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

m Frobenius -> equilateral Euclidean
Improve notation: getting rid of A and B, only V and M now
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where T is the tuning map, J the JIP, and P the projection map.  
where T is the tuning map, J the JIP, and P the projection map.  


The projection map multipled by a [[Temperament mapping matrices|temperament map]] on the left yields its [[Tmonzos and tvals|tempered monzos]]. In particular, if A is the temperament map of P, then
The projection map multipled by a [[Temperament mapping matrices|temperament map]] on the left yields its [[Tmonzos and tvals|tempered monzos]]. In particular, if V is the temperament map of P, then


<math>\displaystyle AP = A</math>
<math>\displaystyle VP = V</math>


Second, the projection map multipled by a monzo list on the right yields the tunings of the list in terms of monzos. In particular, if B is the [[comma list]] of P, then
Second, the projection map multipled by a monzo list on the right yields the tunings of the list in terms of monzos. In particular, if M is the [[comma list]] of P, then


<math>\displaystyle PB = O</math>
<math>\displaystyle PM = O</math>


For any Euclidean aka ''L''<sup>2</sup> tunings, the weighted projection map is
For any Euclidean aka ''L''<sup>2</sup> tuning without constraints, the weight-skewed projection map is


<math>\displaystyle P_W = V^+V</math>
<math>\displaystyle P_{WX} = V_{WX}^+ V_{WX}</math>


where V = AW is the weighted val list of the temperament. Removing the weight, it is
where <sup>+</sup> is the [[Wikipedia:Moore–Penrose inverse|pseudoinverse]], and V<sub>WX</sub> = VWX is the weight-skewed val list of the temperament. Removing the transformation, it is


<math>\displaystyle P = WV^+VW^{-1} = W(AW)^+A</math>
<math>\displaystyle P = WXV_{WX}^+ V_{WX}(WX)^+ = WX(VWX)^+V</math>


== CEE tuning ==
== CEE tuning ==
Let us start with CEE tuning (constrained equilateral-Euclidean tuning): there is no weight or skew, and the constraint is the octave.  
Let us start with CEE tuning (constrained equilateral-Euclidean tuning): both weight and skew are identity matrices, which will be omitted below, and the constraint is the octave.  


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


<math>\displaystyle B_{\rm C} = [ \begin{matrix} 1 & 0 & \ldots & 0 \end{matrix} \rangle</math>
<math>\displaystyle M_{\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 matrices|subgroup basis]].  
but it works as long as it is the first ''r'' elements of the [[Subgroup basis matrices|subgroup basis]].  
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<math>\displaystyle  
<math>\displaystyle  
AP_{\rm C} = A \\
VP_{\rm C} = V \\
P_{\rm C}B = O
P_{\rm C}M = O
</math>
</math>


in addition to
in addition to


<math>\displaystyle P_{\rm C} B_{\rm C} = B_{\rm C}</math>
<math>\displaystyle P_{\rm C} M_{\rm C} = M_{\rm C}</math>


Since P is characteristic of the temperament and is independent of the specific tuning, notice
Since P is characteristic of the temperament and is independent of the specific tuning, notice
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<math>\displaystyle P = P_{\rm C}^+P_{\rm C}</math>
<math>\displaystyle P = P_{\rm C}^+P_{\rm C}</math>


where <sup>+</sup> is the [[Wikipedia:Moore–Penrose inverse|pseudoinverse]]. That makes the pseudoinverse of P<sub>C</sub> easier to work with than P<sub>C</sub> itself, as
That makes the pseudoinverse of P<sub>C</sub> easier to work with than P<sub>C</sub> itself, as


<math>\displaystyle  
<math>\displaystyle  
P_{\rm C}^+ B_{\rm C}
P_{\rm C}^+ M_{\rm C}
= P_{\rm C}^+P_{\rm C} B_{\rm C}
= P_{\rm C}^+P_{\rm C} M_{\rm C}
= P B_{\rm C}
= P M_{\rm C}
</math>
</math>


Both P<sub>C</sub><sup>+</sup>B<sub>C</sub> and PB<sub>C</sub> are the same slice of the first ''r'' columns of P.
Both P<sub>C</sub><sup>+</sup>M<sub>C</sub> and PM<sub>C</sub> are the same slice of the first ''r'' columns of P.


With the first ''r'' rows and columns removed, the remaining part in the mapping will be dubbed the minor matrix, denoted A<sub>M</sub>. The minor matrix of the projection map
With the first ''r'' rows and columns removed, the remaining part in the mapping will be dubbed the minor matrix, denoted V<sub>M</sub>. The minor matrix of the projection map


<math>\displaystyle P_{\rm M} = A_{\rm M}^+ A_{\rm M} </math>
<math>\displaystyle P_{\rm M} = V_{\rm M}^+ V_{\rm M} </math>


forms an orthogonal projection map filling the bottom-right section of P<sub>C</sub><sup>+</sup>.  
forms an orthogonal projection map filling the bottom-right section of P<sub>C</sub><sup>+</sup>.  


