{"id":7800,"date":"2022-09-24T07:22:02","date_gmt":"2022-09-24T05:22:02","guid":{"rendered":"https:\/\/www.glc.us.es\/~jalonso\/vestigium\/?p=7800"},"modified":"2022-09-24T07:22:15","modified_gmt":"2022-09-24T05:22:15","slug":"pfh-la-semana-en-exercitium-23-de-septiembre-de-2022","status":"publish","type":"post","link":"https:\/\/www.glc.us.es\/~jalonso\/vestigium\/pfh-la-semana-en-exercitium-23-de-septiembre-de-2022\/","title":{"rendered":"PFH: La semana en Exercitium (23 de septiembre de 2022)"},"content":{"rendered":"<p>Esta semana he publicado en <a href=\"http:\/\/bit.ly\/2sqPtGs\">Exercitium<\/a> las soluciones de los siguientes problemas:<\/p>\n<ul>\n<li><a href=\"#ej1\">1. Igualdad de conjuntos<\/a><\/li>\n<li><a href=\"#ej2\">2. Uni\u00f3n conjuntista de listas<\/a><\/li>\n<li><a href=\"#ej3\">3. Intersecci\u00f3n conjuntista de listas<\/a><\/li>\n<li><a href=\"#ej4\">4. Diferencia conjuntista de listas<\/a><\/li>\n<li><a href=\"#ej5\">5. Divisores de un n\u00famero<\/a><\/li>\n<li><a href=\"#ej6\">6. Divisores primos<\/a><\/li>\n<\/ul>\n<p>A continuaci\u00f3n se muestran las soluciones.<br \/>\n<!--more--><br \/>\n<a name=\"ej1\"><\/a><\/p>\n<h3>1. Igualdad de conjuntos<\/h3>\n<p>Definir la funci\u00f3n<\/p>\n<pre lang=\"text\">\n   iguales :: Ord a => [a] -> [a] -> Bool\n<\/pre>\n<p>tal que <code>iguales xs ys<\/code> se verifica si <code>xs<\/code> e <code>ys<\/code> son iguales como conjuntos. Por ejemplo,<\/p>\n<pre lang=\"text\">\n   iguales [3,2,3] [2,3]    ==  True\n   iguales [3,2,3] [2,3,2]  ==  True\n   iguales [3,2,3] [2,3,4]  ==  False\n   iguales [2,3] [4,5]      ==  False\n<\/pre>\n<p><b>Soluciones en Haskell<\/b><\/p>\n<pre lang=\"haskell\">\nimport Data.List (nub, sort)\nimport Data.Set (fromList)\nimport Test.QuickCheck\n\n-- 1\u00aa soluci\u00f3n\n-- ===========\n\niguales1 :: Ord a => [a] -> [a] -> Bool\niguales1 xs ys =\n  subconjunto xs ys && subconjunto ys xs\n\n-- (subconjunto xs ys) se verifica si xs es un subconjunto de ys. Por\n-- ejemplo,\n--    subconjunto [3,2,3] [2,5,3,5]  ==  True\n--    subconjunto [3,2,3] [2,5,6,5]  ==  False\nsubconjunto :: Ord a => [a] -> [a] -> Bool\nsubconjunto xs ys =\n  [x | x <- xs, x `elem` ys] == xs\n\n-- 2\u00aa soluci\u00f3n\n-- ===========\n\niguales2 :: Ord a => [a] -> [a] -> Bool\niguales2 xs ys =\n  nub (sort xs) == nub (sort ys)\n\n-- 3\u00aa soluci\u00f3n\n-- ===========\n\niguales3 :: Ord a => [a] -> [a] -> Bool\niguales3 xs ys =\n  fromList xs == fromList ys\n\n-- Comprobaci\u00f3n de equivalencia\n-- ============================\n\n-- La propiedad es\nprop_iguales :: [Int] -> [Int] -> Bool\nprop_iguales xs ys =\n  all (== iguales1 xs ys)\n      [iguales2 xs ys,\n       iguales3 xs ys]\n\n-- La comprobaci\u00f3n es\n--    \u03bb> quickCheck prop_iguales\n--    +++ OK, passed 100 tests.\n\n-- Comparaci\u00f3n de eficiencia\n-- =========================\n\n-- La comparaci\u00f3n es\n--    \u03bb> iguales1 [1..2*10^4] [1..2*10^4]\n--    True\n--    (4.05 secs, 8,553,104 bytes)\n--    \u03bb> iguales2 [1..2*10^4] [1..2*10^4]\n--    True\n--    (4.14 secs, 9,192,768 bytes)\n--    \u03bb> iguales3 [1..2*10^4] [1..2*10^4]\n--    True\n--    (0.01 secs, 8,552,232 bytes)\n<\/pre>\n<p>El c\u00f3digo se encuentra en <a href=\"https:\/\/github.com\/jaalonso\/Exercitium\/blob\/main\/src\/Igualdad_de_conjuntos.hs\">GitHub<\/a>.<\/p>\n<p><b>Soluciones en Python<\/b><\/p>\n<pre lang=\"python\">\nfrom typing import Any\nfrom timeit import Timer, default_timer\nfrom hypothesis import given, strategies as st\n\n# 1\u00aa soluci\u00f3n\n# ===========\n\ndef subconjunto(xs: list[Any],\n                ys: list[Any]) -> bool:\n    return [x for x in xs if x in ys] == xs\n\ndef iguales1(xs: list[Any],\n             ys: list[Any]) -> bool:\n    return subconjunto(xs, ys) and subconjunto(ys, xs)\n\n# 2\u00aa soluci\u00f3n\n# ===========\n\ndef iguales2(xs: list[Any],\n             ys: list[Any]) -> bool:\n    return set(xs) == set(ys)\n\n# Equivalencia de las definiciones\n# ================================\n\n# La propiedad es\n@given(st.lists(st.integers()),\n       st.lists(st.integers()))\ndef test_iguales(xs, ys):\n    assert iguales1(xs, ys) == iguales2(xs, ys)\n\n# La comprobaci\u00f3n es\n#    src> poetry run pytest -q igualdad_de_conjuntos.py\n#    1 passed in 0.28s\n\n# Comparaci\u00f3n de eficiencia\n# =========================\n\ndef tiempo(e):\n    \"\"\"Tiempo (en segundos) de evaluar la expresi\u00f3n e.\"\"\"\n    t = Timer(e, \"\", default_timer, globals()).timeit(1)\n    print(f\"{t:0.2f} segundos\")\n\n# La comparaci\u00f3n es\n#    >>> xs = list(range(20000))\n#    >>> tiempo('iguales1(xs, xs)')\n#    2.71 segundos\n#    >>> tiempo('iguales2(xs, xs)')\n#    0.01 segundos\n<\/pre>\n<p>El c\u00f3digo se encuentra en <a href=\"https:\/\/github.com\/jaalonso\/Exercitium-Python\/blob\/main\/src\/igualdad_de_conjuntos.py\">GitHub<\/a><\/p>\n<p><a name=\"ej2\"><\/a><\/p>\n<h3>2. Uni\u00f3n conjuntista de listas<\/h3>\n<p>Definir la funci\u00f3n<\/p>\n<pre lang=\"text\">\n   union :: Ord a => [a] -> [a] -> [a]\n<\/pre>\n<p>tal que <code>union xs ys<\/code> es la uni\u00f3n de las listas, sin elementos repetidos, <code>xs<\/code> e <code>ys<\/code>. Por ejemplo,<\/p>\n<pre lang=\"text\">\n   union [3,2,5] [5,7,3,4]  ==  [3,2,5,7,4]\n<\/pre>\n<p>Comprobar con QuickCheck que la uni\u00f3n es conmutativa.