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Complexity formulae

🔗Dave Keenan <d.keenan@xx.xxx.xxx>

3/8/1999 9:57:08 PM

[Keenan:]
>>Certainly the next step is to get the complexity of a ratio from its
>>numerator and denominator, but I'm unclear whether to multiply them, or
>>take the maximum value (as is normally done with odd or prime limits).

[Lumma:]
>In my experience their product can be very useful. I learned this from
>Denny Genovese, who calls it the DF. The idea is based on Partch's
>(perhaps dubious) assumption that the length of the period of the
>composite wave determines its consonance. ...
>I'd like to learn more about how this approach overlaps or contradicts
>with the maximum value approach.

If linear complexity measures are multiplied, logarithmic ones should add.
Max value can be done for both, but clearly it throws away some
information. Just another point on the accuracy vs. user-friendliness
tradeoff.

[Keenan:]
>>If you multiply them you should then take the square-root (i.e. you should
>>find the geometric mean). This is to keep them commensurate with the
>>odd->limit or prime-limit where the max value is taken.

[Lumma:]
>I don't see how taking the square root hurts anything, but I don't see what
>it adds either. What do you mean by "commensurate" with the max value
>approaches?

No it doesn't hurt or add, but instead of being a linear measure it becomes
a quadratic one. "Commensurate" means literally "belonging to the same
system of measurement". Hertz are not comensurate with cents, but semitones
or schismas are. More specifically I mean that you can directly compare two
measures (metrics) with at most a constant multiplier required to bring
them into some kind of agreement.

[Monzo:]
>Since the ratio can be represented by the
>same series you use in your formula (the prime
>factors), it seems to me that you should just
>use negative exponents for the factors of the
>denominator, and multiply all the factors together.

Sure. Good idea. Just have to take absolute values of exponents when
calculating complexities.

[Wolf:]
>In chapter 5 of his _Divisions of the Tetrachord_ (Frog Peak 1993), John
>Chalmers gives an excellent summary of methods for the analysis of
>tetrachords which might be usefully applicable to individual intervals and
>systems other than tetrachords. The presentation of Klarenz Barlow's
>indigestability formula is, however, not so useful out of the context of
>Bawrlough's landmark _Bus Journey to Parametron_ (Feedback Papers 21-23,
>available in the states from Frog Peak).

I'm not sure what's fact and what's fiction here :-) but thanks for the
formulae.

[Wolf:]
>(5) Euler's _gradus suavatis_ function:
>
> (a) for prime numbers: the prime number itself
>
> (b) for composite numbers:
> the sum of the prime factors minus one less than
> the number of factors.
>
> (c) for ratios:
> convert a ratio to a segment of the harmonic series, then
> compute the least common multiple of the terms.

I don't understand part (c) above. What does it mean to "convert a ratio to
a segment of the harmonic series"? I just think of the harmonic series as
the series of whole numbers (positive integers).

[Keenan:]
>> I've switched to referring
>> to them as "prime exponent weights" now.
>> I suppose they represent something about
>> how the human brain processes combinations
>> of tones. The relative importance human evolution
>> has given to the various primes. ...

[Monzo:]
>This is very interesting to me. I've already concluded
>regarding the study of meters that we break all meters
>down into simpler and simpler subdivisions, until
>ultimately everything can be expressed by combinations
>of 2s and 3s.
...
>I'm certain that our difficulty of understanding
>more complex meters as anything but 2s and 3s
>has a lot to do with "The relative importance human
>evolution has given to the various primes". We
>can comprehend 1, 2 or 3 of anything *right away*,
>and from my knowledge of how evolution works,
>speed of recognition or comprehension ranks
>near the top of the list of importance.

I wish to make it clear that I don't expect evolution to have given the
same relative weights to the primes in relation to ryhthm (or cake cutting)
as it did in regard to frequency ratios.

