{"id":7427,"date":"2022-10-11T06:00:18","date_gmt":"2022-10-11T04:00:18","guid":{"rendered":"http:\/\/www.glc.us.es\/~jalonso\/exercitium\/?p=7427"},"modified":"2022-12-14T12:35:59","modified_gmt":"2022-12-14T10:35:59","slug":"puntos-en-el-circulo","status":"publish","type":"post","link":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/puntos-en-el-circulo\/","title":{"rendered":"Puntos en el c\u00edrculo"},"content":{"rendered":"<p>En el c\u00edrculo de radio 2 hay 6 puntos cuyas coordenadas son puntos naturales:<\/p>\n<pre lang=\"text\">\n   (0,0),(0,1),(0,2),(1,0),(1,1),(2,0)\n<\/pre>\n<p>y en de radio 3 hay 11:<\/p>\n<pre lang=\"text\">\n   (0,0),(0,1),(0,2),(0,3),(1,0),(1,1),(1,2),(2,0),(2,1),(2,2),(3,0)\n<\/pre>\n<p>Definir la funci\u00f3n<\/p>\n<pre lang=\"text\">\n   circulo :: Int -> Int\n<\/pre>\n<p>tal que <code>circulo n<\/code> es el la cantidad de pares de n\u00fameros naturales (x,y) que se encuentran en el c\u00edrculo de radio <code>n<\/code>. Por ejemplo,<\/p>\n<pre lang=\"text\">\n   circulo 1    ==  3\n   circulo 2    ==  6\n   circulo 3    ==  11\n   circulo 4    ==  17\n   circulo 100  ==  7955\n<\/pre>\n<p><a name=\"haskell\"><\/a><br \/>\n<b>Soluciones en Haskell<\/b><\/p>\n<pre lang=\"haskell\">\nimport Test.QuickCheck\n\n-- 1\u00aa soluci\u00f3n\n-- ===========\n\ncirculo1 :: Int -> Int\ncirculo1 n = length (enCirculo1 n)\n\nenCirculo1 :: Int -> [(Int, Int)]\nenCirculo1 n = [(x,y) | x <- [0..n],\n                        y <- [0..n],\n                        x*x+y*y <= n*n]\n\n-- 2\u00aa soluci\u00f3n\n-- ===========\n\ncirculo2 :: Int -> Int\ncirculo2 0 = 1\ncirculo2 n =\n  2 * length (enSemiCirculo n) + ceiling(fromIntegral n \/ sqrt 2)\n\nenSemiCirculo :: Int -> [(Int, Int)]\nenSemiCirculo n =\n  [(x,y) | x <- [0..floor (sqrt (fromIntegral (n * n)))],\n           y <- [x+1..truncate (sqrt (fromIntegral (n*n - x*x)))]]\n\n-- Comprobaci\u00f3n de equivalencia\n-- ============================\n\n-- La propiedad es\nprop_circulo :: Positive Int -> Bool\nprop_circulo (Positive n) =\n  circulo1 n == circulo2 n\n\n-- La comprobaci\u00f3n es\n--    \u03bb> quickCheck prop_circulo\n--    +++ OK, passed 100 tests.\n\n-- Comparaci\u00f3n de eficiencia\n-- =========================\n\n-- La comparaci\u00f3n es\n--    \u03bb> circulo1 (2*10^3)\n--    3143587\n--    (3.58 secs, 1,744,162,600 bytes)\n--    \u03bb> circulo2 (2*10^3)\n--    3143587\n--    (0.41 secs, 266,374,208 bytes)\n<\/pre>\n<p>El c\u00f3digo se encuentra en <a href=\"https:\/\/github.com\/jaalonso\/Exercitium\/blob\/main\/src\/Puntos_dentro_del_circulo.hs\">GitHub<\/a>.<\/p>\n<p><a name=\"python\"><\/a><br \/>\n<b>Soluciones en Python<\/b><\/p>\n<pre lang=\"python\">\nfrom math import sqrt, trunc, ceil\nfrom timeit import Timer, default_timer\n\nfrom hypothesis import given\nfrom hypothesis import strategies as st\n\n# 1\u00aa soluci\u00f3n\n# ===========\n\ndef circulo1(n: int) -> int:\n    return len([(x, y)\n                for x in range(0, n + 1)\n                for y in range(0, n + 1)\n                if x * x + y * y <= n * n])\n\n# 2\u00aa soluci\u00f3n\n# ===========\n\ndef enSemiCirculo(n: int) -> list[tuple[int, int]]:\n    return [(x, y)\n            for x in range(0, ceil(sqrt(n**2)) + 1)\n            for y in range(x+1, trunc(sqrt(n**2 - x**2)) + 1)]\n\ndef circulo2(n: int) -> int:\n    if n == 0:\n        return 1\n    return (2 * len(enSemiCirculo(n)) + ceil(n \/ sqrt(2)))\n\n# 3\u00aa soluci\u00f3n\n# ===========\n\ndef circulo3(n: int) -> int:\n    r = 0\n    for x in range(0, n + 1):\n        for y in range(0, n + 1):\n            if x**2 + y**2 <= n**2:\n                r = r + 1\n    return r\n\n# 4\u00aa soluci\u00f3n\n# ===========\n\ndef circulo4(n: int) -> int:\n    r = 0\n    for x in range(0, ceil(sqrt(n**2)) + 1):\n        for y in range(x + 1, trunc(sqrt(n**2 - x**2)) + 1):\n            if x**2 + y**2 <= n**2:\n                r = r + 1\n    return 2 * r + ceil(n \/ sqrt(2))\n\n# Comprobaci\u00f3n de equivalencia\n# ============================\n\n# La propiedad es\n@given(st.integers(min_value=1, max_value=100))\ndef test_circulo(n: int) -> None:\n    r = circulo1(n)\n    assert circulo2(n) == r\n    assert circulo3(n) == r\n    assert circulo4(n) == r\n\n# La comprobaci\u00f3n es\n#    src> poetry run pytest -q puntos_dentro_del_circulo.py\n#    1 passed in 0.60s\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('circulo1(2000)')\n#    0.71 segundos\n#    >>> tiempo('circulo2(2000)')\n#    0.76 segundos\n#    >>> tiempo('circulo3(2000)')\n#    2.63 segundos\n#    >>> tiempo('circulo4(2000)')\n#    1.06 segundos\n<\/pre>\n<p>El c\u00f3digo se encuentra en <a href=\"https:\/\/github.com\/jaalonso\/Exercitium-Python\/blob\/main\/src\/puntos_dentro_del_circulo.py\">GitHub<\/a>.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>En el c\u00edrculo de radio 2 hay 6 puntos cuyas coordenadas son puntos naturales: (0,0),(0,1),(0,2),(1,0),(1,1),(2,0) y en de radio 3 hay 11: (0,0),(0,1),(0,2),(0,3),(1,0),(1,1),(1,2),(2,0),(2,1),(2,2),(3,0) Definir la funci\u00f3n circulo :: Int -> Int tal que circulo n es el la cantidad de pares de n\u00fameros naturales (x,y) que se encuentran en el c\u00edrculo de radio n. Por&#8230;<\/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":[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\/7427"}],"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=7427"}],"version-history":[{"count":3,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts\/7427\/revisions"}],"predecessor-version":[{"id":7674,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts\/7427\/revisions\/7674"}],"wp:attachment":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/media?parent=7427"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/categories?post=7427"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/tags?post=7427"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}