Image P shows a portion of a larger photograph. Some operations have been done on that photograph using an image manipulation software. Image Q shows a portion of the resulting photograph. Identify the option(s) that specifies/specify the correct operations.
Step 1: Understanding the Concept:
This question tests spatial reasoning and understanding of 2D image transformations (rotation and reflection/flip). The key is that the order of operations matters, and different sequences can sometimes lead to the same result. Since this is a multiple-select question, we must check all options.
Step 2: Key Formula or Approach:
We can represent 2D transformations using matrices to prove equivalence between different sequences of operations. A clockwise rotation by an angle \(\theta\) is given by the matrix:
\[ R(\theta) = \begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix} \]
A horizontal flip is:
\[ F_H = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} \]
and a vertical flip is:
\[ F_V = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \]
We apply matrices from right to left.
Step 3: Detailed Explanation:
Let's analyze the sequences for options (A) and (C) to see if they are equivalent.
Analysis of (C): Rotate clockwise by 135 degrees, flip vertically.
The combined transformation matrix \(T_C\) is \(F_V \cdot R(135^{\circ})\).
\[ T_C = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} \cos 135^{\circ} & \sin 135^{\circ} \\ -\sin 135^{\circ} & \cos 135^{\circ} \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} -1/\sqrt{2} & 1/\sqrt{2} \\ -1/\sqrt{2} & -1/\sqrt{2} \end{pmatrix} = \begin{pmatrix} -1/\sqrt{2} & 1/\sqrt{2} \\ 1/\sqrt{2} & 1/\sqrt{2} \end{pmatrix} \]
Analysis of (A): Rotate CW by 45°, flip horizontally, then rotate CW by 90°.
The combined transformation matrix \(T_A\) is \(R(90^{\circ}) \cdot F_H \cdot R(45^{\circ})\).
\[ F_H \cdot R(45^{\circ}) = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} \cos 45^{\circ} & \sin 45^{\circ} \\ -\sin 45^{\circ} & \cos 45^{\circ} \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1/\sqrt{2} & 1/\sqrt{2} \\ -1/\sqrt{2} & 1/\sqrt{2} \end{pmatrix} = \begin{pmatrix} -1/\sqrt{2} & -1/\sqrt{2} \\ -1/\sqrt{2} & 1/\sqrt{2} \end{pmatrix} \]
Now, multiply by \(R(90^{\circ})\):
\[ T_A = \begin{pmatrix} \cos 90^{\circ} & \sin 90^{\circ} \\ -\sin 90^{\circ} & \cos 90^{\circ} \end{pmatrix} \begin{pmatrix} -1/\sqrt{2} & -1/\sqrt{2} \\ -1/\sqrt{2} & 1/\sqrt{2} \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} -1/\sqrt{2} & -1/\sqrt{2} \\ -1/\sqrt{2} & 1/\sqrt{2} \end{pmatrix} = \begin{pmatrix} -1/\sqrt{2} & 1/\sqrt{2} \\ 1/\sqrt{2} & 1/\sqrt{2} \end{pmatrix} \]
The resulting matrices for (A) and (C) are identical. This means both sequences of operations produce the exact same final image.
Visual Verification: We need to confirm that this transformation maps P to Q. Let's consider the general orientation. The stones in P are laid somewhat horizontally. In Q, they are oriented diagonally, pointing up and to the right. This visual change is consistent with the derived transformation matrix. Therefore, both (A) and (C) are correct specifications for the operations.
Step 4: Final Answer:
Since the mathematical transformations for options (A) and (C) are equivalent and the resulting transformation correctly maps the visual features of image P to image Q, both are correct answers.
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