Diamond tradeoff: Difference between revisions

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Example: explain edge case
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Here is the mapping, <math>M</math>:
Here is the mapping, <math>M</math>:


<math>
<math>
Line 36: Line 37:
</math>
</math>


This is a 5-limit temperament, so we consider the 5-limit tonality diamond: [1/1, 6/5, 5/4, 4/3, 3/2, 8/5, 5/3]. Of these seven pitches, there are only three we care about. We don't care about the unison, and half of the remaining pitches are octave-complements of the others are thus irrelevant. So, we'll only look at [4/3, 5/4, 6/5].
 
This is a 5-limit temperament, so we consider the 5-limit tonality diamond: {1/1, 6/5, 5/4, 4/3, 3/2, 8/5, 5/3}. Of these seven pitches, there are only three we care about. We don't care about the unison, and half of the remaining pitches are octave-complements of the others are thus irrelevant. So, we'll only look at {4/3, 5/4, 6/5}.


For each of these three diamond consonances, we want to find what generator is required in order that this pitch remains pure after tempering, or in other words, that it is an unchanged interval (sometimes called an [[eigenmonzo]]). And we want to know this for the situation where octaves are pure.  
For each of these three diamond consonances, we want to find what generator is required in order that this pitch remains pure after tempering, or in other words, that it is an unchanged interval (sometimes called an [[eigenmonzo]]). And we want to know this for the situation where octaves are pure.  


Let's do it for 4/3 first. So, we prepare a matrix out of these two unchanged intervals, 2/1 and 4/3, and call it <math>U</math>:
==== First diamond extrema ====
 
Let's do it for 4/3 first, {{vector|2 -1 0}}. So, we prepare a matrix out of these two unchanged intervals, 2/1 and 4/3, and call it <math>U</math>:
 


<math>
<math>
Line 50: Line 55:
</math>
</math>


One property of an unchanged interval of a tuning is that it is [https://mathworld.wolfram.com/Eigenvector.html eigenvector] of the projection matrix<ref>For now, the best explanation of projection matrices seems to be on the [[fractional monzo]]s page.</ref> for the tuning where the eigenvalue <math>λ</math> is 1. If the projection matrix is <math>P</math>, by the definition of eigenvectors, that means <math>P⋅U = λ⋅U</math>, or <math>P⋅U = 1⋅U</math>, or simply <math>P⋅U = U</math>. In other words, the projection matrix maps the interval to itself; it is unchanged by the tuning. Because we know what <math>U</math> is, we could solve for <math>P</math> now. But we don't want <math>P</math>; we want the generators. Fortunately, <math>P</math> is defined in terms of our desired generators, <math>G</math>, and our mapping, <math>M</math>, like this: <math>P = GM</math>. So if <math>P = G⋅M</math>, then we can substitute that in for <math>P</math>, and our equation will now be <math>G⋅M⋅U = U</math>. But we want to solve for our generators, so that's <math>G</math>. So if we right multiply both sides by the inverse of <math>(M⋅U)</math>, we get
<math>
G⋅M⋅U(MU)^{-1} = U(MU)^{-1} \\
G⋅\cancel{M⋅U}\cancel{(MU)^{-1}} = U(MU)^{-1} \\
G = U(MU)^{-1}
</math>


So with <math>MU</math> on the left side canceled out, the rest is busywork.
Next, we find the generators matrix <math>G</math> corresponding to the tuning where these two intervals are unchanged. The formula for this generators matrix is <math>G = U(MU)^{-1}</math> (for a full explanation of this formula, see [[Dave Keenan and Douglas Blumeyer's guide to RTT tuning#Unchanged interval tuning strategy]]). Here, let's work it out for our chosen <math>U</math>. First, we multiply <math>MU</math>:


We multiply <math>M⋅U</math>:


<math>
<math>
\begin{array} {c}
M \\
\left[ \begin{array} {rrr}
\left[ \begin{array} {rrr}
1 & 0 & -4 \\
1 & 0 & -4 \\
0 & 1 & 4 \\
0 & 1 & 4 \\
\end{array} \right]
\end{array} \right]
\end{array}
\begin{array} {c}
U \\
\left[ \begin{array} {rrr}
\left[ \begin{array} {rrr}
1 & 2 \\
1 & 2 \\
0 & -1 \\
0 & -1 \\
0 & 0 \\
0 & 0 \\
\end{array} \right] =
\end{array} \right]
\end{array}
 
\begin{array} {c} \\ = \end{array}
 
\begin{array} {c}
MU \\
\left[ \begin{array} {rrr}
\left[ \begin{array} {rrr}
1 & 1 \\
1 & 1 \\
0 & -1 \\
0 & -1 \\
\end{array} \right]
\end{array} \right]
\end{array}
</math>
</math>


We take the inverse <math>(M⋅U)^{-1}</math> (which in this case is the same):
 
We take the inverse (which in this case is the same) (the best way to find a matrix inverse is to use a mathematical calculation tool such as Wolfram Language, but it can be done by hand; for a step-by-step demonstration of how to take the inverse, see [[Defactoring algorithms#Inversion by hand]]):
 


<math>
<math>
\begin{array} {c}
\\
\left[ \begin{array} {rrr}
1 & 1 \\
0 & -1 \\
\end{array} \right]^{-1}
\end{array}
\begin{array} {c} \\ = \end{array}
\begin{array} {c}
(MU)^{-1} \\
\left[ \begin{array} {rrr}
\left[ \begin{array} {rrr}
1 & 1 \\
1 & 1 \\
0 & -1 \\
0 & -1 \\
\end{array} \right]
\end{array} \right]
\end{array}
</math>
</math>


Then find <math>G</math> which is <math>U⋅(M⋅U)^{-1}</math>:
 
Then find <math>G</math> which is <math>U(MU)^{-1}</math>:
 


<math>
<math>
\begin{array} {c}
U \\
\left[ \begin{array} {rrr}
\left[ \begin{array} {rrr}
1 & 2 \\
1 & 2 \\
Line 95: Line 129:
0 & 0 \\
0 & 0 \\
\end{array} \right]
\end{array} \right]
\end{array}
\begin{array} {c}
(MU)^{-1} \\
\left[ \begin{array} {rrr}
\left[ \begin{array} {rrr}
1 & 1 \\
1 & 1 \\
0 & -1 \\
0 & -1 \\
\end{array} \right] =
\end{array} \right]
\end{array}
 
\begin{array} {c} \\ = \end{array}
 
\begin{array} {c}
G \\
\left[ \begin{array} {rrr}
\left[ \begin{array} {rrr}
1 & -1 \\
1 & -1 \\
Line 104: Line 148:
0 & 0 \\
0 & 0 \\
\end{array} \right]
\end{array} \right]
\end{array}
</math>
</math>


Reading the columns from <math>G</math>, the first one confirms our period of 2/1, and the second column gives our generator 3/2. Which is unsurprising. In cents, that's 1200¢ × log₂(3/2) ≈ 701.955¢. The next unchanged interval will give a more interesting result.


So let's do 5/4 now. We prepare a matrix out of these two unchanged intervals, 2/1 and 5/4, and call it <math>U</math>:
Reading the columns from <math>G</math>, the first one {{vector|1 0 0}} confirms our period of 2/1, and the second column {{vector|-1 1 0}} gives our generator 3/2. Which is unsurprising. In cents, that's 1200¢ × log₂(3/2) ≈ 701.955¢. The next unchanged interval will give a more interesting result.
 
==== Second diamond extrema ====
 
So let's do 5/4 now, {{vector|-2 0 1}}. We prepare a matrix out of these two unchanged intervals, 2/1 and 5/4, and call it <math>U</math>:
 


<math>
<math>
\left[ \begin{array} {rrr}
\left[ \begin{array} {rrr}
1 & -2 \\
1 & -2 \\
Line 116: Line 167:
0 & 1 \\
0 & 1 \\
\end{array} \right]
\end{array} \right]
</math>
</math>


We multiply <math>M⋅U</math>:
 
We multiply <math>MU</math>:
 


