SB Tree Equivalent for Commas

Yesterday I was talking with Mike about some concepts involving octave-equivalent monzo space, hodge duals, tempering, and Stern-Brocot Trees. There isn't enough room in one post for me to describe each of the concepts in detail, so I'll just stick to one for now.

Lets say we have a 5-limit comma, such as 81/80. If we rewrite it as a monzo, it is |-4 4 -1>. However, for what I am about to do next, I don't really care about octaves. So lets just use an octave-equivalent monzo, |4 -1>.

Now if we want to turn these numbers into positions on a generator circle (otherwise known as a val, or a "mapping" to be precise), we must take the Hodge Dual. So then it will give us something like <1 4], if I am correct.

Now here is the fun part. Divide the first element in the val by the second element. This will yield the fraction 1/4. Now take out your handy-dandy Stern-Brocot tree, and assign the comma 81:80 to the position 1/4 on the tree.

If we do this in reverse, by taking the fractions of the SB-Tree and finding the corresponding monzos for each branch, we can map commas to different positions on the tree. Each comma has two "parent" commas and two "children" commas. Once I plotted these commas on the tree, I noticed a weird pattern - whenever a parent and its child are tempered out at the same time, it always results in 1-EDO. Mike was able to figure out the pattern right away.

Here is the precise pattern we discovered:
Comma(parent) = Generator(child)

Lets again consider Meantone as our example. Meantone has two parents on this "comma" tree - 4/3 and 27/20. Notice how both of these commas are also generators in Meantone. As for why this pattern exists, Mike and I could not figure it out.

Anyways, maybe Mike can explain some more things we discussed. I am not very good with coherent thoughts, so sometimes my explanations of things can seem very garbled. If anything I said needs clarification, just let me know.

-Ryan

It's a well known property of the Stern Brocot tree that a certain product of neighboring ratios always comes out as one. I think that's what you're getting here.

Graham

------Original message------
From: Ryan Avella <domeofatonement@yahoo.com>
To: <tuning-math@yahoogroups.com>
Date: Saturday, December 31, 2011 8:03:22 AM GMT-0000
Subject: [tuning-math] SB Tree Equivalent for Commas

Yesterday I was talking with Mike about some concepts involving octave-equivalent monzo space, hodge duals, tempering, and Stern-Brocot Trees. There isn't enough room in one post for me to describe each of the concepts in detail, so I'll just stick to one for now.

Lets say we have a 5-limit comma, such as 81/80. If we rewrite it as a monzo, it is |-4 4 -1>. However, for what I am about to do next, I don't really care about octaves. So lets just use an octave-equivalent monzo, |4 -1>.

Now if we want to turn these numbers into positions on a generator circle (otherwise known as a val, or a "mapping" to be precise), we must take the Hodge Dual. So then it will give us something like <1 4], if I am correct.

Now here is the fun part. Divide the first element in the val by the second element. This will yield the fraction 1/4. Now take out your handy-dandy Stern-Brocot tree, and assign the comma 81:80 to the position 1/4 on the tree.

If we do this in reverse, by taking the fractions of the SB-Tree and finding the corresponding monzos for each branch, we can map commas to different positions on the tree. Each comma has two "parent" commas and two "children" commas. Once I plotted these commas on the tree, I noticed a weird pattern - whenever a parent and its child are tempered out at the same time, it always results in 1-EDO. Mike was able to figure out the pattern right away.

Here is the precise pattern we discovered:
Comma(parent) = Generator(child)

Lets again consider Meantone as our example. Meantone has two parents on this "comma" tree - 4/3 and 27/20. Notice how both of these commas are also generators in Meantone. As for why this pattern exists, Mike and I could not figure it out.

Anyways, maybe Mike can explain some more things we discussed. I am not very good with coherent thoughts, so sometimes my explanations of things can seem very garbled. If anything I said needs clarification, just let me know.

-Ryan

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Alright, I figured out part of the pattern.

Lets say we have two parents on our Stern-Brocot Tree, A/B and C/D (which are a farey pair). Their child is the freshman sum (A+C)/(B+D). These fractions represent mappings, where the numerator is the amount of generators for 3/1 and the denominator is the amount of generators for 5/1.

Since these fractions represent mappings, we can turn them into the monzos M1 = |* B -A>, M2 = |* D -C> and M3 = |* (B+D) (-A-C)>, where * represents an unknown power of two.

Now if we subtract the child monzo from one of the parents, this is the equivalent of dividing one fraction by the other (due to the product-property of exponents).

If we subtract M3 from M1, for example, we get |* -D C>, which is actually -M2. However, in the world of commas, the sign of a monzo doesn't matter to us (81/80 and 80/81 are interchangeable as commas). Therefore, we know that if M3 is tempered out, then M2 and M1 will be made equivalent.

-Ryan