Reverse the isometric projection algorithm - algorithm

Question

I've got this code:
const a = 2; // always > 0 and known in advance
const b = 3; // always > 0 and known in advance
const c = 4; // always > 0 and known in advance
for (let x = 0; x <= a; x++) {
for (let y = 0; y <= b; y++) {
for (let z = 0; z <= c; z++) {
for (let p = 0; p <= 1; p++) {
for (let q = 0; q <= 2; q++) {
let u = b + x - y + p;
let v = a + b + 2 * c - x - y - 2 * z + q;
let w = c + x + y - z;
}
}
}
}
}
The code generates (a+1)*(b+1)*(c+1)*2*3 triplets of (u,v,w), each of them is unique. And because of that fact, I think it should be possible to write reversed version of this algorithm that will calculate x,y,z,p,q based on u,v,w. I understand that there are only 3 equations and 5 variables to get, but known boundaries for x,y,z,p,q and the fact that all variables are integers should probably help.
for (let u = ?; u <= ?; u++) {
for (let v = ?; v <= ?; v++) {
for (let w = ?; w <= ?; w++) {
x = ?;
y = ?;
z = ?;
p = ?;
q = ?;
}
}
}
I even managed to produce the first line: for (let u = 0; u <= a + b + 1; u++) by taking the equation for u and finding min and max but I'm struggling with moving forward. I understand that min and max values for v are depending on u, but can't figure out the formulas.
Examples are in JS, but I will be thankful for any help in any programming language or even plain math formulas.
If anyone is interested in what this code is actually about - it projects voxel 3d model to triangles on a plain. u,v are resulting 2d coordinates and w is distance from the camera. Reversed algorithm will be actually a kind of raytracing.
UPDATE: Using line equations from 2 points I managed to create minmax conditions for v and code now looks like this:
for (let u = 0; u <= a + b + 1; u++) {
let minv = u <= a ? a - u : -a + u - 1;
let maxv = u <= b ? a + 2 * c + u + 2 : a + 2 * b + 2 * c - u + 3;
for (let v = minv; v <= maxv; v++) {
...
}
}
I think I know what to do with x, y, z, p, q on the last step so the problem left is minw and maxw. As far as I understand those values should depend both on u and v and I must use plane equations?

Answer 1

If the triplets are really unique (didn't check that) and if p and q always go up to 1 and 2 (respectively), then you can "group" triplets together and go up the loop chain.
We'll first find the 3 triplets that where made in the same "q loop" : the triplets make with the same x,y,z,p. As only q change, the only difference will be v, and it will be 3 consecutive numbers.
For that, let's group triplets such that, in a group, all triplets have the same u and same w. Then we sort triplets in groups by their v parameters, and we group them 3 by 3. Inside each group it's easy to assign the correct q variable to each triplet.
Then reduce the groups of 3 into the first triplet (the one with q == 0). We start over to assign the p variable : Group all triplets such that they have same v and w inside a group. Then sort them by the u value, and group them 2 by 2. This let's us find their p value. Remember that each triplet in the group of 3 (before reducing) has that same p value.
Then, for each triplet, we have found p and q. We solve the 3 equation for x,y,z :
z = -1 * ((v + w) - a - b - 3c -q)/3
y = (w - u + z + b - c - p)/2
x = u + y - b - p

Answer 2

After spending some time with articles on geometry and with the huge help from Wolfram Alpha, I managed to write needed equations myself. And yes, I had to use plane equations.
const a = 2; // always > 0 and known in advance
const b = 3; // always > 0 and known in advance
const c = 4; // always > 0 and known in advance
const minu = 0;
const maxu = a + b + 1;
let minv, maxv, minw, maxw;
let x, y, z, p, q;
for (let u = minu; u <= maxu; u++) {
if (u <= a) {
minv = a - u;
} else {
minv = -a + u - 1;
}
if (u <= b) {
maxv = a + 2 * c + u + 2;
} else {
maxv = a + 2 * b + 2 * c - u + 3;
}
for (let v = minv; v <= maxv; v++) {
if (u <= b && v >= a + u + 1) {
minw = (-a + 2 * b - 3 * u + v - 2) / 2;
} else if (u > b && v >= a + 2 * b - u + 2) {
minw = (-a - 4 * b + 3 * u + v - 5) / 2;
} else {
minw = a + b - v;
}
if (u <= a && v <= a + 2 * c - u + 1) {
maxw = (-a + 2 * b + 3 * u + v - 1) / 2;
} else if (u > a && v <= -a + 2 * c + u) {
maxw = (5 * a + 2 * b - 3 * u + v + 2) / 2;
} else {
maxw = a + b + 3 * c - v + 2;
}
minw = Math.round(minw);
maxw = Math.round(maxw);
for (let w = minw; w <= maxw; w++) {
z = (a + b + 3 * c - v - w + 2) / 3;
q = Math.round(2 - (z % 1) * 3);
z = Math.floor(z);
y = (a + 4 * b + q - 3 * u - v + 2 * w + 3) / 6;
p = 1 - (y % 1) * 2;
y = Math.floor(y);
x = (a - 2 * b - 3 * p + q + 3 * u - v + 2 * w) / 6;
x = Math.round(x);
}
}
}
This code passes my tests, but if someone can create better solution, I would be very interested.