<math>
<math>
\begin{array} {c}
M \\
\left[ \begin{array} {rrr}
\left[ \begin{array} {rrr}
1 & 0 & -4 \\
1 & 0 & -4 \\
0 & 1 & 4 \\
0 & 1 & 4 \\
\end{array} \right]
\end{array} \right]
\end{array}
\begin{array} {c}
U \\
\left[ \begin{array} {rrr}
\left[ \begin{array} {rrr}
1 & -2 \\
1 & -2 \\
0 & 0 \\
0 & 0 \\
0 & 1 \\
0 & 1 \\
\end{array} \right] =
\end{array} \right]
\end{array}
 
\begin{array} {c} \\ = \end{array}
 
\begin{array} {c}
MU \\
\left[ \begin{array} {rrr}
\left[ \begin{array} {rrr}
1 & -2 \\
1 & -2 \\
0 & 4 \\
0 & 4 \\
\end{array} \right]
\end{array} \right]
\end{array}
</math>
</math>


We take the inverse <math>(M⋅U)^{-1}</math>:
 
We take the inverse:
 


<math>
<math>
\begin{array} {c}
\\
\left[ \begin{array} {rrr}
1 & -2 \\
0 & 4 \\
\end{array} \right]^{-1}
\end{array}
\begin{array} {c} \\ = \end{array}
\begin{array} {c}
(MU)^{-1} \\
\left[ \begin{array} {rrr}
\left[ \begin{array} {rrr}
1 & \frac12 \\
1 & \frac12 \\
0 & \frac14 \\
0 & \frac14 \\
\end{array} \right]
\end{array} \right]
\end{array}
</math>
</math>


Then find <math>G</math> which is <math>U⋅(M⋅U)^{-1}</math>:
 
Then find <math>G</math> which is <math>U(MU)^{-1}</math>:
 


<math>
<math>
\begin{array} {c}
U \\
\left[ \begin{array} {rrr}
\left[ \begin{array} {rrr}
1 & -2 \\
1 & -2 \\
Line 153: Line 244:
0 & 1 \\
0 & 1 \\
\end{array} \right]
\end{array} \right]
\end{array}
\begin{array} {c}
(MU)^{-1} \\
\left[ \begin{array} {rrr}
\left[ \begin{array} {rrr}
1 & \frac12 \\
1 & \frac12 \\
0 & \frac14 \\
0 & \frac14 \\
\end{array} \right] =
\end{array} \right]  
\end{array}
 
\begin{array} {c} \\ = \end{array}
 
\begin{array} {c}
G \\
\left[ \begin{array} {rrr}
\left[ \begin{array} {rrr}
1 & 0 \\
1 & 0 \\
Line 162: Line 263:
0 & \frac14 \\
0 & \frac14 \\
\end{array} \right]
\end{array} \right]
\end{array}
</math>
</math>


This tells us our generator is 5^(1/4). In cents, that's 1200¢ × log₂(5¹⸍⁴) ≈ 696.578¢.


Okay, one more unchanged interval to check: 6/5. We prepare a matrix out of these two unchanged intervals, 2/1 and 6/5, and call it <math>U</math>:
This tells us our generator is {{vector|0 0 <math>\frac14</math>}}, or 5^(1/4). In cents, that's 1200¢ × log₂(5¹⸍⁴) ≈ 696.578¢.
 
==== Third diamond extrema ====
 
Okay, one more unchanged interval to check: 6/5, which is {{vector|1 1 -1}}. We prepare a matrix out of these two unchanged intervals, 2/1 and 6/5, and call it <math>U</math>:
 


<math>
<math>
\left[ \begin{array} {rrr}
\left[ \begin{array} {rrr}
1 & 1 \\
1 & 1 \\
Line 174: Line 282:
0 & -1 \\
0 & -1 \\
\end{array} \right]
\end{array} \right]
</math>
</math>


We multiply <math>M⋅U</math>:
 
We multiply <math>MU</math>:
 


