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→multicommas: redo this so that people can use the table first and then only read the explanation of how to find further cells of it later
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Now let’s see how to do the complement operation.
Now let’s see how to do the complement operation.


# Find the rank, halved, rounded up. In our case, <span><math>\lceil \frac{r}{2} \rceil = \lceil \frac{2}{2} \rceil = \lceil 1 \rceil = 1</math></span>. Save that result for later. Let’s call it <span><math>x</math></span>.
If your temperament's dimensionality <span><math>d</math></span> is 6 or less (within the 13-limit), you can take advantage of this table I've prepared, and use this simplified method:
# Find the lexicographic combinations of <span><math>r</math></span> primes again: (2,3), (2,5), (3,5). Except this time we don’t want the primes themselves, but their indices in the list of primes. So: <span><math>(1,2)</math></span>, <span><math>(1,3)</math></span>, <span><math>(2,3)</math></span>.
# Take the sums of these sets of indices, and to each sum, also add <span><math>x</math></span>. So <span><math>1+2+x</math></span>, <span><math>1+3+x</math></span>, <span><math>2+3+x</math></span> = <span><math>1+2+1</math></span>, <span><math>1+3+1</math></span>, <span><math>2+3+1</math></span> = <span><math>4</math></span>, <span><math>5</math></span>, <span><math>6</math></span>.
# Raise -1 to the exponent of these values. -1⁴, -1⁵, -1⁶ = 1, -1, 1.
# Multiply each term of the multimap by these values. 1×1, 4×-1, 4×1, = 1, -4, 4.
# Reverse the order: 4, -4, 1.
# Set the result in the proper count of brackets: {{vector|4 -4 1}}.
 
Ta-da! Both operations get us to the same result: {{vector|4 -4 1}}.


What’s the proper count of brackets though? Well, the total count of brackets on the multicomma and multimap for a temperament must always sum to the dimensionality of the system from which you tempered. It’s the same thing as <span><math>d - n = r</math></span>, just phrased as <span><math>r + n = d</math></span>, and where <span><math>r</math></span> should be the bracket count for the multimap and <span><math>n</math></span> should be the bracket count for the multicomma. So with 5-limit meantone, with dimensionality 3, there should be 3 total pairs of brackets. If 2 are on the multimap, then only 1 are on the multicomma.
# Find the correct cell in Table 6a below using your temperament's <span><math>d</math></span> and <span><math>r</math></span> (rank). This cell should contain the same number of symbols as there are terms of your multimap.
# Match up the terms of your multimap with these symbols. If the symbol is +, do nothing. If the symbol is -, negate the sign.
# Reverse the order.
# Set the result in the proper count of brackets.


An important observation to make about multicommas and multimaps is that — for a given temperament — they always have the same count of terms. This may surprise you, since the rank and nullity for a temperament are often different, and the length of the multimap comes from the rank while the length of the multicomma comes from the nullity. But there’s a simple explanation for this. In either case, the length is not directly equal to the rank or nullity, but to the dimensionality choose the rank or nullity. And there’s a pattern to combinations that can be visualized in the symmetry of rows of Pascal’s triangle: <span><math>{d \choose n}</math></span> is always equal to <span><math>{d \choose {d - n}}</math></span>, or in other words, <span><math>{d \choose n}</math></span> is always equal to <span><math>{d \choose r}</math></span>. Here are some examples:
{| class="wikitable"
{| class="wikitable"
|+'''Table 5a.''' Multi(co)vector prime combinations (<math>r</math> can be switched for <math>n</math>)
|+ '''Table 6a.''' Complement sign flipping sequences by rank and dimensionality
!<math>d</math>
!<math>r</math>
!<math>d - r</math>
!<math>{d \choose r}</math>
!<math>{d \choose {d - r}}</math>
!count
|-
|3
|2
|1
|<math>(2,3) (2,5) (3,5)</math>
|<math>(2) (3) (5)</math>
|3
|-
|4
|3
|1
|<math>(2,3,5) (2,3,7) (2,5,7) (3,5,7)</math>
|<math>(2) (3) (5) (7)</math>
|4
|-
|5
|3
|2
|<math>(2,3,5) (2,3,7) (2,3,11) (2,5,7) (2,5,11) (2,7,11) (3,5,7) (3,5,11) (3,7,11) (5,7,11)</math>
|<math>(2,3) (2,5) (2,7) (2,11) (3,5), (3,7) (3,11) (5,7) (5,11) (7,11)</math>
|10
|}
Each set of one side corresponds to a set in the other side which has the exact opposite elements.
 