<\/p>\n<p><b>Soluciones en Haskell<\/b><\/p>\n<pre lang=\"haskell\">\nimport Data.List (nub, sort, union)\nimport qualified Data.Set as S (fromList, toList, union)\nimport Test.QuickCheck\n\n-- 1\u00aa soluci\u00f3n\n-- ===========\n\nunion1 :: Ord a => [a] -> [a] -> [a]\nunion1 xs ys = xs ++ [y | y <- ys, y `notElem` xs]\n\n-- 2\u00aa soluci\u00f3n\n-- ===========\n\nunion2 :: Ord a => [a] -> [a] -> [a]\nunion2 [] ys = ys\nunion2 (x:xs) ys\n  | x `elem` ys = xs `union2` ys\n  | otherwise   = x : xs `union2` ys\n\n-- 3\u00aa soluci\u00f3n\n-- ===========\n\nunion3 :: Ord a => [a] -> [a] -> [a]\nunion3 = union\n\n-- 4\u00aa soluci\u00f3n\n-- ===========\n\nunion4 :: Ord a => [a] -> [a] -> [a]\nunion4 xs ys =\n  S.toList (S.fromList xs `S.union` S.fromList ys)\n\n-- Comprobaci\u00f3n de equivalencia\n-- ============================\n\n-- La propiedad es\nprop_union :: [Int] -> [Int] -> Bool\nprop_union xs ys =\n  all (== sort (xs' `union1` ys'))\n      [sort (xs' `union2` ys'),\n       sort (xs' `union3` ys'),\n       xs' `union4` ys']\n  where xs' = nub xs\n        ys' = nub ys\n\n-- La comprobaci\u00f3n es\n--    \u03bb> quickCheck prop_union\n--    +++ OK, passed 100 tests.\n\n-- Comparaci\u00f3n de eficiencia\n-- =========================\n\n-- La comparaci\u00f3n es\n--    \u03bb> length (union1 [0,2..3*10^4] [1,3..3*10^4])\n--    30001\n--    (2.37 secs, 7,153,536 bytes)\n--    \u03bb> length (union2 [0,2..3*10^4] [1,3..3*10^4])\n--    30001\n--    (2.38 secs, 6,553,752 bytes)\n--    \u03bb> length (union3 [0,2..3*10^4] [1,3..3*10^4])\n--    30001\n--    (11.56 secs, 23,253,553,472 bytes)\n--    \u03bb> length (union4 [0,2..3*10^4] [1,3..3*10^4])\n--    30001\n--    (0.04 secs, 10,992,056 bytes)\n\n-- Comprobaci\u00f3n de la propiedad\n-- ============================\n\n-- La propiedad es\nprop_union_conmutativa :: [Int] -> [Int] -> Bool\nprop_union_conmutativa xs ys =\n  iguales (xs `union1` ys) (ys `union1` xs)\n\n-- (iguales xs ys) se verifica si xs e ys son iguales. Por ejemplo,\n--    iguales [3,2,3] [2,3]    ==  True\n--    iguales [3,2,3] [2,3,2]  ==  True\n--    iguales [3,2,3] [2,3,4]  ==  False\n--    iguales [2,3] [4,5]      ==  False\niguales :: Ord a => [a] -> [a] -> Bool\niguales xs ys =\n  S.fromList xs == S.fromList ys\n\n-- La comprobaci\u00f3n es\n--    \u03bb> quickCheck prop_union_conmutativa\n--    +++ OK, passed 100 tests.\n<\/pre>\n<p>El c\u00f3digo se encuentra en <a href=\"https:\/\/github.com\/jaalonso\/Exercitium\/blob\/main\/src\/Union_conjuntista_de_listas.hs\">GitHub<\/a>.<\/p>\n<p><b>Soluciones en Python<\/b><\/p>\n<pre lang=\"python\">\nfrom typing import TypeVar\nfrom timeit import Timer, default_timer\nfrom sys import setrecursionlimit\nfrom hypothesis import given, strategies as st\n\nsetrecursionlimit(10**6)\n\nA = TypeVar('A')\n\n# 1\u00aa soluci\u00f3n\n# ===========\n\ndef union1(xs: list[A], ys: list[A]) -> list[A]:\n    return xs + [y for y in ys if y not in xs]\n\n# 2\u00aa soluci\u00f3n\n# ===========\n\ndef union2(xs: list[A], ys: list[A]) -> list[A]:\n    if not xs:\n        return ys\n    if xs[0] in ys:\n        return union2(xs[1:], ys)\n    return [xs[0]] + union2(xs[1:], ys)\n\n# 3\u00aa soluci\u00f3n\n# ===========\n\ndef union3(xs: list[A], ys: list[A]) -> list[A]:\n    zs = ys[:]\n    for x in xs:\n        if x not in ys:\n            zs.append(x)\n    return zs\n\n# 4\u00aa soluci\u00f3n\n# ===========\n\ndef union4(xs: list[A], ys: list[A]) -> list[A]:\n    return list(set(xs) | set(ys))\n\n# Comprobaci\u00f3n de equivalencia\n# ============================\n#\n# La propiedad es\n@given(st.lists(st.integers()),\n       st.lists(st.integers()))\ndef test_union(xs, ys):\n    xs1 = list(set(xs))\n    ys1 = list(set(ys))\n    assert sorted(union1(xs1, ys1)) ==\\\n           sorted(union2(xs1, ys1)) ==\\\n           sorted(union3(xs1, ys1)) ==\\\n           sorted(union4(xs1, ys1))\n\n# La comprobaci\u00f3n es\n#    src> poetry run pytest -q union_conjuntista_de_listas.py\n#    1 passed in 0.36s\n\n# Comparaci\u00f3n de eficiencia\n# =========================\n\ndef tiempo(e):\n    \"\"\"Tiempo (en segundos) de evaluar la expresi\u00f3n e.\"\"\"\n    t = Timer(e, \"\", default_timer, globals()).timeit(1)\n    print(f\"{t:0.2f} segundos\")\n\n# La comparaci\u00f3n es\n#    >>> tiempo('union1(list(range(0,30000,2)), list(range(1,30000,2)))')\n#    1.30 segundos\n#    >>> tiempo('union2(list(range(0,30000,2)), list(range(1,30000,2)))')\n#    2.84 segundos\n#    >>> tiempo('union3(list(range(0,30000,2)), list(range(1,30000,2)))')\n#    1.45 segundos\n#    >>> tiempo('union4(list(range(0,30000,2)), list(range(1,30000,2)))')\n#    0.00 segundos\n\n# Comprobaci\u00f3n de la propiedad\n# ============================\n\n# iguales(xs, ys) se verifica si xs e ys son iguales como conjuntos. Por\n# ejemplo,\n#    iguales([3,2,3], [2,3])    ==  True\n#    iguales([3,2,3], [2,3,2])  ==  True\n#    iguales([3,2,3], [2,3,4])  ==  False\n#    iguales([2,3], [4,5])      ==  False\ndef iguales(xs: list[A], ys: list[A]) -> bool:\n    return set(xs) == set(ys)\n\n# La propiedad es\n@given(st.lists(st.integers()),\n       st.lists(st.integers()))\ndef test_union_conmutativa(xs, ys):\n    xs1 = list(set(xs))\n    ys1 = list(set(ys))\n    assert iguales(union1(xs1, ys1), union1(ys1, xs1))\n\n# La comprobaci\u00f3n es\n#    src> poetry run pytest -q union_conjuntista_de_listas.py\n#    2 passed in 0.49s\n<\/pre>\n<p>El c\u00f3digo se encuentra en <a href=\"https:\/\/github.com\/jaalonso\/Exercitium-Python\/blob\/main\/src\/union_conjuntista_de_listas.py\">GitHub<\/a><\/p>\n<p><a name=\"ej3\"><\/a><\/p>\n<h3>3. Intersecci\u00f3n conjuntista de listas<\/h3>\n<p>Definir la