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Cross-domain temperament merging: Difference between revisions

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→Examples: map-merge example, and improve subscripts for R
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Then remove the columns of all zeros, go back to number list form, and make sure everything's greater than 1, which it is already. And so the merged interval basis <math>B_1|B_2</math> is 2.3.5.7.
Then remove the columns of all zeros, go back to number list form, and make sure everything's greater than 1, which it is already. And so the merged interval basis <math>B_1|B_2</math> is 2.3.5.7.


Now we change the interval basis for each comma basis to <math>B_1|B_2</math>. Let's do meantone first. First we need to find our interval rebase <math>R_{B_1↔B_1|B_2}</math>.
Now we change the interval basis for each comma basis to <math>B_1|B_2</math>. Let's do meantone first. First we need to find our interval rebase <math>R_{B_1|B_2↔B_1}</math>.


Actually, <math>R_{1↔1|2}</math> is easier enough to read, and still clear enough, so we'll use that notation moving forward. And here it is itself:
Actually, <math>R_{1|2↔1}</math> is easier enough to read, and still clear enough, so we'll use that notation moving forward. And here it is itself:




Line 127: Line 127:




Now we take our comma basis <math>C_1</math> and left-multiply it by this <math>R_{1↔1|2}</math>, just like we would left-multiply by a mapping:
Now we take our comma basis <math>C_1</math> and left-multiply it by this <math>R_{1|2↔1}</math>, just like we would left-multiply by a mapping:




Line 191: Line 191:
\begin{array} {ccc}
\begin{array} {ccc}


R_{1↔1|2} \\
R_{1|2↔1} \\


\begin{array} {ccc}
\begin{array} {ccc}
Line 255: Line 255:




Now we find our other <math>R</math>, the one for archytas, i.e. <math>R_{2↔1|2}</math>:
Now we find our other <math>R</math>, the one for archytas, i.e. <math>R_{1|2↔2}</math>:




Line 358: Line 358:
\begin{array} {ccc}
\begin{array} {ccc}


R_{2↔1|2} \\
R_{1|2↔2} \\


\begin{array} {ccc}
\begin{array} {ccc}
Line 476: Line 476:




And so {{bra|{{vector|4 -4 1}}}}|(2.3.7){{bra|{{vector|6 -1 -1}}}} = {{bra|{{vector|4 -4 1 0}} {{vector|6 -2 0 -1}}}}.
And so {{bra|{{vector|4 -4 1}}}} | (2.3.7){{bra|{{vector|6 -1 -1}}}} = {{bra|{{vector|4 -4 1 0}} {{vector|6 -2 0 -1}}}}.


== Map-merge ==
== Map-merge ==


(WIP)
Now, let's work through an example of a cross-interval basis map-merge: 22 equal temperament <math>T_1</math> with 17 equal temperament <math>T_2</math>, where 22-ET is in the interval basis 2.3.5.11, which we'll call <math>B_1</math>, and 17-ET is in the 2.9.7.11 interval basis, which we'll call <math>B_2</math>.


<nowiki>
First we must intersect these two temperaments' interval bases. The first step of that is to convert them to matrices with the same basis elements:
t1 = {{{22, 35, 51, 76}}, "co", {2, 3, 5, 11}};
 
t2 = {{{17, 54, 48, 59}}, "co", {2, 9, 7, 11}};
 
expectedT = {{{1, 0, 13}, {0, 1, -3}}, "co", {2, 9, 11}};(* {{{22,70,76},{17,54,59}},"co",{2,9,11}}; before canonicalization *)
<math>
test2args[mapMergeWithB, t1, t2, expectedT];
 
</nowiki>
\begin{array} {ccc}
 
\begin{array} {ccc} \\ \end{array} \\
 
\begin{array} {rrr}
\scriptsize{2} \\
\scriptsize{3} \\
\scriptsize{5} \\
\scriptsize{7} \\
\scriptsize{11} \\
\end{array}
 
\end{array}
 
\begin{array} {ccc}
 
\begin{array} {ccc}
\scriptsize{2} & \scriptsize{3} & \scriptsize{5} & \scriptsize{11} \\
\end{array} \\
 
\left[ \begin{array} {rrr}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 \\
\end{array} \right]
 
\end{array}
 
 
\hspace{1cm}
 
 
\begin{array} {ccc}
 
\begin{array} {ccc} \\ \end{array} \\
 
\begin{array} {rrr}
\scriptsize{2} \\
\scriptsize{3} \\
\scriptsize{5} \\
\scriptsize{7} \\
\scriptsize{11} \\
\end{array}
 
