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{
"cells": [
{
"cell_type": "markdown",
"id": "9785abf6",
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"source": [
"We have:\n",
"$$\n",
"\\nabla f = (I + \\beta A) \\vec s - \\vec y\n",
" = \\vec s - \\vec y + \\beta A \\vec s\n",
"$$ \n",
"We can substitute in the partial derivatives:\n",
"$$ \n",
"\\left(\\begin{matrix}\n",
"s_0 - y_0 + \\beta (s_0 - s_1) \\\\\n",
"s_1 - y_1 + \\beta (2s_1 - s_0 - s_2) \\\\\n",
"\\vdots \\\\\n",
"s_N - y_N + \\beta (s_N - s_{N - 1})\n",
"\\end{matrix}\\right) \n",
"= \\vec s - \\vec y + \\beta A \\vec s\n",
"$$\n",
"\n",
"Adding $\\vec y - \\vec s$ to both sides and then dividing by $\\beta$ yields:\n",
"$$ \n",
"\\left(\\begin{matrix}\n",
"s_0 - s_1 \\\\\n",
"2s_1 - s_0 - s_2 \\\\\n",
"\\vdots \\\\\n",
"s_N - s_{N - 1}\n",
"\\end{matrix}\\right) \n",
"= A \\vec s\n",
"$$\n",
"\n",
"Solving for $A$ is trivial and yields:\n",
"$$ \n",
"A = \\left(\\begin{matrix}\n",
"1 & -1 & & & & & \\\\\n",
"-1 & 2 & -1 & & & & \\\\\n",
" & -1 & 2 & -1 & & & \\\\\n",
" & & \\ddots & \\ddots & \\ddots & & \\\\\n",
" & & & -1 & 2 & -1 & \\\\\n",
" & & & & -1 & 2 & -1\\\\\n",
" & & & & & -1 & 1\n",
"\\end{matrix}\\right) \n",
"$$"
]
},
{
"cell_type": "code",
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"execution_count": 17,
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"id": "e7d69a97",
"metadata": {},
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"outputs": [],
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"source": [
"import numpy as np\n",
"import modules.tridiagonal as tridiagonal\n",
"\n",
"def sample_with_noise(points, vectorized_g):\n",
" \"\"\"\n",
" Sample a function at a given set of points,\n",
" introducing noise in the process.\n",
"\n",
" Returns a tuple of the real values of the function\n",
" and the values with introduced noise.\n",
"\n",
" For performance reasons, this function takes in\n",
" a vectorized version of `g`. If you want to pass\n",
" a non vectorized function use `np.vectorize`. Eg:\n",
" ```py\n",
" sample_with_noise(points, np.vectorize(g))\n",
" ```\n",
" \"\"\"\n",
" gs = vectorized_g(points)\n",
" ys = gs + 0.05 * np.random.randn(len(gs))\n",
"\n",
" return (gs, ys)\n",
"\n",
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"def p5(x):\n",
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" \"\"\"\n",
" Vectorized version of the function\n",
" described in the pdf.\n",
" \"\"\"\n",
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" n = 5\n",
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" result = 0\n",
"\n",
" for k in range(n + 1):\n",
" result += x**k\n",
"\n",
" return result/(n + 1)\n",
"\n",
"def smooth(data, beta):\n",
" \"\"\"\n",
" Smooth noisy data (y) by means of solving \n",
" a minimization problem.\n",
"\n",
" Args:\n",
" data (numpy.array): noisy data to be smoothed (y)\n",
" beta (float): parameter >= 0 that balances fit and smoothing\n",
"\n",
" Returns:\n",
" numpy array of smoothed data (s)\n",
" \"\"\"\n",
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" x,y = data\n",
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"\n",
" # We need to solve for s in:\n",
" # (I + βA)s - y = 0\n",
" # <=> (I + βA)s = y\n",
" #\n",
" # Because A and I are tridiagonal, I + βA\n",
" # is tridiagonal as well (linear combinations\n",
" # of tridiagonal matrices are themselves\n",
" # tridiagonal)\n",
" #\n",
" # This means we can use our existing implementation\n",
" # of tridiagonal linear equation system solving.\n",
" # \n",
" # We start by constructing \n",
" # iba := I + βA\n",
" n = len(x) - 1\n",
"\n",
" # The a component of the a-c-e tridiagonal representation of iba\n",
" iba_a = np.ones(n + 1) * (2 * beta + 1)\n",
" iba_a[0] = 1 + beta\n",
" iba_a[-1] = 1 + beta\n",
"\n",
" # I + βA\n",
" iba = (\n",
" iba_a,\n",
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" -beta*np.ones(n),\n",
" -beta*np.ones(n)\n",
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" )\n",
"\n",
" data_smoothed = tridiagonal.solve(*iba, y)\n",
"\n",
" return data_smoothed"
]
},
{
"cell_type": "code",
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"execution_count": 25,
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"id": "2b9f2b79",
"metadata": {},
"outputs": [
{
"data": {
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"text/plain": [
"<Figure size 640x480 with 1 Axes>"
]
},
"metadata": {},
"output_type": "display_data"
}
],
"source": [
"import matplotlib.pyplot as plot\n",
"\n",
"# Reusable function to plot the data\n",
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"def plot_denoise_results(beta, N, ax):\n",
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" # Create the data\n",
" x = np.linspace(-1, 1, num=N+1)\n",
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" g, y = sample_with_noise(x, p5)\n",
