{"id":8279,"date":"2023-09-14T06:00:47","date_gmt":"2023-09-14T04:00:47","guid":{"rendered":"http:\/\/www.glc.us.es\/~jalonso\/exercitium\/?p=8279"},"modified":"2024-05-17T18:48:25","modified_gmt":"2024-05-17T16:48:25","slug":"14-sep-23","status":"publish","type":"post","link":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/14-sep-23\/","title":{"rendered":"La funci\u00f3n de Fibonacci por programaci\u00f3n din\u00e1mica"},"content":{"rendered":"

Los primeros t\u00e9rminos de la sucesi\u00f3n de Fibonacci son<\/p>\n

\n   0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, ...\n<\/pre>\n
Escribir dos definiciones (una recursiva y otra con programaci\u00f3n din\u00e1mica) de la funci\u00f3n<\/p>\n
\n   fib :: Integer -> Integer\n<\/pre>\n
tal que fib n<\/code> es el n<\/code>-\u00e9simo t\u00e9rmino de la sucesi\u00f3n de Fibonacci. Por ejemplo,<\/p>\n
\n   fib 6 == 8\n<\/pre>\n
Comparar la eficiencia de las dos definiciones.
\n<\/p>\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
\nmodule La_funcion_de_Fibonacci_por_programacion_dinamica where\n\nimport Data.Array\nimport Test.Hspec (Spec, hspec, it, shouldBe)\n\n-- 1\u00aa definici\u00f3n (por recursi\u00f3n)\n-- =============================\n\nfib1 :: Integer -> Integer\nfib1 0 = 0\nfib1 1 = 1\nfib1 n = fib1 (n-1) + fib1 (n-2)\n\n-- 2\u00aa definici\u00f3n (con programaci\u00f3n din\u00e1mica)\n-- =========================================\n\nfib2 :: Integer -> Integer\nfib2 n = vectorFib2 n ! n\n\n-- (vectorFib2 n) es el vector con \u00edndices de 0 a n tal que el valor\n-- de la posici\u00f3n i es el i-\u00e9simo n\u00famero de Finonacci. Por ejemplo,\n--    \u03bb> vectorFib2 7\n--    array (0,7) [(0,0),(1,1),(2,1),(3,2),(4,3),(5,5),(6,8),(7,13)]\nvectorFib2 :: Integer -> Array Integer Integer\nvectorFib2 n = v where\n  v = array (0,n) [(i,f i) | i <- [0..n]]\n  f 0 = 0\n  f 1 = 1\n  f m = v!(m-1) + v!(m-2)\n\n-- Comparaci\u00f3n de eficiencia\n-- =========================\n\n-- La comparaci\u00f3n es\n--    \u03bb> fib1 34\n--    5702887\n--    (11.82 secs, 3,504,944,704 bytes)\n--    \u03bb> fib2 34\n--    5702887\n--    (0.01 secs, 587,808 bytes)\n\n-- Verificaci\u00f3n\n-- ============\n\nverifica :: IO ()\nverifica = hspec spec\n\nspec :: Spec\nspec = do\n  it \"e1\" $\n    fib1 6 `shouldBe` 8\n  it \"e2\" $\n    fib2 6 `shouldBe` 8\n  it \"e3\" $\n    map fib1 [0..9] `shouldBe` map fib2 [0..9]\n\n-- La verificaci\u00f3n es\n--    \u03bb> verifica\n--\n--    e1\n--    e2\n--    e3\n--\n--    Finished in 0.0007 seconds\n--    3 examples, 0 failures\n--    (0.01 secs, 788,952 bytes)\n<\/pre>\n
<\/a>
\nSoluciones en Python<\/b><\/p>\n
\nfrom sys import setrecursionlimit\nfrom timeit import Timer, default_timer\n\nimport numpy as np\nimport numpy.typing as npt\n\nsetrecursionlimit(10**6)\n\n# 1\u00aa definici\u00f3n (por recursi\u00f3n)\n# =============================\n\ndef fib1(n: int) -> int:\n    if n == 0:\n        return 0\n    if n == 1:\n        return 1\n    return fib1(n - 1) + fib1(n - 2)\n\n# 2\u00aa definici\u00f3n (con programaci\u00f3n din\u00e1mica)\n# =========================================\n\ndef fib2(n: int) -> int:\n    return vectorFib2(n)[n]\n\n# (vectorFib2 n) es el vector con \u00edndices de 0 a n tal que el valor\n# de la posici\u00f3n i es el i-\u00e9simo n\u00famero de Finonacci. Por ejemplo,\n#    >>> vectorFib2(7)\n#    [0, 1, 1, 2, 3, 5, 8, 13]\ndef vectorFib2(n: int) -> list[int]:\n    v = [0] * (n + 1)\n    v[0] = 0\n    v[1] = 1\n    for i in range(2, n + 1):\n        v[i] = v[i - 1] + v[i - 2]\n    return v\n\n# 2\u00aa definici\u00f3n (con programaci\u00f3n din\u00e1mica y array)\n# =================================================\n\ndef fib3(n: int) -> int:\n    return vectorFib3(n)[n]\n\n# (vectorFib3 n) es el vector con \u00edndices de 0 a n tal que el valor\n# de la posici\u00f3n i es el i-\u00e9simo n\u00famero de Finonacci. Por ejemplo,\n#    >>> vectorFib3(7)\n#    array([ 0,  1,  1,  2,  3,  5,  8, 13])\ndef vectorFib3(n: int) -> npt.NDArray[np.complex64]:\n    v = np.zeros(n + 1, dtype=int)\n    v[0] = 0\n    v[1] = 1\n    for i in range(2, n + 1):\n        v[i] = v[i - 1] + v[i - 2]\n    return v\n\n# Comparaci\u00f3n de eficiencia\n# =========================\n\ndef tiempo(e: str) -> None:\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('fib1(34)')\n#    2.10 segundos\n#    >>> tiempo('fib2(34)')\n#    0.00 segundos\n#    >>> tiempo('fib3(34)')\n#    0.00 segundos\n#\n#    >>> tiempo('fib2(100000)')\n#    0.37 segundos\n#    >>> tiempo('fib3(100000)')\n#    0.08 segundos\n\n# Verificaci\u00f3n\n# ============\n\ndef test_fib() -> None:\n    assert fib1(6) == 8\n    assert fib2(6) == 8\n    assert fib3(6) == 8\n    print(\"Verificado\")\n\n# La verificaci\u00f3n es\n#    >>> test_fib()\n#    Verificado\n<\/pre>\n","protected":false},"excerpt":{"rendered":"
Los primeros t\u00e9rminos de la sucesi\u00f3n de Fibonacci son 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, … Escribir dos definiciones (una recursiva y otra con programaci\u00f3n din\u00e1mica) de la funci\u00f3n fib :: Integer -> Integer tal que fib n es el n-\u00e9simo t\u00e9rmino de la sucesi\u00f3n…<\/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":"default","_kad_post_title":"default","_kad_post_layout":"default","_kad_post_sidebar_id":"","_kad_post_content_style":"default","_kad_post_vertical_padding":"default","_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":[581],"tags":[],"jetpack_featured_media_url":"","jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts\/8279"}],"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=8279"}],"version-history":[{"count":2,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts\/8279\/revisions"}],"predecessor-version":[{"id":8579,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts\/8279\/revisions\/8579"}],"wp:attachment":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/media?parent=8279"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/categories?post=8279"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/tags?post=8279"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}