmentos "abc"  ==  ["", "a","ab","abc","b","bc","c"]
segmentos :: [a] -> [[a]]
segmentos xs =
    [] : concat [tail (inits ys) | ys <- init (tails xs)]

-- 4ª definición
-- =============

sumaMaxima4 :: [Integer] -> Integer
sumaMaxima4 [] = 0
sumaMaxima4 xs = 
    maximum (concat [scanl (+) 0 ys | ys <- tails xs])

-- Comprobación
-- ============

-- La propiedad es
prop_sumaMaxima :: [Integer] -> Bool
prop_sumaMaxima xs =
    sumaMaxima2 xs == n &&
    sumaMaxima3 xs == n &&
    sumaMaxima4 xs == n
    where n = sumaMaxima1 xs

-- La comprobación es
--    λ> quickCheck prop_sumaMaxima
--    +++ OK, passed 100 tests.

-- Comparación de eficiencia
-- =========================

--    λ> let n = 10^2 in sumaMaxima1 [-n..n] 
--    5050
--    (2.10 secs, 390,399,104 bytes)
--    λ> let n = 10^2 in sumaMaxima2 [-n..n] 
--    5050
--    (0.02 secs, 0 bytes)
--    λ> let n = 10^2 in sumaMaxima3 [-n..n] 
--    5050
--    (0.27 secs, 147,705,184 bytes)
--    λ> let n = 10^2 in sumaMaxima4 [-n..n] 
--    5050
--    (0.04 secs, 11,582,520 bytes)

prop_inversa :: [Integer] -> Bool
prop_inversa xs =
    sumaMaxima2 xs == sumaMaxima2 (reverse xs)

import Data.List (inits, tails)

import Test.QuickCheck

-- 1ª definición

-- =============

sumaMaxima1 :: [Integer] -> Integer

sumaMaxima1 [] = 0

sumaMaxima1 xs =

maximum (0 : map sum [sublista xs i j | i <- [0..length xs - 1],

j <- [i..length xs - 1]])

sublista :: [Integer] -> Int -> Int -> [Integer]

sublista xs i j =

[xs!!k | k <- [i..j]]

-- 2ª definición

-- =============

sumaMaxima2 :: [Integer] -> Integer

sumaMaxima2 [] = 0

sumaMaxima2 xs = sumaMaximaAux 0 0 xs

where m = maximum xs

sumaMaximaAux :: Integer -> Integer -> [Integer] -> Integer

sumaMaximaAux m v [] = max m v

sumaMaximaAux m v (x:xs)

| x >= 0 = sumaMaximaAux m (v+x) xs

| v+x > 0 = sumaMaximaAux (max m v) (v+x) xs

| otherwise = sumaMaximaAux (max m v) 0 xs

-- 3ª definición

-- =============

sumaMaxima3 :: [Integer] -> Integer

sumaMaxima3 [] = 0

sumaMaxima3 xs = maximum (map sum (segmentos xs))

-- (segmentos xs) es la lista de los segmentos de xs. Por ejemplo

-- segmentos "abc" == ["", "a","ab","abc","b","bc","c"]

segmentos :: [a] -> [[a]]

segmentos xs =

[] : concat [tail (inits ys) | ys <- init (tails xs)]

-- 4ª definición

-- =============

sumaMaxima4 :: [Integer] -> Integer

sumaMaxima4 [] = 0

sumaMaxima4 xs =

maximum (concat [scanl (+) 0 ys | ys <- tails xs])

-- Comprobación

-- ============

-- La propiedad es

prop_sumaMaxima :: [Integer] -> Bool

prop_sumaMaxima xs =

sumaMaxima2 xs == n &&

sumaMaxima3 xs == n &&

sumaMaxima4 xs == n

where n = sumaMaxima1 xs

-- La comprobación es

-- λ> quickCheck prop_sumaMaxima

-- +++ OK, passed 100 tests.

-- Comparación de eficiencia

-- =========================

-- λ> let n = 10^2 in sumaMaxima1 [-n..n]

-- 5050

-- (2.10 secs, 390,399,104 bytes)

-- λ> let n = 10^2 in sumaMaxima2 [-n..n]

-- 5050

-- (0.02 secs, 0 bytes)

-- λ> let n = 10^2 in sumaMaxima3 [-n..n]

-- 5050

-- (0.27 secs, 147,705,184 bytes)

-- λ> let n = 10^2 in sumaMaxima4 [-n..n]

-- 5050

-- (0.04 secs, 11,582,520 bytes)

prop_inversa :: [Integer] -> Bool

prop_inversa xs =

sumaMaxima2 xs == sumaMaxima2 (reverse xs)