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" s = smooth((x, y), beta)\n",
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"\n",
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" ax.set_title(f\"β={beta} and {N=}\")\n",
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" ax.set_xlabel(\"x\")\n",
" ax.set_ylabel(\"y\")\n",
" ax.plot(x, y, \".\", label=\"raw data points\")\n",
" ax.plot(x, g, \"-\", label=\"original function\")\n",
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" ax.plot(x, s, \"-\", label=\"denoised values\")\n",
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" ax.legend()\n",
"\n",
"\n",
"N = 100\n",
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"beta = 10\n",
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"\n",
"fig, ax = plot.subplots()\n",
"fig.suptitle(f\"Data denoising results\")\n",
"\n",
"# Plot the data\n",
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"plot_denoise_results(beta, N, ax)"
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]
},
{
"cell_type": "markdown",
"id": "f2e5c860",
"metadata": {},
"source": [
"We notice that finding a point where $\\nabla f(s) = 0$ is the same as finding a point where $\\frac 1 \\beta \\nabla f(s) = \\nabla \\frac 1 \\beta f(s) = 0$. We can use this to our advantage, by defining $h(s) = \\frac 1 \\beta f(s)$ and finding the points where $\\nabla h(s) = 0$ instead. When $\\beta \\to \\infty$, we have:\n",
"$$\n",
"\\lim_{\\beta \\to \\inf} h _\\beta (s) =\n",
"\\lim_{\\beta \\to \\inf} \n",
" \\frac 1 {2\\beta} \\sum_{k=0}^n (y_k - s_k)^2\n",
" +\\frac 1 2 \\sum_{k=1}^n (s_k - s_{k - 1})^2\n",
" = \\frac 1 2 \\sum_{k=1}^n (s_k - s_{k - 1})^2\n",
"$$\n",
"\n",
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"It now becomes trivial to verify that $\\nabla f(s) = As$. Looking at the value of $A$ we notice that any vector with all values equal to some constant are solutions for this equation. This means the result will be a constant function (which might differ based on the linear equation solver we are using)."
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]
},
{
"cell_type": "code",
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"execution_count": 29,
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"id": "d7f7dbdf",
"metadata": {
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"scrolled": false
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},
"outputs": [
{
"data": {
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"image/png": "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"text/plain": [
"<Figure size 1280x960 with 9 Axes>"
]
},
"metadata": {},
"output_type": "display_data"
}
],
"source": [
"fig, axs = plot.subplots(\n",
" 3,\n",
" 3,\n",
" layout=\"constrained\",\n",
" figsize=(12.8, 9.6)\n",
")\n",
"\n",
"fig.suptitle(r\"Data denoising results as $\\beta \\to \\infty$\")\n",
"\n",
"N = 100\n",
"\n",
"for i in range(len(axs.flat)):\n",
" ax = axs.flat[i]\n",
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" plot_denoise_results(4**i, N, ax)"
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]
},
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{
"cell_type": "markdown",
"id": "3c9f50cc",
"metadata": {},
"source": [
"When $\\beta \\to 0$ we have:\n",
"$$\n",
"\\lim_{\\beta \\to 0} \\nabla f(s) =\n",
"\\lim_{\\beta \\to 0} (I + \\betaA)s - y = s - y\n",
"$$\n",
"\n",
"This means when solving for $\\nabla f(s) = 0$, we are solving the equation $s - y = 0$, which has the solution $s = y$. The resulting function will then be one equal to the data containing the noise."
]
},
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{
"cell_type": "code",
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"execution_count": 31,
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"id": "dfc485d5",
"metadata": {},
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"outputs": [
{
"data": {
"image/png": "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
"text/plain": [
"<Figure size 1280x960 with 6 Axes>"
]
},
"metadata": {},
"output_type": "display_data"
}
],
"source": [
"fig, axs = plot.subplots(\n",
" 2,\n",
" 3,\n",
" layout=\"constrained\",\n",
" figsize=(12.8, 9.6)\n",
")\n",
"\n",
"fig.suptitle(r\"Data denoising results as $\\beta \\to 0$\")\n",
"\n",
"N = 100\n",
"\n",
"for i in range(len(axs.flat)):\n",
" ax = axs.flat[i]\n",
" plot_denoise_results(2**-i, N, ax)"
]
},
{
"cell_type": "code",
"execution_count": null,
"id": "6ffa20a6",
"metadata": {},
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"outputs": [],
"source": []
}
],
"metadata": {
"kernelspec": {
"display_name": "Python 3",
"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",
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"version": "3.10.9"
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}
},
"nbformat": 4,
"nbformat_minor": 5
}
