{
 "cells": [
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {},
   "outputs": [],
   "source": [
    "import numpy as np"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {},
   "outputs": [],
   "source": [
    "def max_descida_armijo( f, grad, x, step = 1.0, sigma = 0.1, eta = 0.1, epsilon = 1e-6, return_niter = False ):\n",
    "\n",
    "    g_k = grad( x )\n",
    "\n",
    "    niter = 0\n",
    "\n",
    "    while np.linalg.norm( g_k ) > epsilon:\n",
    "\n",
    "        step_k = step\n",
    "\n",
    "        f_k = f( x )\n",
    "        ns_g_k = np.linalg.norm( g_k ) ** 2\n",
    "\n",
    "        x_try = x - step_k * g_k\n",
    "        while f( x_try ) > f_k - sigma * step_k * ns_g_k:\n",
    "            step_k = step_k * eta\n",
    "            x_try = x - step_k * g_k\n",
    "\n",
    "        x = x_try\n",
    "        g_k = grad( x )\n",
    "\n",
    "        niter = niter + 1\n",
    "\n",
    "    if return_niter:\n",
    "        return ( x, niter )\n",
    "    else:\n",
    "        return x"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {},
   "outputs": [],
   "source": [
    "def rosenbrock( x, a = 100, b = 1 ):\n",
    "\n",
    "    return a * ( x[ 1 ] - x[ 0 ] ** 2 ) ** 2 + ( x[ 0 ] - b ) ** 2\n",
    "\n",
    "def rosenbrock_grad( x, a = 100, b = 1 ):\n",
    "\n",
    "    return np.array(\n",
    "        (\n",
    "            -400 * ( x[ 1 ] - x[ 0 ] ** 2 ) * x[ 0 ] + 2 * ( x[ 0 ] - 1 ),\n",
    "            200 * ( x[ 1 ] - x[ 0 ] ** 2 )\n",
    "        )\n",
    "    )"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "(array([1., 1.]), 7473)\n"
     ]
    }
   ],
   "source": [
    "print( max_descida_armijo( rosenbrock, rosenbrock_grad, np.zeros( ( 2, ) ), 1e-2, epsilon = 1e-10, return_niter = True ) )"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Exercício\n",
    "\n",
    "Instrumente a função acima para retornar o número de avaliações da função objetivo que o algoritmo realiza. Compare esse número com o custo computacional das iterações extra realizadas pelo método com passo fixo. Tire conclusões a respeito da eficiência relativa dentre os dois métodos.\n",
    "\n",
    "Lembre-se de alterar o método com passo fixo para utilizar o mesmo critério de parada que o da rotina acima."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": []
  }
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