What you need is the De Casteljau algorithm. This will allow you to split your curve into whatever segments you’d like. However, since you’re dealing with just cubic curves, I’d like to suggest a slightly easier to use formulation that’ll give you a segment from t0 to t1 where 0 <= t0 <= t1 <= … Read more
The C# Language Specification says this: For non-constant expressions (expressions that are evaluated at run-time) that are not enclosed by any checked or unchecked operators or statements, the default overflow checking context is unchecked unless external factors (such as compiler switches and execution environment configuration) call for checked evaluation. The reason for this choice is … Read more
O forgot to say I found the answer eventually. The application is to determine compass direction of a transit vehicle and its destination. Essentially, fancy math for acquiring curvature of Earth, finding an angle/compass reading, and then matching that angle with a generic compass value. You could of course just keep the compassReading and apply … Read more
RESOLVED Here’s how to do it exactly, step by step. Obtain your mouse coordinates within the client area Get your Projection matrix and View matrix if no Model matrix required. Multiply Projection * View Inverse the results of multiplication Construct a vector4 consisting of x = mouseposition.x within a range of window x transform to … Read more
The math.js library supports complex numbers, matrices, and more. The library is compatible with JavaScript’s built-in Math library, so quite easy to use. http://mathjs.org You can just do things like: math.i; // i math.sqrt(-4) // 2i var a = math.complex(‘2 + 3i’); // 2 + 3i var b = math.complex(4, 5); // 4 + 5i … Read more
In Ruby, the prime library gives you the factorization: require ‘prime’ 4800.prime_division #=> [[2, 6], [3, 1], [5, 2]] To get that list of yours, you take the cartesian product of the possible powers: require ‘prime’ def factors_of(number) primes, powers = number.prime_division.transpose exponents = powers.map{|i| (0..i).to_a} divisors = exponents.shift.product(*exponents).map do |powers| primes.zip(powers).map{|prime, power| prime ** … Read more
Your last approach seems promising. You can improve on that by manually considering the lowest bits of a and b: unsigned int average = (a / 2) + (b / 2) + (a & b & 1); This gives the correct results in case both a and b are odd.
Try the Math.Round method. Here’s how: Math.Round(76d / 100d, 0) * 100; Math.Round(121d / 100d, 0) * 100; Math.Round(9660d / 100d, 0) * 100;
FWIW, with gcc-4.2 on MacOS X 10.6 and -O3 compiler flags, x = x * x; and y = pow(y, 2); result in the same assembly code: #include <cmath> void test(double& x, double& y) { x = x * x; y = pow(y, 2); } Assembles to: pushq %rbp movq %rsp, %rbp movsd (%rdi), %xmm0 … Read more
There are three options: >>> 4 / float(100) 0.04 >>> 4 / 100.0 0.04 which is the same behavior as the C, C++, Java etc, or >>> from __future__ import division >>> 4 / 100 0.04 You can also activate this behavior by passing the argument -Qnew to the Python interpreter: $ python -Qnew >>> … Read more