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Using three Euler angles we can represent any 3-d rotation matrix.
Let's assume that allowable Euler angles must be equal to $i\dfrac{\pi}{2}$ $\,$, $i=0.. 3$ $\,$ i.e.
$\alpha = \{0,\dfrac{\pi}{2},{\pi}, \dfrac{3\pi}{2}$ (the same effect as in case of $-\dfrac{\pi}{2}$ rotation) $\}$ (such Euler angles constrain generated rotation matrices to matrices with only $-1,0,1$ entries).

and we have chosen representation XYZ for a rotation matrix (so called Tait–Bryan case) $\,$ i.e.

$R= R_x(\alpha_1)R_y(\alpha_2)R_z(\alpha_3)$

or in the more detailed form

$R= \begin{bmatrix} c_2 c_3 & - c_2 s_3 & s_2 \\ c_1 s_3 + c_3 s_1 s_2 & c_1 c_3 - s_1 s_2 s_3 & - c_2 s_1 \\ s_1 s_3 - c_1 c_3 s_2 & c_3 s_1 + c_1 s_2 s_3 & c_1 c_2 \end{bmatrix}$

where $c_j=\cos(\alpha_j)$ and $s_j=\sin(\alpha_j)$.

Theoretically these conditions allow to get $4\times{4}\times{4}= 64$ different Euler representations for rotation matrix, but it is known (for example from How many $3 \times 3$ integer matrices are orthogonal?) that only $24$ unique rotations are possible, so for sure some Euler representations don't generate unique matrices.

  • How to check (without calculating all possibilities) which of the Euler representations of rotation matrix for this kind $i\dfrac{\pi}{2}$ regular rotation generate unique representation and which don't generate such uniqueness?
  • if some are not unique can we say (also without detailed calculations) how many other representations could generate the same rotation matrix as with the selected particular Euler representation?

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  • $\begingroup$ You don't mention any constraints on the numbers in your matrices, unlike the question that you linked to that is constrained to integer matrices. Without such constraints, the answer is that there are uncountably infinitely many rotation matrices: any $3 \times 3$ matrix such that each column (considered as a vector) has dot product with itself equal to 1, and any two distinct columns have dot product with each other equal to zero. $\endgroup$ – Lee Mosher Dec 14 '17 at 17:54
  • $\begingroup$ You could possibly use (finite) group theory: pick three non-identity “generators” $X=R_x(\pi/2), Y=R_y(\pi/2), Z=R_z(\pi/2)$. You could look at their pairwise products (e. g. geometrically, without multiplying any matrices), then use that knowledge for your eulerian products. $\endgroup$ – arseniiv Dec 14 '17 at 18:42
  • $\begingroup$ @Leo Mosher I mentioned constraints i.e. $\alpha_j = \{0,\dfrac{\pi}{2},{\pi}, \dfrac{3\pi}{2}$ (the same effect as in case of $-\dfrac{\pi}{2}$ rotation) $\}$ what results necessary in entries of matrix being only $-1,0,1$. Henceforth connection with this other question. $\endgroup$ – Widawensen Dec 15 '17 at 9:57
  • $\begingroup$ @arseniiv To say truth I don't see clearly your idea. Maybe it is really hard to devise some clever condition which would discern unique matrices from these which are not unique. I'm staring to believe that the best way would be however to list all $64$ matrices and from their form to try to classify them ( I need such classification for robotic application).. $\endgroup$ – Widawensen Dec 15 '17 at 10:03
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    $\begingroup$ Wait...robotics? You have a computer in the loop here? Just compute all 64 and compare. You can do that, effectively instantly, on about \$0.15 of hardware. I don′t see why anyone (including you) should spend more than \$0.15 worth of time trying to think of a more "principled" approach. As a broad point: Euler angles are almost always the wrong answer for robotics, because even when you address the gimbal-lock problem (which is what's hiding beneath here), you haven't solved the far worse "near gimbal lock" problem which is associated with high wear, uncertainty, and uncontrollability. YMMV. $\endgroup$ – John Hughes Dec 15 '17 at 13:21

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