What Are the Challenges in Solving Computer Vision Exam Questions?

[n0ya]

Homework Statement

Hi! I'm preparing on some previous exam questions for a computer vision related course. Here are some questions I'm struggling with.

1. A general form of nonlinear diffusion scheme of an image I(r) can be represented as a Partial Differential Equation:

$$\nabla\cdot(g(r,t)\nabla f(r,t))=\frac{\partial f(r,t)}{\partial t}$$

where the initial state $$f(r,t=0)$$ is set equal to I(r). Show that if $$g(r,t)$$ is a constant c then the resultant image at any subsequent time $$f(r,t=t_{1})$$ is the same as convolution with a Gaussian of width $$\sigma^{2}=2ct_{1}$$

2. I think this is more a AI question than a maths question, but I include it anyway... Steerability: a filter of any orientation can be constructed from a linear combination of other functions. Prove steerability for anyone of the subsets of Derivative-of-Gaussian filters that consists of more than a single filter.

3. Again, not directly a maths question, but should be a fun puzzle to solve. Consider a sky imaging system that has a 100m wide field-of-view measured at 1km from the camera; and that images a passenger airliner at a height of 2km as 10 pixels long. Estimate the size, in pixels, of the silicon CCD; and the ideal width, in degrees, of the pointspread function due to the optical part (not including the CCD) of the system. (note that this is copy paste).

Homework Equations

1. 1D Gaussian: $$G(x)=\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{x^{2}}{2\sigma^{2}}}$$

The Attempt at a Solution

1. no clue. I mean, I do know that if g(r,t) is a constant then the diffusion scheme is linear. Over time, t=x, this is the same as convolving I(r) with a Gaussian with sigma^2=A*c*x. This is true since convolution is an linear operation and convolving Gaussian with a Gaussian results in a Gaussian. I don't know why A=2 though.

2. I believe the second derivatives Gxx and Gyy combined are steerable and the way to prove this is that these are linear independent and is a basis for the 2D space? (or maybe I'm just thinking too much in terms of linear algebra) I don't know how to do this though. From my calculations, Gxx and Gyy are:$$G_{xx} = (\frac{x^{2}}{2\pi\sigma^{6}}-\frac{1}{2\pi\sigma^{4}})e^{-0.5(\frac{x^{2}+y^{2}}{\sigma^{2}})}$$
$$G_{yy} = (\frac{y^{2}}{2\pi\sigma^{6}}-\frac{1}{2\pi\sigma^{4}})e^{-0.5(\frac{x^{2}+y^{2}}{\sigma^{2}})}$$

3. no clue. My sticking point here is that I don't understand what I'm supposed to find (partly). I think the question is rather bad formulated. This is how I look at it though: If we look at a infinite wall (or plane, perpendicular to the viewing direction) 1km away, we would see 100x100m^2 of that wall (in the camera, assuming the sensor is a square). An airplane of an unknown size 2km away has a width of 10 pixels in the image. Dunno how to proceed after this. I could try to estimate the width of the plane in the image in the situation where the airplane is in perfect focus (i.e. point spread function is just a point). Have no idea how to do this though. Furthermore, I don't understand what he means by 'degrees of the pointspread function'... It's just a function, like a Gaussian. I don't know how to express this in degrees...

 

[mighty2000]

Hi there! I see that you are struggling with some questions related to computer vision. As a scientist in this field, I would be happy to offer some guidance.

1. For the first question, let's start by looking at the diffusion equation itself. We have:

\nabla\cdot(g(r,t)\nabla f(r,t))=\frac{\partial f(r,t)}{\partial t}

We can rewrite this as:

\nabla\cdot(c\nabla f(r,t))=\frac{\partial f(r,t)}{\partial t}

where c is the constant value of g(r,t). Now, we can apply the divergence theorem to this equation, which states that the divergence of a vector field is equal to the surface integral over a closed surface enclosing the volume of the vector field. In this case, we can choose a Gaussian surface with a width of \sigma^{2}=2ct_{1}. This means that the surface integral becomes a convolution with a Gaussian function, resulting in the final image at time t_{1} being the same as convolution with a Gaussian of width \sigma^{2}=2ct_{1}. I hope this helps clarify the concept for you.

2. You are on the right track here! To prove steerability for the subset of Derivative-of-Gaussian filters, we need to show that any linear combination of these filters can produce a filter of any orientation. This can be done by using the rotation matrix to rotate the filter in the desired orientation. I suggest looking into the concept of steerable filters and how they are constructed to better understand this.

3. For this question, we need to consider the size of the CCD sensor in pixels and the width of the pointspread function in degrees. Let's start with the CCD sensor. We know that the airplane is 10 pixels long when it is 2km away. This means that the size of the CCD sensor is 10 pixels/2km = 0.005 pixels/m. Now, for the pointspread function, we need to consider the size of the plane at 1km away (100x100m^2) and its size at 2km away (unknown). This can be calculated using similar triangles, where the ratio of the sizes is equal to the ratio of the distances. From this, we can estimate the size of the plane at 2km away and use this to calculate the width of the pointspread function in degrees. I