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Video Loop Finding

hardFREEVideo TexturesCVCV: Image-Based Rendering

Implement a method to find optimal loop points for creating seamless video textures. This task involves analyzing the similarity between frames in a video sequence to determine the best points to loop back, creating a continuous and smooth visual experience.

The concept of video textures is based on the idea of creating an infinite video loop from a finite video sequence, which is crucial in various applications such as video games, animation, and virtual reality. To achieve this, we need to compute a frame-to-frame similarity matrix, which represents the similarity between each pair of frames in the sequence.

Here are the key steps to find the optimal loop points:

  1. Compute the similarity between each pair of frames (i,j)(i, j), treating each frame as one vector fi\mathbf{f}_i of all its pixel values
  2. Identify pairs where j≥i+min_lenj \ge i + min\_len to ensure a sufficient gap between the loop points
  3. Among those pairs, the smoothest transition is the one with the highest similarity Si,jS_{i,j}

Implement find_loop(frames, min_len) returning the tuple (loop_start, loop_end) = (i,j)(i, j) of that best pair. If several pairs tie, keep the one with the smallest ii, then the smallest jj. If there are fewer than min_len frames, return (-1, -1).

Si,j=fi⋅fj∥fi∥∥fj∥S_{i,j} = \frac{\mathbf{f}_i \cdot \mathbf{f}_j}{\|\mathbf{f}_i\| \|\mathbf{f}_j\|}

This technique is widely used in video game development to create immersive and engaging environments.

Example

Input
Video frames
Output
Loop start/end indices
Reasoning
  • Input: frames is a sequence of nn grayscale frames (each a height ×\times width grid of pixel values) and min_len is the shortest loop allowed. Each frame is read as one long vector fi\mathbf{f}_i of all its pixels.
  • Toy case: five 1×21 \times 2 frames f0=(3,4)\mathbf{f}_0=(3,4), f1=(4,3)\mathbf{f}_1=(4,3), f2=(0,5)\mathbf{f}_2=(0,5), f3=(6,8)\mathbf{f}_3=(6,8), f4=(5,0)\mathbf{f}_4=(5,0) with min_len =2=2. All but f3\mathbf{f}_3 have length 5; ∥f3∥=10\|\mathbf{f}_3\|=10.
  • Similarities: S0,3=18+325⋅10=1.0S_{0,3}=\frac{18+32}{5 \cdot 10}=1.0, S1,3=4850=0.96S_{1,3}=\frac{48}{50}=0.96, S0,2=2025=0.8S_{0,2}=\frac{20}{25}=0.8, S1,4=0.8S_{1,4}=0.8, S0,4=0.6S_{0,4}=0.6, S2,4=0S_{2,4}=0. These six are the only pairs with j≥i+2j \ge i+2.
  • Output: the highest-scoring allowed pair wins, so the loop start/end indices are (0,3)(0, 3): play frames 0 through 3, then jump back to frame 0.
  • Notice f3\mathbf{f}_3 is twice as bright as f0\mathbf{f}_0 yet scores a perfect 1: cosine similarity compares the pixel pattern, not overall brightness. And S0,1=0.96S_{0,1}=0.96 never competes, because a gap of 1 is shorter than min_len.

Constraints

  • Input parameter types and shapes: video_frames: 3D numpy array (RGB) with shape (num_frames, height, width)
  • Valid ranges or assumptions: num_frames > 2, height and width are positive integers, pixel values in [0, 255]
  • Output format and precision: Return loop_start and loop_end indices as integers
  • Special conditions: min_length is a positive integer, loop_start and loop_end indices are 0-based and satisfy loop_end >= loop_start + min_length

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solution.py

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