{"id":7398,"date":"2022-09-28T09:19:12","date_gmt":"2022-09-28T07:19:12","guid":{"rendered":"http:\/\/www.glc.us.es\/~jalonso\/exercitium\/?p=7398"},"modified":"2022-12-14T14:21:07","modified_gmt":"2022-12-14T12:21:07","slug":"suma-de-los-cuadrados-de-los-primeros-numeros-naturales","status":"publish","type":"post","link":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/suma-de-los-cuadrados-de-los-primeros-numeros-naturales\/","title":{"rendered":"Suma de los cuadrados de los primeros n\u00fameros naturales"},"content":{"rendered":"

Definir la funci\u00f3n<\/p>\n

\n   sumaDeCuadrados :: Integer -> Integer\n<\/pre>\n
tal que sumaDeCuadrados n<\/code> es la suma de los cuadrados de los primeros n<\/code> n\u00fameros; es decir, 1\u00b2 + 2\u00b2 + … + n\u00b2. Por ejemplo,<\/p>\n
\n   sumaDeCuadrados 3    ==  14\n   sumaDeCuadrados 100  ==  338350\n   length (show (sumaDeCuadrados (10^100)))  ==  300\n<\/pre>\n
Soluciones<\/b><\/p>\n
A continuaci\u00f3n se muestran las soluciones en Haskell<\/a> y las soluciones en Python<\/a>.<\/p>\n
<\/a>
\nSoluciones en Haskell<\/b><\/p>\n
\nimport Data.List (foldl')\nimport Test.QuickCheck\n\n-- 1\u00aa soluci\u00f3n\n-- ===========\n\nsumaDeCuadrados1 :: Integer -> Integer\nsumaDeCuadrados1 n = sum [x^2 | x <- [1..n]]\n\n-- 2\u00aa soluci\u00f3n\n-- ===========\n\nsumaDeCuadrados2 :: Integer -> Integer\nsumaDeCuadrados2 n = n*(n+1)*(2*n+1) `div` 6\n\n-- 3\u00aa soluci\u00f3n\n-- ===========\n\nsumaDeCuadrados3 :: Integer -> Integer\nsumaDeCuadrados3 1 = 1\nsumaDeCuadrados3 n = n^2 + sumaDeCuadrados3 (n-1)\n\n-- 4\u00aa soluci\u00f3n\n-- ===========\n\nsumaDeCuadrados4 :: Integer -> Integer\nsumaDeCuadrados4 n = foldl (+) 0 (map (^2) [0..n])\n\n-- 5\u00aa soluci\u00f3n\n-- ===========\n\nsumaDeCuadrados5 :: Integer -> Integer\nsumaDeCuadrados5 n = foldl' (+) 0 (map (^2) [0..n])\n\n-- Comprobaci\u00f3n de equivalencia\n-- ============================\n\n-- La propiedad es\nprop_sumaDeCuadrados :: Positive Integer -> Bool\nprop_sumaDeCuadrados (Positive n) =\n  all (== sumaDeCuadrados1 n)\n      [sumaDeCuadrados2 n,\n       sumaDeCuadrados3 n,\n       sumaDeCuadrados4 n,\n       sumaDeCuadrados5 n]\n\n-- La comprobaci\u00f3n es\n--    \u03bb> quickCheck prop_sumaDeCuadrados\n--    +++ OK, passed 100 tests.\n\n-- Comparaci\u00f3n de eficiencia\n-- =========================\n\n-- La comparaci\u00f3n es\n--    \u03bb> sumaDeCuadrados1 (2*10^6)\n--    2666668666667000000\n--    (1.90 secs, 1,395,835,576 bytes)\n--    \u03bb> sumaDeCuadrados2 (2*10^6)\n--    2666668666667000000\n--    (0.01 secs, 563,168 bytes)\n--    \u03bb> sumaDeCuadrados3 (2*10^6)\n--    2666668666667000000\n--    (2.37 secs, 1,414,199,400 bytes)\n--    \u03bb> sumaDeCuadrados4 (2*10^6)\n--    2666668666667000000\n--    (1.33 secs, 1,315,836,128 bytes)\n--    \u03bb> sumaDeCuadrados5 (2*10^6)\n--    2666668666667000000\n--    (0.71 secs, 1,168,563,384 bytes)\n<\/pre>\n
El c\u00f3digo se encuentra en GitHub<\/a>.<\/p>\n
<\/a>
\nSoluciones en Python<\/b><\/p>\n
