{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Batman, ecuaciones y python"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "*Esta notebook fue creada originalmente como un blog post por [Raúl E. López Briega](http://relopezbriega.com.ar/) en [Mi blog sobre Python](http://relopezbriega.github.io). El contenido esta bajo la licencia BSD.*"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<img alt=\"Ecuación Batman\" title=\"Ecuación Batman\" src=\"http://relopezbriega.github.io/images/ecnbatmancolor.png\">"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "[Batman](https://es.wikipedia.org/wiki/Batman) siempre fue mi superhéroe favorito porque es uno de los pocos héroes que no posee ningún superpoder, sino que debe recurrir a su intelecto y a la ciencia para construir las bati-herramientas que utiliza para combatir al crimen. Además posee ese toque de oscuridad producto de la dualidad entre realizar el bien, protegiendo a la gente de ciudad gótica, y su sed de venganza contra el crimen y la corrupción que acabó con la vida de su familia.  \n",
    "\n",
    "Es un personaje con muchos recursos, en cada nueva aparición podemos verlo utilizar nuevas y muy modernas bati-herramientas; su intelecto es tan agudo que incluso escondió una ecuación matemática en su bati-señal!!\n",
    "\n",
    "La [ecuación de batman](http://www.wolframalpha.com/input/?i=batman+equation) fue creada por el profesor de matemáticas [Matthew Register](http://www.quora.com/J-Matthew-Register) y se popularizó a través de un post de uno de sus alumnos en la red social [reddit](https://www.reddit.com/r/pics/comments/j2qjc/do_you_like_batman_do_you_like_math_my_math/); su expresión matemática es la siguiente:"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "$$\n",
    "((\\frac{x}{7})^2 \\cdot \\sqrt{\\frac{||x|-3|}{(|x|-3)}}+ (\\frac{y}{3})^2 \\cdot \\sqrt{\\frac{|y+3 \\cdot \\frac{\\sqrt{33}}{7}|}{y+3 \\cdot \\frac{\\sqrt{33}}{7}}}-1) \\cdot (|\\frac{x}{2}|-((3 \\cdot \\frac{\\sqrt{33}-7)}{112}) \\cdot x^2-3+\\sqrt{1-(||x|-2|-1)^2}-y) \\cdot (9 \\cdot \\sqrt{\\frac{|(|x|-1) \\cdot (|x|-0.75)|}{((1-|x|)*(|x|-0.75))}}-8 \\cdot |x|-y) \\cdot (3 \\cdot |x|+0.75 \\cdot \\sqrt{\\frac{|(|x|-0.75) \\cdot (|x|-0.5)|}{((0.75-|x|) \\cdot (|x|-0.5))}}-y) \\cdot (2.25 \\cdot \\sqrt{\\frac{|(x-0.5) \\cdot (x+0.5)|}{((0.5-x) \\cdot (0.5+x))}}-y) \\cdot (\\frac{6 \\cdot \\sqrt{10}}{7}+(1.5-0.5 \\cdot |x|) \\cdot \\sqrt{\\frac{||x|-1|}{|x|-1}}-(\\frac{6 \\cdot \\sqrt{10}}{14}) \\cdot \\sqrt{4-(|x|-1)^2}-y) =0\n",
    "$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Si bien a simple vista la ecuación parece sumamente compleja e imposible de graficar, la misma se puede descomponer en seis curvas distintas, mucho más simples.\n",
    "\n",
    "La primera de estas curvas, es la función del elipse $(\\frac{x}{7})^2 + (\\frac{y}{3})^2 = 1$, restringida a la región $\\sqrt{\\frac{||x|-3|}{(|x|-3)}}$ y $\\sqrt{\\frac{|y+3 \\cdot \\frac{\\sqrt{33}}{7}|}{y+3 \\cdot \\frac{\\sqrt{33}}{7}}}$ para cortar la parte central.\n",
    "\n",
    "Los cinco términos siguientes pueden ser entendidos como simples funciones de x, tres de los cuales son lineales.\n",
    "Por ejemplo, la siguiente función es la que grafica las curvas de la parte inferior de la bati-señal.\n",
    "\n",
