Yahoo Tuning Groups Ultimate Backup tuning Another way to get to infinite limit temperaments

An idea I just had:

1) start with a mapping matrix of dimension RxC in which R <= C.

[5 8]
[7 11]

2) add the next prime by just selecting the coefficients that generate
the temperament lowest in ________ badness

2.3:
[5 8 12]
[7 11 16]

3) goto 2

As you can see, this will go on forever. So we have two ways to
terminate this algorithm:

1) arbitrarily at some "limit," but which is now being chosen for
purely computational rather than theoretical reasons
2) if ________ badness is a type of badness which is designed to
converge as the limit increases, after the change in badness with each
next successive prime falls below some threshold of delta that you
don't care about

But there's one question that's most pertinent of all: is there a
linear-time or constant-time function which will generate the above
sequence of digits given any initial "seed?" Or, if that's too
specific, will the sequence of digits produced by the resultant matrix
be some well-known sequence, or be related to some well known
function, perhaps with an official mathematical-sounding name? Or is
there some such badness function that will make that the case?

Any ideas come to mind?

-Mike

🔗gbreed@...

I have an algorithm for finding all equal temperaments for a given octave division below a given error. For good temperaments with a strict cutoff it probably approaches linear time. The worst case is exponential.
I don't thing there's a general linear time algorithm. It's possible for the optimal mapping to change when you consider a new prime so you have to back track. What you can do is find each new prime according to the optimal octave stretch for the ones you already have. For classic TOP that will give an optimal mapping once you get beyond the point where half a step is good enough. It also gives you a good cutoff to use with the strict algorithm.

Graham

------Original message------
From: Mike Battaglia <battaglia01@gmail.com>
To: <tuning@yahoogroups.com>
Date: Wednesday, December 21, 2011 5:04:41 AM GMT-0500
Subject: [tuning] Another way to get to infinite limit temperaments

An idea I just had:

1) start with a mapping matrix of dimension RxC in which R <= C.

[5 8]
[7 11]

2) add the next prime by just selecting the coefficients that generate
the temperament lowest in ________ badness

2.3:
[5 8 12]
[7 11 16]

3) goto 2

As you can see, this will go on forever. So we have two ways to
terminate this algorithm:

Any ideas come to mind?

-Mike

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🔗Mike Battaglia <battaglia01@...>

On Wed, Dec 21, 2011 at 5:17 PM, gbreed@... <gbreed@...> wrote:
>
> I don't thing there's a general linear time algorithm. It's possible for the optimal mapping to change when you consider a new prime so you have to back track.

This is a good point that I should have realized. Damn. So if I just
do some cascaded TOP thing, where I continually discover new
matrices/vals that minimize the maximum Tenney-weighted error, then
this will fail, because I'll miss certain "best" matrices then? That's
terrible.

> What you can do is find each new prime according to the optimal octave stretch for the ones you already have.

I don't understand, what do you mean according to the optimal octave
stretch? So let's say my "mapping matrix" is just one row, and is <12
19 28|. What happens now? My optimal octave stretch is a bit flat of
JI, and then how do I use this to work out the next prime?

-Mike

🔗Carl Lumma <carl@...>

🔗gbreed@...

You have 12 19 28 as a five limit mapping. That leads to an optimal octave stretch and step size. You can then find the nearest match to 7:4 as a multiple of that step size. The result is more likely to be optimal than keeping on taking the nearest approximations to equal divisions of the 2:1

Graham

------Original message------
From: Mike Battaglia <battaglia01@...>
To: <tuning@yahoogroups.com>
Date: Thursday, December 22, 2011 10:17:24 AM GMT-0500
Subject: Re: [tuning] Another way to get to infinite limit temperaments

> What you can do is find each new prime according to the optimal octave stretch for the ones you already have.

-Mike

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