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{
"cells": [
{
"cell_type": "code",
"execution_count": 1,
"metadata": {},
"outputs": [],
"source": [
"import numpy as np\n",
"import scipy as sp"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Arnoldi"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Defined as Algorithm 2 in the paper."
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {},
"outputs": [],
"source": [
"def Arnoldi(A,v0,m):\n",
" v = v0\n",
" beta = sp.linalg.norm(v)\n",
" v = v/beta\n",
" H = sp.sparse.lil_matrix((m+1,m)) \n",
" V = sp.sparse.lil_matrix((A.shape[0],m+1))\n",
" V[:,0] = v # each column of V is a vector v\n",
"\n",
" for j in range(m):\n",
" w = A @ v \n",
" for i in range(j):\n",
" H[i,j] = v.T @ w # tmp is a 1x1 matrix, so it's O(1) in memory\n",
" w = w - H[i,j]*v \n",
" \n",
" H[j+1,j] = np.linalg.norm(w)\n",
"\n",
" if H[j+1,j] == 0:\n",
" print(\"Arnoldi breakdown\")\n",
" m = j\n",
" v = 0\n",
" break\n",
" else:\n",
" if j < m-1:\n",
" v = w/H[j+1,j]\n",
" V[:,j+1] = v\n",
"\n",
" return V, H, beta, j "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Small test case\n",
"\n",
"The final implementation will be using all sparse arrays and matrices, no numpy arrays"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {},
"outputs": [],
"source": [
"n = 110\n",
"\n",
"# start with a random sparse matrix\n",
"A = sp.sparse.rand(n,n, density=0.1, format='lil')\n",
"\n",
"# Starting vector\n",
"v = np.repeat(1/n,n)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Now we can run the Arnoldi Process"
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The number of iterations is: 109\n",
"The matrix H is: (111, 110)\n",
"The matrix V is: (110, 111)\n"
]
}
],
"source": [
"V, H, beta, j = Arnoldi(A,v,110)\n",
"print(\"The number of iterations is: \", j)\n",
"print(\"The matrix H is: \", H.shape)\n",
"print(\"The matrix V is: \", V.shape)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"As we can see, it returns $H_{m+1}$ that is an upper-hassemberg matrix, with one extra row."
]
}
],
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"mimetype": "text/x-python",
"name": "python",
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"pygments_lexer": "ipython3",
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