@ -2,7 +2,7 @@
"cells": [
{
"cell_type": "code",
"execution_count": 3 ,
"execution_count": 2 ,
"metadata": {},
"outputs": [],
"source": [
@ -33,7 +33,7 @@
},
{
"cell_type": "code",
"execution_count": 4 ,
"execution_count": 28 ,
"metadata": {},
"outputs": [],
"source": [
@ -63,10 +63,11 @@
" v = w/H[j+1,j]\n",
" V[:,j+1] = v\n",
"\n",
" return V, H, v, beta, j "
" return V, H, beta, j "
]
},
{
"attachments": {},
"cell_type": "markdown",
"metadata": {},
"source": [
@ -74,264 +75,172 @@
"\n",
"This algorithm is based on the \"Algorithm 4\" of the paper, the pseudocode provided by the authors is the following \n",
"\n",
"\n",
"\n",
"Line 14 is particularly tricky to understand, not working for now. Need to figure out how to solve that linear system. My idea was to do something like that\n",
"\n",
"\n",
"\n",
"And use the `sp.sparse.linalg.spsolve` function to solve the linear system as $Ax=0$ where $A$ is $[\\bar H_m^i ~ | ~ z]$ but it returns an array of zeros. So the idea it's wrong"
""
]
},
{
"cell_type": "code",
"execution_count": 5 ,
"execution_count": 1,
"metadata": {},
"outputs": [],
"source": [
"# def Algo4_OLD(Pt, v, m, a: list, tau, maxit: int, x):\n",
" \n",
"# # I'm using a non declared variable n here , it's declared in the next cell when I call this function. This will be fixed later in the main.py file\n",
"\n",
"# iter , mv = 1, 0\n",
"# res = np.zeros(len(a)) \n",
"\n",
"# # I'm defining 3 canonical vectors of different sizes. It's probably stupid, will be fixed once the algorithm actually works\n",
"\n",
"# H_e1 = np.zeros((m+1,1)) # canonical basis vector of size H.shape[0]\n",
"# H_e1[0] = 1\n",
"\n",
"# V_e1 = np.zeros((n,1)) # canonical basis vector of size V.shape[0]\n",
"# V_e1[0] = 1\n",
"\n",
"# s_e1 = np.zeros((len(a),1)) # canonical basis vector of size s.shape[0]\n",
"# s_e1[0] = 1\n",
"\n",
"# def find_k(res): # function to find the index of the largest element in res\n",
"# k = 0\n",
"# for i in range(len(a)):\n",
"# if res[i] == max(res):\n",
"# k = i\n",
"# break\n",
"# return k\n",
"\n",
"# def compute_gamma(res, a, k): # function to compute gamma\n",
"# gamma = np.zeros(len(a))\n",
"# for i in range(len(a)):\n",
"# if i != k:\n",
"# gamma[i] = (res[i]*a[k])/(res[k]*a[i])\n",
"# else:\n",
"# gamma[i] = 0\n",
"# return gamma\n",
"\n",
"# # compute the residual vector\n",
"\n",
"# # create a sp.sparse.eye matrix of size Pt\n",
"# I = sp.sparse.eye(Pt.shape[0], Pt.shape[1], format='lil')\n",
"# for i in range(len(a)):\n",
"# r = ((1-a[i])/a[i])*v - ((1/a[i])*I - Pt) @ x\n",
"# res[i] = a[i]*norm(r)\n",
"\n",
"\n",
"# while max(res) >= tau and iter <= maxit:\n",
"# k = find_k(res)\n",
"# gamma = compute_gamma(res, a, k)\n",
"# V, H, v, beta, j = Arnoldi((1/a[k])*I - Pt, r, m)\n",
"\n",
"# mv = mv + j\n",
"\n",
"# # compute y as the minimizer of || beta*e1 - Hy ||_2 using the least squares method\n",
"# y = sp.sparse.linalg.lsqr(H, beta*H_e1)[0]\n",
"\n",
"# # reshape y to be a column vector\n",
"# y = y.reshape(y.shape[0],1)\n",
"\n",
"# # update x \n",
"# x += V[:,0:y.shape[0]] @ y\n",
"\n",
"# # compute the residual vector\n",
