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CS549-Homework 8 Solved

Smallest Optical Flow (4 pts): What velocity ๐‘‰โƒ— ๐‘€๐ผ๐‘ that satisfies the Optical Flow
Constraint Equation ๐ผ๐‘ฅ๐‘ข+๐ผ๐‘ฆ๐‘ฃ+๐ผ๐‘ก = 0 has the smallest magnitude |๐‘‰โƒ— |? Hint: This can be solved geometrically as was outlined in class by considering the OFCE in ๐‘ข,๐‘ฃ space.

                                                                                                  ๐ผ๐‘ฅ๐‘ข +๐ผ๐‘ฆ๐‘ฃ+๐ผ๐‘ก = 0

 

Moving Gaussian Blob (6 pts): A Gaussian blob is observed over time to have brightness
 2 ๐‘ฅ ๐‘ฆ ๐‘ฅ ๐‘ฆ 2 ๐ผ(๐‘ฅ,๐‘ฆ,๐‘ก) = ๐‘’−(๐‘ก −2(๐‘˜1+๐‘˜2)๐‘ก+(๐‘˜1+๐‘˜2) )

 

What are ๐ผ๐‘ฅ, ๐ผ๐‘ฆ, and ๐ผ๐‘ก? Hint: You should find that these derivatives have a simple form.
The Optical Flow Constraint Equation is ๐ผ๐‘ฅ๐‘ข +๐ผ๐‘ฆ๐‘ฃ+๐ผ๐‘ก = 0. Write this out using the results of Part a. and simplify it as much as possible. For example, you should be able to cancel terms that occur in each of ๐ผ๐‘ฅ, ๐ผ๐‘ฆ, and ๐ผ๐‘ก.
 

Quadratic Optical Flow (8 pts): Suppose the image brightness is given by
๐ผ(๐‘ฅ,๐‘ฆ,๐‘ก) = ๐ผ0 +[(๐‘ฅ−๐‘1๐‘ก)2 +(๐‘ฆ−๐‘2๐‘ก)2]

What are Ix, Iy, and It? Hint: You should find that these derivatives have a simple form.
 

Express the Optical Flow Constraint Equation ๐ผ๐‘ฅ๐‘ข +๐ผ๐‘ฆ๐‘ฃ+๐ผ๐‘ก = 0 in the simplest terms possible for this image sequence.
 

The equation from b. must hold for all x, y, and t. Find a constant solution for u and v that makes this true, that is, such that u and v do not depend on x, y, and t.
 

Iterative Optical Flow(8 pts): We saw in class an iterative method for computing optical flow, where at each iteration, the optical flow ๐‘ข is updated according to
                              [             ]new = [๐œ†๐ผ๐‘ฅ2 +4                 ๐œ†๐ผ๐‘ฅ๐ผ๐‘ฆ ]neighbors(๐‘ฅ,๐‘ฆ)๐‘ขold(๐‘›)−๐œ†๐ผ๐‘ฅ๐ผ๐‘ก   

๐‘ข(๐‘ฅ,๐‘ฆ)

                                 ๐‘ฃ(๐‘ฅ,๐‘ฆ)                ๐œ†๐ผ๐‘ฅ๐ผ๐‘ฆ              ๐œ†๐ผ                       ๐‘ฃold(๐‘›)−๐œ†๐ผ๐‘ฆ๐ผ๐‘ก

                                                                                                                [      ๐‘›neighbors(๐‘ฅ,๐‘ฆ)                                                   ]

Show that this is equivalent to
                 [๐‘ข new                                         ๐œ†๐ผ๐‘ฅ๐ผ๐‘ฆ                         neighbors(๐‘ฅ,๐‘ฆ)๐‘ขold(๐‘›)−๐œ†๐ผ๐‘ฅ๐ผ๐‘ก   

]

                  ๐‘ฃ                                                         ๐œ†๐ผ๐‘ฅ๐ผ๐‘ฆ            ๐œ†๐ผ                          ๐‘ฃold(๐‘›)−๐œ†๐ผ๐‘ฆ๐ผ๐‘ก

                                                                                                                                      [      ๐‘›neighbors(๐‘ฅ,๐‘ฆ)                                                   ]

Show that this is equivalent to update equations
                                                          new(๐‘ฅ,๐‘ฆ) = ๐‘ขฬ…old −          ๐ผ๐‘ฅ               4(๐ผ๐‘ฅ๐‘ขฬ…old +๐ผ๐‘ฆ๐‘ฃฬ… old +๐ผ๐‘ก)

๐‘ข

                                                                                                  ๐ผ๐‘ฅ2 +๐ผ๐‘ฆ2 +๐œ†                                   

๐ผ

                                                      ๐‘ฃnew(๐‘ฅ,๐‘ฆ) = ๐‘ฃฬ… old − ๐‘ฆ 4(๐ผ๐‘ฅ๐‘ขฬ…old +๐ผ๐‘ฆ๐‘ฃฬ… old +๐ผ๐‘ก)

๐ผ๐‘ฅ2 +๐ผ๐‘ฆ2 +๐œ†

where ๐‘ขฬ…old,๐‘ฃฬ… old are the averages of the 4 neighbors of ๐‘ข(๐‘ฅ,๐‘ฆ),๐‘ฃ(๐‘ฅ,๐‘ฆ). Hint: You only need to show this for ๐‘ขnew because ๐‘ฃnew follows an identical derivation.

In the case that ๐œ† = 0, what do the update equations reduce to?

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