Máxima suma en una matriz

PorJosé A. Alonso 11-marzo-201628-marzo-2016

Las matrices puede representarse mediante tablas cuyos índices son pares de números naturales:

   type Matriz = Array (Int,Int) Int

1	type Matriz = Array (Int,Int) Int

Definir la función

   maximaSuma :: Matriz -> Int

1	maximaSuma :: Matriz -> Int

tal que (maximaSuma p) es el máximo de las sumas de las listas de elementos de la matriz p tales que cada elemento pertenece sólo a una fila y a una columna. Por ejemplo,

   ghci> maximaSuma (listArray ((1,1),(3,3)) [1,2,3,8,4,9,5,6,7])
   17

1 2	ghci> maximaSuma (listArray ((1,1),(3,3)) [1,2,3,8,4,9,5,6,7]) 17

ya que las selecciones, y sus sumas, de la matriz

   |1 2 3|
   |8 4 9|
   |5 6 7|

|1 2 3|

|8 4 9|

|5 6 7|

son

   [1,4,7] --> 12
   [1,9,6] --> 16
   [2,8,7] --> 17
   [2,9,5] --> 16
   [3,8,6] --> 17
   [3,4,5] --> 12

[1,4,7] --> 12

[1,9,6] --> 16

[2,8,7] --> 17

[2,9,5] --> 16

[3,8,6] --> 17

[3,4,5] --> 12

Hay dos selecciones con máxima suma: [2,8,7] y [3,8,6].

Soluciones

import Data.Array
import Data.List (permutations)

type Matriz = Array (Int,Int) Int

maximaSuma :: Matriz -> Int
maximaSuma p = maximum [sum xs | xs <- selecciones p]

-- (selecciones p) es la lista de las selecciones en las que cada
-- elemento pertenece a un única fila y a una única columna de la matriz
-- p. Por ejemplo,
--    ghci> selecciones (listArray ((1,1),(3,3)) [1,2,3,8,4,9,5,6,7])
--    [[1,4,7],[2,8,7],[3,4,5],[2,9,5],[3,8,6],[1,9,6]]
selecciones :: Matriz -> [[Int]]
selecciones p = 
    [[p!(i,j) | (i,j) <- ijs] | 
     ijs <- [zip [1..n] xs | xs <- permutations [1..n]]] 
    where (_,(m,n)) = bounds p

-- 2ª solución (mediante submatrices):
maximaSuma2 :: Matriz -> Int
maximaSuma2 p 
    | (m,n) == (1,1) = p!(1,1)
    | otherwise = maximum [p!(1,j) 
                  + maximaSuma2 (submatriz 1 j p) | j <- [1..n]]
    where (_,(m,n)) = bounds p

-- (submatriz i j p) es la matriz obtenida a partir de la p eliminando
-- la fila i y la columna j. Por ejemplo, 
--    ghci> submatriz 2 3 (listArray ((1,1),(3,3)) [1,2,3,8,4,9,5,6,7])
--    array ((1,1),(2,2)) [((1,1),1),((1,2),2),((2,1),5),((2,2),6)]
submatriz :: Int -> Int -> Matriz -> Matriz
submatriz i j p = 
    array ((1,1), (m-1,n -1))
          [((k,l), p ! f k l) | k <- [1..m-1], l <- [1.. n-1]]
    where (_,(m,n)) = bounds p
          f k l | k < i  && l < j  = (k,l)
                | k >= i && l < j  = (k+1,l)
                | k < i  && l >= j = (k,l+1)
                | otherwise        = (k+1,l+1)

import Data.Array

import Data.List (permutations)

type Matriz = Array (Int,Int) Int

maximaSuma :: Matriz -> Int

maximaSuma p = maximum [sum xs | xs <- selecciones p]

-- (selecciones p) es la lista de las selecciones en las que cada

-- elemento pertenece a un única fila y a una única columna de la matriz

-- p. Por ejemplo,

-- ghci> selecciones (listArray ((1,1),(3,3)) [1,2,3,8,4,9,5,6,7])

-- [[1,4,7],[2,8,7],[3,4,5],[2,9,5],[3,8,6],[1,9,6]]

selecciones :: Matriz -> [[Int]]

selecciones p =

[[p!(i,j) | (i,j) <- ijs] |

ijs <- [zip [1..n] xs | xs <- permutations [1..n]]]

where (_,(m,n)) = bounds p

-- 2ª solución (mediante submatrices):

maximaSuma2 :: Matriz -> Int

maximaSuma2 p

| (m,n) == (1,1) = p!(1,1)

| otherwise = maximum [p!(1,j)