"# res[k] = a[k]*np.linalg.norm(beta*V_e1 - V[:,0:y.shape[0]] @ y)\n",
" \n",
"# # for i in range(len(a)) but not k\n",
"# for i in range(len(a)):\n",
"# if i != k and res[i] >= tau:\n",
"# # Compute H as described in the paper\n",
"# H = H + ((1-a[i])/a[i] - (1-a[k])/a[k])*sp.sparse.eye(H.shape[0], H.shape[1], format='lil') \n",
"\n",
"# z = beta*H_e1 - H @ y # define z as in the paper (page 9)\n",
"# A_tmp = sp.sparse.hstack([H, z]) # stack H and z, as in the paper, to solve the linear system (?)\n",
"# A_tmp = A_tmp.tocsc() # Convert A to CSC format for sparse solver\n",
"\n",
"# # What should I put here? What does it mean in the paper the line 14 of the pseudocode?\n",
"# result = sp.sparse.linalg.spsolve(A_tmp, np.zeros(A_tmp.shape[0])) # if I solve this, I get a vector of zeros.\n",
"# print(result)\n",
" \n",
"# # I don't know if the code below is correct since I don't get how to solve the linear system above, so I'm unsure about what y and gamma should be. For now it's commented out.\n",
"\n",
"# # # update x\n",
"# # x += V[:,0:y.shape[0]] @ y\n",
"# # # update the residual vector\n",
"# # res[i] = (a[i]/a[k])*gamma[k]*res[k] \n",
"\n",
"# iter = iter + 1\n",
"\n",
"# return x, iter, mv"
"def compute_gamma(res, a, k): # function to compute gamma\n",
" gamma = np.ones(len(a))\n",
" for i in range(len(a)):\n",
" if i != k:\n",
" gamma[i] = (res[i]*a[k])/(res[k]*a[i])\n",
" else:\n",
" gamma[i] = 0\n",
" return gamma"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Basic test case with random numbers to test the algorithm."
]
},
{
"cell_type": "code",
"execution_count": 109 ,
"execution_count": 35,
"metadata": {},
"outputs": [],
"source": [
"def find_k(res): # function to find the index of the largest element in res\n",
" k = 0\n",
" for i in range(len(a)):\n",
" if res[i] == max(res):\n",
" k = i\n",
" break\n",
" return k\n",
"n = 1000\n",
"m = 1100\n",
"tau = 1e-6\n",
"a = [0.85, 0.88, 0.9, 0.95]\n",
"\n",
"def compute_gamma(res, a, k): # function to compute gamma\n",
" gamma = np.zeros(len(a))\n",
" for i in range(len(a)):\n",
" if i != k:\n",
" gamma[i] = (res[i]*a[k])/(res[k]*a[i])\n",
" else:\n",
" gamma[i] = 0\n",
" return gamma"
"x = sp.sparse.lil_matrix((n,1))\n",
"x[0,0] = 1\n",
"\n",
"# generate a random graph\n",
"G = nx.gnp_random_graph(n, 0.1)\n",
"v = np.repeat(1.0 / 1000, 1000) # p is the personalization vector\n",
"v = v.reshape(v.shape[0],1)\n",
"\n",
"A = nx.to_scipy_sparse_array(G, dtype=float)\n",
"S = A.sum(axis=1) # S[i] is the sum of the weights of edges going out of node i\n",
"S[S != 0] = 1.0 / S[S != 0] # S[i] is now the sum of the weights of edges going into node i\n",
"Q = sp.sparse.csr_array(sp.sparse.spdiags(S.T, 0, *A.shape)) # Q is the matrix of edge weights"
]
},
{
"cell_type": "code",
"execution_count": 12 3,
"execution_count": 83,
"metadata": {},
"outputs": [],
"source": [
"def Algo4(Pt, v, m, a: list, tau, maxit: int, x):\n",
"\n",
" mv, iter = 0, 1 # mv is the number of matrix-vector products, iter is the number of iterations\n",
" \n",
