--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> This is what Browne calls the Euclidean compliment, and there is a
> somewhat nice and standard formula for this. I'm going to apply it to
> the 13-limit case for some examples.

Thanks very much for these. Please don't be offended by my attempts to
translate some parts in a way that I imagine might be more easily
comprehended by Monz and others.

> The basic idea is that we find
> the signs, for which we can use the formula, apply them, reverse the
> ordering, change the m-monzo to an (n-m)-val, and we have the
> compliment of a monzo. The same applies the other way around for vals.
>
> If we have a 13-limit 2-monzo, it will have dimension 6*5/2 = 15

It think it will cause less confusion if we call them bi-monzos and
tri-monzos (with or without hyphens). There are enough numbers flying
around as it is. But by all means start using the numbers once we get
beyond grade 3. And why not use "g" for a variable grade instead of
"m"? And call them multi-monzos when the grade is unspecified and
doesn't need to be referred to in any formula. This is the sort of
stuff that doesn't matter to someone who already understands it, but
for someone trying to break into it, every little bit helps.

Now if the "dimension" of a multi-monzo is the number of coefficients
in it, then what do you call the thing that I've been calling the
dimension, which in our application is the index of the highest prime
that's represented? That seems more like "dimension" to me, since it
is the dimension of the underlying simple space. The other I have been
calling simply "number of coefficients".

To say that in 4D-space a plane has dimension 6 wouldn't make a lot of
sense to most people. To say that the bi-vector representing it has 6
coefficients, is more likely to be understood.

Translation:

A 13-limit monzo (prime-exponent-vector) has 6 coefficients so a
13-limit bi-monzo, will have 6*5/2 = 15 coefficients. In general, if
the monzo has n coefficients, the corresponding bi-monzo will have
n*(n-1)/2 coefficients. The same goes of course for maps and bi-maps
(vals and bi-vals).

> Let
>
> [[c1 c2 c3 c4 c5 c6 c7 c6 c7 c8 c9 c10 c11 c12 c13 c14 c15>>
>
> be that monzo.

You mean "bi-monzo".

> Now take the ordered combinations of six things, taken
> two at a time, in alphabetical order:
>
> [[1, 2], [1, 3], [1, 4], [1, 5], [1, 6], [2, 3], [2, 4], [2, 5],
> [2, 6], [3, 4], [3, 5], [3, 6], [4, 5], [4, 6], [5, 6]]

It might be clearer to use compound indices in the bimonzo as follows
(this trick will work up to the 23-limit, beyond which we have more
than 9 coefficients in the monzo or map, but we hardly ever go there).

Graham raised the question, why talk about alphabetical order when
it's numerical order. Well, up to simple indices of 9, they are the
same thing, but beyond that if you want it to still be numerical you
have to use a number base greater than the maximum simple index.

The monzo [c1 c2 c3 c4 c5 c6> represents the 13-limit ratio
2^c1 * 3^c2 * 5^c3 * 7^c4 * 1^c5 * 13^c6.

The bimonzo [[c12 c13 c14 c15 c16 c23 c24 c25 c26 c34 c35 c36 c45 c46
c56>>

represents a two 13-limit ratios simultaneously, but in such a way
that the bimonzo for any other pair which are linear combinations of
this pair would give the same bimonzo. For example, if these are two
commas which vanish in some temperament, then any other pair of commas
whose vanishing would define the same temperament will have the same
bimonzo. There's no redundancy as there would be in a matrix
representation of this (which would have 6x6 = 36 coefficients).

Notice that _within_ each compound index for a single coefficient, the
digits are in numerical/alphabetical order, and the same digit does
not occur more than once.

Notice also that the coefficients themselves are listed in
numerical/alphabetical order of their compound index.

> Sum the numbers in each and put down a 1 if the sum is even, and a -1
> if it is odd, giving
>
> [-1, 1, -1, 1, -1, -1, 1, -1, 1, -1, 1, -1, -1, 1, -1]
>
> Now note that m=2,

The grade g is 2 since it's a _bi_monzo. This is what the number of
brackets shows us.

> so m(m+1)/2 is 2*3/2 = 3, an odd number.

so Ceiling(g/2) = 1, an odd number.

> Hence we
> must multiply the above by -1; at the same time we will reverse the
> ordering, giving us now
>
> [1, -1, 1, 1, -1, 1, -1, 1, -1, 1, 1, -1, 1, -1, 1]
>
> We now use these signs, together with the coefficients in descending
> order, to get the complimentary 4-val:
>
> <<<<c15 -c14 c13 c12 -c11 c10 -c9 c8 -c7 c6 c5 -c4 c3 -c2 c1]]]]

That's "compl_e_mentary", although it's true Gene didn't charge any
money for it. ;-)

Because the 13-prime-limit is a 6-dimensional space (2 3, 5 7 11, 13),
the complement of a bi-monzo is a 6-2 = 4-map (4-val) and where the
bi-monzo has 2 brackets pointing to the right, its complement has 4
brackets pointing to the left.

