"This function is needed in the algorithm. Note that this is a NON-functioning version, for now it's just a place holder. When algo2_testing will be completed, this will be updated and I'll work on algo4_testing."
"## Arnoldi \n",
"\n",
"This is a copy of the algorithm defined and tested in the notebook `algo2_testing`. It's an implementation of the Algorithm 2 from the paper. It's needed in this notebook since this function is called by the `algo4` function. It's implemented to return exactly what's needed in the `algo4` function.\n",
"\n",
"Everything will be reorganized in the main.py file once everything is working."
]
},
{
"cell_type": "code",
"execution_count": 2,
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
@ -42,7 +46,6 @@
" V[:,0] = v # each column of V is a vector v\n",
"\n",
" for j in range(m):\n",
" # print(\"j = \", j)\n",
" w = A @ v \n",
" for i in range(j):\n",
" tmp = v.T @ w # tmp is a 1x1 matrix, so it's O(1) in memory\n",
@ -61,12 +64,6 @@
" v = w/H[j+1,j]\n",
" V[:,j+1] = v\n",
"\n",
" # print(j, \" iterations completed\")\n",
" # print(\"V = \", V.shape)\n",
" # print(\"H = \", H.shape)\n",
" # print(\"v = \", v.shape)\n",
" # print(\"beta = \", beta)\n",
"\n",
" return V, H, v, beta, j "
]
},
@ -74,36 +71,47 @@
"cell_type": "markdown",
"metadata": {},
"source": [
"## Algorithm 4 testing\n",
"# Algorithm 4 testing\n",
"\n",
"This algorithm is based on the \"Algorithm 4\" of the paper, the pseudocode provided by the authors is the following \n",
"\n",
"Still a complete mess. Conceptually and technically wrong. I'll work on it when algo2_testing will be completed."
"\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": 15,
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"def Algo4(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 = 1\n",
" mv = 0\n",
" I = sp.sparse.eye(n, n, format='lil')\n",
" r = sp.sparse.lil_matrix((n,1))\n",
" res = np.zeros(len(a)) \n",
"\n",
" H_e1 = np.zeros((m+1,1))\n",
" # I'm defining 3 canonical vectors of different sizes. It's probably stupid, will be fixed once the algorithm actually works\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",
" # solve the system, without using the least squares method\n",
" # CONVERT TO CSC FORMAT TO USE SPARSE SOLVERS\n",
" A_tmp = A_tmp.tocsc()\n",
"\n",
" result = sp.sparse.linalg.spsolve(A_tmp, b_tmp)[0]\n",
" print(\"result:\", result.shape)\n",
" \n",
" # What should I put here? What does it mean in the paper the 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",
" # split the result into y and gamma\n",
" y = result[0:y.shape[0]]\n",
" gamma = result[y.shape[0]:]\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",
" # update x\n",
" x = x + tmp @ y\n",
" # update residual\n",
" res = (a[i]/a[k])*res*gamma[k]\n",
" iter = iter + 1\n",
"\n",
" iter = iter + 1"
" return x, iter, mv"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Basic test case with random numbers to test the algorithm."
]
},
{
"cell_type": "code",
"execution_count": 16,
"execution_count": null,
"metadata": {},
"outputs": [
{
"name": "stderr",
"output_type": "stream",
"text": [
"/tmp/ipykernel_264197/102804730.py:12: FutureWarning: adjacency_matrix will return a scipy.sparse array instead of a matrix in Networkx 3.0.\n",
"\u001b[0;32m/tmp/ipykernel_264197/102804730.py\u001b[0m in \u001b[0;36m<module>\u001b[0;34m\u001b[0m\n\u001b[1;32m 16\u001b[0m \u001b[0mPt\u001b[0m \u001b[0;34m=\u001b[0m \u001b[0mA\u001b[0m\u001b[0;34m.\u001b[0m\u001b[0mT\u001b[0m\u001b[0;34m\u001b[0m\u001b[0;34m\u001b[0m\u001b[0m\n\u001b[1;32m 17\u001b[0m \u001b[0;31m# run the algorithm\u001b[0m\u001b[0;34m\u001b[0m\u001b[0;34m\u001b[0m\u001b[0m\n\u001b[0;32m---> 18\u001b[0;31m \u001b[0mAlgo4\u001b[0m\u001b[0;34m(\u001b[0m\u001b[0mPt\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[0mmaxit\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",
"\u001b[0;32m/tmp/ipykernel_264197/2668036735.py\u001b[0m in \u001b[0;36mAlgo4\u001b[0;34m(Pt, v, m, a, tau, maxit, x)\u001b[0m\n\u001b[1;32m 83\u001b[0m \u001b[0;34m\u001b[0m\u001b[0m\n\u001b[1;32m 84\u001b[0m \u001b[0;31m# split the result into y and gamma\u001b[0m\u001b[0;34m\u001b[0m\u001b[0;34m\u001b[0m\u001b[0m\n\u001b[0;32m---> 85\u001b[0;31m \u001b[0my\u001b[0m \u001b[0;34m=\u001b[0m \u001b[0mresult\u001b[0m\u001b[0;34m[\u001b[0m\u001b[0;36m0\u001b[0m\u001b[0;34m:\u001b[0m\u001b[0my\u001b[0m\u001b[0;34m.\u001b[0m\u001b[0mshape\u001b[0m\u001b[0;34m[\u001b[0m\u001b[0;36m0\u001b[0m\u001b[0;34m]\u001b[0m\u001b[0;34m]\u001b[0m\u001b[0;34m\u001b[0m\u001b[0;34m\u001b[0m\u001b[0m\n\u001b[0m\u001b[1;32m 86\u001b[0m \u001b[0mgamma\u001b[0m \u001b[0;34m=\u001b[0m \u001b[0mresult\u001b[0m\u001b[0;34m[\u001b[0m\u001b[0my\u001b[0m\u001b[0;34m.\u001b[0m\u001b[0mshape\u001b[0m\u001b[0;34m[\u001b[0m\u001b[0;36m0\u001b[0m\u001b[0;34m]\u001b[0m\u001b[0;34m:\u001b[0m\u001b[0;34m]\u001b[0m\u001b[0;34m\u001b[0m\u001b[0;34m\u001b[0m\u001b[0m\n\u001b[1;32m 87\u001b[0m \u001b[0;34m\u001b[0m\u001b[0m\n",
"\u001b[0;31mIndexError\u001b[0m: invalid index to scalar variable."