\end{array}
 
\begin{array} {ccc}
 
\begin{array} {ccc}
\scriptsize{2} & \scriptsize{9} & \scriptsize{7} & \scriptsize{11} \\
\end{array} \\
 
\left[ \begin{array} {rrr}
1 & 0 & 0 & 0 \\
0 & 2 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
\end{array} \right]
 
\end{array}
 
</math>
 
 
Build our block matrix:
 
 
<math>
\left[ \begin{array} {rrrr|rrrr}
1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 2 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 \\
\hline
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\
\end{array} \right]
</math>
 
 
HNF it:
 
 
<math>
\left[ \begin{array} {rrrrrrrr}
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 & 2 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
\end{array} \right]
</math>
 
 
Identify what we want:
 
 
<math>
\left[ \begin{array} {rrrrrrrr}
1 & 0 & 0 & 0 & 0 & \colorbox{pink}0 & \colorbox{pink}0 & \colorbox{pink}0 \\
0 & 1 & 0 & 0 & 0 & \colorbox{pink}0 & \colorbox{pink}0 & \colorbox{pink}0 \\
0 & 0 & 1 & 0 & 0 & \colorbox{pink}0 & \colorbox{pink}0 & \colorbox{pink}0 \\
0 & 0 & 0 & 1 & 0 & \colorbox{pink}0 & \colorbox{pink}0 & \colorbox{pink}0 \\
0 & 0 & 0 & 0 & 1 & \colorbox{pink}0 & \colorbox{pink}0 & \colorbox{pink}0 \\
0 & 0 & 0 & 0 & 0 & \colorbox{yellow}1 & \colorbox{yellow}0 & \colorbox{yellow}0 \\
0 & 1 & 0 & 0 & 0 & \colorbox{yellow}0 & \colorbox{yellow}2 & \colorbox{yellow}0 \\
0 & 0 & 1 & 0 & 0 & \colorbox{yellow}0 & \colorbox{yellow}0 & \colorbox{yellow}0 \\
0 & 0 & 0 & 0 & 0 & \colorbox{yellow}0 & \colorbox{yellow}0 & \colorbox{yellow}0 \\
0 & 0 & 0 & 0 & 0 & \colorbox{yellow}0 & \colorbox{yellow}0 & \colorbox{yellow}1 \\
\end{array} \right]
</math>
 
 
Take just that:
 
 
<math>
\left[ \begin{array} {rrrrrrrr}
1 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 1 \\
\end{array} \right]
</math>
 
 
Canonicalize (HNF, remove all-zero columns, back to number list form, and make super). So that tells us that the intersected interval basis <math>B1∩B2</math> is 2.9.11.
 
Now we change the interval basis for each mapping to <math>B1∩B2</math>. Let's do 22-ET first. First we need to find our interval rebase <math>R_{1↔1∩2}</math>:
 
 
<math>
 
\begin{array} {ccc}
 
\begin{array} {rrr}
\\
\end{array} \\
 
\begin{array} {rrr}
\scriptsize{2} \\
\scriptsize{3} \\
\scriptsize{5} \\
\scriptsize{11} \\
\end{array}
 
\end{array}
 
 
\begin{array} {ccc}
 
\begin{array} {ccc}
\scriptsize{2} & \scriptsize{9} & \scriptsize{11} \\
\end{array} \\
 
\left[ \begin{array} {rrr}
1 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 0 \\
0 & 0 & 1 \\
\end{array} \right]
 
\end{array}
 
</math>
 
 
Now we take our mapping <math>M_1</math> and right-multiply it by this <math>R_{1∩2↔1}</math>, just like we would right-multiply it by a list of vectors:
 
 
<math>
 
\begin{array} {ccc}
 
M_1 \\
 
\begin{array} {ccc}
\scriptsize{2} & \!\!\! & \scriptsize{3} & \!\!\! & \scriptsize{5} & \!\!\! & \scriptsize{11}
\end{array} \\
 
\left[ \begin{array} {rrr}
22 & 35 & 51 & 76 \\
\end{array} \right] \\
 
\begin{array} {rrr}
\\
\\
\end{array}
 
\end{array}
 
 
\hspace{0.5cm}
\large{×} \normalsize{}
\hspace{0.5cm}
 
 
\begin{array} {ccc}
 
\\
 
\begin{array} {rrr}
\\
\end{array} \\
 
\begin{array} {rrr}
\scriptsize{2} \\
\scriptsize{3} \\
\scriptsize{5} \\
\scriptsize{11} \\
\end{array}
 
\end{array}
 
 
\begin{array} {ccc}
 
R_{1↔1∩2} \\
 
\begin{array} {ccc}
\scriptsize{2} & \scriptsize{9} & \scriptsize{11} \\
\end{array} \\
 