[Monzo:]
>With your original approximated weights
>(0.3, 0.8, 0.9, 1.0, 1.0, ...), which I thought were
>OK at first, it's evident that there is a sharp
>increase in complexity after 2. Perhaps with
>a different weighting your formula is a mathematical
>validation and explanation of my idea.

Since you are free to choose the weights to fit your own judgement it can't
be a validation or explanation of that judgement, merely a description of
it. But tell us what weights you favour and we can compare and maybe home
in on a rough agreement.

Please use the generalised/parameterised Barlow-type formula I mentioned in
my previous post to this thread (i.e. multiply weights by absolute
exponents then add them up), or at least make it clear which version the
weights are for (including whether log or linear).

Actually, this could more simply be called a parameterised Wilson's
harmonic complexity (rather than "the absolute value of the reciprocal of a
parameterised Barlow's harmonicity"). Wilson's choice of weights
corresponds to the primes themselves, except for 2 whose weight is zero.

>> Give me a better term. [Keenan]
>
>"Prime importance"? [Monzo]

"Prime importance" could be used instead of "prime exponent weight", but I
don't think it could be used instead of "harmonic complexity" (of a ratio).
I now favour "harmonic complexity" over "musical complexity". It reminds us
it only works for (near) harmonic timbres and "musical" was too broad since
it could apply to rhythm as well. (Thanks for mentioning rhythm).

[Monzo:]
>I gave it a weight of as low as .05 before it looked
>like something I agreed with, but the weight of 3
>was much lower also - I don't remember what now.
>
>As I said above, my inclination would be to have
>low weights for 2 and 3 and then increase sharply
>after that. What do you think?

I might well agree with such a weighting. Try me.

[Monzo:]
>How about the
>question of 15? I've always thought that 15/8
>is a pretty consonance in a chord - is it more,
>or less, consonant than 7/4?

Ah but we're not (yet) talking about "in a chord". I agree with Paul Erlich
that the 15/8 just happens to appear in a chord with other highly consonant
intervals, but anyway, on its own I think 15/8 is more dissonant than 7/4.

>[Monzo (replying to Erlich):]
>OK, Paul, you got me there. So apparently the
>effect of octave-equivalence has more to do with
>prime-base 2 have an extremely low weight
>in Keenan's "musical complexity" prime-weights
>formula than anything else. But to account
>for the fact of octave-equivalency when no
>other prime equivalency occurs anywhere near
>as strongly, the curve of the weights must be
>very low at 2 then rise sharply for 3 and above.

I agree. You might say that the second harmonic so often accompanied a
fundamental that it carried very little additional information. It has very
low importance. But why *prime* importance. Why is 4 (and 8 and 16) also so
unimportant? Maybe it was just an accident of economical "wiring" that
treats four as a power of two. Or maybe it is just that the 4th harmonic is
just treated as the 2nd harmonic of the 2nd harmonic.

[Wolf:]
>Although I cannot invite the list to dinner in Cologne, I propose that we
>duplicate the experiment. Since most of us on the list are into higher
>primes, let's try all divisions through 19. Please send me, off list ,
>your own ranking of the difficulties in slicing tortes (cakes, pies) from 2
>to 19 equal parts. I'll compile the results and report back.

As I said, I don't think one can draw any musical conclusions from this,
but it sounds like fun and will be interesting to compare.

[Morrison:]
> Am I correct in inferring that, by how much they "cost", you mean how
much taking it out affects the high-level harmonic character?

Yes. Or equivalently how much we over or underestimate the dissonance of an
interval by ignoring *how many* factors of 2 (or 3 etc.) it has.

Regards,

-- Dave Keenan
http://dkeenan.com

🔗Daniel Wolf <DJWOLF_MATERIAL@xxxxxxxxxx.xxxx>

3/9/1999 1:40:12 PM

Message text written by Dave Keenan
>
I'm not sure what's fact and what's fiction here :-) but thanks for the
formulae.<

In case you're wondering, the spelling of Barlow's name is highly variable.
Although he has lived in Europe for the past 30 years, he comes from the
Anglophone minority community in Calcutta, and I suspect that he intends
this as a humorous slant on both Indian English orthography and the
decaying remains of colonialism. To the best of my knowlege, everthing in
the passage you cited was true.