    "$y = |\\frac{x}{2}|-(\\frac{3 \\cdot \\sqrt{33} -7}{112})\\cdot x^2 - 3 + \\sqrt{1-(||x|-2| -1)^2}$\n",
    "\n",
    "Las restantes ecuaciones de las curvas que completan el gráfico, son las siguientes:\n",
    "\n",
    "$y = \\frac{6\\cdot\\sqrt{10}}{7} + (-0.5|x| + 1.5) - \\frac{3\\cdot\\sqrt{10}}{7}\\cdot\\sqrt{4 - (|x|-1)^2}, |x| > 1$\n",
    "\n",
    "$y = 9 -8|x|, 0.75 < |x| < 1$\n",
    "\n",
    "$y = 3|x| + 0.75, 0.5 < |x| < 0.75$\n",
    "\n",
    "$y = 2.25, |x| < 0.5$\n",
    "\n",
    "La [ecuación de batman](http://www.wolframalpha.com/input/?i=batman+equation) puede ser fácilmente graficada utilizando [Matplotlib](http://matplotlib.org/) del siguiente modo:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {
    "collapsed": false
   },
   "outputs": [],
   "source": [
    "# graficos embebidos\n",
    "%matplotlib inline  "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "# Importando lo necesario para los cálculos\n",
    "import matplotlib.pyplot as plt\n",
    "from numpy import sqrt \n",
    "from numpy import meshgrid\n",
    "from numpy import arange"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
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      "text/plain": [
       "<matplotlib.figure.Figure at 0x7f17f61722e8>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "# Graficando la ecuación de Batman.\n",
    "xs = arange(-7.25, 7.25, 0.01)\n",
    "ys = arange(-5, 5, 0.01)\n",
    "x, y = meshgrid(xs, ys)\n",
    "\n",
    "eq1 = ((x/7)**2*sqrt(abs(abs(x)-3)/(abs(x)-3))+(y/3)**2*sqrt(abs(y+3/7*sqrt(33))/(y+3/7*sqrt(33)))-1)\n",
    "eq2 = (abs(x/2)-((3*sqrt(33)-7)/112)*x**2-3+sqrt(1-(abs(abs(x)-2)-1)**2)-y)\n",
    "eq3 = (9*sqrt(abs((abs(x)-1)*(abs(x)-.75))/((1-abs(x))*(abs(x)-.75)))-8*abs(x)-y)\n",
    "eq4 = (3*abs(x)+.75*sqrt(abs((abs(x)-.75)*(abs(x)-.5))/((.75-abs(x))*(abs(x)-.5)))-y)\n",
    "eq5 = (2.25*sqrt(abs((x-.5)*(x+.5))/((.5-x)*(.5+x)))-y)\n",
    "eq6 = (6*sqrt(10)/7+(1.5-.5*abs(x))*sqrt(abs(abs(x)-1)/(abs(x)-1))-(6*sqrt(10)/14)*sqrt(4-(abs(x)-1)**2)-y)\n",
    "\n",
    "for f, c in [(eq1, \"red\"), (eq2, \"purple\"), (eq3, \"green\"),\n",
    "             (eq4, \"blue\"), (eq5, \"orange\"), (eq6, \"black\")]:\n",
    "    plt.contour(x, y, f, [0], colors=c)\n",
    "\n",
    "plt.show()"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": true
   },
   "source": [
    "Ahora ya saben...si están en algún apuro y necesitan la ayuda del bati-héroe, solo necesitan graficar una ecuación para llamarlo con la bati-señal!\n",
    "\n",
    "Saludos!\n",
    "\n",
    "*Este post fue escrito utilizando IPython notebook. Pueden descargar este [notebook](https://github.com/relopezbriega/relopezbriega.github.io/blob/master/downloads/Batman.ipynb) o ver su version estática en [nbviewer](http://nbviewer.ipython.org/github/relopezbriega/relopezbriega.github.io/blob/master/downloads/Batman.ipynb).*"
   ]
  }
 ],
 "metadata": {
  "kernelspec": {
   "display_name": "Python 3",
   "language": "python",
   "name": "python3"
  },
  "language_info": {
   "codemirror_mode": {
    "name": "ipython",
    "version": 3
   },
   "file_extension": ".py",
   "mimetype": "text/x-python",
   "name": "python",
   "nbconvert_exporter": "python",
   "pygments_lexer": "ipython3",
   "version": "3.4.3+"
  }
 },
 "nbformat": 4,
 "nbformat_minor": 0
}