Indexing in equilateral triangle grid given simple 2D Cartesian coordinates - grid

Question

I have an equilateral triangle grid constructed like this:
Now, given that 2D coordinate origin point is red dot at very bottom left of image, i need to find index into this triangle grid.
Given Input: X and Y coordinates of point (floating point) of interest, triangle side length and height
Need Output: X and Y index of triangle. (sample indexes seen in image)
Getting Y (row) coordinate of triangle is simple as it is just integer part of (GivenY / TRIANGLE_HEIGHT), but i cant get around to calculate Needed X coordinate without doing to much of operations.

Answer 1

Here is a more explicit answer, for people having issues (or too lazy) to figure out the geometry.
First scale your coordinates to a convenient basis
vec2 pos=(input-origin)/vec2(edge/2,median);
Split your coordinates in integer and fractional part
int x=pos.x, y=pos.y; float u=pos.x-x, v=pos.y-y;
Test for the diagonal (x,y of different parity) or anti-diagonal edge (x,y of same parity)
if(x%2 ^ y%2) { if(v+u<1) x--; } else { if(v-u>0) x--; }
That's it (x,y) is now your face index.
Finding vertex index for each face is a bit more involved.
You have four cases, Here is a list of CCW vertex indices for each face:
face vertices
xy xy xy xy
00 -> 00 10 01
10 -> 11 01 10
01 -> 01 12 02
11 -> 12 01 11
It's easier to see the pattern if you don't add 1 to the y vertex indices, So your final table is:
00 -> 00 10 01
10 -> 11 01 10
01 -> 00 11 01
11 -> 11 00 10
Each column is respectively: x, x, !x, x^y, x, !x. Alternatively, you can simply use a lookup table.
It works with arbitrary face indices, you just need to add (x/2, y) and do your lookup on (x%2, y%2).
In the end, the triangle vertex indices are:
x/2 + x%2, y + x%2; x/2 + !(x%2), y + (x%2^y%2); x/2 + (x%2), y + !(x%2)
with vertex coordinates in your original cartesian space:
origin+vec2(2*x+y%2),y)*vec2(edge/2,median)

Answer 2

More triangles
If you draw a line from the bottom left of 0,0, to the top right of 1,0, the shaded are of the two triangles is now made up of four right angled triangles. Drop a line down from the top right of 1,0 you end up with another right angled triangle and a triangle composed of three right angled triangles height is your original triangle height, width is 1.5 * triangle length, hypotenuse I leave for an exercise for the class.
So then whether it's in the left or right triangle is where your point intercepts that line.
You can precaclc most of this based on triangle length.
Another option is to work out the parallelogram, calc it's centre, move that to the origin, rotatee 30 degrees clockwise, move back, Precalc the transform, you have a diamond with a width of the above hypotenuse, Which side of the middle is which triangle, height of the diamond is triangle length, you'll still need to work or the parallegram as 1.0 and 2,0 will overlap.