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[EmptyNull]

Oh noes! It's math time again  

I'd like some help to replace the inefficient hacks I'm using to see if a point is inside an isometric tile.

I'm currently looping trough every tile to check if the point is inside it's bounding box.
If it is I do a check between the tile's mask and a pixel at the place's point to check if the point is actually inside the tile.

This is the code I use to calculate the place where my tiles must be displayed on the screen.
x = ((tile.width/2)*rowx+((tile.width/2)*rowy))
y = ((tile.height*4)+((tile.height/2)*rowy)-((tile.height/2)*rowx)) + (300-tile.eight*1.5)



I've also looked around a bit to see how other people fixed their math isometric problem but most of their solutions were above me.

One option I heard of was using matrices to rotate the tiles so that they form squares and can be used fully in a isinRect check.
Though I have no idea how to do that.

So anyone have any advice?

[Christian Knudsen]

This might be helpful:

http://laserbrainstudios.com/2010/08/the-basics-of-isometric-programming/

[EmptyNull]

This might be helpful:

http://laserbrainstudios.com/2010/08/the-basics-of-isometric-programming/

Sadly this does not help me with my issue.
Since photobucket refused to replace my image for some reason it wasn't updated and I posted the old example picture here.
And as you can see on that picture my 0.0 origin is at the west side.
And the one in your example sadly uses a top origin.

[Polly]

It's super easy. First divide the point coordinates by the tile size, and then simply add / subtract the coordinates for both axes.

Code:

float x = position_x / tile_width;
float y = position_y / tile_height;

int tile_x = x + y - origin_y;
int tile_y = x - y - origin_y;

Should give you this

[EmptyNull]

That is remarkable. Such elegant math.
Simply brilliant! Hats off to you good sir! 

[Christian Knudsen]

Which is pretty much the exact algorithm in the blog post I linked to...