<math>
<math>
\begin{array} {c}
M \\
\left[ \begin{array} {rrr}
\left[ \begin{array} {rrr}
1 & 0 & -4 \\
1 & 0 & -4 \\
0 & 1 & 4 \\
0 & 1 & 4 \\
\end{array} \right]
\end{array} \right]
\end{array}
\begin{array} {c}
U \\
\left[ \begin{array} {rrr}
\left[ \begin{array} {rrr}
1 & 1 \\
1 & 1 \\
0 & 1 \\
0 & 1 \\
0 & -1 \\
0 & -1 \\
\end{array} \right] =
\end{array} \right]  
\end{array}
 
\begin{array} {c} \\ = \end{array}
 
\begin{array} {c}
MU \\
\left[ \begin{array} {rrr}
\left[ \begin{array} {rrr}
1 & 2 \\
1 & 2 \\
0 & -3 \\
0 & -3 \\
\end{array} \right]
\end{array} \right]
\end{array}
</math>
</math>


We take the inverse <math>(M⋅U)^{-1}</math>:
 
We take the inverse:
 


<math>
<math>
\begin{array} {c}
\\
\left[ \begin{array} {rrr}
1 & 2 \\
0 & -3 \\
\end{array} \right]^{-1}
\end{array}
\begin{array} {c} \\ = \end{array}
\begin{array} {c}
(MU)^{-1} \\
\left[ \begin{array} {rrr}
\left[ \begin{array} {rrr}
1 & \frac23 \\
1 & \frac23 \\
0 & -\frac13 \\
0 & -\frac13 \\
\end{array} \right]
\end{array} \right]
\end{array}
</math>
</math>


Then find <math>G</math> which is <math>U⋅(M⋅U)^{-1}</math>:
 
Then find <math>G</math> which is <math>U(MU)^{-1}</math>:
 


<math>
<math>
\begin{array} {c}
U \\
\left[ \begin{array} {rrr}
\left[ \begin{array} {rrr}
1 & 1 \\
1 & 1 \\
Line 211: Line 359:
0 & -1 \\
0 & -1 \\
\end{array} \right]
\end{array} \right]
\end{array}
\begin{array} {c}
(MU)^{-1} \\
\left[ \begin{array} {rrr}
\left[ \begin{array} {rrr}
1 & \frac23 \\
1 & \frac23 \\
0 & -\frac13 \\
0 & -\frac13 \\
\end{array} \right] =
\end{array} \right]  
\end{array}
 
\begin{array} {c} \\ = \end{array}
 
\begin{array} {c}
G \\
\left[ \begin{array} {rrr}
\left[ \begin{array} {rrr}
1 & \frac13 \\
1 & \frac13 \\
Line 220: Line 378:
0 & \frac13 \\
0 & \frac13 \\
\end{array} \right]
\end{array} \right]
\end{array}
</math>
</math>


This tells us our generator is (10/3)^(1/3). In cents, that's 1200¢ × log₂((10/3)¹⸍³) ≈ 694.786¢.  
 
This tells us our generator is {{vector|<math>\frac13</math> <math>-\frac13</math> <math>\frac13</math>}}, or expressed another way, (10/3)^(1/3). In cents, that's 1200¢ × log₂((10/3)¹⸍³) ≈ 694.786¢.  
 
==== Determining the final range ====


We now have our generator sizes that give us pure consonances in the tonality diamond: 701.955¢, 696.578¢, and 694.786¢. The minimum of those is 694.786¢ and the maximum is 701.955¢, so that's our diamond tradeoff range. Anywhere inside that range, we are making at least one of our diamond consonances purer; outside it, we're making them all less pure.
We now have our generator sizes that give us pure consonances in the tonality diamond: 701.955¢, 696.578¢, and 694.786¢. The minimum of those is 694.786¢ and the maximum is 701.955¢, so that's our diamond tradeoff range. Anywhere inside that range, we are making at least one of our diamond consonances purer; outside it, we're making them all less pure.
This method works as long as you do not choose for your unchanged intervals any intervals that are separated by commas of the temperament. In that case, <math>MU</math> will come out singular, that is, determinant 0, with no inverse, which is math's way of telling you that there's no way to not change both intervals when one of them must be tempered to the other. For example, you couldn't set both 3/2 and 40/27 to be unchanged intervals of meantone, because they're off from each other by 81/80. But then again, you wouldn't try to do something weird like that in the first place, would you.