" # initialize x as a random sparse matrix. Each col is the pagerank vector for a different alpha\n",
" x = sp.sparse.lil_matrix((Pt.shape[0], len(a)))\n",
" mv = 0\n",
"\n",
" # compute the residual vector\n",
"\n",
" # compute the residual vector, it is a matrix of size (n, len(a)). Each col is the residual vector for a different alpha. \n",
" I = sp.sparse.eye(Pt.shape[0], Pt.shape[1], format='lil')\n",
" r = sp.sparse.lil_matrix((Pt.shape[0], len(a)))\n",
" res = np.zeros(len(a))\n",
"\n",
" for i in range(len(a)):\n",
" r = sp.sparse.linalg.spsolve(I - a[i]*Pt, v)\n",
" res[i] = a[i]*np.linalg.norm(r)\n",
" r[:,[i]] = sp.sparse.linalg.spsolve(I - a[i]*Pt, v)\n",
" col = r[:,[i]].toarray()\n",
" res[i] = np.linalg.norm(col)\n",
"\n",
" iter = 0\n",
" for _ in range(maxit):\n",
" print(\"\\niter: \", iter)\n",
" k = find_k(res)\n",
" # check if we have converged\n",
" err = np.absolute(np.amax(res))\n",
" if err < tau:\n",
" print(\"Computation ended successfully in \", iter, \" iterations and \", mv, \" matrix-vector products.\")\n",
" return x, iter, mv\n",
"\n",
" print(\"\\niter = \", iter)\n",
" print(\"res: \", res)\n",
" print(\"err = \", err)\n",
"\n",
"\n",
" k = int(np.argmax(res))\n",
" print(\"k = \", k)\n",
" gamma = compute_gamma(res, a, k)\n",
" \n",
" # Run Arnoldi\n",
" V, H, v, beta, j = Arnoldi((1/a[k])*I - Pt, r, m)\n",
"\n",
" r_k = r[:,[k]].toarray()\n",
" A_arnoldi = (1/a[k])*I - Pt\n",
" V, H, beta, j = Arnoldi((1/a[k])*I - Pt, r_k, m)\n",
" H = H[:-1,:]\n",
" V = V[:,:-1]\n",
" mv = mv + j\n",
"\n",
" # compute y as the minimizer of || beta*e1 - Hy ||_2 using the least squares method\n",
" H_e1 = np.zeros(H.shape[0])\n",
" H_e1[0] = 1\n",
" y = sp.sparse.linalg.lsqr(H, beta*H_e1)[0]\n",
"\n",
" # compute y as the minimizer of || beta*e1 - Hy ||_2 using the least squares method\n",
" y = sp.sparse.lil_matrix((H.shape[1],len(a)))\n",
" y[:,[k]] = sp.sparse.linalg.lsqr(H, beta*H_e1)[0]\n",
" y_k = y[:,[k]].toarray()\n",
"\n",
" # # Update x\n",
" x_new = x\n",
" x_new[:,[k]] = x[:,[k]] + V @ y_k\n",
"\n",
" # Update res[k]\n",
" y = y.reshape(y.shape[0],1)\n",
" V_e1 = np.zeros(V.shape[0])\n",
" V_e1[0] = 1\n",
"\n",
" res[k] = a[k]*np.linalg.norm(beta*V_e1 - V @ y) \n",
" print(\"res[\", k, \"] = \", res[k]) \n",
" norm_k =np.linalg.norm(beta*V_e1 - V @ y_k) \n",
" res[k] = a[k]*norm_k \n",
"\n",
" # multi shift\n",
" for i in range(len(a)):\n",
" print(\"res = \", res)\n",
" if i != k and res[i] >= tau:\n",
" print(\"res[\", i, \"] is greater than tau = \", tau, \" at iteration \", iter, \" entering the for loop\")\n",
" if res[i] >= tau:\n",
" # print(\"res[\", i, \"] is larger than tau = \", tau)\n",
"\n",
" # Compute H as described in the paper\n",
" H_new = H + ((1-a[i])/a[i] - (1-a[k])/a[k])*sp.sparse.eye(H.shape[0], H.shape[1], format='lil')\n",
" # # Compute H as described in the paper\n",
" # H_k = H[:,[k]].toarray()\n",
" # H_i = H_k + ((1-a[i])/a[i] - (1-a[k])/a[k])\n",
" # H[:,[i]] = H_i\n",
" H = H + ((1-a[i])/a[i] - (1-a[k])/a[k])*sp.sparse.eye(H.shape[0], H.shape[1], format='lil')\n",
"\n",
" # Compute z as described in the paper\n",
