C++ DirectX11 2d Game: Stopping enemy sprites moving over each other? - c++

Question

I am using IntersectsWith(this->boundingBox)) method to detect collisions between sprites and player. I want to somehow be able to use this method in detecting my enemy sprites that collide with each other, and when they do to make sure they don't move over one another.
All of the enemy sprites follow the player.
MainGame.cpp
Loops over each enemy in the vector and does the update loop:
for (auto &enemyMobsObj : this->enemyMobs)
{
enemyMobsObj->Update(tickTotal, tickDelta, timeTotal, timeDelta, windowBounds, this->ship, this->firstBoss,
this->enemyMobs, this->bullets, this->missiles, NULL, "NULL", "NULL");
}
Here is what I tried before to stop each sprite moving over each other:
EnemyMobOne::Update:
int nextEnemy;
for (int i = 0; i < enemyMobOne.size(); i++)
{
nextEnemy = i + 1;
if (nextEnemy < enemyMobOne.size())
{
//Deal with mobs collision
if (enemyMobOne[i].boundingBox.IntersectsWith(enemyMobOne[nextEnemy].boundingBox))
{
enemyMobOne[i].position.x = enemyMobOne[nextEnemy].position.x - enemyMobOne[i].boundingBox.Width;
}
}
}
However this makes each enemy sprite obviously stick to each other, which doesn't look right, it also makes them teleport.
Anyone know the correct code to stop them moving over each other? Thanks.

Answer 1

When you detect an intersection between two collision objects, you need to make a decision about how you're going to counteract the overlap (as I'm sure you figured out). However, how one does this is a bit trickier than simply "pushing" them to one side (as you've done in your code). What you likely want to do is to have a counter-force to the applied force, so to speak. Basically, you want to calculate the minimum translation, or the direction by which the least amount of movement would be required, to get at least ONE of the boxes to move OUT of the other one.
This is a bit more complicated than simply "put me on the right (or left, depending on how you set up your coordinates, I suppose) side of the other guy," which is more or less what your code does, now.
For a simple solution, just check if one of the colliders is closer to the left, right, top, or bottom of the other. To do this, you can simply take the collision intersection position and check the relative distance between that point and the minimum and maximum x and y coordinates relative to one of the colliders, then move the one or both of the sprites, accordingly.
Ex:
[Edit] After reviewing my previous answer for this, I realized you would need to calculate the overlap of the boxes, which would make it much easier to accomplish this all like so:
float minX = min(sprite0.boundingBox.maxX, sprite1.boundingBox.maxX);// Minimum of the boxes' right-side points (top right and bottom right) x coordinates
float minY = min(sprite0.boundingBox.maxY, sprite1.boundingBox.maxY);// Minimum of the boxes' top-side points (top left and top right) y coordinates
float maxX = max(sprite0.boundingBox.minX, sprite1.boundingBox.minX);// Maximum of the boxes' left-side points (top left and bottom left) x coordinates
float maxY = max(sprite0.boundingBox.minY, sprite1.boundingBox.minY);// Maximum of the boxes' bottom-side points (bottom left and bottom right) y coordinates
float distHoriz = minX - maxX;// The horizontal intersection distance
float distVert = minY - maxY;// The vertical instersection distance
// If the boxes are overlapping less on the horizontal axis than the vertical axis,
// move one of the sprites (in this case, sprite0) in the opposite direction of the
// x-axis overlap
if(abs(distHoriz) < abs(distVert))
{
sprite0.x -= distHoriz;
}
// Else, move one of the sprites (again, I just decided to use sprite0 here,
// arbitrarily) in the opposite direction of the y-axis overlap
else
{
sprite0.y -= distVert;
}
To further clarify (beyond the comments), what we're basically doing here is checking the distance between the overlapping lines. For example:
Box 0 x-axis xmin0|------------------|xmax0
Box 1 x-axis xmin1|----------------------|xmax1
|----|<-- Overlap (xmax0 - xmin1)
Notice that the minimum from the two bounding boxes that is used for the overlap is the maximum among the two minima (xmin0 and xmin1), and the maximum that is used for the overlap is the minimum among the two maxima (xmax0 and xmax1).
The y-axis calculation works exactly the same way. Once we have both axes, we simply check to see which one has a lower absolute value (which distance is shorter) and move along that distance to counteract the intersection.