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{
"cells": [
{
"cell_type": "markdown",
"metadata": {
"lang": "fr"
},
"source": [
"## La méthode du gradient pour résoudre A x = b\n",
"\n",
"Le but de ce TP est de vous laisser avancer tout seul. Reprenez les cours et programmez la méthode du gradient\n",
"pour résoudre le système matriciel $A {\\bf x} = {\\bf b}$ avec A symétrique et à diagonale dominante\n",
"($a_{ii} > \\sum_{k \\ne i} |a_{ik}|$).\n",
"\n",
"* Commencez en 2D avec une matrice 2x2, vérifier que le résultat est bon et tracer la courbe de convergence\n",
"* Passez en nxn (on montrera que cela marche avec une matrice 9x9)\n",
"\n",
"Il peut être intéressant de normaliser la matrice A pour éviter que les calculs explosent."
]
},
{
"cell_type": "code",
"execution_count": 25,
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"array([[0.15, 0.15],\n",
" [0.15, 0.55]])"
]
},
"execution_count": 25,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"import numpy as np\n",
"A = np.random.randint(10, size=(2,2))\n",
"A = A + A.T\n",
"A += np.diag(A.sum(axis=1))\n",
"A = A / np.sum(np.abs(A))\n",
"A"
]
},
{
"cell_type": "code",
"execution_count": 26,
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"array([0.3, 0.7])"
]
},
"execution_count": 26,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"b = A.sum(axis=1)\n",
"b"
]
},
{
"cell_type": "code",
"execution_count": 36,
"metadata": {},
"outputs": [],
"source": [
"def solve_with_gradiant(A, b):\n",
" mu = 0.01\n",
" xi = np.array(np.zeros(b.shape))\n",
" while True:\n",
" xn = xi - mu * (A @ xi - b)\n",
" if np.square(xn - xi).sum() < 1e-6 ** 2:\n",
" break\n",
" xi = xn\n",
" return xi"
]
},
{
"cell_type": "code",
"execution_count": 37,
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"array([0.9990519 , 1.00031603])"
]
},
"execution_count": 37,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"solve_with_gradiant(A,b)"
]
},
{
"cell_type": "markdown",
"metadata": {
"lang": "fr"
},
"source": [
"## Introduire de l'inertie\n",
"\n",
"Introduire de l'inertie dans la méthode du gradient. Que constate-t-on ?"
]
},
{
"cell_type": "code",
"execution_count": 39,
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"array([0.9990519 , 1.00031603])"
]
},
"execution_count": 39,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"def solve_with_gradiant_inertie(A, b):\n",
" mu = 0.01\n",
" i = 0.5\n",
" xi = np.array(np.zeros(b.shape))\n",
" while True:\n",
" xn = xi - mu * (A @ xi - b)\n",
" if np.square(xn - xi).sum() < 1e-6 ** 2:\n",
" break\n",
" xi = xn\n",
" return xi\n",
"solve_with_gradiant_inertie(A, b)"
]
},
{
"cell_type": "markdown",
"metadata": {
"lang": "fr"
},
"source": [
"## Valeur optimale de µ\n",
"\n",
"On note que deux directions de pente sucessives sont orthogonales si le point suivant est le minumum dans\n",
"la direction donnée ($\\nabla J ({\\bf x}^k$)).\n",
"\n",
"$$\\nabla J ({\\bf x}^{k+1})^T \\; \\nabla J ({\\bf x}^k) = 0$$\n",
"\n",
"#### Démonstration \n",
"\n",
"\n",
"<table><tr>\n",
" <th>\n",
"On veut régler µ pour arriver au minimum de J lorsqu'on avance dans la direction $- \\nabla J({\\bf x}^{k})$.\n",
"Cela veut dire que la dérivée partielle de $J({\\bf x}^{k+1})$ par rapport à µ doit être\n",
"égale à 0 ou bien en faisant apparaitre µ dans l'équation :\n",
"\n",
"$$\n",
"\\frac{\\partial J ({\\bf x}^k - µ \\; \\nabla J ({\\bf x}^k))}{\\partial µ} = 0\n",
"$$\n",
"\n",
"En développant on a\n",
"\n",
"$$\n",
"\\begin{align}\n",
"\\frac{\\partial J ({\\bf x}^{k+1})}{\\partial {\\bf x}^{k+1}} \\; \n",
"\\frac{\\partial {\\bf x}^{k+1}}{\\partial µ} & = 0 \\\\\n",
"J'({\\bf x}^{k+1}) \\, . \\, (- \\nabla J ({\\bf x}^k)) & = 0 \\\\\n",
"(A\\, {\\bf x}^{k+1} - b) \\, . \\, \\nabla J ({\\bf x}^k) & = 0 \\quad \\textrm{puisque A est symétrique}\\\\\n",
"\\nabla J ({\\bf x}^{k+1}) \\, . \\, \\nabla J ({\\bf x}^k) & = 0 \\quad \\textrm{CQFD}\n",
"\\end{align}\n",
"$$\n",
"\n",
"\n",
"</th><th>\n",
"<img src=\"images/gradient.png\" width = \"400px\"/>\n",
"</th></tr></table>\n",
"\n",
"En utilisant cette propriété, évaluer la valeur optimale de µ pour atteindre le minimum dans la direction de\n",
"$\\nabla J ({\\bf x}^k)$.\n",
"\n",
"Écrire le méthode du gradient avec le calcul du µ optimal à chaque itération pour résoudre $A {\\bf x} = {\\bf b}$."
]
},
{
"cell_type": "code",
"execution_count": 40,
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"array([0.99999941, 0.99999941])"
]
},
"execution_count": 40,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"def solve_with_gradiant_inertie_opti(A, b):\n",
" mu = 0.01\n",
" i = 0.5\n",
" xi = np.array(np.zeros(b.shape))\n",
" while True:\n",
" J = A @ xi - b\n",
" mu = np.dot(J, J) / np.dot(A @ J, J)\n",
" xn = xi - mu * (A @ xi - b)\n",
" if np.square(xn - xi).sum() < 1e-6 ** 2:\n",
" break\n",
" xi = xn\n",
" return xi\n",
"solve_with_gradiant_inertie_opti(A,b)"
]
}
],
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|