funci\u00f3n<\/p>\n<pre lang=\"text\">\n   interseccion :: Eq a => [a] -> [a] -> [a]\n<\/pre>\n<p>tal que <code>interseccion xs ys<\/code> es la intersecci\u00f3n de las listas sin elementos repetidos <code>xs<\/code> e <code>ys<\/code>. Por ejemplo,<\/p>\n<pre lang=\"text\">\n   interseccion [3,2,5] [5,7,3,4]  ==  [3,5]\n   interseccion [3,2,5] [9,7,6,4]  ==  []\n<\/pre>\n<p><b>Soluciones en Haskell<\/b><\/p>\n<pre lang=\"haskell\">\nimport Data.List (nub, sort, intersect)\nimport qualified Data.Set as S (fromList, toList, intersection )\nimport Test.QuickCheck\n\n-- 1\u00aa soluci\u00f3n\n-- ===========\n\ninterseccion1 :: Eq a => [a] -> [a] -> [a]\ninterseccion1 xs ys =\n  [x | x <- xs, x `elem` ys]\n\n-- 2\u00aa soluci\u00f3n\n-- ===========\n\ninterseccion2 :: Ord a => [a] -> [a] -> [a]\ninterseccion2 [] _ = []\ninterseccion2 (x:xs) ys\n  | x `elem` ys = x : xs `interseccion2` ys\n  | otherwise   = xs `interseccion2` ys\n\n-- 3\u00aa soluci\u00f3n\n-- ===========\n\ninterseccion3 :: Ord a => [a] -> [a] -> [a]\ninterseccion3 = intersect\n\n-- 4\u00aa soluci\u00f3n\n-- ===========\n\ninterseccion4 :: Ord a => [a] -> [a] -> [a]\ninterseccion4 xs ys =\n  S.toList (S.fromList xs `S.intersection` S.fromList ys)\n\n-- Comprobaci\u00f3n de equivalencia\n-- ============================\n\n-- La propiedad es\nprop_interseccion :: [Int] -> [Int] -> Bool\nprop_interseccion xs ys =\n  all (== sort (xs' `interseccion1` ys'))\n      [sort (xs' `interseccion2` ys'),\n       sort (xs' `interseccion3` ys'),\n       xs' `interseccion4` ys']\n  where xs' = nub xs\n        ys' = nub ys\n\n-- La comprobaci\u00f3n es\n--    \u03bb> quickCheck prop_interseccion\n--    +++ OK, passed 100 tests.\n\n-- Comparaci\u00f3n de eficiencia\n-- =========================\n\n-- La comparaci\u00f3n es\n--    \u03bb> length (interseccion1 [0..3*10^4] [1,3..3*10^4])\n--    15000\n--    (2.94 secs, 6,673,360 bytes)\n--    \u03bb> length (interseccion2 [0..3*10^4] [1,3..3*10^4])\n--    15000\n--    (3.04 secs, 9,793,440 bytes)\n--    \u03bb> length (interseccion3 [0..3*10^4] [1,3..3*10^4])\n--    15000\n--    (5.39 secs, 6,673,472 bytes)\n--    \u03bb> length (interseccion4 [0..3*10^4] [1,3..3*10^4])\n--    15000\n--    (0.04 secs, 8,593,176 bytes)\n<\/pre>\n<p>El c\u00f3digo se encuentra en <a href=\"https:\/\/github.com\/jaalonso\/Exercitium\/blob\/main\/src\/Interseccion_conjuntista_de_listas.hs\">GitHub<\/a>.<\/p>\n<p><b>Soluciones en Python<\/b><\/p>\n<pre lang=\"python\">\nfrom typing import TypeVar\nfrom timeit import Timer, default_timer\nfrom sys import setrecursionlimit\nfrom hypothesis import given, strategies as st\n\nsetrecursionlimit(10**6)\n\nA = TypeVar('A')\n\n# 1\u00aa soluci\u00f3n\n# ===========\n\ndef interseccion1(xs: list[A], ys: list[A]) -> list[A]:\n    return [x for x in xs if x in ys]\n\n# 2\u00aa soluci\u00f3n\n# ===========\n\ndef interseccion2(xs: list[A], ys: list[A]) -> list[A]:\n    if not xs:\n        return []\n    if xs[0] in ys:\n        return [xs[0]] + interseccion2(xs[1:], ys)\n    return interseccion2(xs[1:], ys)\n\n# 3\u00aa soluci\u00f3n\n# ===========\n\ndef interseccion3(xs: list[A], ys: list[A]) -> list[A]:\n    zs = []\n    for x in xs:\n        if x in ys:\n            zs.append(x)\n    return zs\n\n# 4\u00aa soluci\u00f3n\n# ===========\n\ndef interseccion4(xs: list[A], ys: list[A]) -> list[A]:\n    return list(set(xs) & set(ys))\n\n# Comprobaci\u00f3n de equivalencia\n# ============================\n#\n# La propiedad es\n@given(st.lists(st.integers()),\n       st.lists(st.integers()))\ndef test_interseccion(xs, ys):\n    xs1 = list(set(xs))\n    ys1 = list(set(ys))\n    assert sorted(interseccion1(xs1, ys1)) ==\\\n           sorted(interseccion2(xs1, ys1)) ==\\\n           sorted(interseccion3(xs1, ys1)) ==\\\n           sorted(interseccion4(xs1, ys1))\n\n# La comprobaci\u00f3n es\n#    src> poetry run pytest -q interseccion_conjuntista_de_listas.py\n#    1 passed in 0.33s\n\n# Comparaci\u00f3n de eficiencia\n# =========================\n\ndef tiempo(e):\n    \"\"\"Tiempo (en segundos) de evaluar la expresi\u00f3n e.\"\"\"\n    t = Timer(e, \"\", default_timer, globals()).timeit(1)\n    print(f\"{t:0.2f} segundos\")\n\n# La comparaci\u00f3n es\n#    >>> tiempo('interseccion1(list(range(0,20000)), list(range(1,20000,2)))')\n#    0.98 segundos\n#    >>> tiempo('interseccion2(list(range(0,20000)), list(range(1,20000,2)))')\n#    2.13 segundos\n#    >>> tiempo('interseccion3(list(range(0,20000)), list(range(1,20000,2)))')\n#    0.87 segundos\n#    >>> tiempo('interseccion4(list(range(0,20000)), list(range(1,20000,2)))')\n#    0.00 segundos\n<\/pre>\n<p>El c\u00f3digo se encuentra en <a href=\"https:\/\/github.com\/jaalonso\/Exercitium-Python\/blob\/main\/src\/interseccion_conjuntista_de_listas.py\">GitHub<\/a>.<\/p>\n<p><a name=\"ej4\"><\/a><\/p>\n<h3>4. Diferencia conjuntista de listas<\/h3>\n<p>Definir la funci\u00f3n<\/p>\n<pre lang=\"text\">\n   diferencia :: Eq a => [a] -> [a] -> [a]\n<\/pre>\n<p>tal que <code>diferencia xs ys<\/code> es la diferencia de las listas sin elementos repetidos <code>xs<\/code> e <code>ys<\/code>. Por ejemplo,<\/p>\n<pre lang=\"text\">\n   diferencia [3,2,5,6] [5,7,3,4]  ==  [2,6]\n   diferencia [3,2,5] [5,7,3,2]    ==  []\n<\/pre>\n<p><b>Soluciones en Haskell<\/b><\/p>\n<pre lang=\"haskell\">\nimport Data.List (nub, sort, (\\\\))\nimport qualified Data.Set as S (fromList, toList, (\\\\) )\nimport Test.QuickCheck\n\n-- 1\u00aa soluci\u00f3n\n-- ===========\n\ndiferencia1 :: Eq a => [a] -> [a] -> [a]\ndiferencia1 xs ys =\n  [x | x <- xs, x `notElem` ys]\n\n-- 2\u00aa soluci\u00f3n\n-- ===========\n\ndiferencia2 :: Ord a => [a] -> [a] -> [a]\ndiferencia2 [] _ = []\ndiferencia2 (x:xs) ys\n  | x `elem` ys = xs `diferencia2` ys\n  | otherwise   = x : xs `diferencia2` ys\n\n-- 3\u00aa soluci\u00f3n\n-- ===========\n\ndiferencia3 :: Ord a => [a] -> [a] -> [a]\ndiferencia3 = (\\\\)\n\n-- 4\u00aa soluci\u00f3n\n-- ===========\n\ndiferencia4 :: Ord a => [a] -> [a] -> [a]\ndiferencia4 xs ys =\n  S.toList (S.fromList xs S.