🔗Dave Keenan <d.keenan@xx.xxx.xxx>

3/10/1999 11:41:08 PM

[Erlich:]
>P.S. Dave: I don't like it if 15/8 is given the same complexity as 6/5,
>assuming factors of 2 are ignored.

No. Nor do I. But first notice that I'm specifically saying that factors of
2 should *not* be ignored, just given a low weight. But no matter what the
weighting of 2, 15/4 would have the same complexity as 12/5. I certainly
don't thing they have the same dissonance (15/4 is greater). So maybe
you've just blown away all formulae where we ignore whether various factors
are on the same or different sides of the vinculum (the line between
numerator and denominator). If so, good work.

Maybe the tolerance function can take care of it. But it seems unlikely.
Can you establish that it won't?

More cases which would have the same complexity under any such scheme (same
complexity on the same line).
6/1 3/2
10/1 5/2
12/1 4/3
14/1 7/2
15/1 5/3 *
18/1 9/2
20/1 5/4
21/1 7/3 *
22/1 11/2
24/1 8/3
26/1 13/2
28/1 7/4
30/1 15/2 10/3 6/5 *
33/1 11/3 *
34/1 17/2
35/1 7/5 *
36/1 9/4
38/1 19/2
39/1 13/3
40/1 8/5
42/1 21/2 14/3 7/6 *
44/1 11/4
45/1 9/5 *
46/1 23/2
48/1 16/3
50/1 25/2
51/1 17/3
52/1 13/4
54/1 27/2
55/1 11/5 *
56/1 8/7 *
58/1 29/2
60/1 15/4 20/3 12/5 *
:
120/1 15/8 40/3 24/5 *

Those where only powers-of-two change sides, are not too objectionable. Nor
are those that involve primes higher than 11, since the tolerance function
should take care of them. I've flagged the remaining ones with *.

Surely oo/1 is as consonant as 1/1. ("oo" is meant to look like the lazy-8
infinity symbol). I'd even say n/1, where n>16, are almost as consonant as
1/1. Can we arrange for the tolerance function (e.g. Harmonic Entropy) to
give that result?

But it doesn't blow away all formulae based on separate prime
factorisations of numerator and denominator. There must be ways of
combining them that avoid this problem. So Paul (or anyone), how might we
modify odd-limit to include some consideration of 2's? e.g. so 6/5 is not
forced to have the same complexity as 5/3 (or 3/2 same as 4/3). (I think
Dan Wolf asked you the same question).

Do we agree that the superparticular series 2/1, 3/2, 4/3, 5/4, 6/5, 7/6,
8/7, 9/8, 10/9, 11/10 ... is progressively more dissonant until it gets
close enough to 1/1 for the tolerance function to take over? If someone
were to ask me "What is dissonance?". I don't think I could do better than
to say "It is that quality of the sound (independent of pitch and loudness)
which you hear increasing as you go up this series of just intervals
(assuming a near harmonic timbre)".

Here's are more series of intervals which IMHO increase in dissonance
(until getting too near some low complexity interval). Since the tolerance
function can take care of the "until too near ..." bit, these can be taken
as unlimited series of increasing *complexity*. I've arbitrarily stayed
within 2 octaves.