<<<<c56 -c46 c45 c36 -c35 c34 -c26 c25 -c24 c23 c16 -c15 c14 -c13 c12]]]]

So you don't actually have to write out the 1's and -1's. Just
calculate Ceiling(g/2) and add it to the sum of the digits making up
each compound index and put a minus sign if the result is odd. You
ask, how can I remember whether I put the minus sign when it's odd or
even? Easy. You're actually multiplying them by
-1^(sum_of_indices+Ceiling(g/2)) and minus-one to an even power is 1,
while to an odd power it is -1.

For example, consider c13 in our bimonzo. Ceiling(g/2) = 1 and 1 + 1 +
3 = 5, an odd number so it becomes -c13 in the complement.

> This isn't what I got before, where I presumably goofed.
>
> A 13-limit 3-monzo has dimension 6*5*4/6 = 20, and may be written
>
> [[[c1 c2 c3 c4 c5 c6 c6 c8 c9 c10
> c11 c12 c13 c14 c15 c16 c17 c18 c19 c20>>>
>
> The ordered combinations of [1,2,3,4,5,6] taken three at a time are
>
> [[1, 2, 3], [1, 2, 4], [1, 2, 5], [1, 2, 6], [1, 3, 4], [1, 3, 5],
> [1,3, 6], [1, 4, 5], [1, 4, 6], [1, 5, 6], [2, 3, 4], [2, 3, 5],
> [2, 3, 6], [2, 4, 5], [2, 4, 6], [2, 5, 6], [3, 4, 5], [3, 4, 6],
> [3, 5, 6], [4, 5, 6]]
>
> Taking the sums of the coefficients of these and putting 1 for even,
> -1 for odd gives us
>
> [1, -1, 1, -1, 1, -1, 1, 1, -1, 1, -1, 1, -1, -1, 1, -1, 1, -1, 1, -1]
>
> We are taking the compliment of a 3-monzo, so m=3 and m(m+1)/2 = 6, an
> even number; hence we don't change sign. However, we do as always
> reverse the ordering and take the coefficients in reverse order:
>
> [-1, 1, -1, 1, -1, 1, -1, -1, 1, -1, 1, -1, 1, 1, -1, 1, -1, 1, -1, 1]
>
> <<<-c20 c19 -c18 c17 -c16 c15 -c14 -c13 c12 -c11
> c10 -c9 c8 c7 -c6 c5 -c4 c3 -c2 c1]]]

In other words the tri-monzo is:

[[[c123 c124 c125 c126 c134 c135 c136 c145 c146 c156 c234 c235 c236
c245 c246 c256 c345 c346 c356 c456>>>

and we find Ceiling(g/2) = Ceiling(3/2) = Ceiling(1.5) = 2. So we add
2 to the sum of the digits in each case, which of course is the same
as adding zero, since all we care about is whether the result is odd
or even. In fact we don't even have to add them up. We can just count
how many are odd, and if this count is itself odd, then we negate.

So we negate all those coefficients with an odd number of odd digits
(in their compound index), and we reverse their order, to get the tri-map:

[[[-c456 c356 -c346 c345 -c256 c246 -c245 -c236 c235 -c234 c156 -c146
c145 c136 -c135 c134 -c126 c125 -c124 c123>>>

>
> Now for the compliment of a 4-monzo, which is a 6*5*4*3/24 = 15
> dimensional object again
>
> [[[[c1 c2 c3 c4 c5 c6 c7 c8 c9 c10 c11 c12 c13 c14 c15>>>>
>
> The combinations of [1,2,3,4,5,6] taken four at a time in order are
>
> [[1, 2, 3, 4], [1, 2, 3, 5], [1, 2, 3, 6], [1, 2, 4, 5], [1, 2, 4, 6],
> [1, 2, 5, 6], [1, 3, 4, 5], [1, 3, 4, 6], [1, 3, 5, 6], [1, 4, 5, 6],
> [2, 3, 4, 5], [2, 3, 4, 6], [2, 3, 5, 6], [2, 4, 5, 6], [3, 4, 5, 6]]
>
> Even and odd for this gives
>
> [1, -1, 1, 1, -1, 1, -1, 1, -1, 1, 1, -1, 1, -1, 1]
>
> m=4, and 4*5/2 = 10, an even number, so we don't change sign but as
> always do reverse the order
>
> [1, -1, 1, -1, 1, 1, -1, 1, -1, 1, -1, 1, 1, -1, 1]
>
> Taking these signs with descending coefficients gives
>
> <<c15 -c14 c13 -c12 c11 c10 -c9 c8 -c7 c6 -c5 c4 c3 -c2 c1]]
>
> as the compliment.

Take the 4-monzo
[[[[c1234 c1235 c1236 c1245 c1246 c1256 c1345 c1346 c1356 c1456 c2345
c2346 c2356 c2456 c3456>>>>

Ceiling(g/2) = Ceiling(4/2) = 2 which is even, so we don't need to add
anything more to our count of the odd digits in each index, and the
complement of the above 4-monzo is the following bi-map:

<<c3456 -c2456 c2356 -c2346 c2345 c1456 -c1356 c1346 -c1345 c1256
-c1246 c1245 c1236 -c1235 c1234]]

It should be mentioned that taking the complement of the complement
doesn't always give you back what you started with, sometimes it's the
negative of what you started with. So in those cases it's analogous to
multiplying by i (the square root of -1). This depends on the
dimension and the grade. But taking the complement four-times always
gives you back exactly what you started with.