\\\\ S.fromList ys)\n\n-- Comprobaci\u00f3n de equivalencia\n-- ============================\n\n-- La propiedad es\nprop_diferencia :: [Int] -> [Int] -> Bool\nprop_diferencia xs ys =\n  all (== sort (xs' `diferencia1` ys'))\n      [sort (xs' `diferencia2` ys'),\n       sort (xs' `diferencia3` ys'),\n       xs' `diferencia4` ys']\n  where xs' = nub xs\n        ys' = nub ys\n\n-- La comprobaci\u00f3n es\n--    \u03bb> quickCheck prop_diferencia\n--    +++ OK, passed 100 tests.\n\n-- Comparaci\u00f3n de eficiencia\n-- =========================\n\n-- La comparaci\u00f3n es\n--    \u03bb> length (diferencia1 [0..3*10^4] [1,3..3*10^4])\n--    15001\n--    (3.39 secs, 9,553,528 bytes)\n--    \u03bb> length (diferencia2 [0..3*10^4] [1,3..3*10^4])\n--    15001\n--    (2.98 secs, 9,793,528 bytes)\n--    \u03bb> length (diferencia3 [0..3*10^4] [1,3..3*10^4])\n--    15001\n--    (3.61 secs, 11,622,502,792 bytes)\n--    \u03bb> length (diferencia4 [0..3*10^4] [1,3..3*10^4])\n--    15001\n--    (0.02 secs, 10,092,832 bytes)\n<\/pre>\n<p>El c\u00f3digo se encuentra en <a href=\"https:\/\/github.com\/jaalonso\/Exercitium\/blob\/main\/src\/Diferencia_conjuntista_de_listas.hs\">GitHub<\/a>.<\/p>\n<p><b>Soluciones en Python<\/b><\/p>\n<pre lang=\"python\">\nfrom typing import TypeVar\nfrom timeit import Timer, default_timer\nfrom sys import setrecursionlimit\nfrom hypothesis import given, strategies as st\n\nsetrecursionlimit(10**6)\n\nA = TypeVar('A')\n\n# 1\u00aa soluci\u00f3n\n# ===========\n\ndef diferencia1(xs: list[A], ys: list[A]) -> list[A]:\n    return [x for x in xs if x not in ys]\n\n# 2\u00aa soluci\u00f3n\n# ===========\n\ndef diferencia2(xs: list[A], ys: list[A]) -> list[A]:\n    if not xs:\n        return []\n    if xs[0] in ys:\n        return diferencia2(xs[1:], ys)\n    return [xs[0]] + diferencia2(xs[1:], ys)\n\n# 3\u00aa soluci\u00f3n\n# ===========\n\ndef diferencia3(xs: list[A], ys: list[A]) -> list[A]:\n    zs = []\n    for x in xs:\n        if x not in ys:\n            zs.append(x)\n    return zs\n\n# 4\u00aa soluci\u00f3n\n# ===========\n\ndef diferencia4(xs: list[A], ys: list[A]) -> list[A]:\n    return list(set(xs) - set(ys))\n\n# Comprobaci\u00f3n de equivalencia\n# ============================\n#\n# La propiedad es\n@given(st.lists(st.integers()),\n       st.lists(st.integers()))\ndef test_diferencia(xs, ys):\n    xs1 = list(set(xs))\n    ys1 = list(set(ys))\n    assert sorted(diferencia1(xs1, ys1)) ==\\\n           sorted(diferencia2(xs1, ys1)) ==\\\n           sorted(diferencia3(xs1, ys1)) ==\\\n           sorted(diferencia4(xs1, ys1))\n\n# La comprobaci\u00f3n es\n#    src> poetry run pytest -q diferencia_conjuntista_de_listas.py\n#    1 passed in 0.39s\n\n# Comparaci\u00f3n de eficiencia\n# =========================\n\ndef tiempo(e):\n    \"\"\"Tiempo (en segundos) de evaluar la expresi\u00f3n e.\"\"\"\n    t = Timer(e, \"\", default_timer, globals()).timeit(1)\n    print(f\"{t:0.2f} segundos\")\n\n# La comparaci\u00f3n es\n#    >>> tiempo('diferencia1(list(range(0,20000)), list(range(1,20000,2)))')\n#    0.89 segundos\n#    >>> tiempo('diferencia2(list(range(0,20000)), list(range(1,20000,2)))')\n#    2.11 segundos\n#    >>> tiempo('diferencia3(list(range(0,20000)), list(range(1,20000,2)))')\n#    1.06 segundos\n#    >>> tiempo('diferencia4(list(range(0,20000)), list(range(1,20000,2)))')\n#    0.01 segundos\n<\/pre>\n<p>El c\u00f3digo se encuentra en <a href=\"https:\/\/github.com\/jaalonso\/Exercitium-Python\/blob\/main\/src\/diferencia_conjuntista_de_listas.py\">GitHub<\/a>.<\/p>\n<p><a name=\"ej5\"><\/a><\/p>\n<h3>5. Divisores de un n\u00famero<\/h3>\n<p>Definir la funci\u00f3n<\/p>\n<pre lang=\"text\">\n   divisores :: Integer -> [Integer]\n<\/pre>\n<p>tal que <code>divisores n<\/code> es la lista de los divisores de <code>n<\/code>. Por ejemplo,<\/p>\n<pre lang=\"text\">\n  divisores 30  ==  [1,2,3,5,6,10,15,30]\n  length (divisores (product [1..10]))  ==  270\n  length (divisores (product [1..25]))  ==  340032\n<\/pre>\n<p><b>Soluciones en Haskell<\/b><\/p>\n<pre lang=\"haskell\">\nimport Data.List (group, inits, nub, sort, subsequences)\nimport Data.Numbers.Primes (primeFactors)\nimport Data.Set (toList)\nimport Math.NumberTheory.ArithmeticFunctions (divisors)\nimport Test.QuickCheck\n\n-- 1\u00aa soluci\u00f3n\n-- ===========\n\ndivisores1 :: Integer -> [Integer]\ndivisores1 n = [x | x <- [1..n], n `rem` x == 0]\n\n-- 2\u00aa soluci\u00f3n\n-- ===========\n\ndivisores2 :: Integer -> [Integer]\ndivisores2 n = [x | x <- [1..n], x `esDivisorDe` n]\n\n-- (esDivisorDe x n) se verifica si x es un divisor de n. Por ejemplo,\n--    esDivisorDe 2 6  ==  True\n--    esDivisorDe 4 6  ==  False\nesDivisorDe :: Integer -> Integer -> Bool\nesDivisorDe x n = n `rem` x == 0\n\n-- 3\u00aa soluci\u00f3n\n-- ===========\n\ndivisores3 :: Integer -> [Integer]\ndivisores3 n = filter (`esDivisorDe` n) [1..n]\n\n-- 4\u00aa soluci\u00f3n\n-- ===========\n\ndivisores4 :: Integer -> [Integer]\ndivisores4 = filter <$> flip