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Lecture 7: Intro to Geometry, Splines, and Bezier Curves (40)

Is "B" going to be an invertible matrix regardless of the control points that we choose?


If I'm interpreting this correctly, B is the same basis matrix every time (and thus guaranteed to be invertible). The values of c, x, and d change in accordance to the parameters/inputs we pass in.


I don't think B is the same matrix every time. If you look, on this slide we are assuming that u0=0u_0 = 0 and u1=0u_1 = 0, which is what produces the specific values for BB. When u0=u1u_0 = u_1, you'll end up with two identical rows in your BB, which will give you a non-invertible matrix. This makes sense because then your curve would be degenerate. I am not too sure of how you could prove that any distinct u0,u1u_0, u_1 would produce an invertible matrix though


If this system of equations is solvable, then B should be incertible. It is a way to solve the equation to get our coefficients. I think if your initial assumption is wrong, then maybe your B won't be invertible since your equations won't be solvable.


The matrix does not depend on the value of the control points, but it does depend on u0 and u1. If u0=u1 then the matrix would be rank deficient, but a curve from 0 to 0 would not make sense so that’s okay. Most of the time there is no reason not to leave u0=0 and u1=1.


Just remind me of a linear algebra trick for less computational cost. Instead of inverting a matrix, QR decomposition would be better for solving this function, i.e. B = QR -> QRc = h -> Rc = Q^Th. Time complexity: O(N^3) -> O(N^2)


I'm confused on how evaluating the points and having the tangent line (derivative) allows us to construct this curved line. Like the example that we are doing, how does evaluating at only u0 and u1 allow us to construct a curved line between the points when the only other information we have is the tangent lines. Intuitively, I would think this would not reveal anything about how the line curves so I'm curious how we can get this with only two points and no intermediary point.


Since this is a degree 3 polynomial we are trying to fit, we need 4 points to uniquely determine it. In this case, it seems that rather than having 4 points on the line, we use 2 points and 2 derivative points. This results in 4 systems of equations we can then solve.

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