{
 "cells": [
  {
   "cell_type": "code",
   "execution_count": null,
   "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.sparse.linalg import norm"
   ]
  },
  {
   "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": [
    "# this should return mv (the number of iteration needed for the convergence), and two vector called x and r. Where x is the vector of the pagerank and r is the residual vector\n",
    "\n",
    "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 iteration\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 # should it be the same u of the beginning?\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": "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) \n",
    "\n"
   ]
  },
  {
   "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['iterations'].tolist()\n",
    "\n",
    "fig1 = go.Figure(data=go.Scatter(x=x, y=y, mode='lines+markers'), \n",
    "                                layout=go.Layout(title='Iterations needed for the convergence', xaxis_title='tau', yaxis_title='iterations'))\n",
    "                                                                                \n",
    "# save the plot in a html file\n",
    "fig1.write_html(\"../data/results/algo1/taus_over_iterations.html\")\n",
    "\n",
    "##### RESULTS OVER TIME #####\n",
    "\n",
    "x1 = df['tau'][::-1].tolist()\n",
    "y1 = df['time'].tolist()\n",
    "\n",
    "fig2 = 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",
    "fig2.write_html(\"../data/results/algo1/taus_over_time.html\")\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "To view the graph just use the command\n",
    "\n",
    "```bash\n",
    "firefox taus_over_iterations.html \n",
    "```\n",
    "or \n",
    "\n",
    "```bash\n",
    "firefox taus_over_time.html\n",
    "```\n",
    "\n",
    "_In the right folder_"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "def Arnoldi(A, v, m): #  defined ad algorithm 2 in the paper\n",
    "    beta = norm(v)\n",
    "    print(\"A\")\n",
    "    v = v/beta\n",
    "    print(\"B\")\n",
    "    h = sp.sparse.lil_matrix((m,m))\n",
    "    print(\"C\")\n",
    "\n",
    "    for j in range(1,m):\n",
    "        w = A.dot(v)\n",
    "        for i in range(1,j):\n",
    "            h[i,j] = v.T.dot(w)\n",
    "            w = w - h[i,j]*v\n",
    "\n",
    "        h[j+1,j] = norm(w)\n",
    "\n",
    "        if h[j+1,j] == 0:\n",
    "            m = j\n",
    "            v[m+1] = 0\n",
    "            break\n",
    "        else:\n",
    "            v = w**h[j+1,j]\n",
    "\n",
    "    return v, h, m, beta, j"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "A = sp.sparse.rand(100,100, density=0.5, format='lil')\n",
    "v = sp.sparse.rand(100,1, density=1, format='lil')"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "v, h, m, beta, j = Arnoldi(A, v, 100)"
   ]
  }
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