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I'm dealing with objects on 15x15 chunks in my game.
I'd like the chunks to rotate, for example, rotation version 0 is the original position of 2,5;
(EDIT: Oops, rotation 2 and 3 names need switching in the image)
How can I get the other positions in the simplest and quickest way? It only needs to rotate in 90 degrees increments, having 4 version (so 0, 90, 180, 270)
If horizontal axis is x, vertical is y, and chunk size is n, then clockwise rotation can be performed as follows:
(x2, y2) = (n - 1 - y1, x1)
and counter-clockwise rotation would be:
(x2, y2) = (y1, n - 1 - x1)
If you number the grid from the middle as (0,0), rather than the edge, rotating becomes as simple as
(x2, y2) = (-y1, x1)
You can still provide the grid to your player with conventional numbering, of course.
If you use Weckar's suggestion to number the grid with (0,0) in the center, this is specialized case of the general rotation formula. You could get any rotation using these equations
new_x = old_x * cos(angle) - old_y * sin(angle)
new_y = old_x * sin(angle) + old_y * cos(angle)
In your case, angle can only be 0, 90, 180, and 270, which means the sins and cosines can only produce values of -1, 0, and +1.
y), but in rotation 1 it's 5 positions from the right (x). So you need to define the newxin terms of the oldy. Putting this in more mathematical terms, we go fromy=5tox=14-5orx_new=14-y_old. Do something similar fory_newand double-check it and you have the answer. \$\endgroup\$ – Bernhard Barker yesterdayfor(x = 0; x <= n - 1; x++){ for(y = 0; y <= n - 1; y++){ plot(x,y) } }displays Rotation 0, thenfor(y = n - 1; y >= 0; y--){ for(x = 0; x <= n - 1; x++){ plot(x,y) } }will be Rotation 1. (Basically you'll notice the changes to the loops are similar to what you get in the accepted answer) \$\endgroup\$ – Kaddath yesterday