Complexity formulae: Reply to Paul Hahn

>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.

I never thought of "renege" as a necessarily strong word. I would apply
it to idle conversation. Sorry if I misused it in a way that blew the
importance of any of this out of proportion. If I appear to get
vitriolic on the list sometimes, I'm sorry; Paul, I think of you as a
kindred spirit since we think so much alike (and have the same name :)
and I really enjoy these intellectual discussions -- if I was hurtful, I
hope you can think of it as a boxing match between friends. Maybe my
tone on this list comes from that of Harry Partch in _Genesis_, which
was my primary influence and some of whose ideas I like to defend with
the same fervor as Harry himself.

>> 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-)> )

Yeah, that was me, and we still have to come back to that discussion at
a later date.

But let's look at the algorithm you posted on Fri, 12 Mar 1999 10:30:47
(TD 95, message 12), which is what I was referring to:

>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)

>[....]

The rest of the post dealt only with these 3-component (3,5,7) vectors.
Now a Fokker-style vector implies (to me) one with an entry for each
prime above 2. Manuel in fact referred to primes in his post on the
algorithm. Now I admit that you did not specifically mention primes; as
your examples only went up to 7, there's no way to know whether you
meant primes or odds. But another reason I assumed you meant primes is
that if you meant odds, you would need an additional step in the
algorithm where the user specifies the odd limit (dimensionality) of the
lattice. That step is clearly needed to apply the algorithm
unambiguously. As the prime-based complexity measures do not require
this step, I proposed an interpretation where the dimensionality was
infinite, and the complexity of the interval turns out to be just the
(log of the) odd limit of the interval. But that makes the whole lattice
kind of superfluous and I would definitely prefer a means of limiting
the dimensionality. Unfortunately, that cannot be done from the ratios
themselves; a 9:5 could occur as a secondary interval if the odd limit
is 5, or a primary interval if the odd limit is 9 or higher. So the
dimensionality of the lattice must also be specified.

>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.

You didn't mention any such conversion in the post I quoted above, which
appeared to be an exposition of your algorithm. You see why I thought we
disagreed?

>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?

I think we both understand now that both are the right answer, under
different specifications of the dimensionality of the lattice. Right?

Manuel mentioned primes, not odds; and did not mention a step which
specifies the dimensionality of the lattice. So I'm pretty sure my
objections still apply to his algorithm. Since you appeared to agree
with his algorithm, I included you as one of the authors of the
algorithm I disagreed with. Sorry if I misinterpreted you.

So what do you think, Paul? (regarding my posting of Fri, 19 Mar 1999
17:09:55 -0500, message 23 in digest 111)