\nfrom operator import add\nfrom functools import reduce\nfrom sys import setrecursionlimit\nfrom timeit import Timer, default_timer\nfrom hypothesis import given, strategies as st\nsetrecursionlimit(10**6)\n\n# 1\u00aa soluci\u00f3n\n# ===========\n\ndef sumaDeCuadrados1(n: int) -> int:\n    return sum(x**2 for x in range(1, n + 1))\n\n# 2\u00aa soluci\u00f3n\n# ===========\n\ndef sumaDeCuadrados2(n: int) -> int:\n    return n * (n + 1) * (2 * n + 1) \/\/ 6\n\n# 3\u00aa soluci\u00f3n\n# ===========\n\ndef sumaDeCuadrados3(n: int) -> int:\n    if n == 1:\n        return 1\n    return n**2 + sumaDeCuadrados3(n - 1)\n\n# 4\u00aa soluci\u00f3n\n# ===========\n\ndef sumaDeCuadrados4(n: int) -> int:\n    return reduce(add, (x**2 for x in range(1, n + 1)))\n\n# 5\u00aa soluci\u00f3n\n# ===========\n\ndef sumaDeCuadrados5(n: int) -> int:\n    x, r = 1, 0\n    while x <= n:\n        r = r + x**2\n        x = x + 1\n    return r\n\n# 6\u00aa soluci\u00f3n\n# ===========\n\ndef sumaDeCuadrados6(n: int) -> int:\n    r = 0\n    for x in range(1, n + 1):\n        r = r + x**2\n    return r\n\n# Comprobaci\u00f3n de equivalencia\n# ============================\n\n# La propiedad es\n@given(st.integers(min_value=1, max_value=1000))\ndef test_sumaDeCuadrados(n):\n    r = sumaDeCuadrados1(n)\n    assert sumaDeCuadrados2(n) == r\n    assert sumaDeCuadrados3(n) == r\n    assert sumaDeCuadrados4(n) == r\n    assert sumaDeCuadrados5(n) == r\n    assert sumaDeCuadrados6(n) == r\n\n# La comprobaci\u00f3n es\n#    src> poetry run pytest -q suma_de_los_cuadrados_de_los_primeros_numeros_naturales.py\n#    1 passed in 0.19s\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('sumaDeCuadrados1(20000)')\n#    0.01 segundos\n#    >>> tiempo('sumaDeCuadrados2(20000)')\n#    0.00 segundos\n#    >>> tiempo('sumaDeCuadrados3(20000)')\n#    0.02 segundos\n#    >>> tiempo('sumaDeCuadrados4(20000)')\n#    0.02 segundos\n#    >>> tiempo('sumaDeCuadrados5(20000)')\n#    0.02 segundos\n#    >>> tiempo('sumaDeCuadrados6(20000)')\n#    0.02 segundos\n#\n#    >>> tiempo('sumaDeCuadrados1(10**7)')\n#    2.19 segundos\n#    >>> tiempo('sumaDeCuadrados2(10**7)')\n#    0.00 segundos\n#    >>> tiempo('sumaDeCuadrados4(10**7)')\n#    2.48 segundos\n#    >>> tiempo('sumaDeCuadrados5(10**7)')\n#    2.53 segundos\n#    >>> tiempo('sumaDeCuadrados6(10**7)')\n#    2.22 segundos\n<\/pre>\n
El c\u00f3digo se encuentra en GitHub<\/a>.<\/p>\n","protected":false},"excerpt":{"rendered":"
Definir la funci\u00f3n sumaDeCuadrados :: Integer -> Integer tal que sumaDeCuadrados n es la suma de los cuadrados de los primeros n n\u00fameros; es decir, 1\u00b2 + 2\u00b2 + … + n\u00b2. Por ejemplo, sumaDeCuadrados 3 == 14 sumaDeCuadrados 100 == 338350 length (show (sumaDeCuadrados (10^100))) == 300 Soluciones A continuaci\u00f3n se muestran las soluciones…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","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":[2],"tags":[],"jetpack_featured_media_url":"","jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts\/7398"}],"collection":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/comments?post=7398"}],"version-history":[{"count":2,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts\/7398\/revisions"}],"predecessor-version":[{"id":7682,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts\/7398\/revisions\/7682"}],"wp:attachment":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/media?parent=7398"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/categories?post=7398"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/tags?post=7398"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}