{ "cells": [ { "cell_type": "code", "execution_count": 1, "metadata": {}, "outputs": [], "source": [ "import numpy as np\n", "import networkx as nx\n", "import time\n", "import math\n", "import pandas as pd\n", "import scipy as sp\n", "import plotly.express as px\n", "import plotly.graph_objs as go\n", "from scipy.sparse import *\n", "from scipy import linalg\n", "from scipy.sparse.linalg import norm\n", "from scipy.optimize import least_squares" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "# ALGORITHM 1 TESTING\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Let's create two graphs from the list of edges downloaded from the Snap database. " ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [ "G1 = nx.read_edgelist('../data/web-Stanford.txt', create_using=nx.DiGraph(), nodetype=int)\n", "\n", "G2 = nx.read_edgelist('../data/web-BerkStan.txt', create_using=nx.DiGraph(), nodetype=int)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Creating the transition probability matrix" ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [ "# square matrix of size n x n, where n is the number of nodes in the graph. The matrix is filled with zeros and the (i,j) element is x if the node i is connected to the node j. Where x is 1/(number of nodes connected to i).\n", "\n", "def create_matrix(G):\n", " n = G.number_of_nodes()\n", " P = sp.sparse.lil_matrix((n,n))\n", " for i in G.nodes():\n", " for j in G[i]: #G[i] is the list of nodes connected to i, it's neighbors\n", " P[i-1,j-1] = 1/len(G[i])\n", " return P" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "To ensure that the random process has a unique stationary distribution and it will not stagnate, the transition matrix P is usually modified to be an irreducible stochastic matrix A (called the Google matrix) as follows\n", "\n", "$$ A = \\alpha \\tilde{P} + (1-\\alpha)v e^T$$\n", "\n", "Where $\\tilde{P}$ is defined as \n", "\n", "$$ \\tilde{P} = P + v d^T$$\n", "\n", "Where $d \\in \\mathbb{N}^{n \\times 1}$ s a binary vector tracing the indices of dangling web-pages with no hyperlinks, i.e., $d(i ) = 1$ if the `ith` page has no hyperlink, $v \\in \\mathbb{R}^{n \\times 1}$ is a probability vector, $e = [1, 1, . . . , 1]^T$ , and $0 < \\alpha < 1$ is the so-called damping factor that represents the probability in the model that the surfer transfer by clicking a hyperlink rather than other ways" ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [ "n = G1.number_of_nodes()\n", "P = create_matrix(G1) " ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "the vector `d` solves the dangling nodes problem" ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [ "# define d as a nx1 sparse matrix, where n is the number of nodes in the graph. The vector is filled with d(i) = 1 if the i row of the matrix P is filled with zeros, other wise is 0\n", "\n", "# d is the vector of dangling nodes\n", "d = sp.sparse.lil_matrix((n,1))\n", "for i in range(n):\n", " if P[i].sum() == 0:\n", " d[i] = 1" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "The vector v is a probability vector, the sum of its elements bust be one" ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [ "# define v as the probability vector of size n x 1, where n is the number of nodes in the graph. The vector is filled with 1/n\n", "# https://en.wikipedia.org/wiki/Probability_vector\n", "\n", "v = sp.sparse.lil_matrix((n,1))\n", "for i in range(n):\n", " v[i] = 1/n " ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Now we can compute the transition matrix\n" ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [ "Pt = P + v.dot(d.T)\n", "\n", "# Pt is a sparse matrix too" ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [ "# e is a nx1 sparse matrix filled with ones\n", "e = sp.sparse.lil_matrix((1,n))\n", "for i in range(n):\n", " e[0,i] = 1" ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [ "# # v*eT is a nxn sparse matrix filled all with 1/n, let's call it B\n", "\n", "# B = sp.sparse.lil_matrix((n,n))\n", "# for i in range(n):\n", "# for j in range(n):\n", "# B[i,j] = 1/n\n", "\n", "# A = alpha*Pt + (1-alpha)*B" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Algorithm 1 Shifted-Power method for PageRank with multiple damping factors:" ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [ "# pandas dataframe to store the results\n", "df = pd.DataFrame(columns=['alpha', 'iterations', 'tau', 'time'])" ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [ "def Algorithm1(Pt, v, tau, max_mv, a: list):\n", " \n", " start_time = time.time()\n", "\n", " u = Pt.dot(v) - v \n", " mv = 1 # number of matrix vector products\n", " r = sp.sparse.lil_matrix((n,1)) \n", " Res = sp.sparse.lil_matrix((len(a),1))\n", " x = sp.sparse.lil_matrix((n,1)) \n", "\n", " for i in range(len(a)):\n", " r = a[i]*(u) \n", " normed_r = norm(r)\n", " Res[i] = normed_r \n", "\n", " if Res[i] > tau:\n", " x = r + v \n", "\n", " while max(Res) > tau and mv < max_mv:\n", " u = Pt*u \n", " mv += 1 \n", "\n", " for i in range(len(a)):\n", " if Res[i] >= tau: \n", " r = (a[i]**(mv+1))*(u)\n", " Res[i] = norm(r)\n", "\n", " if Res[i] > tau:\n", " x = r + x\n", "\n", " if mv == max_mv:\n", " print(\"The algorithm didn't converge in \", max_mv, \" iterations\")\n", " else:\n", " print(\"The algorithm converged in \", mv, \" iterations\")\n", "\n", " total_time = time.time() - start_time\n", " print(\"The algorithm took \", total_time, \" seconds\")\n", " \n", " return mv, x, r, total_time " ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Testing setup" ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [ "# list of alpha values, from 0.85 to 0.99 with step 0.01\n", "a = []\n", "for i in range(85,100):\n", " a.append(i/100)\n", "\n", "max_mv = 1000\n", "\n", "# run the algorithm for different values of tau from 10^-5 to 10^-9 with step 10^-1\n", "for i in range(5,10):\n", " tau = 10**(-i)\n", " print(\"\\ntau = \", tau)\n", " mv, x, r, total_time = Algorithm1(Pt, v, tau, max_mv, a)\n", " df = df.append({'alpha': a, 'iterations': mv, 'tau': tau, 'time': total_time}, ignore_index=True) " ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [ "# save the results in a csv file\n", "df.to_csv('../data/results/algo1/different_tau.csv', index=False)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Plotting the results of the algorithm for different values of tau, and fixed alpha" ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [ "x = df['tau'][::-1].tolist()\n", "y = df['products m-v'].tolist()\n", "\n", "fig = go.Figure(data=go.Scatter(x=x, y=y, mode='lines+markers'),\n", " layout=go.Layout(title='products needed for the convergence', xaxis_title='tau', yaxis_title='products matrix vector'))\n", "\n", "# save the figure as a html file\n", "fig.write_html(\"../data/results/algo1/taus_over_prods.html\")\n", "print(\"The plot has been saved in the folder data/results/algo1\")\n", "\n", "##### RESULTS OVER TIME #####\n", "\n", "x1 = df['tau'][::-1].tolist()\n", "y1 = df['time'].tolist()\n", "\n", "fig = go.Figure(data=go.Scatter(x=x1, y=y1, mode='lines+markers'),\n", " layout=go.Layout(title='Time needed for the convergence', xaxis_title='tau', yaxis_title='time (seconds)'))\n", "\n", "# save the plot in a html file\n", "fig.write_html(\"../data/results/algo1/taus_over_time.html\")\n", "print(\"The plot has been saved in the folder data/results/algo1\")\n" ] } ], "metadata": { "kernelspec": { "display_name": "Python 3.10.6 64-bit", "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": "3.10.6" }, "orig_nbformat": 4, "vscode": { "interpreter": { "hash": "916dbcbb3f70747c44a77c7bcd40155683ae19c65e1c03b4aa3499c5328201f1" } } }, "nbformat": 4, "nbformat_minor": 2 }