" tmp = H_e1*beta \n",
" z1, z2 = tmp.reshape(tmp.shape[0],1), H @ y\n",
" z1 = H_e1*beta\n",
" z1 = z1.reshape(z1.shape[0],1)\n",
" z2 = H @ y[:,[1]]\n",
" z2 = z2.reshape(z2.shape[0],1)\n",
" z = z1 - z2\n",
"\n",
" # Solve the linear system \n",
" A = sp.sparse.hstack([H_new , z])\n",
" b = gamma[i]* (beta*H_e1)\n",
" A = sp.sparse.hstack([H, z])\n",
" b = (beta*H_e1)\n",
" b = b.reshape(b.shape[0],1)\n",
" \n",
" # use the least squares method to solve the linear system\n",
" y_to_split = sp.sparse.linalg.lsqr(A, b)[0]\n",
" \n",
" # the last element of y_to_split is the last element of gamma[i], the other elements are the elements of y\n",
" y = y_to_split[:-1].reshape(y_to_split[:-1].shape[0],1)\n",
" gamma[i] = y_to_split[-1]\n",
" to_split = sp.sparse.linalg.lsqr(A, b)[0]\n",
" \n",
" x_new = x\n",
" x_new[:,[k]] = x[:,[k]] + V @ y\n",
" # the last element of y_to_split is the last element of gamma[i], the other elements are the elements of y[:[i]]\n",
" y[:,[i]] = to_split[:-1]\n",
" gamma[i] = to_split[-1]\n",
"\n",
" # update x\n",
" x_new[:,i] = x[:,i] + V @ y\n",
" x_new[:,i] = x[:,i] + V @ y[:,[i]] \n",
"\n",
" # update the residual vector\n",
" res[i] = (a[i]/a[k])*gamma[k]*res[k]\n",
" print(\"res[\", i, \"] = \", res[i])\n",
" res[i] = (a[i]/a[k])*pow(gamma[i], i)*res[k]\n",
"\n",
" else:\n",
" if res[i] < tau:\n",
" print(\"res[\", i, \"] = \", res[i], \" is smaller than tau = \", tau, \" at iteration \", iter)\n",
"\n",
" err = np.absolute(res).max()\n",
" print(\"err = \", err)\n",
" if err < tau:\n",
" return x_new, iter, mv\n",
" print(\"res[\", i, \"] is smaller than tau = \", tau, \" at iteration \", iter)\n",
"\n",
" iter = iter + 1\n",
" x = x_new\n",
"\n",
" raise Exception('Maximum number of iterations reached')\n",
" \n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Basic test case with random numbers to test the algorithm."
]
},
{
"cell_type": "code",
"execution_count": 124,
"metadata": {},
"outputs": [],
"source": [
"n = 1000\n",
"m = 1100\n",
"tau = 1e-5\n",
"a = [0.85, 0.9, 0.95, 0.99]\n",
"\n",
"x = sp.sparse.lil_matrix((n,1))\n",
"x[0,0] = 1\n",
"\n",
"# generate a random graph\n",
"G = nx.gnp_random_graph(n, 0.01)\n",
"v = np.repeat(1.0 / 1000, 1000) # p is the personalization vector\n",
"v = v.reshape(v.shape[0],1)\n",
"\n",
"A = nx.to_scipy_sparse_array(G, dtype=float)\n",
"S = A.sum(axis=1) # S[i] is the sum of the weights of edges going out of node i\n",
"S[S != 0] = 1.0 / S[S != 0] # S[i] is now the sum of the weights of edges going into node i\n",
"Q = sp.sparse.csr_array(sp.sparse.spdiags(S.T, 0, *A.shape)) # Q is the matrix of edge weights"
" raise Exception('Maximum number of iterations reached')"
]
},
{
"cell_type": "code",
"execution_count": 125 ,
"execution_count": 84,
"metadata": {},
"outputs": [
{
@ -339,43 +248,39 @@
"output_type": "stream",
"text": [
"\n",
"iter: 0\n",
"res[ 3 ] = 1.9500347533542841\n",
"res = [0.02974037 0.03169717 0.03368163 1.95003475]\n",
"res[ 0 ] is greater than tau = 1e-05 at iteration 0 entering the for loop\n",
"res[ 0 ] = 0.0\n",
"res = [0. 0.03169717 0.03368163 1.95003475]\n",