2/1, 3/2, 4/3, 5/4, 6/5, 7/6, 8/7, 9/8, 10/9, 11/10 ... until too near 1/1
3/2, 5/3, 7/4, 9/5, 11/6, 13/7, 15/8 ... until too near 2/1
3/1, 5/2, 7/3, 9/4, 11/5, 13/6, 15/7 ... until too near 2/1
2/1, 5/2, 8/3, 11/4, 13/5, ... until too near 3/1
4/1, 7/2, 10/3, 13/4, 14/5, ... until too near 3/1
3/1, 7/2, 11/3, 15/4, ... until too near 4/1
1/1, 4/3, 7/5, 10/7, 13/8, ... until too near 3/2
2/1, 5/3, 8/5, 11/7, 14/9, ... until too near 3/2
2/1, 7/3, 12/5, ... until too near 5/2
3/1, 8/3, 13/5, ... until too near 5/2
3/2, 7/5, 11/8, ... until too near 4/3
1/1, 5/4, 9/7, ... until too near 4/3
3/2, 8/5, 13/8, ... until too near 5/3
2/1, 7/4, 12/7, ... until too near 5/3
4/3, 9/7, ... until too near 5/4
1/1, 6/5, 11/9, ... until too near 5/4
3/1, 10/3, ... until too near 7/2
4/1, 11/3, ... until too near 7/2

Anyone disagree with any of these? Next we need to decide how they interleave.

[Rosati:]
> During a discussion with Paul Erlich awhile back, I mentioned that 13/8
> sounded more consonant to me than 11/8. I attributed this to 11/8 falling in
> the tritone "hump", while 13/8 sits between 5/3 and 8/5. After Paul gave us
> David Canright's new web address I was reading his "Tour up the Harmonic
> Series" (http://www.mbay.net/~anne/david/harmser/index.htm) and noticed that
> he also hears 13/8 as more consonant than 11/8. So, I was wondering how
> other distinguished ears in this forum might weigh in on this and what
> implications it might have for theories of odd-limit (or prime limit, for
> that matter) consonance indexing.

I don't think it has any implications for any *complexity* measure, since
11/8 and 13/8 are in regions where the lower complexity of notes either
side will dominate, i.e. this result will be taken care of by the tolerance
or blurring function. I agree totally with what you said. The dissonance
hump between 4/3 and 7/5 (where 11/8 is) is higher than the one between 8/5
and 5/3 (where 13/8 is).

[Keenan:]
>>The second function has been called tolerance. This is some kind of
>>blurring function where the dissonance of a complex ratio will depend more
>>on its proximity to nearby simpler ratios, thus limiting the significance
>>of higher primes.

[Erlich replied:]
>Have you noticed that this argument is far more valid if you replace
>"primes" with "odds" at the end of the sentence,

Actually, I'd prefer to just say "limiting the significance of higher
*numbers*", whether they be prime, odd or otherwise.

[Erlich:]
>showing that a strict
>lattice approach like all those we've been discussing is much more
>likely to be meaningful if we restrict ourselves to, say, the
>11-odd-limit, than if we restrict ourselves to the 11-prime-limit, the
>7-prime-limit, the 5-prime-limit, or even the 3-prime-limit?

I didn't think we needed to limit the complexity measure in this way
because the tolerance function should take care of it in a less arbitrary
way. Isn't what-you-are-talking-about a more user-friendly but less
accurate way of modelling tolerance? Certainly a useful point in that
tradeoff, and I agree odd-limit is much better than prime-limit for this
purpose of "limiting the limit".

[Erlich:]
>The 11-odd-limit seems about right given ideal conditions for pitch
>discrimination.

I agree. And before anyone jumps on us, remember we're talking about bare
intervals (not in chords).

[Erlich:]
>And again, a triangular lattice would be more appropriate.

So what does the corresponding equation (or algorithm) look like, to get
complexity from n/d, without ignoring factors of 2.

I'm convinced that n+d isn't a good enough complexity measure *in chords*,
but remind me what is wrong with it for intervals?

Regards,
-- Dave Keenan
http://dkeenan.com

🔗Paul H. Erlich <PErlich@xxxxxxxxxxxxx.xxxx>

3/11/1999 3:45:42 PM

Dave Keenan wrote,

>So maybe
>you've just blown away all formulae where we ignore whether various
factors
>are on the same or different sides of the vinculum (the line between
>numerator and denominator). If so, good work.

Thanks. I like to blow things away.

>Surely oo/1 is as consonant as 1/1. ("oo" is meant to look like the
lazy-8
>infinity symbol). I'd even say n/1, where n>16, are almost as consonant
as
>1/1. Can we arrange for the tolerance function (e.g. Harmonic Entropy)
to
>give that result?

