Task 1. ChatGPT¶
User prompt: Any interesting riddle for 3d computer vision?
ChatGPT: Nice — love this! Here are five short, fun riddles (easy → hard) themed around 3D computer vision. Each riddle has a small hint and a brief answer/explanation so you can try them and then check.
1 — The Epipolar Sneak
Riddle: Two calibrated cameras look at the same scene. You pick a pixel in the left image. Your friend points to a line in the right image and says “the matching pixel must be somewhere along this line.” Why is your friend right?
Hint: Think about where the 3D ray from the left camera can land relative to the right camera’s image plane.
Answer:
Epipolar line / epipolar geometry. The 3D point traced by the ray from the left pixel projects to the right image along a line (the epipolar line) determined by the cameras’ relative pose. That line contains all possible correspondences for that left pixel.⸻
2 — The Invisible Depth
Riddle: A single pinhole camera rotates in place (no translation). A distant point projects onto the image and—after rotation—projects to a different pixel. Can you recover the absolute depth of that point from only those two images?
Hint: Rotation changes orientation but doesn’t create parallax.
Answer:
No. Pure rotation produces no parallax; the image motion is explainable by camera orientation only, so absolute depth (scale along the viewing ray) is unobservable from rotation-only motion.⸻
3 — The Three-Point Detective
Riddle: You have two unlabeled 3D point clouds (same object) captured in different rigid poses. What is the minimum number of non-collinear point correspondences you need to uniquely determine the rotation and translation that aligns the first cloud to the second (assuming no scale change)?
Hint: Think about degrees of freedom: rotation (3) + translation (3).
Answer:
Three non-collinear points. Three corresponding, non-collinear 3D points give enough constraints to solve for rotation and translation uniquely (up to measurement noise). (Two points leave an infinite set of rotations around the baseline; collinear triples fail.)⸻
4 — The Mirror Trick
Riddle: You’re tracking a moving object and see its reflection in a planar mirror in the scene. Treat the mirror as a plane — how can you convert the reflected observations into equivalent observations of a “virtual” camera or virtual object so you can use standard triangulation?
Hint: Reflections across a plane are equivalent to mirroring the world or the camera across that plane.
Answer:
Use a virtual camera (or virtual points) mirrored across the plane. Reflect the camera center (or reflect the observed point rays) across the mirror plane to create a virtual camera viewpoint; then triangulate between real and virtual views (or between virtual points) as usual. Reflection acts like a symmetric transformation across the mirror plane.⸻
5 — The Stereo Baseline Puzzle (harder, brain-teaser)
Riddle: Two identical cameras are placed some distance apart (baseline). A single point in space projects to exactly the same pixel coordinates in both images. What does that imply about the 3D location of the point relative to the cameras? (Assume perfect pinhole cameras, identical intrinsics, and no occlusion.)
Hint: Each pixel corresponds to a ray through the camera center; “same pixel” means rays have the same direction in each camera’s coordinate frame.
Answer:
If the cameras have identical intrinsics and the pixel is at the same image coordinates, then in each camera’s own coordinate system the ray direction (in camera coordinates) is identical. The two rays are therefore lines through two different camera centers but with the same direction vector. Two parallel rays either never meet (point at infinity) or—if the point lies in front of both—this can only happen when the point is at infinity in that direction (i.e., effectively the same direction vector but no finite intersection), unless the cameras are colocated or the direction points along the baseline so the rays intersect. In short: for finite non-coincident camera centers, identical pixel coordinates imply the point is on the plane at infinity (no finite depth) or special degenerate geometry — practically this is a degenerate case for triangulation.(Shorter: same pixel in two different camera centers is a degenerate case — triangulation fails or indicates a point at infinity or special alignment.)
⸻
Want more like these? I can: • give a challenge riddle with a small simulated image pair and ask you to compute depth, • turn one into a short coding exercise (e.g., derive epipolar line equation, or implement 3-point registration), • or give a one-question mini-quiz with increasing difficulty for a class/meetup.
Which would you like? 😄
