April 23, 2024: Dwarf Fortress 50.13 has been released.
News: February 3, 2024: The February '24 Report is up.
News: February 4, 2021: Dwarf Fortress Talk #28 has been posted.
News: November 21, 2018: A new Threetoe story has been posted.
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What kind of server?
That really really doesn't make any sense. What you actually need to do: If your velocity vector is v and your unit normal vector is n, then zeroing v along n is done by taking v-n*(v.n).Also, I assume that zeroing the vector along an axis involves vector multiplication (dot product or cross product), but I'm not sure of the exact operations I should be doing here. It would be extremely useful if I could get this working with slopes.You can multiply it by the "swap" of the normal. This is usually what we mean when we talk about transposing a vector. You can go into the math, but in programming terms, it boils down to this: if the normal vector is {x, y}, then you want to multiply the velocity vector by {y, x}.
OK yeah, I see what the issue is now. You need to do the collision detection by ray tracing the velocity vector, finding the nearest collision intersection, then zeroing the velocity along that normal and ray tracing the new velocity vector (once again checking for collisions).The major problem with that method is that you can't walk uphill.
Your example shouldn't happen under this scheme -- the box is already colliding with the ground on the y-axis so any ray trace would find that immediately, zero your y velocity, then try to move you along the x-axis instead.
Supose I have found the points of intersection of the lines that confine the area. How do I identify which one of those points is the point where the objective function reaches its maximum or minimum? It also can be that there is either no maximum or no minimum, or both no maximum and no minimum.
Suppose we have n constraints (n lines) and less-than-n points of their intersection with one another. That means the area is unbounded in the direction of the gradient of the objective function. How do I find it out not knowing how the polygon (which is shaped by the area) looks like?To find a potential maximum, just evaluate the objective function f at all intersections of two lines and take the intersection P with the highest objective value f(P) such that P still fulfills all inequalities. To determine whether that point is actually maximal, take the line {Q : f(Q) = f(P) + 1} and intersect it with all the given boundary lines. If none of the resulting points is valid, then P is indeed maximal, otherwise P is not maximal.
Hmmm... I got my own implementation of quadtrees working, and it successfully reduced the number of collision comparisons to 0.166 what it was before, using about 3.50% of my (quad-core) CPU while moving around and drawing 1000 AABBs at 60fps. But one of the code examples I looked at looks like this:Code: [Select]public List retrieve(List returnObjects, Rectangle pRect) {
int index = getIndex(pRect);
if (index != -1 && nodes[0] != null) {
nodes[index].retrieve(returnObjects, pRect);
}
returnObjects.addAll(objects);
return returnObjects;
}
This function returns a list of objects that the given rect could potentially be intersecting. When getIndex returns -1, it means the given rect doesn't fit within one of the child nodes, so must be placed within the current one. The problem I found with this function is that it's not symmetrical. Say two collision rectangles are added to the quadtree, and one is placed into one of the child nodes as it fits entirely within one quadrant. But the other is halfway between two of the quadrants, so it gets placed in the parent node. Using the function above, the former will register as colliding with the latter, but not vice versa.
I can't even tell if that's intended behavior here.
My robotics team just won like all the thing.Wow, congrats! Want to share some more details?
Top scoring, Excellence award (AKA the top award), and of course the tournament.
There's aa bit in the chunk file that states whether or not you need to generate ores for it. I don't know what it's set as in that map.This.
Now, it reads as almost-English: "if compChoice beats userChoice print you lose" and "if userChoice beats compChoice print you win".
