$25
Smallest Optical Flow (4 pts): What velocity πβ ππΌπ that satisfies the Optical Flow
Constraint Equation πΌπ₯π’+πΌπ¦π£+πΌπ‘ = 0 has the smallest magnitude |πβ |? Hint: This can be solved geometrically as was outlined in class by considering the OFCE in π’,π£ space.
πΌπ₯π’ +πΌπ¦π£+πΌπ‘ = 0
Moving Gaussian Blob (6 pts): A Gaussian blob is observed over time to have brightness
2 π₯ π¦ π₯ π¦ 2 πΌ(π₯,π¦,π‘) = π−(π‘ −2(π1+π2)π‘+(π1+π2) )
What are πΌπ₯, πΌπ¦, and πΌπ‘? Hint: You should find that these derivatives have a simple form.
The Optical Flow Constraint Equation is πΌπ₯π’ +πΌπ¦π£+πΌπ‘ = 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 πΌπ₯, πΌπ¦, and πΌπ‘.
Quadratic Optical Flow (8 pts): Suppose the image brightness is given by
πΌ(π₯,π¦,π‘) = πΌ0 +[(π₯−π1π‘)2 +(π¦−π2π‘)2]
What are Ix, Iy, and It? Hint: You should find that these derivatives have a simple form.
Express the Optical Flow Constraint Equation πΌπ₯π’ +πΌπ¦π£+πΌπ‘ = 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 π’ is updated according to
[ ]new = [ππΌπ₯2 +4 ππΌπ₯πΌπ¦ ]neighbors(π₯,π¦)π’old(π)−ππΌπ₯πΌπ‘
π’(π₯,π¦)
π£(π₯,π¦) ππΌπ₯πΌπ¦ ππΌ π£old(π)−ππΌπ¦πΌπ‘
[ πneighbors(π₯,π¦) ]
Show that this is equivalent to
[π’ new ππΌπ₯πΌπ¦ neighbors(π₯,π¦)π’old(π)−ππΌπ₯πΌπ‘
]
π£ ππΌπ₯πΌπ¦ ππΌ π£old(π)−ππΌπ¦πΌπ‘
[ πneighbors(π₯,π¦) ]
Show that this is equivalent to update equations
new(π₯,π¦) = π’Μ
old − πΌπ₯ 4(πΌπ₯π’Μ
old +πΌπ¦π£Μ
old +πΌπ‘)
π’
πΌπ₯2 +πΌπ¦2 +π
πΌ
π£new(π₯,π¦) = π£Μ
old − π¦ 4(πΌπ₯π’Μ
old +πΌπ¦π£Μ
old +πΌπ‘)
πΌπ₯2 +πΌπ¦2 +π
where π’Μ
old,π£Μ
old are the averages of the 4 neighbors of π’(π₯,π¦),π£(π₯,π¦). Hint: You only need to show this for π’new because π£new follows an identical derivation.
In the case that π = 0, what do the update equations reduce to?