Find Optimal Loop Point

mediumPROVideo TexturesVideoCV: Image-Based Rendering

Implement a solution to find the optimal loop point for creating a seamless video texture. The goal is to determine the best pair of frames to loop back to, ensuring a smooth transition.

The concept of video textures involves creating an infinite loop of a video sequence, which is crucial for applications like computer vision and image-based rendering. A key challenge is finding the optimal loop point, where the transition from the end frame back to the start frame is as seamless as possible. This is measured using a similarity matrix, where $S[i][j]$ represents the dissimilarity between frames $i$ and $j$ .

To find the optimal loop point, consider the following steps:

Iterate over all possible frame pairs $(i, j)$ with $j \geq i + min\_length$ , where $min\_length$ is the minimum required loop length.
For each pair, calculate the dissimilarity $S[j][i]$ when transitioning from frame $j$ back to frame $i$ .

S[j][i] = dissimilarity(j, i)

This technique is widely used in video game development and virtual reality applications.

Example

Input

find_loop([[0, 100, 50], [100, 0, 100], [50, 100, 0]], 1)

Output

(0, 2)

Reasoning

Finding best loop with min length 1: Check j→i transitions where j > i by at least 1:

(i=0, j=1): similarities[1][0] = 100
(i=0, j=2): similarities[2][0] = 50 ← best!
(i=1, j=2): similarities[2][1] = 100 Best loop: play frames 0→2, then jump back to 0 (cost=50)

Constraints

similarities: NxN matrix where similarities[j][i] is cost of j→i transition
min_loop_length: minimum number of frames in the loop
Return (start_frame, end_frame) tuple

More from CV: Image-Based Rendering

Differentiable Renderer Forwardhard Video Loop Findinghard Light Field Refocusinghard Depth-Based View Synthesishard Front-to-Back Splat Accumulationmedium

Concept primerPREVIEW

You can think of this as a constrained “best pair of indices in a matrix” problem, motivated by making a video loop cut where the end frame looks as close as possible to the start frame while keeping the loop at least some minimum length.

1. Background Knowledge

In video textures, the goal is to take a short video of some quasi-repetitive motion (e.g., fire, water, waving flag) and turn it into a seamless loop. Conceptually, you pick a starting frame $i$ , play forward to frame $j$ , and then jump back to frame $i$ . For that jump not to be noticeable, the appearance and motion between frames $j$ and $i$ should be as similar as possible.

The similarity (or dissimilarity) matrix $S$ encodes how good that jump is: $S[i][j]$ (or $S[j][i]$ depending on convention) is a scalar measure of how different it looks if you jump from frame $j$ back to frame $i$ . A low value means a good, smooth transition; a high value means an obvious jump. On top of that, we want the loop to be sufficiently long to be visually interesting, so we constrain the loop length $L = j - i$ to be at least some minimum.

So the problem reduces to: given a 2D matrix of costs $S$ , find the pair $(i, j)$ with $j > i$ , $j - i \ge \text{minLength}$ , and minimum cost $S[j][i]$ . This is a pure algorithmic selection problem over a constrained subset of matrix entries.

2. Algorithm / General Approach

This is essentially a brute-force search with constraints:

You are not asked to optimize a sequence, only to choose one best pair $(i, j)$ .
The objective is a simple scalar: minimize $S[j][i]$ .
The constraint is simple: $j - i \ge \text{minLength}$ .

Because of that, the natural pattern is:

Iterate over all valid pairs $(i, j)$ that satisfy the length constraint.
Track the minimum value of $S[j][i]$ and its indices.
Return the indices with the best (smallest) value.

If the number of frames $n$ is not huge (e.g., typical coding challenge constraints), a straightforward $O(n^2)$ double loop is usually acceptable and easiest to implement correctly.

3. Step-by-Step Strategy

Understand indices and constraints

Frames are typically numbered $0 \dots n-1$ (or $1 \dots n$ ); check the problem statement.
You must enforce $j - i \ge \text{minLength}$ and $j > i$ .

PRO

Continue the full explanation

You're reading the free preview. Unlock the complete walkthrough, the code editor, test runner and reference solution with Premium.

PRO

Editor locked

The code editor is locked for Pro problems. It is only available for free problems. Please upgrade to gain access to the code editor for all problems.

solution.py

Python 3.13

Test results