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March 8, 2022 08:35
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SPL_LU_doolittle.ipynb
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| { | |
| "nbformat": 4, | |
| "nbformat_minor": 0, | |
| "metadata": { | |
| "colab": { | |
| "name": "SPL_LU_doolittle.ipynb", | |
| "provenance": [], | |
| "authorship_tag": "ABX9TyNFZdugVwG2k/2a3dLFt/z8", | |
| "include_colab_link": true | |
| }, | |
| "kernelspec": { | |
| "name": "python3", | |
| "display_name": "Python 3" | |
| }, | |
| "language_info": { | |
| "name": "python" | |
| } | |
| }, | |
| "cells": [ | |
| { | |
| "cell_type": "markdown", | |
| "metadata": { | |
| "id": "view-in-github", | |
| "colab_type": "text" | |
| }, | |
| "source": [ | |
| "<a href=\"https://colab.research.google.com/gist/DianSano/7a953614b4719b25f0f587364c97d874/spl_lu_doolittle.ipynb\" target=\"_parent\"><img src=\"https://colab.research.google.com/assets/colab-badge.svg\" alt=\"Open In Colab\"/></a>" | |
| ] | |
| }, | |
| { | |
| "cell_type": "code", | |
| "execution_count": 1, | |
| "metadata": { | |
| "colab": { | |
| "base_uri": "https://localhost:8080/" | |
| }, | |
| "id": "raRVZTciqk-q", | |
| "outputId": "549a086d-212e-4548-d623-3e83234de9d9" | |
| }, | |
| "outputs": [ | |
| { | |
| "output_type": "stream", | |
| "name": "stdout", | |
| "text": [ | |
| "[[ 1. 0. 0. ]\n", | |
| " [ 1. 1. 0. ]\n", | |
| " [ 2. -4.5 1. ]] [[ 1. 4. 1.]\n", | |
| " [ 0. 2. -2.]\n", | |
| " [ 0. 0. -9.]]\n", | |
| "[ 5. 1. -2.]\n" | |
| ] | |
| } | |
| ], | |
| "source": [ | |
| "import numpy as np\n", | |
| "\n", | |
| "a = np.array([[1, 4, 1],\n", | |
| " [1, 6, -1],\n", | |
| " [2, -1, 2]], float)\n", | |
| "#the b matrix constant terms of the equations \n", | |
| "b = np.array([7, 13, 5], float)\n", | |
| "\n", | |
| "def doolittle(A):\n", | |
| " #source https://johnfoster.pge.utexas.edu/numerical-methods-book/LinearAlgebra_LU.html\n", | |
| " n = a.shape[0]\n", | |
| " U = np.zeros((n, n), dtype=np.double)\n", | |
| " L = np.eye(n, dtype=np.double)\n", | |
| " # Note: @ numpy operator for matrix multiplication\n", | |
| " for k in range(n):\n", | |
| " U[k, k:] = a[k, k:] - L[k,:k] @ U[:k,k:]\n", | |
| " L[(k+1):,k] = (a[(k+1):,k] - L[(k+1):,:] @ U[:,k]) / U[k, k]\n", | |
| " return L, U\n", | |
| "\n", | |
| "L, U = doolittle(a)\n", | |
| "print(L, U)\n", | |
| "\n", | |
| "\n", | |
| "def forward_substitution(L, b):\n", | |
| " #source https://johnfoster.pge.utexas.edu/numerical-methods-book/LinearAlgebra_LU.html\n", | |
| " #Get number of rows\n", | |
| " n = L.shape[0]\n", | |
| " #Allocating space for the solution vector\n", | |
| " y = np.zeros_like(b, dtype=np.double);\n", | |
| " #Here we perform the forward-substitution. \n", | |
| " #Initializing with the first row.\n", | |
| " y[0] = b[0] / L[0, 0]\n", | |
| " #Looping over rows in reverse (from the bottom up),\n", | |
| " #starting with the second to last row, because the \n", | |
| " #last row solve was completed in the last step.\n", | |
| " for i in range(1, n):\n", | |
| " y[i] = (b[i] - np.dot(L[i,:i], y[:i])) / L[i,i] \n", | |
| " return y\n", | |
| "\n", | |
| "\n", | |
| "def back_substitution(U, y):\n", | |
| " #source https://johnfoster.pge.utexas.edu/numerical-methods-book/LinearAlgebra_LU.html\n", | |
| " #Number of rows\n", | |
| " n = U.shape[0]\n", | |
| " #Allocating space for the solution vector\n", | |
| " x = np.zeros_like(y, dtype=np.double);\n", | |
| " #Here we perform the back-substitution. \n", | |
| " #Initializing with the last row.\n", | |
| " x[-1] = y[-1] / U[-1, -1]\n", | |
| " #Looping over rows in reverse (from the bottom up), \n", | |
| " #starting with the second to last row, because the \n", | |
| " #last row solve was completed in the last step.\n", | |
| " for i in range(n-2, -1, -1):\n", | |
| " x[i] = (y[i] - np.dot(U[i,i:], x[i:])) / U[i,i]\n", | |
| " return x\n", | |
| "\n", | |
| "y = forward_substitution(L, b)\n", | |
| "x = back_substitution(U, y)\n", | |
| "\n", | |
| "print(x)\n" | |
| ] | |
| } | |
| ] | |
| } |
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