+ maximaSuma2 (submatriz 1 j p) | j <- [1..n]]

where (_,(m,n)) = bounds p

-- (submatriz i j p) es la matriz obtenida a partir de la p eliminando

-- la fila i y la columna j. Por ejemplo,

-- ghci> submatriz 2 3 (listArray ((1,1),(3,3)) [1,2,3,8,4,9,5,6,7])

-- array ((1,1),(2,2)) [((1,1),1),((1,2),2),((2,1),5),((2,2),6)]

submatriz :: Int -> Int -> Matriz -> Matriz

submatriz i j p =

array ((1,1), (m-1,n -1))

[((k,l), p ! f k l) | k <- [1..m-1], l <- [1.. n-1]]

where (_,(m,n)) = bounds p

f k l | k < i && l < j = (k,l)

| k >= i && l < j = (k+1,l)

| k < i && l >= j = (k,l+1)

| otherwise = (k+1,l+1)

Máxima longitud de las sublistas comunes

PorJosé A. Alonso 9-marzo-201616-marzo-2016

Las sublistas comunes de «1325» y «36572» son «», «3»,»32″, «35», «2» y «5». El máximo de sus longitudes es 2.

Definir la función

   maximo :: Eq a => [a] -> [a] -> Int

1	maximo :: Eq a => [a] -> [a] -> Int

tal que (maximo xs ys) es el máximo de las longitudes de las sublistas comunes de xs e ys. Por ejemplo,

   maximo "1325" "36572"       == 2
   maximo [1,4..33] [2,4..33]  == 5
   maximo [1..10^6] [1..10^6]  == 100000

maximo "1325" "36572" == 2

maximo [1,4..33] [2,4..33] == 5

maximo [1..10^6] [1..10^6] == 100000

Soluciones

import Data.List (subsequences, intersect)

-- 1ª definición
-- =============

maximo1 :: Eq a => [a] -> [a] -> Int
maximo1 xs ys = 
    maximum (map length (sublistasComunes xs ys))

-- (sublistasComunes xs ys) es la lista de las sublistas comunes de xs e
-- ys. Por ejemplo,
sublistasComunes :: Eq a => [a] -> [a] -> [[a]]
sublistasComunes xs ys =
    subsequences xs `intersect` subsequences ys

-- 2ª definición
-- =============

maximo2 :: Eq a => [a] -> [a] -> Int
maximo2 l1@(x:xs) l2@(y:ys) 
    | x == y    = 1 + maximo2 xs ys
    | otherwise = max (maximo2 xs l2) (maximo2 l1 ys)  
maximo2 _ _ = 0

-- Comparación de eficiencia
-- =========================

--    λ> maximo1 [1,4..30] [2,4..30]
--    5
--    (3.60 secs, 0 bytes)
--    λ> maximo2 [1,4..30] [2,4..30]
--    5
--    (1.58 secs, 216,347,472 bytes)
--    
--    λ> maximo2 [1..10^6] [1..10^6]
--    1000000
--    (2.44 secs, 407,051,096 bytes)

import Data.List (subsequences, intersect)

-- 1ª definición

-- =============

maximo1 :: Eq a => [a] -> [a] -> Int

maximo1 xs ys =

maximum (map length (sublistasComunes xs ys))

-- (sublistasComunes xs ys) es la lista de las sublistas comunes de xs e

-- ys. Por ejemplo,

sublistasComunes :: Eq a => [a] -> [a] -> [[a]]

sublistasComunes xs ys =

subsequences xs `intersect` subsequences ys

-- 2ª definición

-- =============

maximo2 :: Eq a => [a] -> [a] -> Int

maximo2 l1@(x:xs) l2@(y:ys)

| x == y = 1 + maximo2 xs ys

| otherwise = max (maximo2 xs l2) (maximo2 l1 ys)

maximo2 _ _ = 0

-- Comparación de eficiencia

-- =========================

-- λ> maximo1 [1,4..30] [2,4..30]

-- 5

-- (3.60 secs, 0 bytes)

-- λ> maximo2 [1,4..30] [2,4..30]

-- 5

-- (1.58 secs, 216,347,472 bytes)

-- λ> maximo2 [1..10^6] [1..10^6]

-- 1000000

-- (2.44 secs, 407,051,096 bytes)

Regresión lineal

PorJosé A. Alonso 22-febrero-201626-marzo-2022

Dadas dos listas de valores

   xs = [x(1), x(2), ..., x(n)]
   ys = [y(1), y(2), ..., y(n)]