== See also ==
== See also ==

Revision as of 00:28, 27 May 2022

A tuning for a rank-r p-limit regular temperament is diamond tradeoff, or diamond strict, if it fits the following definition: we may define the diamond tradeoff tuning range by finding the convex hull in tuning space of the set of all tunings with r eigenmonzos chosen as follows: one eigenmonzo 2 (pure octaves tunings) and the rest of the eigenmonzos any set of r - 1 members of the p-odd limit tonality diamond, whenever such a tuning is defined (this definition is based on Gene Ward Smith's one).

Original name

In the original work by Andrew Milne, Bill Sethares and James Plamondon this tuning range was known as the "purer" range. On the wiki and in the regular temperament community, for many years this tuning range was referred to simply as the "nice" tuning range.

Vs. diamond monotone

Diamond tradeoff tunings are always guaranteed to occur, but diamond monotone tunings are not.

Additional notes from Andrew Milne

The diamond tradeoff tuning range marks tuning boundaries inside of which the temperament's approximations to simple low-ratio frequency ratios can be "traded" against each other; that is, if I make the 3/2 more accurate, the 5/4 will suffer. However, outside this range, you will improve the tunings of all such intervals by moving back inside. This range therefore makes sense when one is concerned with approximating JI as closely as possible (without asserting a priori which specific consonances are the most important) because, under that criterion, it makes no logical sense to choose a tuning outside that range.

However, it is quite clear that tunings outside of this diamond tradeoff range can function perfectly well as less accurate (and arguably more characterful) representations of the JI intervals specified by the temperament. That is, they are likely to be correctly recognized (whatever that actually means). For example, a 17-TET rendition of a standard piece of meantone music still makes complete musical sense, and major and minor chords still sound like major and minor chords, even though this tuning is outside the diamond tradeoff tuning range.

Explanation adapted from Keenan Pepper

For meantone temperament, there are three specific tunings that are special: one that tunes 4/3 and 3/2 pure, another that tunes 5/4 and 8/5 pure, and the third that tunes 6/5 and 5/3 pure. The tradeoff tuning range consists of these three points in tuning space and everything in between. In this case, the three points fall along a line, where the pure-5/4 tuning is in between the pure-4/3 tuning and the pure-6/5 tuning.

The next thing to understand is that tunings in the middle between the pure-4/3 tuning and the pure-6/5 tuning (including the pure-5/4 tuning, because that's in the middle) are in a sense "reasonable compromises" because whenever you're making one interval of the diamond worse, you're always making another one better. You're in the realm of tradeoffs.

On the other hand, if you go outside these boundaries - for example, if you make 4/3 even flatter than pure - then you're making some intervals in the 5-limit diamond worse without making any of them better. You're past the realm of compromises and now you're just damaging things for no reason.

Example

Here we will demonstrate the calculation of the diamond tradeoff tuning range for meantone.

Here is the mapping, [math]\displaystyle{ M }[/math]:


[math]\displaystyle{ \left[ \begin{array} {rrr} 1 & 0 & -4 \\ 0 & 1 & 4 \\ \end{array} \right] }[/math]


This is a 5-limit temperament, so we consider the 5-limit tonality diamond: {1/1, 6/5, 5/4, 4/3, 3/2, 8/5, 5/3}. Of these seven pitches, there are only three we care about. We don't care about the unison, and half of the remaining pitches are octave-complements of the others are thus irrelevant. So, we'll only look at {4/3, 5/4, 6/5}.

For each of these three diamond consonances, we want to find what generator is required in order that this pitch remains pure after tempering, or in other words, that it is an unchanged interval (sometimes called an eigenmonzo). And we want to know this for the situation where octaves are pure.

