There is in fact no real reason to partition the sphere into a regular non-overlapping mesh, try this: partition your sphere into semi-overlapping circles see here for generating uniformly distributed points (your circle centers) Dispersing n points uniformly on a sphere you can identify the points in each circle very fast by a simple dot … Read more
2D case Just like the dot product is proportional to the cosine of the angle, the determinant is proportional to its sine. So you can compute the angle like this: dot = x1*x2 + y1*y2 # Dot product between [x1, y1] and [x2, y2] det = x1*y2 – y1*x2 # Determinant angle = atan2(det, dot) … Read more
Known data: bounding box width W, height H, rotation angle Fi Wanted: coordinates of rotated rectangle vertices. Unknown source rectangle size: w x h Bounding box size for this dimension and rotation angle: H = w * Abs(Sin(Fi)) + h * Abs(Cos(Fi)) W = w * Abs(Cos(Fi)) + h * Abs(Sin(Fi)) denote as = Abs(Sin(Fi)) … Read more
Try this, much simpler than Haversine! Location me = new Location(“”); Location dest = new Location(“”); me.setLatitude(myLat); me.setLongitude(myLong); dest.setLatitude(destLat); dest.setLongitude(destLong); float dist = me.distanceTo(dest);
Indeed, it is a very low number, awfully close to zero. Here’s a great article which can help you understand the common challenges and pitfalls of floats: “What Every Computer Scientist Should Know About Floating-Point Arithmetic“
The complete rotation matrices are derived and given at https://sites.google.com/site/glennmurray/glenn-murray-ph-d/rotation-matrices-and-formulas/rotation-about-an-arbitrary-axis-in-3-dimensions. From the paper: 5.2 The simplified matrix for rotations about the origin Note this assumes that (u, v, w) is a direction vector for the axis of rotation and that u^2 + v^2 + w^2 = 1. If you have a point (x, y, z) … Read more
Look at the powerset() recipe in the itertools docs. from itertools import chain, combinations def powerset(iterable): “powerset([1,2,3]) –> () (1,) (2,) (3,) (1,2) (1,3) (2,3) (1,2,3)” s = list(iterable) return chain.from_iterable(combinations(s, r) for r in range(len(s)+1)) def subsets(s): return map(set, powerset(s))
Here is an implementation of a streaming algorithm described by Jeremy Gibbons in Unbounded Spigot Algorithms for the Digits of Pi (2004), Chaper 6: function * generateDigitsOfPi() { let q = 1n; let r = 180n; let t = 60n; let i = 2n; while (true) { let digit = ((i * 27n – 12n) … Read more
I’ve had a chance to look at fdlibm’s implementation. The comments describe the algorithm used: * n * Method: Let x = 2 * (1+f) * 1. Compute and return log2(x) in two pieces: * log2(x) = w1 + w2, * where w1 has 53-24 = 29 bit trailing zeros. * 2. Perform y*log2(x) = … Read more
private int ReverseModulus(int div, int a, int remainder) { if(remainder >= div) throw new ArgumentException(“Remainder cannot be greater than or equal to divisor”); if(a < remainder) return remainder – a; return div + remainder – a; } e.g. : // (53 + x) % 62 = 44 var res = ReverseModulus(62,53,44); // res = 53 … Read more