## Checking to see if 3 points are on the same line

You can check if the area of the ABC triangle is 0: [ Ax * (By – Cy) + Bx * (Cy – Ay) + Cx * (Ay – By) ] / 2 Of course, you don’t actually need to divide by 2.

You can check if the area of the ABC triangle is 0: [ Ax * (By – Cy) + Bx * (Cy – Ay) + Cx * (Ay – By) ] / 2 Of course, you don’t actually need to divide by 2.

No your algorithm is not scalable. What I’ve done before is to issue numbers serially (+1 each time) and then pass them through an XOR operation to jumble the bits thus giving me a seemingly random numbers. Of course they aren’t really random, but they look so to users eyes. [Edit] Additional information This algorithm’s … Read more

I am assuming that you store the frequencies as floating point numbers between 0 and 1 that total to make 1. First you should prepare a table of cumulative frequencies, i.e. the sum of the frequency of that letter and all letters before it. To simplify, if you start with this frequency distribution: A 0.1 … Read more

It works! In an earlier iteration of the Johnson algorithm, I had supposed that A was an adjacency matrix. Instead, it appears to represent an adjacency list. In that example, implemented below, the vertices {a, b, c} are numbered {0, 1, 2}, yielding the following circuits. Addendum: As noted in this proposed edit and helpful … Read more

Very interesting question. I wanted to implement this into mine 4D rendering engine as I was curious how would it look like but I was too lazy and incompetent to handle ND transcendent problems from the math side. Instead I come up with different solution to this problem. Its not a Fibonaci Latice !!! Instead … Read more

Breadth-first search traverses a graph and in fact finds all paths from a starting node. Usually, BFS doesn’t keep all paths, however. Instead, it updates a prededecessor function π to save the shortest path. You can easily modify the algorithm so that π(n) doesn’t only store one predecessor but a list of possible predecessors. Then … Read more

if (X1+W1<X2 or X2+W2<X1 or Y1+H1<Y2 or Y2+H2<Y1): Intersection = Empty else: Intersection = Not Empty If you have four coordinates – ((X,Y),(A,B)) and ((X1,Y1),(A1,B1)) – rather than two plus width and height, it would look like this: if (A<X1 or A1<X or B<Y1 or B1<Y): Intersection = Empty else: Intersection = Not Empty

This problem reduces to the 0-1 Knapsack Problem, where you are trying to find a set with an exact sum. The solution depends on the constraints, in the general case this problem is NP-Complete. However, if the maximum search sum (let’s call it S) is not too high, then you can solve the problem using … Read more

Dmitriy is right that you’ll want the Sieve of Atkin to generate the prime list but I don’t believe that takes care of the whole issue. Now that you have a list of primes you’ll need to see how many of those primes act as a divisor (and how often). Here’s some python for the … Read more

Choosing a random pivot minimizes the chance that you will encounter worst-case O(n2) performance (always choosing first or last would cause worst-case performance for nearly-sorted or nearly-reverse-sorted data). Choosing the middle element would also be acceptable in the majority of cases. Also, if you are implementing this yourself, there are versions of the algorithm that … Read more