### sandwiching a spline

The strong convex hull property says that for a spline of degree , the convex hull of at most control points of a spline contains the point of a spline at parameter . Intersection methods can use this property to rule out certain interections. Union of a convex hulls of consecutive points completely encompass the … More sandwiching a spline

### tightly bounding a bspline

In the post on tessellation step length where we discussed ways to get a bound for step length, the trickiness of computing maxima forced us to take a different approach to getting a bound. We had to look at integrals as alternatives. In this post we take a closer look at computing points where maxima … More tightly bounding a bspline

### Tessellation step length

A common step in graphics processing and in geometry processing is to is to discretize(tessellate) curves into smaller segments that closely approximate the curve. Let be a curve defined on . In this post we will compute a value for global step size such that the chord from to is not farther from than for … More Tessellation step length

### arc length

Computing arc length, and arc length parametrization using a queer looking power series … More arc length

### Waist on a torus

Last year I worked on a interesting problem which asks about the possibility of partitioning a convex set in a plane in to internally disjoint convex pieces of equal area and perimeter. The problem has since been proved when is a prime power, by Aronov and Hubard and Roman Karasev for all dimensions. Here in … More Waist on a torus

### q-binomials

The binomial like coefficients that we saw in earlier posts: are called –binomial coefficients. In keeping with the existing convention we will use as the indeterminate instead of and write for It was established in an earlier post that: Using this identity we may reduce the -binomial coefficient to: We also made use of the … More q-binomials

### ponder problem now on last legs

This post culminates our efforts in solving the problem: If are positive integers with and satisfying the following is divisible by for all positive integers , then is a power of . First recap a few facts from this post: We began by showing that: is divisible by for all and therefore by the lcm, … More ponder problem now on last legs