Actually, Harmonic Entropy already works that way! Some time ago I
posted an old derivation of mine that the certainty with which an
interval is perceived as a just ratio is inversely proportional to the
DENOMINATOR of the ratio (when ratios are expressed with a larger
numerator than denominator). This derivation is a corrected version of
one originally done by Van Eck and is based on the exact same model I've
used for all the harmonic entropy work I've described so far (including
some graphs I just set to Joe Monzo). Now notice that the odd limit is
an upper bound for the denominator, and you'll see why I find the odd
limit useful even in octave-specific situations!

>But it doesn't blow away all formulae based on separate prime
>factorisations of numerator and denominator. There must be ways of
>combining them that avoid this problem. So Paul (or anyone), how might
we
>modify odd-limit to include some consideration of 2's? e.g. so 6/5 is
not
>forced to have the same complexity as 5/3 (or 3/2 same as 4/3). (I
think
>Dan Wolf asked you the same question).

Use just the denominator. That's based on the derivation I described
above. One assumption in that derivation is that the pitch of the upper
note is fixed when comparing different intervals. If instead you fix the
center of the interval, I think my derivation can be modified to show
that the certainty is inversely proportional to the sum of the numerator
and the denominator. According to you, that sum models critical-band
roughness very well up to ratios of 17, so it seems _both_ components of
dissonance can be modeled by the sum of the numerator and the
denominator!

>>showing that a strict
>>lattice approach like all those we've been discussing is much more
>>likely to be meaningful if we restrict ourselves to, say, the
>>11-odd-limit, than if we restrict ourselves to the 11-prime-limit, the
>>7-prime-limit, the 5-prime-limit, or even the 3-prime-limit?

>I didn't think we needed to limit the complexity measure in this way
>because the tolerance function should take care of it in a less
arbitrary
>way. Isn't what-you-are-talking-about a more user-friendly but less
>accurate way of modelling tolerance?

Yes. I have a more explicit provision for tolerance in this context
which I'll dig up soon.

>Certainly a useful point in that
>tradeoff, and I agree odd-limit is much better than prime-limit for
this
>purpose of "limiting the limit".

Yup.

>>And again, a triangular lattice would be more appropriate.

>So what does the corresponding equation (or algorithm) look like, to
get
>complexity from n/d, without ignoring factors of 2.

If you mean in a lattice context, I haven't thought about
octave-specific lattices. As for the usual octave-invariant lattice, it
appears Paul Hahn found an error in his original algorithm, but I'm sure
if anyone can correct it, he can.

>I'm convinced that n+d isn't a good enough complexity measure *in
chords*,
>but remind me what is wrong with it for intervals?

Nothing!

🔗Dave Keenan <d.keenan@xx.xxx.xxx>

3/11/1999 10:12:15 PM

[Paul Erlich:]
>Re-reading Dave's e-mail to me, I see that the sum [of numerator and
>denominator] only models
>critical-band roughness well if the _sum_ is 17 or less. So intervals
>like 11/8 and 13/8 are already too complex for this type of formula, as
>they are too close to other ratios of similar complexity to be clearly
>distinguished from them.

Agreed.

>Note that my harmonic entropy model, to the
>accuracy that I've evaluated it so far, did not show local minima at
>11/8 or 13/8, but there was a very tiny one at 11/6.

Neato!

I'd be interested to see your harmonic entropy applied to, not a Farey
series but, all ratios n/d, where n>d, n+d <= N, where N is large, say 40
or 80. Will this still predict low dissonance for n/1 where n>=16?

Regards,
-- Dave Keenan
http://dkeenan.com

🔗Paul Hahn <Paul-Hahn@xxxxxxx.xxxxx.xxxx>

3/12/1999 8:30:47 AM

On Thu, 11 Mar 1999, Paul H. Erlich wrote:
> As for the usual octave-invariant lattice, it
> appears Paul Hahn found an error in his original algorithm, but I'm sure
> if anyone can correct it, he can.

