Woche 2 Start
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DotNaos
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code/part-2/Linear Algebra Assignment 3.ipynb
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175
code/part-2/Linear Algebra Assignment 3.ipynb
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"cells": [
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{
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"cell_type": "code",
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"execution_count": 1,
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"metadata": {},
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"outputs": [],
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"source": [
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"import numpy as np\n",
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"import matplotlib.pyplot as plt\n"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## Übung\n",
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"\n",
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"Gegeben ist das folgende Array:"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 2,
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"metadata": {},
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"outputs": [],
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"source": [
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"n = np.array([\n",
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" [0. , 0. , 0.5, 1. , 1. , 0. , 1. , 0. , 1. ],\n",
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" [0. , 1. , 1.5, 1. , 0. , 1. , 1. , 0. , 0. ]\n",
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"])"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"Die erste Zeile sind dabei x-Werte, die zweite Zeile y-Werte von Punkten. Plotten Sie die Punkte als Linienzug mit `plt.plot(...)`."
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]
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},
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{
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"cell_type": "code",
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"execution_count": 3,
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"metadata": {},
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"outputs": [],
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"source": [
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"# Linienzug plotten\n",
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"# plt.plot(...)\n"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"Als nächstes sollen die Punkte rotiert werden. Eine Matrix, die eine Rotation um einen Winkel $\\theta$ durchführt, lautet wie folgt:\n",
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"\n",
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"\\begin{equation}\n",
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"m_1 =\n",
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"\\begin{bmatrix}\n",
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"\\cos \\theta & -\\sin \\theta \\\\ \n",
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"\\sin \\theta & \\cos \\theta \\\\\n",
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"\\end{bmatrix}\n",
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"\\end{equation}\n",
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"\n",
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"Implementieren Sie eine Funktion, die für einen gegeben Winkel diese Matrix als `ndarray` erzeugt:\n"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 4,
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"metadata": {},
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"outputs": [],
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"source": [
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"# angle soll in Grad angegeben werden\n",
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"def rotation_matrix(angle):\n",
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" pass\n",
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"\n",
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"# mögliche Tests\n",
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"# assert(np.allclose(rotation_matrix(90).dot(np.array([1,0])), np.array([0,1])))\n",
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"# assert(np.allclose(rotation_matrix(-90).dot(np.array([1,0])), np.array([0,-1])))"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"Zum Rotieren aller Punkte muss jeder Punkt mit der Matrix `m` multipliziert werden. Rotieren Sie alle Punkte um 60 Grad und plotten Sie das Ergebnis erneut (Bonus: Versuchen Sie ohne `for`-Schleife auszukommen):"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 5,
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"metadata": {},
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"outputs": [],
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"source": [
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"# n_rot soll die rotierten Punkte enthalten, n_rot.shape == n.shape\n",
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"n_rot = ...\n",
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"\n",
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"# und erneut plotten..."
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"Als nächstes sollen die Punkte bezüglich des Ursprungs skaliert werden werden. Eine Matrix, die eine Skalierung in x-Richtung mit Faktor $s_x$ und in y-Richtung mit Faktor $s_y$ durchführt, lautet:\n",
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"\n",
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"\\begin{equation}\n",
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"m_2 =\n",
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"\\begin{bmatrix}\n",
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"s_x & 0 \\\\ \n",
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"0 & s_y \\\\\n",
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"\\end{bmatrix}\n",
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"\\end{equation}\n",
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"\n",
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"Implementieren Sie eine Funktion, die für einen gegeben Winkel diese Matrix als `ndarray` erzeugt:\n"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 6,
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"metadata": {},
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"outputs": [],
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"source": [
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"def scale_matrix(sx, sy):\n",
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" pass\n",
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"\n",
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"\n",
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"# todo test schreiben?"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"Skalieren Sie die gegebenen Punkte um beliebige Faktoren und plotten Sie die ursprünglichen und skalierten Punkte im selben Diagramm:"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 7,
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"metadata": {},
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"outputs": [],
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"source": [
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"# Punkte skalieren\n",
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"\n",
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"\n",
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"\n",
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"# und plotten\n",
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"\n"
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]
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}
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],
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"metadata": {
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"kernelspec": {
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"display_name": "Python 3 (ipykernel)",
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"language": "python",
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"name": "python3"
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},
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"language_info": {
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"codemirror_mode": {
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"name": "ipython",
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"version": 3
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},
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"file_extension": ".py",