In general, if B<sub>C</sub> is the first ''r'' elements of the subgroup basis, then P<sub>C</sub> is of the form
In general, if M<sub>C</sub> is the first ''r'' elements of the subgroup basis, then P<sub>C</sub> is of the form


<math>\displaystyle  
<math>\displaystyle  
P_{\rm C} =  
P_{\rm C} =  
\begin{bmatrix}  
\begin{bmatrix}  
A^+AB_{\rm C} & \begin{matrix} O \\ A_{\rm M}^+A_{\rm M} \end{matrix}
V^+VM_{\rm C} & \begin{matrix} O \\ V_{\rm M}^+V_{\rm M} \end{matrix}
\end{bmatrix}^+
\end{bmatrix}^+
</math>
</math>
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<math>\displaystyle  
<math>\displaystyle  
\begin{align}
\begin{align}
V &= AWX \\
V_{WX} &= VWX \\
M_{\rm C} &= (WX)^+ B_{\rm C}
(M_{\rm C})_{WX} &= (WX)^+ M_{\rm C}
\end{align}
\end{align}
</math>
</math>
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(P_{\rm C})_{WX} =  
(P_{\rm C})_{WX} =  
\begin{bmatrix}  
\begin{bmatrix}  
V^+VM_{\rm C} & \begin{matrix} O \\ V_{\rm M}^+V_{\rm M} \end{matrix}
V_{WX}^+ V_{WX} (M_{\rm C})_{WX} & \begin{matrix} O \\ (V_{WX})_{\rm M}^+ (V_{WX})_{\rm M} \end{matrix}
\end{bmatrix}^+
\end{bmatrix}^+
</math>
</math>
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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.  
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.  


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


<math>\displaystyle  
<math>\displaystyle  
B_{\rm C} =  
M_{\rm C} =  
\begin{bmatrix}
\begin{bmatrix}
1 & 0 \\
1 & 0 \\
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0 & 0
0 & 0
\end{bmatrix},  
\end{bmatrix},  
B_{\rm C}^\perp =  
M_{\rm C}^\perp =  
\begin{bmatrix}
\begin{bmatrix}
0 & 0 \\
0 & 0 \\
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<math>\displaystyle  
<math>\displaystyle  
\begin{align}
\begin{align}
A_S &= AS \\
V_S &= VS \\
(B_{\rm C})_S &= S^{-1}B_{\rm C}
(M_{\rm C})_S &= S^{-1}M_{\rm C}
\end{align}
\end{align}
</math>
</math>
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<math>\displaystyle  
<math>\displaystyle  
\begin{align}
\begin{align}
V &= AWX \\
V_{WX} &= VWX \\
M_{\rm C} &= (WX)^+ B_{\rm C}
(M_{\rm C})_{WX} &= (WX)^+ M_{\rm C}
\end{align}
\end{align}
</math>
</math>
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<math>\displaystyle  
<math>\displaystyle  
\begin{align}
\begin{align}
V_S &= VS \\
V_{WXS} &= VWXS \\
(M_{\rm C})_S &= S^{-1} M_{\rm C}
(M_{\rm C})_{WXS} &= (WXS)^+ M_{\rm C}
\end{align}
\end{align}
</math>
</math>
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Proceed as before. The projection map found this way will be weight-skewed and in the working basis. To reconstruct the original projection map, apply
Proceed as before. The projection map found this way will be weight-skewed and in the working basis. To reconstruct the original projection map, apply


<math>\displaystyle P_{\rm C} = WXS (P_{\rm C})_{WXS} S^{-1} (WX)^+</math>
<math>\displaystyle P_{\rm C} = WXS (P_{\rm C})_{WXS} (WXS)^+</math>


== Example ==
== Example ==
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<math>\displaystyle  
<math>\displaystyle  
A =
V =
\begin{bmatrix}
\begin{bmatrix}
1 & 0 & -4 & -13 \\
1 & 0 & -4 & -13 \\
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<math>\displaystyle  
<math>\displaystyle  
\begin{align}
\begin{align}
P &= A^+A \\
P &= V^+V \\
&= \frac{1}{446}
&= \frac{1}{446}
\begin{bmatrix}
\begin{bmatrix}
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The minor matrix of the mapping is
The minor matrix of the mapping is


<math>\displaystyle A_{\rm M} = \begin{bmatrix} 1 & 4 & 10\end{bmatrix}</math>
<math>\displaystyle V_{\rm M} = \begin{bmatrix} 1 & 4 & 10\end{bmatrix}</math>


and the minor matrix of the projection map is
and the minor matrix of the projection map is
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<math>\displaystyle  
<math>\displaystyle  
\begin{align}
\begin{align}
P_{\rm M} &= A_{\rm M}^+A_{\rm M} \\
P_{\rm M} &= V_{\rm M}^+V_{\rm M} \\
&= \frac{1}{117}
&= \frac{1}{117}
\begin{bmatrix}  
\begin{bmatrix}