How to avoid Gimbal Lock on Rubik's Cube [duplicate]

Question

I am making a Rubik's cube program. That will simulate a Rubik's cube. I am having problems with gimbal lock. I need to know how to avoid it. To my understanding I can either multiply the x, y, z rotation matrices in a specific order to acquire the rotation I am trying to get. Or I can use a quaternion which is like rotating around a single user defined axis instead of the axis x, y, or z. I can't really understand visually how I can get the resulting axis from two other axis. (x, y, z) How can I get the resulting axis for my quaternion from my Euler rotations? Will it just result in gimbal lock again?

One of the big problems is. I don't know what rotations are going to be made. I need to make it general purpose.

Edit: Little bit more documentation of my problem:

My Euler matrix multiplication looks like this

    rotate = XMMatrixRotationX(content.mesh[meshToRotate[i]].rotation.x) * XMMatrixRotationY(content.mesh[meshToRotate[i]].rotation.y) * XMMatrixRotationZ(radians);
    content.mesh[meshToRotate[i]].rotation.z = radians;

As a gimbal lock would suggest the last axis in the rotation gets the problem. When you attempt to rotate the front or back face you get problems.

Answer 1

In the case of a puzzle like this you can animate the rotation then reset the model back to its original position but with the cube's stickers rotated.

Look at it like cheating with a real puzzle by moving the stickers.

You'll have to make it possible to change the stickers' colors. This can be done with material colors or a dynamic texture.

And this way you can reduce it to 2 animations: center 90' rotation & end 90' rotation.

You flip/rotate the cube and stickers to animate the opposite end or on a different axis, reverse the animation to turn in the other direction.

The model (Mesh) always comes back to its original position after animating the 90 degree rotation, it's the stickers that change places. This way we completely avoid gimbal locks.

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Another way is to use rotation matrices.

Each cube cells (the 3x3x3 elements of the puzzle) have 6 faces, up to 3 of them with stickers and their own rotation matrix. Using the face normal multiplied by the rotation matrix of the cell you can figure out which sticker points where as only 1 axis in the normal will ever have a non-zero value +/- 1.0 in X, Y or Z.

And there will (should) only ever be 9 stickers pointing to the same direction.

You can use the cell's positions to figure out which ones to rotate.

Since the cube cells will only ever be rotated to 0, 90, 180, or 270 at the end of the animation we can fix the cumulative rotation errors by doing:

if (matrix[r][c] > 0.5) {

    matrix[r][c] = 1; 

} else if (matrix[r][c] < -0.5) {

    matrix[r][c] = -1; 

} else {

    matrix[r][c] = 0;

}

For each cell (r=row, c=column) to eradicate any rounding errors once the animation is done.

Usually comparing float values to exact values is a bad idea but in this case we're fixing the matrices to be exactly -1.0, 0.0, or 1.0, without any translation and anything (N) multiplied by those numbers always end up exactly -N, 0 or +N so we can safely compare them.