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"you can move an endpoint along the curve without changing the remainder of the curve" Is trivial true. Think of a curve as t(0..1)->p as a function. Change (0.

"you can move an endpoint along the curve without changing the remainder of the curve" Is trivial true. Think of a curve as t(0..1)->p as a function. Change (0..1) to (0..a) with a<=1. It's the same function, just scale t by 1/a first.

gizmo686 10y ago
The spline function itself is defined as a function of the points. In general, if you change the points of a spline, then you change the resulting curve. Animats seems to be asking about special cases where, if you change the endpoint of the curve such that it is still on the original curve, the resulting curve happens to be the same as the original (over the interval on which it is defined).
- starmole 10y ago
  I got that. I think I provided a trivial way to do that. I might be wrong but how/why? Pick the new end point at p'(a) = curve(a), a[0..1], multiply the curve by 1/a. Same curve, new endpoint?
  - gizmo686 10y ago
    I am not sure I understand what you are suggesting. Are you suggesting to think of the spline-like curve as a function that with domain [0,1]. For simplicity, let us simply say that original curve is defined between 0 and 1 on the x axis, so t(x) gives you the curve without additional work. If you simply change the domain of t to be [0,a], then you have (trivially) produced an otherwise identical function (call it t') with a different endpoint; but I am not sure where the scaling t' by 1/a comes in. If you literally mean take t'(x)/a, then (unless a is 1) that is clearly a different function. If you mean take t'(x/a) then you have arrived back at the original function, but compressed along the x axis. As I mentioned above, simply changing the domain without scaling does "work". The problem is that it is not obvious that the resulting curve is "spline-like". Admittedly, to my knowledge, "spline-like" has never been formally defined. However, we defined a spline by the fixed points through which it passes. For example, suppose t(x) is the spline-like curve under discussion. We should have t = SplineLike((x1,y1),(x2,y2),...,(xn,yn)). Where SplineLike is a function that takes a set of points, and produces a spline-like curve that passes through said points. Now consider the curve t' = SplineLike((x1,y1),(x2,y2),...,(a,t(a))). That is to say, take the previous set of points, and move the last point to somewhere else on the resulting curve. For simplicities sake, make the additional assumption that (a,t(a)) is still the rightmost point. We are looking for a higher order function SplineLike such that t'(x) = t(x) where t' is defined. Additionally, SplineLike should produce functions that "feel" like a spline. Since the splines we are talking about are quadratic, a reasonable criteria for this is that the resulting functions are continuous, and have a continuous derivative. Providing a way to go straight from t to t' does not provide any insight to the higher order SplineLike function; or even if such a function exists.

gizmo686 10y ago

The spline function itself is defined as a function of the points. In general, if you change the points of a spline, then you change the resulting curve. Animats seems to be asking about special cases where, if you change the endpoint of the curve such that it is still on the original curve, the resulting curve happens to be the same as the original (over the interval on which it is defined).

starmole 10y ago
I got that. I think I provided a trivial way to do that. I might be wrong but how/why? Pick the new end point at p'(a) = curve(a), a[0..1], multiply the curve by 1/a. Same curve, new endpoint?
- gizmo686 10y ago
  I am not sure I understand what you are suggesting. Are you suggesting to think of the spline-like curve as a function that with domain [0,1]. For simplicity, let us simply say that original curve is defined between 0 and 1 on the x axis, so t(x) gives you the curve without additional work. If you simply change the domain of t to be [0,a], then you have (trivially) produced an otherwise identical function (call it t') with a different endpoint; but I am not sure where the scaling t' by 1/a comes in. If you literally mean take t'(x)/a, then (unless a is 1) that is clearly a different function. If you mean take t'(x/a) then you have arrived back at the original function, but compressed along the x axis. As I mentioned above, simply changing the domain without scaling does "work". The problem is that it is not obvious that the resulting curve is "spline-like". Admittedly, to my knowledge, "spline-like" has never been formally defined. However, we defined a spline by the fixed points through which it passes. For example, suppose t(x) is the spline-like curve under discussion. We should have t = SplineLike((x1,y1),(x2,y2),...,(xn,yn)). Where SplineLike is a function that takes a set of points, and produces a spline-like curve that passes through said points. Now consider the curve t' = SplineLike((x1,y1),(x2,y2),...,(a,t(a))). That is to say, take the previous set of points, and move the last point to somewhere else on the resulting curve. For simplicities sake, make the additional assumption that (a,t(a)) is still the rightmost point. We are looking for a higher order function SplineLike such that t'(x) = t(x) where t' is defined. Additionally, SplineLike should produce functions that "feel" like a spline. Since the splines we are talking about are quadratic, a reasonable criteria for this is that the resulting functions are continuous, and have a continuous derivative. Providing a way to go straight from t to t' does not provide any insight to the higher order SplineLike function; or even if such a function exists.