"res[ 1 ] is greater than tau = 1e-05 at iteration 0 entering the for loop\n",
"res[ 1 ] = 0.0\n",
"res = [0. 0. 0.03368163 1.95003475]\n",
"res[ 2 ] is greater than tau = 1e-05 at iteration 0 entering the for loop\n",
"res[ 2 ] = 0.0\n",
"res = [0. 0. 0. 1.95003475]\n",
"err = 1.9500347533542841\n",
"\n",
"iter: 1\n",
"res[ 3 ] = 1.9500347533542841\n",
"res = [0. 0. 0. 1.95003475]\n",
"res[ 0 ] = 0.0 is smaller than tau = 1e-05 at iteration 1\n",
"res = [0. 0. 0. 1.95003475]\n",
"res[ 1 ] = 0.0 is smaller than tau = 1e-05 at iteration 1\n",
"res = [0. 0. 0. 1.95003475]\n",
"res[ 2 ] = 0.0 is smaller than tau = 1e-05 at iteration 1\n",
"res = [0. 0. 0. 1.95003475]\n",
"err = 1.9500347533542841\n",
"\n",
"iter: 2\n",
"res[ 3 ] = 1.9500347533542841\n",
"res = [0. 0. 0. 1.95003475]\n",
"res[ 0 ] = 0.0 is smaller than tau = 1e-05 at iteration 2\n",
"res = [0. 0. 0. 1.95003475]\n",
"res[ 1 ] = 0.0 is smaller than tau = 1e-05 at iteration 2\n",
"res = [0. 0. 0. 1.95003475]\n",
"res[ 2 ] = 0.0 is smaller than tau = 1e-05 at iteration 2\n",
"res = [0. 0. 0. 1.95003475]\n",
"err = 1.9500347533542841\n",
"\n",
"iter: 3\n"
"iter = 1\n",
"res: [0.03189738 0.03190716 0.03191369 0.03193001]\n",
"err = 0.031930006625941795\n",
"k = 0\n",
"\n",
"iter = 2\n",
"res: [1.11728737e+00 8.26005227e-04 5.55288870e-10 4.81520495e-13]\n",
"err = 1.1172873666904701\n",
"k = 3\n",
"res[ 2 ] is smaller than tau = 1e-06 at iteration 2\n",
"\n",
"iter = 3\n",
"res: [1.17714008e+00 1.29941354e-03 5.55288870e-10 1.93969263e-18]\n",
"err = 1.1771400826095457\n",
"k = 3\n",
"res[ 2 ] is smaller than tau = 1e-06 at iteration 3\n",
"\n",
"iter = 4\n",
"res: [1.17714008e+00 1.29941354e-03 5.55288870e-10 1.93969263e-18]\n",
"err = 1.1771400826095457\n",
"k = 3\n",
"res[ 2 ] is smaller than tau = 1e-06 at iteration 4\n",
"\n",
"iter = 5\n",
"res: [1.17714008e+00 1.29941354e-03 5.55288870e-10 1.93969263e-18]\n",
"err = 1.1771400826095457\n",
"k = 3\n",
"res[ 2 ] is smaller than tau = 1e-06 at iteration 5\n",
"\n",
"iter = 6\n",
"res: [1.17714008e+00 1.29941354e-03 5.55288870e-10 1.93969263e-18]\n",
"err = 1.1771400826095457\n",
"k = 3\n"
]
},
{
@ -385,10 +290,10 @@
"traceback": [
"\u001b[0;31m---------------------------------------------------------------------------\u001b[0m",
"\u001b[0;31mKeyboardInterrupt\u001b[0m Traceback (most recent call last)",
"\u001b[0;32m/tmp/ipykernel_85082 /3677688099.py\u001b[0m in \u001b[0;36m<module>\u001b[0;34m\u001b[0m\n\u001b[0;32m----> 1\u001b[0;31m \u001b[0mx\u001b[0m\u001b[0;34m,\u001b[0m \u001b[0miter\u001b[0m\u001b[0;34m,\u001b[0m \u001b[0mmv\u001b[0m \u001b[0;34m=\u001b[0m \u001b[0mAlgo4\u001b[0m\u001b[0;34m(\u001b[0m\u001b[0mQ\u001b[0m\u001b[0;34m,\u001b[0m \u001b[0mv\u001b[0m\u001b[0;34m,\u001b[0m \u001b[0mm\u001b[0m\u001b[0;34m,\u001b[0m \u001b[0ma\u001b[0m\u001b[0;34m,\u001b[0m \u001b[0mtau\u001b[0m\u001b[0;34m,\u001b[0m \u001b[0;36m100\u001b[0m\u001b[0;34m,\u001b[0m \u001b[0mx\u001b[0m\u001b[0;34m)\u001b[0m\u001b[0;34m\u001b[0m\u001b[0;34m\u001b[0m\u001b[0m\n\u001b[0m",
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]
}