esDivisorDe <*> enumFromTo 1\n\n-- 5\u00aa soluci\u00f3n\n-- ===========\n\ndivisores5 :: Integer -> [Integer]\ndivisores5 n = xs ++ [n `div` y | y <- ys]\n  where xs = primerosDivisores1 n\n        (z:zs) = reverse xs\n        ys | z^2 == n  = zs\n           | otherwise = z:zs\n\n-- (primerosDivisores n) es la lista de los divisores del n\u00famero n cuyo\n-- cuadrado es menor o gual que n. Por ejemplo,\n--    primerosDivisores 25  ==  [1,5]\n--    primerosDivisores 30  ==  [1,2,3,5]\nprimerosDivisores1 :: Integer -> [Integer]\nprimerosDivisores1 n =\n   [x | x <- [1..round (sqrt (fromIntegral n))],\n        x `esDivisorDe` n]\n\n-- 6\u00aa soluci\u00f3n\n-- ===========\n\ndivisores6 :: Integer -> [Integer]\ndivisores6 n = aux [1..n]\n  where aux [] = []\n        aux (x:xs) | x `esDivisorDe` n = x : aux xs\n                   | otherwise         = aux xs\n\n-- 7\u00aa soluci\u00f3n\n-- ===========\n\ndivisores7 :: Integer -> [Integer]\ndivisores7 n = xs ++ [n `div` y | y <- ys]\n  where xs = primerosDivisores2 n\n        (z:zs) = reverse xs\n        ys | z^2 == n  = zs\n           | otherwise = z:zs\n\nprimerosDivisores2 :: Integer -> [Integer]\nprimerosDivisores2 n = aux [1..round (sqrt (fromIntegral n))]\n  where aux [] = []\n        aux (x:xs) | x `esDivisorDe` n = x : aux xs\n                   | otherwise         = aux xs\n\n-- 8\u00aa soluci\u00f3n\n-- ===========\n\ndivisores8 :: Integer -> [Integer]\ndivisores8 =\n  nub . sort . map product . subsequences . primeFactors\n\n-- 9\u00aa soluci\u00f3n\n-- ===========\n\ndivisores9 :: Integer -> [Integer]\ndivisores9 = sort\n           . map (product . concat)\n           . productoCartesiano\n           . map inits\n           . group\n           . primeFactors\n\n-- (productoCartesiano xss) es el producto cartesiano de los conjuntos\n-- xss. Por ejemplo,\n--    \u03bb> productoCartesiano [[1,3],[2,5],[6,4]]\n--    [[1,2,6],[1,2,4],[1,5,6],[1,5,4],[3,2,6],[3,2,4],[3,5,6],[3,5,4]]\nproductoCartesiano :: [[a]] -> [[a]]\nproductoCartesiano []       = [[]]\nproductoCartesiano (xs:xss) =\n  [x:ys | x <- xs, ys <- productoCartesiano xss]\n\n-- 10\u00aa soluci\u00f3n\n-- ============\n\ndivisores10 :: Integer -> [Integer]\ndivisores10 = sort\n            . map (product . concat)\n            . mapM inits\n            . group\n            . primeFactors\n\n-- 11\u00aa soluci\u00f3n\n-- ============\n\ndivisores11 :: Integer -> [Integer]\ndivisores11 = toList . divisors\n\n-- Comprobaci\u00f3n de equivalencia\n-- ============================\n\n-- La propiedad es\nprop_divisores :: Positive Integer -> Bool\nprop_divisores (Positive n) =\n  all (== divisores1 n)\n      [ divisores2 n\n      , divisores3 n\n      , divisores4 n\n      , divisores5 n\n      , divisores6 n\n      , divisores7 n\n      , divisores8 n\n      , divisores9 n\n      , divisores10 n\n      , divisores11 n\n      ]\n\n-- La comprobaci\u00f3n es\n--    \u03bb> quickCheck prop_divisores\n--    +++ OK, passed 100 tests.\n\n-- Comparaci\u00f3n de la eficiencia\n-- ============================\n\n-- La comparaci\u00f3n es\n--    \u03bb> length (divisores1 (product [1..11]))\n--    540\n--    (18.55 secs, 7,983,950,592 bytes)\n--    \u03bb> length (divisores2 (product [1..11]))\n--    540\n--    (18.81 secs, 7,983,950,592 bytes)\n--    \u03bb> length (divisores3 (product [1..11]))\n--    540\n--    (12.79 secs, 6,067,935,544 bytes)\n--    \u03bb> length (divisores4 (product [1..11]))\n--    540\n--    (12.51 secs, 6,067,935,592 bytes)\n--    \u03bb> length (divisores5 (product [1..11]))\n--    540\n--    (0.03 secs, 1,890,296 bytes)\n--    \u03bb> length (divisores6 (product [1..11]))\n--    540\n--    (21.46 secs, 9,899,961,392 bytes)\n--    \u03bb> length (divisores7 (product [1..11]))\n--    540\n--    (0.02 secs, 2,195,800 bytes)\n--    \u03bb> length (divisores8 (product [1..11]))\n--    540\n--    (0.09 secs, 107,787,272 bytes)\n--    \u03bb> length (divisores9 (product [1..11]))\n--    540\n--    (0.02 secs, 2,150,472 bytes)\n--    \u03bb> length (divisores10 (product [1..11]))\n--    540\n--    (0.01 secs, 1,652,120 bytes)\n--    \u03bb> length (divisores11 (product [1..11]))\n--    540\n--    (0.01 secs, 796,056 bytes)\n--\n--    \u03bb> length (divisores5 (product [1..17]))\n--    10752\n--    (10.16 secs, 3,773,953,128 bytes)\n--    \u03bb> length (divisores7 (product [1..17]))\n--    10752\n--    (9.83 secs, 4,679,260,712 bytes)\n--    \u03bb> length (divisores9 (product [1..17]))\n--    10752\n--    (0.06 secs, 46,953,344 bytes)\n--    \u03bb> length (divisores10 (product [1..17]))\n--    10752\n--    (0.02 secs, 33,633,712 bytes)\n--    \u03bb> length (divisores11 (product [1..17]))\n--    10752\n--    (0.03 secs, 6,129,584 bytes)\n--\n--    \u03bb> length (divisores10 (product [1..27]))\n--    677376\n--    (2.14 secs, 3,291,277,736 bytes)\n--    \u03bb> length (divisores11 (product [1..27]))\n--    677376\n--    (0.56 secs, 396,042,280 bytes)\n<\/pre>\n<p>El c\u00f3digo se encuentra en <a href=\"https:\/\/github.com\/jaalonso\/Exercitium\/blob\/main\/src\/Divisores_de_un_numero.hs\">GitHub<\/a>.