I'm touched by your faith in me. In any case, I have in fact found a
simpler algorithm, though I think Manuel has beaten me to it already,
since it sounds like he has already coded it up. Nevertheless, here it
is, for the curious:

Given a Fokker-style interval vector (I1, I2, . . . In):

1. Go to the rightmost nonzero exponent; add the product of its
absolute value with the log of its base to the total.

2. Use that exponent to cancel out as many exponents of the opposite
sign as possible, starting to its immediate left and working right;
discard anything remaining of that exponent.

Example: starting with, say, (4 2 -3), we would add 3 lg(7) to
our total, then cancel the -3 against the 2, then the remaining
-1 against the 4, leaving (3 0 0). OTOH, starting with
(-2 3 5), we would add 5 lg(7) to our total, then cancel 2 of
the 5 against the -2 and discard the remainder, leaving (0 3 0).

3. If any nonzero exponents remain, go back to step one, otherwise
stop.

To illustrate on the list of intervals Manuel culled from an earlier
post of mine:

225:224 (2 2 -1):
1st iteration lg(7); (2 1 0)
2nd iteration lg(7) + lg(5); (2 0 0)
3rd iteration lg(7) + lg(5) + 2 lg(3);(0 0 0) Done.

126:125 (2 -3 1):
1st iteration lg(7); (2 -2 0)
2nd iteration lg(7) + 2 lg(5); (0 0 0) Done.

128:125 (0 -3):
1st iteration 3 lg(5); (0 0) Done.

81:80 (4 -1):
1st iteration lg(5); (3 0)
2nd iteration lg(5) + 3 lg(3); (0 0) Done.

64:63 (-2 0 -1):
1st iteration lg(7); (-2 0 0)
2nd iteration lg(7) + 2 lg(3); ( 0 0 0) Done.

50:49 (0 2 -2):
1st iteration 2 lg(7); (0 0 0) Done.

--pH <manynote@lib-rary.wustl.edu> http://library.wustl.edu/~manynote
O
/\ "How about that? The guy can't run six balls,
-\-\-- o and they make him president."

NOTE: dehyphenate node to remove spamblock. <*>

🔗Dave Keenan <d.keenan@xx.xxx.xxx>

3/12/1999 8:40:45 AM

Dear tuning folk,

As promised, I've put up a spreadsheet with a chart comparing 10 different
complexity measures for all ratios n/d up to n+d=21.

http://dkeenan.com/Music/HarmonicComplexity.xls 174kB

Unfortunately the complexity comparison is not complete. I need a third
opinion on how Euler's totient function and gradus suavatis apply to
ratios. Also, for whole numbers, is the totient function equal to the prime
(or one less) in the case of primes?

I'll also take requests to include other dyadic complexity measures.

Remember, complexity alone is not dissonance. You need tolerance when
complexity is high. :-)

I've also put up my spreadsheet implementation of Sethares' dissonance
curve (from timbre) algorithm. Thanks Bill. Unfortunately the number of
points it can plot is limited to 28 so you have to zoom in on areas of
interest by changing the start point and reducing the step size.

http://dkeenan.com/Music/SetharesDissonance.xls 1.5MB

Regards,
-- Dave Keenan
http://dkeenan.com

🔗Paul H. Erlich <PErlich@xxxxxxxxxxxxx.xxxx>

3/12/1999 3:23:56 PM

I've thought about it and I don't agree with the algorithm Manuel and
Paul Hahn have come up with for computing the city-block metric on the
triangular lattice. The problem comes down, as usual, to an
over-reliance on primes. For instance, the algorithm makes 11:10 a
single step of length lg(11), while 9:5 is a step of length lg(3) plus a
step of lg(5), which is longer. But if you have an 11-axis in the
lattice, then that means you're considering 11-limit intervals
consonant, so you should also consider 9-limit intervals consonant, and
you should have a 9-axis too. That would make 9:5 a single step of
length lg(9).