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"mimetype": "text/x-python",
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython3",
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"version": "3.11.0"
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}
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},
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"nbformat": 4,
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"nbformat_minor": 2
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}
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0
code/part-2/Plots.ipynb
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code/part-2/Plots.ipynb
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132
code/part-2/Roboterfahrt Aufgabe 1.ipynb
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code/part-2/Roboterfahrt Aufgabe 1.ipynb
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{
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"cells": [
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{
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"cell_type": "code",
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"execution_count": null,
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"id": "00d20df1",
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"metadata": {},
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"outputs": [],
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"source": [
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"import numpy as np\n",
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"import matplotlib.pyplot as plt"
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]
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},
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{
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"cell_type": "markdown",
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"id": "b986454c",
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"metadata": {},
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"source": [
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"## Übung: Roboterbewegung\n",
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"\n",
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"Ein Roboter steht in einem Raum, seine Position auf dem Boden wird in ein Koordinatensystem eingetragen - die y-Achse ist dabei die Himmelsrichtung _Nord_, die x-Achse die Himmelsrichtung _Ost_. Zu Beginn steht der Roboter auf der Position $x=3$ und $y=5$. Danach führt der Roboter folgende Bewegungen durch.\n",
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"\n",
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"- Fahrt 1 Meter Ost, 2 Meter Nord\n",
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"- Fahrt 2 Meter Ost, 1 Meter Nord\n",
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"- Fahrt 2 Meter Ost, 1 Meter Süd\n",
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"- Fahrt 1 Meter West, 3 Meter Süd\n",
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"\n",
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"a) Drücken Sie die 4 Bewegungen als Arrays aus und speichern Sie diese in Python-Variablen."
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"id": "bec0fab0",
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"metadata": {},
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"outputs": [],
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"source": [
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"start = np.array([3,5])\n",
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"fahrt1 = ...\n",
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"fahrt2 = ...\n",
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"fahrt3 = ...\n",
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"fahrt4 = ..."
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"id": "e55f20bc",
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"metadata": {},
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"outputs": [],
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"source": [
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"# Mit dieser Zelle können Sie die Fahrt des Roboters plotten - einfach ausführen\n",
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"def plot_path(start, moves):\n",
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" path = np.vstack([start] + moves).cumsum(axis=0)\n",
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" plt.plot(path[:,0], path[:,1])\n",
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" plt.xlim((0,10))\n",
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" plt.ylim((0,10))\n",
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"\n",
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"plot_path(start, [fahrt1, fahrt2, fahrt3, fahrt4])"
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]
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},
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{
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"cell_type": "markdown",
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"id": "9f47078c",
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"metadata": {},
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"source": [
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"b) Berechnen Sie den Zielpunkt des Roboters nach seiner Fahrt."
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"id": "90852486",
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"metadata": {},
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"outputs": [],
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"source": []
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},
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{
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"cell_type": "markdown",
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"id": "3c8684d8",
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"metadata": {},
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"source": [
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"c) Welche Strecke hat der Roboter zurückgelegt (nicht der direkte Weg)?"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"id": "6ee1ca50",
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"metadata": {},
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"outputs": [],
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"source": []
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},
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{
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"cell_type": "markdown",
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"id": "3474df24",
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"metadata": {},
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"source": [
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"d) Wie weit wäre der Roboter gefahren, wenn er die direkte Strecke von Start zu Ziel gefahren wäre?"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"id": "04a2e60a",
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"metadata": {},
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"outputs": [],
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"source": []
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}
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],
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"metadata": {
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"kernelspec": {
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"display_name": "Python 3 (ipykernel)",
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"language": "python",
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"name": "python3"
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},
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"language_info": {
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"codemirror_mode": {
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"name": "ipython",
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"version": 3
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},
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"file_extension": ".py",
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"mimetype": "text/x-python",
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython3",
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"version": "3.11.5"
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}
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},
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"nbformat": 4,
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"nbformat_minor": 5
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}
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