<\/p>\n<p><b>Soluciones en Python<\/b><\/p>\n<pre lang=\"python\">\nfrom math import factorial, sqrt\nfrom timeit import Timer, default_timer\nfrom sys import setrecursionlimit\nfrom sympy import divisors\nfrom hypothesis import given, strategies as st\n\nsetrecursionlimit(10**6)\n\n# 1\u00aa soluci\u00f3n\n# ===========\n\ndef divisores1(n: int) -> list[int]:\n    return [x for x in range(1, n + 1) if n % x == 0]\n\n# 2\u00aa soluci\u00f3n\n# ===========\n\n# esDivisorDe(x, n) se verifica si x es un divisor de n. Por ejemplo,\n#    esDivisorDe(2, 6)  ==  True\n#    esDivisorDe(4, 6)  ==  False\ndef esDivisorDe(x: int, n: int) -> bool:\n    return n % x == 0\n\ndef divisores2(n: int) -> list[int]:\n    return [x for x in range(1, n + 1) if esDivisorDe(x, n)]\n\n# 3\u00aa soluci\u00f3n\n# ===========\n\ndef divisores3(n: int) -> list[int]:\n    return list(filter(lambda x: esDivisorDe(x, n), range(1, n + 1)))\n\n# 4\u00aa soluci\u00f3n\n# ===========\n\n# primerosDivisores(n) es la lista de los divisores del n\u00famero n cuyo\n# cuadrado es menor o gual que n. Por ejemplo,\n#    primerosDivisores(25)  ==  [1,5]\n#    primerosDivisores(30)  ==  [1,2,3,5]\ndef primerosDivisores(n: int) -> list[int]:\n    return [x for x in range(1, 1 + round(sqrt(n))) if n % x == 0]\n\ndef divisores4(n: int) -> list[int]:\n    xs = primerosDivisores(n)\n    zs = list(reversed(xs))\n    if zs[0]**2 == n:\n        return xs + [n \/\/ a for a in zs[1:]]\n    return xs + [n \/\/ a for a in zs]\n\n# 5\u00aa soluci\u00f3n\n# ===========\n\ndef divisores5(n: int) -> list[int]:\n    def aux(xs: list[int]) -> list[int]:\n        if xs:\n            if esDivisorDe(xs[0], n):\n                return [xs[0]] + aux(xs[1:])\n            return aux(xs[1:])\n        return xs\n\n    return aux(list(range(1, n + 1)))\n\n# 6\u00aa soluci\u00f3n\n# ============\n\ndef divisores6(n: int) -> list[int]:\n    xs = []\n    for x in range(1, n+1):\n        if n % x == 0:\n            xs.append(x)\n    return xs\n\n# 7\u00aa soluci\u00f3n\n# ===========\n\ndef divisores7(n: int) -> list[int]:\n    x = 1\n    xs = []\n    ys = []\n    while x * x < n:\n        if n % x == 0:\n            xs.append(x)\n            ys.append(n \/\/ x)\n        x = x + 1\n    if x * x == n:\n        xs.append(x)\n    return xs + list(reversed(ys))\n\n# 8\u00aa soluci\u00f3n\n# ============\n\ndef divisores8(n: int) -> list[int]:\n    return divisors(n)\n\n# Comprobaci\u00f3n de equivalencia\n# ============================\n\n# La propiedad es\n@given(st.integers(min_value=2, max_value=1000))\ndef test_divisores(n):\n    assert divisores1(n) ==\\\n           divisores2(n) ==\\\n           divisores3(n) ==\\\n           divisores4(n) ==\\\n           divisores5(n) ==\\\n           divisores6(n) ==\\\n           divisores7(n) ==\\\n           divisores8(n)\n\n# La comprobaci\u00f3n es\n#    src> poetry run pytest -q divisores_de_un_numero.py\n#    1 passed in 0.84s\n\n# Comparaci\u00f3n de eficiencia\n# =========================\n\ndef tiempo(e):\n    \"\"\"Tiempo (en segundos) de evaluar la expresi\u00f3n e.\"\"\"\n    t = Timer(e, \"\", default_timer, globals()).timeit(1)\n    print(f\"{t:0.2f} segundos\")\n\n# La comparaci\u00f3n es\n#    >>> tiempo('divisores5(4*factorial(7))')\n#    1.40 segundos\n#\n#    >>> tiempo('divisores1(factorial(11))')\n#    1.79 segundos\n#    >>> tiempo('divisores2(factorial(11))')\n#    3.80 segundos\n#    >>> tiempo('divisores3(factorial(11))')\n#    5.22 segundos\n#    >>> tiempo('divisores4(factorial(11))')\n#    0.00 segundos\n#    >>> tiempo('divisores6(factorial(11))')\n#    3.51 segundos\n#    >>> tiempo('divisores7(factorial(11))')\n#    0.00 segundos\n#    >>> tiempo('divisores8(factorial(11))')\n#    0.00 segundos\n#\n#    >>> tiempo('divisores4(factorial(17))')\n#    2.23 segundos\n#    >>> tiempo('divisores7(factorial(17))')\n#    3.22 segundos\n#    >>> tiempo('divisores8(factorial(17))')\n#    0.00 segundos\n#\n#    >>> tiempo('divisores8(factorial(27))')\n#    0.28 segundos\n<\/pre>\n<p>El c\u00f3digo se encuentra en <a href=\"https:\/\/github.com\/jaalonso\/Exercitium-Python\/blob\/main\/src\/divisores_de_un_numero.py\">GitHub<\/a>.<\/p>\n<p><a name=\"ej6\"><\/a><\/p>\n<h3>6. Divisores primos<\/h3>\n<p>Definir la funci\u00f3n<\/p>\n<pre lang=\"text\">\n   divisoresPrimos :: Integer -> [Integer]\n<\/pre>\n<p>tal que <code>divisoresPrimos x<\/code> es la lista de los divisores primos de <code>x<\/code>. Por ejemplo,<\/p>\n<pre lang=\"text\">\n   divisoresPrimos 40 == [2,5]\n   divisoresPrimos 70 == [2,5,7]\n   length (divisoresPrimos (product [1..20000])) == 2262\n<\/pre>\n<p><b>Soluciones en Haskell<\/b><\/p>\n<pre lang=\"haskell\">\nimport Data.List (nub)\nimport Data.Set (toList)\nimport Data.Numbers.Primes (isPrime, primeFactors)\nimport Math.NumberTheory.ArithmeticFunctions (divisors)\nimport Test.QuickCheck\n\n-- 1\u00aa soluci\u00f3n\n-- ===========\n\ndivisoresPrimos1 :: Integer -> [Integer]\ndivisoresPrimos1 x = [n | n <- divisores1 x, primo1 n]\n\n-- (divisores n) es la lista de los divisores del n\u00famero n. Por ejemplo,\n--    divisores 25  ==  [1,5,25]\n--    divisores 30  ==  [1,2,3,5,6,10,15,30]\ndivisores1 :: Integer -> [Integer]\ndivisores1 n = [x | x <- [1..n], n `mod` x == 0]\n\n-- (primo n) se verifica si n es primo. Por ejemplo,\n--    primo 30  == False\n--    primo 31  == True\nprimo1 :: Integer -> Bool\nprimo1 n = divisores1 n == [1, n]\n\n-- 2\u00aa soluci\u00f3n\n-- ===========\n\ndivisoresPrimos2 :: Integer -> [Integer]\ndivisoresPrimos2 x = [n | n <- divisores2 x, primo2 n]\n\ndivisores2 :: Integer -> [Integer]\ndivisores2 n = xs ++ [n `div` y | y <- ys]\n  where xs = primerosDivisores2 n\n        (z:zs) = reverse xs\n        ys | z^2 == n  = zs\n           | otherwise = z:zs\n\n-- (primerosDivisores n) es la lista de los divisores del n\u00famero n cuyo\n-- cuadrado es menor o gual que n. Por ejemplo,\n--    primerosDivisores 25  ==  [1,5]\n--    primerosDivisores 30  ==  [1,2,3,5]\nprimerosDivisores2 :: Integer -> [Integer]\nprimerosDivisores2 n =\n   [x | x <- [1..round (sqrt (fromIntegral n))],\n        n `mod` x == 0]\n\nprimo2 :: Integer -> Bool\nprimo2 1 = False\nprimo2 n = primerosDivisores2 n == [1]\n\n-- 3\u00aa soluci\u00f3n\n-- ===========\n\ndivisoresPrimos3 :: Integer -> [Integer]\ndivisoresPrimos3 x = [n | n <- divisores3 x, primo3 n]\n\ndivisores3 :: Integer -> [Integer]\ndivisores3 n = xs ++ [n `div` y | y <- ys]\n  where xs = primerosDivisores3 n\n        (z:zs) = reverse xs\n        ys | z^2 == n  = zs\n           | otherwise = z:zs\n\nprimerosDivisores3 :: Integer -> [Integer]\nprimerosDivisores3 n =\n   filter ((== 0) . mod n) [1..round (sqrt (fromIntegral n))]\n\nprimo3 :: Integer -> Bool\nprimo3 1 = False\nprimo3 n = primerosDivisores3 n == [1]\n\n-- 4\u00aa soluci\u00f3n\n-- ===========\n\ndivisoresPrimos4 :: Integer -> [Integer]\ndivisoresPrimos4 n\n  | even n = 2 : divisoresPrimos4 (reducido n 2)\n  | otherwise = aux n [3,5..n]\n  where aux 1 _  = []\n        aux _ [] = []\n        aux m (x:xs) | m `mod` x == 0 = x : aux (reducido m x) xs\n                     | otherwise      = aux m xs\n\n-- (reducido m x) es el resultado de dividir repetidamente m por x,\n-- mientras sea divisible. Por ejemplo,\n--    reducido 36 2  ==  9\nreducido :: Integer -> Integer -> Integer\nreducido m x | m `mod` x == 0 = reducido (m `div` x) x\n             | otherwise      = m\n\n-- 5\u00aa soluci\u00f3n\n-- ===========\n\ndivisoresPrimos5 :: Integer -> [Integer]\ndivisoresPrimos5 = nub . primeFactors\n\n-- 6\u00aa soluci\u00f3n\n-- ===========\n\ndivisoresPrimos6 :: Integer -> [Integer]\ndivisoresPrimos6 = filter isPrime . toList . divisors\n\n-- Comprobaci\u00f3n de equivalencia\n-- ============================\n\n-- La propiedad es\nprop_divisoresPrimos :: Integer -> Property\nprop_divisoresPrimos n =\n  n > 1 ==>\n  all (== divisoresPrimos1 n)\n      [divisoresPrimos2 n,\n       divisoresPrimos3 n,\n       divisoresPrimos4 n,\n       divisoresPrimos5 n,\n       divisoresPrimos6 n]\n\n-- La comprobaci\u00f3n es\n--    \u03bb> quickCheck prop_divisoresPrimos\n--    +++ OK, passed 100 tests; 108 discarded.\n\n-- Comparaci\u00f3n de eficiencia\n-- =========================\n\n-- La comparaci\u00f3n es\n--    \u03bb> divisoresPrimos1 (product [1..11])\n--    [2,3,5,7,11]\n--    (18.34 secs, 7,984,382,104 bytes)\n--    \u03bb> divisoresPrimos2 (product [1..11])\n--    [2,3,5,7,11]\n--    (0.02 secs, 2,610,976 bytes)\n--    \u03bb> divisoresPrimos3 (product [1..11])\n--    [2,3,5,7,11]\n--    (0.02 secs, 2,078,288 bytes)\n--    \u03bb> divisoresPrimos4 (product [1..11])\n--    [2,3,5,7,11]\n--    (0.02 secs, 565,992 bytes)\n--    \u03bb> divisoresPrimos5 (product [1..11])\n--    [2,3,5,7,11]\n--    (0.01 secs, 568,000 bytes)\n--    \u03bb> divisoresPrimos6 (product [1..11])\n--    [2,3,5,7,11]\n--    (0.00 secs, 2,343,392 bytes)\n--\n--    \u03bb> divisoresPrimos2 (product [1..16])\n--    [2,3,5,7,11,13]\n--    (2.32 secs, 923,142,480 bytes)\n--    \u03bb> divisoresPrimos3 (product [1..16])\n--    [2,3,5,7,11,13]\n--    (0.80 secs, 556,961,088 bytes)\n--    \u03bb> divisoresPrimos4 (product [1..16])\n--    [2,3,5,7,11,13]\n--    (0.01 secs, 572,368 bytes)\n--    \u03bb> divisoresPrimos5 (product [1..16])\n--    [2,3,5,7,11,13]\n--    (0.01 secs, 31,665,896 bytes)\n--    \u03bb> divisoresPrimos6 (product [1..16])\n--    [2,3,5,7,11,13]\n--    (0.01 secs, 18,580,584 bytes)\n--\n--    \u03bb> length (divisoresPrimos4 (product [1..30]))\n--    10\n--    (0.01 secs, 579,168 bytes)\n--    \u03bb> length (divisoresPrimos5 (product [1..30]))\n--    10\n--    (0.01 secs, 594,976 bytes)\n--    \u03bb> length (divisoresPrimos6 (product [1..30]))\n--    10\n--    (3.38 secs, 8,068,783,408 bytes)\n--\n--    \u03bb> length (divisoresPrimos4 (product [1..20000]))\n--    2262\n--    (1.20 secs, 1,940,069,976 bytes)\n--    \u03bb> length (divisoresPrimos5 (product [1..20000]))\n--    2262\n--    (1.12 secs, 1,955,921,736 bytes)\n<\/pre>\n<p>El c\u00f3digo se encuentra en <a href=\"https:\/\/github.com\/jaalonso\/Exercitium\/blob\/main\/src\/Divisores_primos.hs\">GitHub<\/a>.<\/p>\n<p><b>Soluciones en Python<\/b><\/p>\n<pre lang=\"python\">\nfrom math import sqrt\nfrom operator import mul\nfrom functools import reduce\nfrom timeit import Timer, default_timer\nfrom sys import setrecursionlimit\nfrom sympy import divisors, isprime, primefactors\nfrom hypothesis import given, strategies as st\n\nsetrecursionlimit(10**6)\n\n# 1\u00aa soluci\u00f3n\n# ===========\n\n# divisores(n) es la lista de los divisores del n\u00famero n. Por ejemplo,\n#    divisores(30)  ==  [1,2,3,5,6,10,15,30]\ndef divisores1(n: int) -> list[int]:\n    return [x for x in range(1, n + 1) if n % x == 0]\n\n# primo(n) se verifica si n es primo. Por ejemplo,\n#    primo(30)  == False\n#    primo(31)  == True\ndef primo1(n: int) -> bool:\n    return divisores1(n) == [1, n]\n\ndef divisoresPrimos1(x: int) -> list[int]:\n    return [n for n in divisores1(x) if primo1(n)]\n\n# 2\u00aa soluci\u00f3n\n# ===========\n\n# primerosDivisores(n) es la lista de los divisores del n\u00famero n cuyo\n# cuadrado es menor o gual que n. Por ejemplo,\n#    primerosDivisores(25)  ==  [1,5]\n#    primerosDivisores(30)  ==  [1,2,3,5]\ndef primerosDivisores2(n: int) -> list[int]:\n    return [x for x in range(1, 1 + round(sqrt(n))) if n % x == 0]\n\ndef divisores2(n: int) -> list[int]:\n    xs = primerosDivisores2(n)\n    zs = list(reversed(xs))\n    if zs[0]**2 == n:\n        return xs + [n \/\/ a for a in zs[1:]]\n    return xs + [n \/\/ a for a in zs]\n\ndef primo2(n: int) -> bool:\n    return divisores2(n) == [1, n]\n\ndef divisoresPrimos2(x: int) -> list[int]:\n    return [n for n in divisores2(x) if primo2(n)]\n\n# 3\u00aa soluci\u00f3n\n# ===========\n\n# reducido(m, x) es el resultado de dividir repetidamente m por x,\n# mientras sea divisible. Por ejemplo,\n#    reducido(36, 2)  ==  9\ndef reducido(m: int, x: int) -> int:\n    if m % x == 0:\n        return reducido(m \/\/ x, x)\n    return m\n\ndef divisoresPrimos3(n: int) -> list[int]:\n    if n % 2 == 0:\n        return [2] + divisoresPrimos3(reducido(n, 2))\n\n    def aux(m, xs):\n        if m == 1:\n            return []\n        if xs == []:\n            return []\n        if m % xs[0] == 0:\n            return [xs[0]] + aux(reducido(m, xs[0]), xs[1:])\n        return aux(m, xs[1:])\n    return aux(n, range(3, n + 1, 2))\n\n# 4\u00aa soluci\u00f3n\n# ===========\n\ndef divisoresPrimos4(x: int) -> list[int]:\n    return [n for n in divisors(x) if isprime(n)]\n\n# 5\u00aa soluci\u00f3n\n# ===========\n\ndef divisoresPrimos5(n):\n    return primefactors(n)\n\n# Comprobaci\u00f3n de equivalencia\n# ============================\n\n# La propiedad es\n@given(st.integers(min_value=2, max_value=1000))\ndef test_divisoresPrimos(n):\n    assert divisoresPrimos1(n) ==\\\n           divisoresPrimos2(n) ==\\\n           divisoresPrimos3(n) ==\\\n           divisoresPrimos4(n) ==\\\n           divisoresPrimos5(n)\n\n# La comprobaci\u00f3n es\n#    src> poetry run pytest -q divisores_primos.py\n#    1 passed in 0.70s\n\n# Comparaci\u00f3n de eficiencia\n# =========================\n\ndef tiempo(e):\n    \"\"\"Tiempo (en segundos) de evaluar la expresi\u00f3n e.\"\"\"\n    t = Timer(e, \"\", default_timer, globals()).timeit(1)\n    print(f\"{t:0.2f} segundos\")\n\ndef producto(xs: list[int]) -> int:\n    return reduce(mul, xs)\n\n# La comparaci\u00f3n es\n#    >>> tiempo('divisoresPrimos1(producto(list(range(1, 12))))')\n#    11.14 segundos\n#    >>> tiempo('divisoresPrimos2(producto(list(range(1, 12))))')\n#    0.03 segundos\n#    >>> tiempo('divisoresPrimos3(producto(list(range(1, 12))))')\n#    0.00 segundos\n#    >>> tiempo('divisoresPrimos4(producto(list(range(1, 12))))')\n#    0.00 segundos\n#    >>> tiempo('divisoresPrimos5(producto(list(range(1, 12))))')\n#    0.00 segundos\n#\n#    >>> tiempo('divisoresPrimos2(producto(list(range(1, 17))))')\n#    14.21 segundos\n#    >>> tiempo('divisoresPrimos3(producto(list(range(1, 17))))')\n#    0.00 segundos\n#    >>> tiempo('divisoresPrimos4(producto(list(range(1, 17))))')\n#    0.01 segundos\n#    >>> tiempo('divisoresPrimos5(producto(list(range(1, 17))))')\n#    0.00 segundos\n#\n#    >>> tiempo('divisoresPrimos3(producto(list(range(1, 32))))')\n#    0.00 segundos\n#    >>> tiempo('divisoresPrimos4(producto(list(range(1, 32))))')\n#    4.59 segundos\n#    >>> tiempo('divisoresPrimos5(producto(list(range(1, 32))))')\n#    0.00 segundos\n#\n#    >>> tiempo('divisoresPrimos3(producto(list(range(1, 10001))))')\n#    3.00 segundos\n#    >>> tiempo('divisoresPrimos5(producto(list(range(1, 10001))))')\n#    0.24 segundos\n<\/pre>\n<p>El c\u00f3digo se encuentra en <a href=\"https:\/\/github.com\/jaalonso\/Exercitium-Python\/blob\/main\/src\/divisores_primos.py\">GitHub<\/a>.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Esta semana he publicado en Exercitium las soluciones de los siguientes problemas: 1. Igualdad de conjuntos 2. Uni\u00f3n conjuntista de listas 3. Intersecci\u00f3n conjuntista de listas 4. Diferencia conjuntista de listas 5. Divisores de un n\u00famero 6. Divisores primos A continuaci\u00f3n se muestran las soluciones.<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"jetpack_post_was_ever_published":false,"_kad_post_transparent":"","_kad_post_title":"","_kad_post_layout":"","_kad_post_sidebar_id":"","_kad_post_content_style":"","_kad_post_vertical_padding":"","_kad_post_feature":"","_kad_post_feature_position":"","_kad_post_header":false,"_kad_post_footer":false,"_jetpack_newsletter_access":"","_jetpack_dont_email_post_to_subs":false,"_jetpack_newsletter_tier_id":0,"_jetpack_memberships_contains_paywalled_content":false,"footnotes":"","_jetpack_memberships_contains_paid_content":false},"categories":[337],"tags":[],"jetpack_featured_media_url":"","jetpack_sharing_enabled":true,"jetpack_likes_enabled":false,"_links":{"self":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/vestigium\/wp-json\/wp\/v2\/posts\/7800"}],"collection":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/vestigium\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/vestigium\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/vestigium\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/vestigium\/wp-json\/wp\/v2\/comments?post=7800"}],"version-history":[{"count":1,"href":"https:\/\/www.glc.us.es\/~jalonso\/vestigium\/wp-json\/wp\/v2\/posts\/7800\/revisions"}],"predecessor-version":[{"id":7801,"href":"https:\/\/www.glc.us.es\/~jalonso\/vestigium\/wp-json\/wp\/v2\/posts\/7800\/revisions\/7801"}],"wp:attachment":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/vestigium\/wp-json\/wp\/v2\/media?parent=7800"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/vestigium\/wp-json\/wp\/v2\/categories?post=7800"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/vestigium\/wp-json\/wp\/v2\/tags?post=7800"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}