[SOLVED] CS 549 -Computer Vision-HW #11

1. Inexact Match (7 pts): In stereo imaging, if the rays defined by 𝑋𝑋⃗_𝐿𝐿 and 𝑋𝑋⃗_𝑅𝑅 do not intersect, we can find 𝑋𝑋⃗^𝑊𝑊 anyway by minimizing an error measure. One way to do this is to project

𝑋𝑋⃗^𝑊𝑊 into the left and right image planes to give 𝑋𝑋⃗_𝐿𝐿′ and 𝑋𝑋⃗_𝑅𝑅′. The error E is defined as the squared difference between the observed image points 𝑋𝑋⃗_𝐿𝐿 and 𝑋𝑋⃗_𝑅𝑅 and the projected image points 𝑋𝑋⃗_𝐿𝐿′ and 𝑋𝑋⃗_𝑅𝑅′.

1. Show that for the simplified parallel optical axis camera geometry used in class where

𝑓𝑓⃗ ∙ 𝐵𝐵⃗ = 0, 𝑅𝑅 = 𝐼𝐼, 𝑇𝑇⃗     ^𝐵𝐵⃗

𝐸𝐸 ≡ 𝑋𝑋⃗_𝐿𝐿^′ − 𝑋𝑋⃗_𝐿𝐿² + 𝑋𝑋⃗_𝑅𝑅^′ − 𝑋𝑋⃗_𝑅𝑅²

𝑓𝑓            𝑏𝑏            2           𝑓𝑓                 2           𝑓𝑓            𝑏𝑏            2

= 𝑧𝑧_𝑊𝑊 𝑥𝑥𝑊𝑊 + 2 − 𝑥𝑥𝐿𝐿 + 𝑧𝑧𝑊𝑊 𝑦𝑦𝑊𝑊 − 𝑦𝑦𝐿𝐿 + 𝑧𝑧𝑊𝑊 𝑥𝑥𝑊𝑊 − 2 − 𝑥𝑥𝑅𝑅

𝑓𝑓                 2

+ 𝑧𝑧_𝑊𝑊 𝑦𝑦𝑊𝑊 − 𝑦𝑦𝑅𝑅

1. By differentiating E with respect to 𝑥𝑥^𝑊𝑊 and 𝑦𝑦^𝑊𝑊, show that

𝑥𝑥𝑊𝑊 ₌ 𝑥𝑥^𝐿𝐿+𝑥𝑥^𝑅𝑅 𝑧𝑧^𝑊𝑊 and 𝑦𝑦𝑊𝑊 ₌ 𝑦𝑦^𝐿𝐿+𝑦𝑦^𝑅𝑅 𝑧𝑧^𝑊𝑊

2        𝑓𝑓                                  2        𝑓𝑓

1. By differentiating E with respect to 𝑧𝑧^𝑊𝑊 and using the results of b., show that

^𝑊𝑊 = 𝑓𝑓𝑏𝑏

𝑧𝑧

∆_𝑥𝑥

and conclude that

𝐵𝐵⃗2

𝑋𝑋⃗𝑊𝑊 = 𝑋𝑋⃗𝐴𝐴𝐴𝐴𝐴𝐴

𝐵𝐵⃗ ∙ ∆⃗

2. Reference Image (4 pts): Sometimes we treat one image in a stereo pair as a “reference image” aligned with world coordinates. The other image and disparity are taken with respect to the reference image.  Suppose that the right image is the reference image.  Then we have

𝑋𝑋⃗_𝑅𝑅^𝐶𝐶 = 𝑋𝑋⃗^𝑊𝑊 and 𝑋𝑋⃗_𝐿𝐿^𝐶𝐶 = 𝑋𝑋⃗^𝑊𝑊 + 𝐵𝐵⃗

Show that in this formulation, the world point is given by

𝐵𝐵⃗²    𝐵𝐵⃗

𝑋𝑋⃗_𝑊𝑊 = 𝑋𝑋⃗_{𝐴𝐴𝐴𝐴𝐴𝐴}    −

𝐵𝐵⃗ ∙ ∆⃗  2

You may assume that the left and right image rays intersect, although as Problem 1. shows, they don’t have to.

3. Inexact Match, Alternative Version (10 pts): Another way to determine distance when 𝑋𝑋⃗_𝐿𝐿 and 𝑋𝑋⃗_𝑅𝑅 do not intersect is to find the point 𝑋𝑋⃗^𝑊𝑊 closest to the rays defined by 𝑋𝑋⃗_𝐿𝐿 and 𝑋𝑋⃗_𝑅𝑅. The left image ray is defined by

^𝑊𝑊 = − 𝐵𝐵⃗ + 𝑙𝑙𝑋𝑋⃗_𝐿𝐿, 0 ≤ 𝑙𝑙

𝑋𝑋⃗_𝐿𝐿

2

The goal is to find points 𝑋𝑋^⃗_𝐿𝐿^𝑊𝑊 and 𝑋𝑋^⃗_𝑅𝑅^𝑊𝑊 on the left and right image rays that are closest to each other; the point equidistant between them is the desired 𝑋𝑋⃗^𝑊𝑊. We can find these points by minimizing the following:

min𝑋𝑋⃗𝐿𝐿𝑊𝑊 − 𝑋𝑋⃗𝑅𝑅𝑊𝑊2

𝑙𝑙,𝑟𝑟

Show that the solution is given by

𝑋𝑋⃗_𝑅𝑅²𝑋𝑋⃗_𝐿𝐿 ∙ 𝐵𝐵⃗ − 𝑋𝑋⃗_𝐿𝐿 ∙ 𝑋𝑋⃗_𝑅𝑅𝑋𝑋⃗_𝑅𝑅 ∙ 𝐵𝐵⃗𝑋𝑋⃗_𝐿𝐿 + 𝑋𝑋⃗_𝐿𝐿 ∙ 𝑋𝑋⃗_𝑅𝑅𝑋𝑋⃗_𝐿𝐿 ∙ 𝐵𝐵⃗ − 𝑋𝑋⃗_𝐿𝐿²𝑋𝑋⃗_𝑅𝑅 ∙ 𝐵𝐵⃗𝑋𝑋⃗_𝑅𝑅

1

𝑋𝑋⃗𝑊𝑊 = ²                                        𝑋𝑋⃗_𝐿𝐿2𝑋𝑋⃗_𝑅𝑅2 − 𝑋𝑋⃗_𝐿𝐿 ∙ 𝑋𝑋⃗_𝑅𝑅2