1 2	xs = [x(1), x(2), ..., x(n)] ys = [y(1), y(2), ..., y(n)]

la ecuación de la recta de regresión de ys sobre xs es y = a+bx, donde

   b = (nΣx(i)y(i) - Σx(i)Σy(i)) / (nΣx(i)² - (Σx(i))²)
   a = (Σy(i) - bΣx(i)) / n

1 2	b = (nΣx(i)y(i) - Σx(i)Σy(i)) / (nΣx(i)² - (Σx(i))²) a = (Σy(i) - bΣx(i)) / n

Definir la función

   regresionLineal :: [Double] -> [Double] -> (Double,Double)

1	regresionLineal :: [Double] -> [Double] -> (Double,Double)

tal que (regresionLineal xs ys) es el par (a,b) de los coeficientes de la recta de regresión de ys sobre xs. Por ejemplo, para los valores

   ejX, ejY :: [Double]
   ejX = [5,  7, 10, 12, 16, 20, 23, 27, 19, 14]
   ejY = [9, 11, 15, 16, 20, 24, 27, 29, 22, 20]

ejX, ejY :: [Double]

ejX = [5, 7, 10, 12, 16, 20, 23, 27, 19, 14]

ejY = [9, 11, 15, 16, 20, 24, 27, 29, 22, 20]

se tiene

   λ> regresionLineal ejX ejY
   (5.195045748716805,0.9218924347243919)

1 2	λ> regresionLineal ejX ejY (5.195045748716805,0.9218924347243919)

Definir el procedimiento

   grafica :: [Double] -> [Double] -> IO ()

1	grafica :: [Double] -> [Double] -> IO ()

tal que (grafica xs ys) pinte los puntos correspondientes a las listas de valores xs e ys y su recta de regresión. Por ejemplo, con (grafica ejX ejY) se obtiene el siguiente dibujo

Soluciones

import Graphics.Gnuplot.Simple

ejX, ejY :: [Double]
ejX = [5,  7, 10, 12, 16, 20, 23, 27, 19, 14]
ejY = [9, 11, 15, 16, 20, 24, 27, 29, 22, 20]

regresionLineal :: [Double] -> [Double] -> (Double,Double)
regresionLineal xs ys = (a,b)
    where n     = fromIntegral (length xs)
          sumX  = sum xs
          sumY  = sum ys
          sumX2 = sum (zipWith (*) xs xs)
          sumY2 = sum (zipWith (*) ys ys)
          sumXY = sum (zipWith (*) xs ys)
          b     = (n*sumXY - sumX*sumY) / (n*sumX2 - sumX^2)
          a     = (sumY - b*sumX) / n

grafica :: [Double] -> [Double] -> IO ()
grafica xs ys = 
    plotPathsStyle 
      [YRange (0,10+mY)] 
      [(defaultStyle {plotType = Points,
                      lineSpec = CustomStyle [LineTitle "Datos",
                                              PointType 2,
                                              PointSize 2.5]},
                     zip xs ys),
       (defaultStyle {plotType = Lines,
                      lineSpec = CustomStyle [LineTitle "Ajuste",
                                              LineWidth 2]},
                     [(x,a+b*x) | x <- [0..mX]])]
    where (a,b) = regresionLineal xs ys                      
          mX    = maximum xs 
          mY    = maximum ys

import Graphics.Gnuplot.Simple

ejX, ejY :: [Double]

ejX = [5, 7, 10, 12, 16, 20, 23, 27, 19, 14]

ejY = [9, 11, 15, 16, 20, 24, 27, 29, 22, 20]

regresionLineal :: [Double] -> [Double] -> (Double,Double)

regresionLineal xs ys = (a,b)

where n = fromIntegral (length xs)

sumX = sum xs

sumY = sum ys

sumX2 = sum (zipWith (*) xs xs)

sumY2 = sum (zipWith (*) ys ys)

sumXY = sum (zipWith (*) xs ys)

b = (n*sumXY - sumX*sumY) / (n*sumX2 - sumX^2)

a = (sumY - b*sumX) / n

grafica :: [Double] -> [Double] -> IO ()

grafica xs ys =

plotPathsStyle

[YRange (0,10+mY)]

[(defaultStyle {plotType = Points,

lineSpec = CustomStyle [LineTitle "Datos",

PointType 2,

PointSize 2.5]},

zip xs ys),

(defaultStyle {plotType = Lines,

lineSpec = CustomStyle [LineTitle "Ajuste",

LineWidth 2]},

[(x,a+b*x) | x <- [0..mX]])]

where (a,b) = regresionLineal xs ys

mX = maximum xs

mY = maximum ys

Máxima suma de elementos consecutivos

Soluciones

Máxima suma en una matriz

Soluciones

Máxima longitud de las sublistas comunes

Soluciones

Regresión lineal

Soluciones

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