First diamond extrema

Let's do it for 4/3 first, [2 -1 0. So, we prepare a matrix out of these two unchanged intervals, 2/1 and 4/3, and call it [math]\displaystyle{ U }[/math]:


[math]\displaystyle{ \left[ \begin{array} {rrr} 1 & 2 \\ 0 & -1 \\ 0 & 0 \\ \end{array} \right] }[/math]


Next, we find the generators matrix [math]\displaystyle{ G }[/math] corresponding to the tuning where these two intervals are unchanged. The formula for this generators matrix is [math]\displaystyle{ G = U(MU)^{-1} }[/math] (for a full explanation of this formula, see Dave Keenan and Douglas Blumeyer's guide to RTT tuning#Unchanged interval tuning strategy). Here, let's work it out for our chosen [math]\displaystyle{ U }[/math]. First, we multiply [math]\displaystyle{ MU }[/math]:


[math]\displaystyle{ \begin{array} {c} M \\ \left[ \begin{array} {rrr} 1 & 0 & -4 \\ 0 & 1 & 4 \\ \end{array} \right] \end{array} \begin{array} {c} U \\ \left[ \begin{array} {rrr} 1 & 2 \\ 0 & -1 \\ 0 & 0 \\ \end{array} \right] \end{array} \begin{array} {c} \\ = \end{array} \begin{array} {c} MU \\ \left[ \begin{array} {rrr} 1 & 1 \\ 0 & -1 \\ \end{array} \right] \end{array} }[/math]


We take the inverse (which in this case is the same) (the best way to find a matrix inverse is to use a mathematical calculation tool such as Wolfram Language, but it can be done by hand; for a step-by-step demonstration of how to take the inverse, see Defactoring algorithms#Inversion by hand):


[math]\displaystyle{ \begin{array} {c} \\ \left[ \begin{array} {rrr} 1 & 1 \\ 0 & -1 \\ \end{array} \right]^{-1} \end{array} \begin{array} {c} \\ = \end{array} \begin{array} {c} (MU)^{-1} \\ \left[ \begin{array} {rrr} 1 & 1 \\ 0 & -1 \\ \end{array} \right] \end{array} }[/math]


Then find [math]\displaystyle{ G }[/math] which is [math]\displaystyle{ U(MU)^{-1} }[/math]:


[math]\displaystyle{ \begin{array} {c} U \\ \left[ \begin{array} {rrr} 1 & 2 \\ 0 & -1 \\ 0 & 0 \\ \end{array} \right] \end{array} \begin{array} {c} (MU)^{-1} \\ \left[ \begin{array} {rrr} 1 & 1 \\ 0 & -1 \\ \end{array} \right] \end{array} \begin{array} {c} \\ = \end{array} \begin{array} {c} G \\ \left[ \begin{array} {rrr} 1 & -1 \\ 0 & 1 \\ 0 & 0 \\ \end{array} \right] \end{array} }[/math]


Reading the columns from [math]\displaystyle{ G }[/math], the first one [1 0 0 confirms our period of 2/1, and the second column [-1 1 0 gives our generator 3/2. Which is unsurprising. In cents, that's 1200¢ × log₂(3/2) ≈ 701.955¢. The next unchanged interval will give a more interesting result.

Second diamond extrema

So let's do 5/4 now, [-2 0 1. We prepare a matrix out of these two unchanged intervals, 2/1 and 5/4, and call it [math]\displaystyle{ U }[/math]:


[math]\displaystyle{ \left[ \begin{array} {rrr} 1 & -2 \\ 0 & 0 \\ 0 & 1 \\ \end{array} \right] }[/math]


We multiply [math]\displaystyle{ MU }[/math]:


[math]\displaystyle{ \begin{array} {c} M \\ \left[ \begin{array} {rrr} 1 & 0 & -4 \\ 0 & 1 & 4 \\ \end{array} \right] \end{array} \begin{array} {c} U \\ \left[ \begin{array} {rrr} 1 & -2 \\ 0 & 0 \\ 0 & 1 \\ \end{array} \right] \end{array} \begin{array} {c} \\ = \end{array} \begin{array} {c} MU \\ \left[ \begin{array} {rrr} 1 & -2 \\ 0 & 4 \\ \end{array} \right] \end{array} }[/math]


We take the inverse:


[math]\displaystyle{ \begin{array} {c} \\ \left[ \begin{array} {rrr} 1 & -2 \\ 0 & 4 \\ \end{array} \right]^{-1} \end{array} \begin{array} {c} \\ = \end{array} \begin{array} {c} (MU)^{-1} \\ \left[ \begin{array} {rrr} 1 & \frac12 \\ 0 & \frac14 \\ \end{array} \right] \end{array} }[/math]