\left[ \begin{array} {rrr}
1 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 0 \\
0 & 0 & 1 \\
\end{array} \right]
 
\end{array}
 
 
\hspace{0.5cm}
\large{→} \normalsize{}
\hspace{0.5cm}
 
 
\begin{array} {ccc}
 
(1∩2)M_1 \\
 
\begin{array} {ccc}
\scriptsize{2} & \scriptsize{9} & \scriptsize{11} \\
\end{array} \\
 
\left[ \begin{array} {rrr}
22 & 70 & 76 \\
\end{array} \right] \\
 
\begin{array} {rrr}
\\
\\
\end{array}
 
\end{array}
 
</math>
 
 
Now we find our other <math>R</math>, the one for 17-ET, i.e. <math>R_{2↔1∩2}</math>:
 
 
<math>
 
\begin{array} {ccc}
 
\begin{array} {rrr}
\\
\end{array} \\
 
\begin{array} {rrr}
\scriptsize{2} \\
\scriptsize{9} \\
\scriptsize{7} \\
\scriptsize{11} \\
\end{array}
 
\end{array}
 
 
\begin{array} {ccc}
 
\begin{array} {ccc}
\scriptsize{2} & \scriptsize{9} & \scriptsize{11} \\
\end{array} \\
 
\left[ \begin{array} {rrr}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 0 \\
0 & 0 & 1 \\
\end{array} \right]
 
\end{array}
 
</math>
 
 
And change the interval basis for the 17-ET mapping in the same way:
 
 
<math>
 
\begin{array} {ccc}
 
M_2 \\
 
\begin{array} {ccc}
\scriptsize{2} & \!\!\! & \scriptsize{9} & \!\!\! & \scriptsize{7} & \!\!\! & \scriptsize{11}
\end{array} \\
 
\left[ \begin{array} {rrr}
17 & 54 & 48 & 59 \\
\end{array} \right] \\
 
\begin{array} {rrr}
\\
\\
\end{array}
 
\end{array}
 
 
\hspace{0.5cm}
\large{×} \normalsize{}
\hspace{0.5cm}
 
 
\begin{array} {ccc}
 
\\
 
\begin{array} {rrr}
\\
\end{array} \\
 
\begin{array} {rrr}
\scriptsize{2} \\
\scriptsize{9} \\
\scriptsize{7} \\
\scriptsize{11} \\
\end{array}
 
\end{array}
 
 
\begin{array} {ccc}
 
R_{2↔1∩2} \\
 
\begin{array} {ccc}
\scriptsize{2} & \scriptsize{9} & \scriptsize{11} \\
\end{array} \\
 
\left[ \begin{array} {rrr}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 0 \\
0 & 0 & 1 \\
\end{array} \right]
 
\end{array}
 
 
\hspace{0.5cm}
\large{→} \normalsize{}
\hspace{0.5cm}
 
 
\begin{array} {ccc}
 
(1∩2)M_2 \\
 
\begin{array} {ccc}
\scriptsize{2} & \scriptsize{9} & \scriptsize{11} \\
\end{array} \\
 
\left[ \begin{array} {rrr}
17 & 54 & 59 \\
\end{array} \right] \\
 
\begin{array} {rrr}
\\
\\
\end{array}
 
\end{array}
 
</math>
 
 
Finally, we can merge these two mappings as usual:
 
 
<math>
 
\begin{array} {ccc}
 
\left[ \begin{array} {rrr}
22 & 70 & 76 \\
\end{array} \right] \\
 
\& \\
 
\left[ \begin{array} {rrr}
17 & 54 & 59 \\
\end{array} \right] \\
 
↓ \\
 
\left[ \begin{array} {rrr}
22 & 70 & 76 \\
17 & 54 & 59 \\
\end{array} \right] \\
 
\text{which canonicalizes to} \\
 
\left[ \begin{array} {rrr}
1 & 0 & 13 \\
0 & 1 & -3 \\
\end{array} \right] \\
 
\end{array}
 
</math>
 
 
And so (2.3.5.11){{ket|{{map|22 35 51 76}}}} & (2.9.7.11){{ket|{{map|17 54 48 59}}}} = (2.9.11){{ket|{{map|1 0 13}} {{map|0 1 -3}}}}.


[[Category:Regular temperament theory]]
[[Category:Regular temperament theory]]
[[Category:Terms]]
[[Category:Terms]]
[[Category:Math]]
[[Category:Math]]
Retrieved from "https://en.xen.wiki/w/Cross-domain_temperament_merging"