Clearly in this lattice 9:5 occurs in two different places. But, Paul
Hahn himself wrote,

>If you're an odd-limit proponent such as myself, things get a little
>complicated at the 9-limit and above, since 9 is composite, and
>odd-factorization does not necessarily yield unique results. However,
>the minimum complexity is achieved by assigning the 9 exponent as large
>as possible, and the 3 exponent 0 or 1 as appropriate.

So it appears he agreed with me to begin with, then renegged (or
forgot). Another way to think about it is that the lattice with prime
axes is the basic construct, so every pitch appears only once, but then
you have "wormholes" with shorter than the apparent length for intervals
like 9:5.

The correct algorithm is of course much simpler than the last one Paul
H. described. It is: remove all factors of two, then take the log of the
denominator or the numerator, whichever is larger.

🔗Paul Hahn <Paul-Hahn@xxxxxxx.xxxxx.xxxx>

3/13/1999 8:19:13 AM

On Fri, 12 Mar 1999, Paul H. Erlich wrote:
> I've thought about it and I don't agree with the algorithm Manuel and
> Paul Hahn have come up with for computing the city-block metric on the
> triangular lattice. [snip]
> So it appears he agreed with me to begin with, then renegged (or
> forgot).

I do wish you wouldn't use such emotionally charged words, especially in
view of the fact that I have neither "reneged" nor forgotten.
Everything I've written on this subject so far works using vectors
including separate exponents for the odd composites, provided you
reduce them as I described in an earlier article.

> The correct algorithm is of course much simpler than the last one Paul
> H. described. It is: remove all factors of two, then take the log of the
> denominator or the numerator, whichever is larger.

No, it isn't. Consider 225:224, for example: your way (and with my
uncorrected algorithm for the weighted version) the complexity is
2 lg(3) + 2 lg(5), when in fact the correct version, if you work it out
on a lattice or use my corrected algorithm, is 2 lg(3) + lg(5) + lg(7).

--pH <manynote@lib-rary.wustl.edu> http://library.wustl.edu/~manynote
O
/\ "How about that? The guy can't run six balls,
-\-\-- o and they make him president."

NOTE: dehyphenate node to remove spamblock. <*>

🔗Paul H. Erlich <PErlich@xxxxxxxxxxxxx.xxxx>

3/16/1999 12:21:19 PM

I wrote,

>> I've thought about it and I don't agree with the algorithm Manuel and
>> Paul Hahn have come up with for computing the city-block metric on
the
>> triangular lattice. [snip]
>> So it appears he agreed with me to begin with, then renegged (or
>> forgot).

>I do wish you wouldn't use such emotionally charged words, especially
in
>view of the fact that I have neither "reneged" nor forgotten.
>Everything I've written on this subject so far works using vectors
>including separate exponents for the odd composites, provided you
>reduce them as I described in an earlier article.

I didn't think there was any emotion there. But there does appear to be
a conflict between your/Manuel's algorithm and using all odd composites,
as my 9:5 vs. 11:5 example pointed out.

>> The correct algorithm is of course much simpler than the last one
Paul
>> H. described. It is: remove all factors of two, then take the log of
the
>> denominator or the numerator, whichever is larger.

>No, it isn't. Consider 225:224, for example: your way (and with my
>uncorrected algorithm for the weighted version) the complexity is
>2 lg(3) + 2 lg(5), when in fact the correct version, if you work it out
>on a lattice or use my corrected algorithm, is 2 lg(3) + lg(5) + lg(7).

I stand by my way and would rather you addressed my examples of 9:5 and
11:5 instead of 225:224 (for which the lattice evaluation is unlikely to
be very meaningful anyway). If it must be discussed, my value for
225:224 assumes there is a 225-axis or 225-wormholes in the lattice. It
might make more sense to pick an odd limit for the lattice like 7, 9, or
11, but then you couldn't evaluate anything with 13 in it. Perhaps a
separate complexity measure for several choices of odd limit would
satisfy us both (but complicate the matter greatly).