Then find [math]\displaystyle{ G }[/math] which is [math]\displaystyle{ U(MU)^{-1} }[/math]:


[math]\displaystyle{ \begin{array} {c} U \\ \left[ \begin{array} {rrr} 1 & -2 \\ 0 & 0 \\ 0 & 1 \\ \end{array} \right] \end{array} \begin{array} {c} (MU)^{-1} \\ \left[ \begin{array} {rrr} 1 & \frac12 \\ 0 & \frac14 \\ \end{array} \right] \end{array} \begin{array} {c} \\ = \end{array} \begin{array} {c} G \\ \left[ \begin{array} {rrr} 1 & 0 \\ 0 & 0 \\ 0 & \frac14 \\ \end{array} \right] \end{array} }[/math]


This tells us our generator is [0 0 [math]\displaystyle{ \frac14 }[/math], or 5^(1/4). In cents, that's 1200¢ × log₂(5¹⸍⁴) ≈ 696.578¢.

Third diamond extrema

Okay, one more unchanged interval to check: 6/5, which is [1 1 -1. We prepare a matrix out of these two unchanged intervals, 2/1 and 6/5, and call it [math]\displaystyle{ U }[/math]:


[math]\displaystyle{ \left[ \begin{array} {rrr} 1 & 1 \\ 0 & 1 \\ 0 & -1 \\ \end{array} \right] }[/math]


We multiply [math]\displaystyle{ MU }[/math]:


[math]\displaystyle{ \begin{array} {c} M \\ \left[ \begin{array} {rrr} 1 & 0 & -4 \\ 0 & 1 & 4 \\ \end{array} \right] \end{array} \begin{array} {c} U \\ \left[ \begin{array} {rrr} 1 & 1 \\ 0 & 1 \\ 0 & -1 \\ \end{array} \right] \end{array} \begin{array} {c} \\ = \end{array} \begin{array} {c} MU \\ \left[ \begin{array} {rrr} 1 & 2 \\ 0 & -3 \\ \end{array} \right] \end{array} }[/math]


We take the inverse:


[math]\displaystyle{ \begin{array} {c} \\ \left[ \begin{array} {rrr} 1 & 2 \\ 0 & -3 \\ \end{array} \right]^{-1} \end{array} \begin{array} {c} \\ = \end{array} \begin{array} {c} (MU)^{-1} \\ \left[ \begin{array} {rrr} 1 & \frac23 \\ 0 & -\frac13 \\ \end{array} \right] \end{array} }[/math]


Then find [math]\displaystyle{ G }[/math] which is [math]\displaystyle{ U(MU)^{-1} }[/math]:


[math]\displaystyle{ \begin{array} {c} U \\ \left[ \begin{array} {rrr} 1 & 1 \\ 0 & 1 \\ 0 & -1 \\ \end{array} \right] \end{array} \begin{array} {c} (MU)^{-1} \\ \left[ \begin{array} {rrr} 1 & \frac23 \\ 0 & -\frac13 \\ \end{array} \right] \end{array} \begin{array} {c} \\ = \end{array} \begin{array} {c} G \\ \left[ \begin{array} {rrr} 1 & \frac13 \\ 0 & -\frac13 \\ 0 & \frac13 \\ \end{array} \right] \end{array} }[/math]


This tells us our generator is [[math]\displaystyle{ \frac13 }[/math] [math]\displaystyle{ -\frac13 }[/math] [math]\displaystyle{ \frac13 }[/math], or expressed another way, (10/3)^(1/3). In cents, that's 1200¢ × log₂((10/3)¹⸍³) ≈ 694.786¢.

Determining the final range

We now have our generator sizes that give us pure consonances in the tonality diamond: 701.955¢, 696.578¢, and 694.786¢. The minimum of those is 694.786¢ and the maximum is 701.955¢, so that's our diamond tradeoff range. Anywhere inside that range, we are making at least one of our diamond consonances purer; outside it, we're making them all less pure.

See also

For examples and other information, see the topic page Tuning ranges of regular temperaments.