What you want to use is often called the “perp dot product”, that is, find the vector perpendicular to one of the vectors, and then find the dot product with the other vector. if(a.x*b.y – a.y*b.x < 0) angle = -angle; You can also do this: angle = atan2( a.x*b.y – a.y*b.x, a.x*b.x + a.y*b.y …
Why not just: (val, max, min) => (val – min) / (max – min);
Here’s an implementation of the Euclidean algorithm that returns the greatest common divisor without performing any heap allocation. You can substitute ulong for uint if needed. An unsigned type is used, as the technique does not work for signed values. If you know your a and b values are not negative, you can use long …
In every case (including divisible by 2): if (number % n == 0) do(); Anding with a mask of low order bits is just obfuscation, and with a modern compiler will not be any faster than writing the code in a readable fashion. If you have to test all of the cases, you might improve …
“General Theory” of CAS is a pretty huge scope for a question. That being said, I’ll do my best to cover as much as I can in the hopes that something helps you find what you’re looking for 🙂 The proceedings of the ISSAC and SIGSAM groups would no doubt have some good stuff about …
“yep base-10 is what I mean” In that case, yes, they can be extended to base-10 in several ways, though they aren’t nearly as useful as in binary. One idea is that &, |, etc. are the same as doing arithmetic mod-2 to the individual binary digits. If a and b are single binary-digits, then …
In calculus there is a thing called Taylor Series which provides an easy way to calculate many irrational values to arbitrary precision. Pi/4 = 1 – 1/3 + 1/5 – 1/7 + … (from http://www.math.hmc.edu/funfacts/ffiles/30001.1-3.shtml ) Keep adding those terms until the number of digits of precision you want stabilize. Taylor’s theorem is a powerful …
This question has a bureaucratic part and an algorithmic part. A floating point number is stored internally as (2e × m), where e is an exponent (itself in binary) and m is a mantissa. The bureaucratic part of the question is how to access this data, but R. seems more interested in the algorithmic part of the question, …
This constraint t[0] + t[1] = 1 would be an equality (type=”eq”) constraint, where you make a function that must equal zero: def con(t): return t[0] + t[1] – 1 Then you make a dict of your constraint (list of dicts if more than one): cons = {‘type’:’eq’, ‘fun’: con} I’ve never tried it, but …