🔗Paul Hahn <Paul-Hahn@xxxxxxx.xxxxx.xxxx>

3/17/1999 8:34:11 AM

On Tue, 16 Mar 1999, Paul H. Erlich wrote:
>> I do wish you wouldn't use such emotionally charged words, especially in
>> view of the fact that I have neither "reneged" nor forgotten.
>> Everything I've written on this subject so far works using vectors
>> including separate exponents for the odd composites, provided you
>> reduce them as I described in an earlier article.
>
> I didn't think there was any emotion there.

You don't think that someone might show a bit of heat at being (wrongly)
accused of having reneged on something? Aside from that, I find it a
bit presumptuous that sufficient commitment was assumed for the word
"renege" to be used in the first place.

> But there does appear to be
> a conflict between your/Manuel's algorithm and using all odd composites,
> as my 9:5 vs. 11:5 example pointed out.

No, there isn't. (Sheesh! I don't remember the last time I've had to
argue so hard to convince someone I agreed with him. Though I have a
feeling that it was Paul E. that time, too. 8-)> )

Look. Let's look at what I said about using the algorithms in an
odd-limit vs. prime-limit environment, _which you (Paul E.) yourself
quoted_:

| If you're an odd-limit proponent such as myself, things get a little
| complicated at the 9-limit and above, since 9 is composite, and
| odd-factorization does not necessarily yield unique results. However,
^^^^^^^^
| the minimum complexity is achieved by assigning the 9 exponent as large
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
| as possible, and the 3 exponent 0 or 1 as appropriate.
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

Okay. The 11-limit prime vectors for 9:5 and 11:5 are (2 -1 0 0) and
(0 -1 0 1). Converting these to the optimal 11-limit odd-factor vectors
as described above, we get (0 -1 0 1 0) for 9:5 and (0 -1 0 0 1) for
11:5.

Now apply the algorithms:

Simple version:

In each interval, the absolute values of the sums of the
positive and negative exponents are 1. Hence, each of these are
primary/consonant intervals within the 11-limit.

Weighted version:

(0 -1 0 1 0):
1st iteration lg(9) (0 0 0 0 0) done

(0 -1 0 0 1):
1st iteration lg(11) (0 0 0 0 0) done

See? It works.

Incidentally, since you don't like the 225:224 example, let's just
consider the 9:5 within the traditional 5-limit (vector [2 -1]). With
your largest-odd-factor method, we get lg(9). However, the shortest
path through the 5-limit lattice for 9:5 is a 3:2 and a 6:5, or
lg(3) + lg(5). Eh?

>>> The correct algorithm is of course much simpler than the last one Paul
>>> H. described. It is: remove all factors of two, then take the log of the
>>> denominator or the numerator, whichever is larger.
>
>> No, it isn't. Consider 225:224, for example: [snip]
>
> I stand by my way and would rather you addressed my examples of 9:5 and
> 11:5

See above.

> instead of 225:224 (for which the lattice evaluation is unlikely to
> be very meaningful anyway).

Meaningfulness is a separate question. As we have already seen, we all
use these various metric functions for slightly different purposes and
in slightly different ways. But is the function not defined on a large
interval like 225:224? Is not the shortest path through the 7-limit
lattice one septimal interval, one 5-limit interval, and two 3-limit
intervals?

> If it must be discussed, my value for
> 225:224 assumes there is a 225-axis or 225-wormholes in the lattice.

As Carl Lumma has already pointed out, this seems quite strange--it
would seem to imply that you can only apply this metric to intervals
which one considers primary/within a given odd limit/consonant.
IOW, it _is_ (log of) odd-limit. Which would negate the whole point of
using a city-block function in the first place.

--pH <manynote@lib-rary.wustl.edu> http://library.wustl.edu/~manynote
O
/\ "How about that? The guy can't run six balls,
-\-\-- o and they make him president."

NOTE: dehyphenate node to remove spamblock. <*>