Solving the Aperture Problem
Solving the Aperture Problem
This video is part of the Udacity course “Introduction to Computer Vision”. Watch the full course at https://www.udacity.com/course/ud810
Content
0.64 -> So how are we going to
solve the aperture problem?
3.14 -> Well, look,
the basic idea is the same as before,
5.79 -> we have to impose some sort of
constraints, but this time we're
9.67 -> going to impose some local constraints
to get more equations per pixel.
14.09 -> So for example, We talked about
maybe we'll assume that locally
18.69 -> right around this point,
the motion field is very smooth.
22.67 -> In fact how smooth do we want to assume?
25.3 -> How bout we assume that it's
actually the same value.
28.3 -> That is over a window,
30.27 -> we're going to assume that u v
is the same value everywhere.
34.93 -> So if we were using, let's say a five
by five window, so there's 25 pixels,
40.3 -> if we were assuming that there
was one u v for all of these,
45.2 -> the uv for, let's say, the center pixel.
47.89 -> That would give us 25
equations per pixel.
51.91 -> And here they are so,
what we've got here is, here's our u and
55.39 -> v that we're assuming is
the u v of some center pixel.
59.14 -> And here we have 25 rows
of the derivative on x and
63.83 -> the derivative in y at
each of those 25 points.
68.12 -> And th, this, these are the gradients,
dotted with u v equals the negative,
73.09 -> and here is the temporal derivatives
at each of those points.
77.25 -> So I can write that as a d equals b,
where d is this
82.16 -> displacement vector u v and
b is just this 25 by 1 vector,
87.43 -> that is essentially the negative
of all the temporal derivatives.
90.95 -> Okay, so
we have this equation a d equals b and
94.37 -> it's got 25 equations and two unknowns.
96.91 -> All right, well, so
that's a little bit of a problem.
99.77 -> How do we solve a system when we
have more equations than unknowns?
103.6 -> Well We do our standard
least square solution.
106.5 -> When we have an equation like this,
we say,
108.58 -> well, we're going to minimize
the square difference, okay.
112.53 -> And they way you do that is by using
the standard pseudo uh,inverse method.
116.77 -> So we multiply a by a transpose.
119.67 -> And since a is 25 by 2,
a trans-flows is 2 by 25, so
123.856 -> a transpose is just a 2 by 2 and d is
a 2 by 1, and a transpose b is a 2 by 1,
129.348 -> and we end up with this
equation right here, okay?
133.73 -> And these have the terms written out.
136.52 -> And these sums are summed over
whatever the window is it
140.22 -> that you're summing over.
141.1 -> So it's five by five would be over
all those 25 pixels and over here,
144.92 -> you just have the product of the,
the sum of the product of the,
148.89 -> image gradient and
times the temporal gradient.
151.39 -> This is our, what we have to solve,
and by the way,
153.38 -> this technique was first profos,
proposed by Luke Scigotti way back in
157.55 -> the dark ages of computer vision, 1981,
which is not quite the dark, dark age.
161.34 -> It was the dawning of it.
162.93 -> The dark ages were in the 60s.
164.93 -> All right, we just did that.
166.42 -> We can solve our squares and
come up with a solution, but
169.49 -> let's understand exactly what
solving this means, okay.
173.49 -> So you remember the aperture problem,
right?
175.36 -> And so here it's drawn out.
177.04 -> That the idea is if I have
a gradient in some direction, and so
181.19 -> this is my little gradient there.
183.1 -> Right?
So
183.63 -> that any u v that's along this
line would be an acceptable u v.
189.06 -> That is any u v whose component in
the direction of the gradient would be
193 -> acceptable.
193.55 -> And that's what this says.
194.51 -> That there's a single equation for that.
197.02 -> And that's the equation
of a line in uv space.
Source: https://www.youtube.com/watch?v=O1Mc5GAVAiQ