Constrained tuning/Analytical solution to constrained Euclidean tunings: Difference between revisions

Skew!
Cleanup
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== CFE tuning ==
== CFE tuning ==
Let us start with CFE tuning (<u>c</u>onstrained <u>F</u>rob<u>e</u>nius tuning) since its weighter is the identity matrix and the constraint is simply the octave.  
Let us start with CFE tuning (<u>c</u>onstrained <u>F</u>rob<u>e</u>nius tuning): there is no weight or skew, and the constraint is the octave.  


Denote the constraint by B<sub>C</sub>. In the case of CFE, it is the octave:  
Denote the constraint by B<sub>C</sub>. In the case of CFE, it is the octave:  


<math>\displaystyle B_{\rm C} = [ \begin{matrix} 1 & 0 & 0 & 0 \end{matrix} \rangle</math>
<math>\displaystyle B_{\rm C} = [ \begin{matrix} 1 & 0 & \ldots & 0 \end{matrix} \rangle</math>


but it works as long as it is the first ''r'' elements of the subgroup basis.  
but it works as long as it is the first ''r'' elements of the subgroup basis.  
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== Nontrivially constrained tuning ==
== Nontrivially constrained tuning ==
What if the constraint is something more complex, especially when it is not the first ''r'' elements of the subgroup basis? It turns out we can always transform the subgroup basis to encapsulate the constraint. Such a subgroup is formed by the constraint and its orthonormal complement.  
What if the constraint is something more complex, especially when it is not the first ''r'' elements of the subgroup basis? It turns out we can always transform the subgroup basis to encapsulate the constraint. Such a subgroup S is formed by the constraint and its orthonormal complement.  


For example, if the temperament is in the subgroup basis of 2.3.5.7, and the constraint is 2.5/3:
<math>\displaystyle S = [\begin{matrix} B_{\rm C} & B_{\rm C}^\perp \end{matrix}] </math>


<math>\displaystyle B_{\rm C} = [[ \begin{matrix} 1 & 0 & 0 & 0 \end{matrix} \rangle, [ \begin{matrix} 0 & -1 & 1 & 0 \end{matrix} \rangle]</math>
For example, if the temperament is in the subgroup basis of 2.3.5.7, and if the constraint is 2.5/3, then
 
Then the basis transformation matrix will be


<math>\displaystyle  
<math>\displaystyle  
S = [\begin{matrix} B_{\rm C} & B_{\rm C}^\perp \end{matrix}] =  
B_{\rm C} =  
\begin{bmatrix}
1 & 0 \\
0 & -1 \\
0 & 1 \\
0 & 0
\end{bmatrix},
B_{\rm C}^\perp =
\begin{bmatrix}
0 & 0 \\
1/\sqrt{2} & 0 \\
1/\sqrt{2} & 0 \\
0 & 1
\end{bmatrix},
S =  
\begin{bmatrix}
\begin{bmatrix}
1 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 \\
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<math>\displaystyle  
<math>\displaystyle  
S_{WX} = [\begin{matrix} M_{\rm C} & M_{\rm C}^\perp \end{matrix}]
S = [\begin{matrix} M_{\rm C} & M_{\rm C}^\perp \end{matrix}]
</math>
</math>


We should apply this S<sub>WX</sub> to the weight-skewed map and the weight-skewed constraint to convert them into the working basis:  
We should apply this S to the weight-skewed map and the weight-skewed constraint to convert them into the working basis:  


<math>\displaystyle  
<math>\displaystyle  
\begin{align}
\begin{align}
V_S &= VS_{WX} \\
V_S &= VS \\
(M_{\rm C})_S &= S_{WX}^+ M_{\rm C}
(M_{\rm C})_S &= S^{-1} M_{\rm C}
\end{align}
\end{align}
</math>
</math>