Initial commit with project files

2_Bildbearbeitung/ü8/README.md

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+# Übung 8: Geometrische Transformation
+
+In dieser Übung soll die Transformationsforschrift 
+
+<p align="center">
+<img src="https://latex.codecogs.com/svg.image?\begin{pmatrix}&space;x'\\y'\end{pmatrix}&space;=\begin{pmatrix}&space;a&space;&&space;b\\c&space;&&space;d\end{pmatrix}\cdot\begin{pmatrix}&space;x\\y\end{pmatrix}+&space;\begin{pmatrix}&space;e\\f\end{pmatrix}.&space;" title="\begin{pmatrix} x'\\y'\end{pmatrix} =\begin{pmatrix} 0.5 & 0\\0 & 0.5\end{pmatrix}\cdot\begin{pmatrix} x\\y\end{pmatrix} ." />
+</p>
+
+für die Transformation des Bildes I1 zu I2 hergeleitet werden.
+
+
+| I1 | I2 |
+| --- | --- |
+| ![](./data/original.jpg) | ![](./data/new.jpg) |
+
+
+Leiten Sie die Transformationsvorschrift her und testen Sie die Vorschrift, indem Sie ein Skript in die Datei [a.py](a.py)
+programmieren. Die Lösung ist in der Datei [l_a.py](l_a.py) zu finden!
+
+
+
+

2_Bildbearbeitung/ü8/a.py

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+import cv2
+import numpy as np
+
+img = cv2.imread("../../data/car.png")
+img = cv2.resize(img, (500, 500))
+cv2.imshow("Car", img)
+
+print("Image Shape:", img.shape)
+
+
+def T(pos, rotation, translation):
+    """ Transformation Matrix """
+    new_pos = np.matmul(rotation, pos) + translation
+
+    return new_pos
+
+
+new_image = np.zeros_like(img)
+
+rotation = [
+    [..., ...],
+    [..., ...]
+]
+translation = [
+    ...,
+    ...
+]
+
+
+for x in range(img.shape[1]):
+    for y in range(img.shape[0]):
+        old_pos = np.asarray([[x], [y]])
+        new_pos = T(old_pos, rotation, translation)
+        new_pos = new_pos.astype(int)
+        if 0 <= new_pos[0] < img.shape[1] and 0 <= new_pos[1] < img.shape[0]:
+            new_image[new_pos[1], new_pos[0]] = img[y, x]
+
+print(new_image.shape)
+cv2.imshow("After Transformation", new_image)
+
+cv2.waitKey(0)
+

2_Bildbearbeitung/ü8/l_a.py

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+import cv2
+import numpy as np
+
+img = cv2.imread("../../data/car.png")
+img = cv2.resize(img, (500, 500))
+cv2.imshow("Car", img)
+
+print("Image Shape:", img.shape)
+
+
+def T(pos, rotation, translation):
+    """ Transformation Matrix """
+    new_pos = np.matmul(rotation, pos) + translation
+
+    return new_pos
+
+
+new_image = np.zeros_like(img)
+
+alpha = -0.5 * np.pi
+rotation = [
+    [np.cos(alpha), -np.sin(alpha)],
+    [np.sin(alpha), np.cos(alpha)]
+]
+translation = [
+    [0],
+    [img.shape[1]]
+]
+
+
+for x in range(img.shape[1]):
+    for y in range(img.shape[0]):
+        old_pos = np.asarray([[x], [y]])
+        new_pos = T(old_pos, rotation, translation)
+        new_pos = new_pos.astype(int)
+        if 0 <= new_pos[0] < img.shape[1] and 0 <= new_pos[1] < img.shape[0]:
+            new_image[new_pos[1], new_pos[0]] = img[y, x]
+
+print(new_image.shape)
+cv2.imshow("After Transformation", new_image)
+
+""" 
+Naive Solution
+ 
+1. Rotate the image with 90°  -> alpha=90°
+2. Translate in y axis with image length -> [ [0], [height] ] 
+ 
+"""
+
+"""
+Mathematical solution
+
+1. Find correspondences
+
+x,y      -->   x',y'
+
+0,0      -->   0,1
+1,0      -->   0,0
+0,1      -->   1,1
+1,1      -->   1,0
+0.5,0.5  -->   0.5,0.5
+
+
+2. Solve system:
+[a, b] * [x, y]^T  + [e, f]^T = [x', y']^T
+[c, d]
+
+ax + by + e = x'
+cx + dy + f = y'
+
+
+Translation (0,0 --> 0,1):
+0 + 0 + e = 0   -->  e = 0
+0 + 0 + f = 1   -->  f = 1
+
+Rotation (1,0   --> 0,0):
+a + 0 + 0 = 0   -->  a = 0
+c + 0 + 1 = 0   -->  c = -1
+
+Rotation (0,1 --> 1,1):
+0 + b + 0 = 1  -->  b = 1
+0 + d + 1 = 1  -->  d = 0
+
+"""
+
+
+cv2.waitKey(0)
+