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
I recommend looking at the “Introduction to Algorithms” book (by Cormen, Leiserson and Rivest). I’ve always found its pseudo-code description of algorithms very clear and consistent. An example: DIJKSTRA(G, w, s) 1 INITIALIZE-SINGLE-SOURCE(G, s) 2 S ← Ø 3 Q ← V[G] 4 while Q ≠ Ø 5 do u ← EXTRACT-MIN(Q) 6 S ← … Read more
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.
Yeah. Defining distinct is a product of deferring to a color space then when we say maximally distinct colors what we mean to say is colors which are as far from all the other colors as possible. But since the color space doesn’t change the answer isn’t going to change. And implementing something that better … Read more
See Harry Hsu. “An algorithm for finding a minimal equivalent graph of a digraph.”, Journal of the ACM, 22(1):11-16, January 1975. The simple cubic algorithm below (using an N x N path matrix) suffices for DAGs, but Hsu generalizes it to cyclic graphs. // reflexive reduction for (int i = 0; i < N; ++i) … Read more
This is how I see it, after explanation of Ernest Friedman-Hill (accepted answer): So the example from the same book (Figure 1.4). I’ve added names to the vertices to make it clear So at first step all the vertices are single vertex chains, so we connect A-D, B-E and C-F pairs, b/c distance between them … 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
Here is how I would do it in Java: int seconds = (int) (milliseconds / 1000) % 60 ; int minutes = (int) ((milliseconds / (1000*60)) % 60); int hours = (int) ((milliseconds / (1000*60*60)) % 24);
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
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