If you review the seven steps in the process for taking the complement, you may notice that a lot of it is busywork that will never change from one multimap to another. It all amounts to a specific sequence of 1’s and -1’s corresponding to a given rank and dimensionality. In consideration of this, I have gone ahead and prepared a table with the sequences you need to multiply the terms of your multimap by before flipping them:
 
{| class="wikitable"
|+ '''Table 5b.''' Complement sign flipping sequences by rank and dimensionality
!r\d
!r\d
!0
!0
Line 1,382: Line 1,342:
|}
|}


Note the Pascal’s triangle shape to the results. Also note that the mirrored results within each dimensionality are reverses of each other. Sometimes that means they’re identical, like 1 -1 1 -1 1 and 1 -1 1 -1 1; other times not, like 1 -1 1 -1 1 -1 1 1 -1 1 and 1 -1 1 1 -1 1 -1 1 -1 1. (Well, and sometimes they’re reverses of each other, but then flipped signs so that the first time is always 1)
So in this case:
 
# We have d=3, r=2, so the correct cell contains the symbols + - +.
# Matching these symbols up with the terms of our multimap, we don't negate 1, we do negate 4 to -4, and we don't negate the second 4.
# Now we reverse 1 -4 4 to 4 -4 1.
# Now we set the result in the proper count of brackets: {{vector|4 -4 1}}
 
Ta-da! Both operations get us to the same result: {{vector|4 -4 1}}.
 
What’s the proper count of brackets though? Well, the total count of brackets on the multicomma and multimap for a temperament must always sum to the dimensionality of the system from which you tempered. It’s the same thing as <span><math>d - n = r</math></span>, just phrased as <span><math>r + n = d</math></span>, and where <span><math>r</math></span> should be the bracket count for the multimap and <span><math>n</math></span> should be the bracket count for the multicomma. So with 5-limit meantone, with dimensionality 3, there should be 3 total pairs of brackets. If 2 are on the multimap, then only 1 are on the multicomma.


If you’re instead converting a multicomma to a multimap, then you think of it a couple different ways. Either use <span><math>n</math></span> as <span><math>r</math></span> when looking up in this table, and then reverse the result, or find <span><math>r</math></span> by subtracting <span><math>n</math></span> from <span><math>d</math></span> and then look it up.
Note the Pascal’s triangle shape to the numbers in Table 6a. Also note that the mirrored results within each dimensionality are reverses of each other. Sometimes that means they’re identical, like 1 -1 1 -1 1 and 1 -1 1 -1 1; other times not, like 1 -1 1 -1 1 -1 1 1 -1 1 and 1 -1 1 1 -1 1 -1 1 -1 1. (Well, and sometimes they’re reverses of each other, but then flipped signs so that the first time is always 1.)
 
If you’re instead converting a multicomma to a multimap, then you can think of it a couple different ways. Either use <span><math>n</math></span> as <span><math>r</math></span> when looking up in this table, and then reverse the result, or find <span><math>r</math></span> by subtracting <span><math>n</math></span> from <span><math>d</math></span> and then look it up.
 
An important observation to make about multicommas and multimaps is that — for a given temperament — they always have the same count of terms. This may surprise you, since the rank and nullity for a temperament are often different, and the length of the multimap comes from the rank while the length of the multicomma comes from the nullity. But there’s a simple explanation for this. In either case, the length is not directly equal to the rank or nullity, but to the dimensionality choose the rank or nullity. And there’s a pattern to combinations that can be visualized in the symmetry of rows of Pascal’s triangle: <span><math>{d \choose n}</math></span> is always equal to <span><math>{d \choose {d - n}}</math></span>, or in other words, <span><math>{d \choose n}</math></span> is always equal to <span><math>{d \choose r}</math></span>. Here are some examples:
{| class="wikitable"
|+'''Table 6b.''' Multi(co)vector prime combinations (<math>r</math> can be switched for <math>n</math>)
!<math>d</math>
!<math>r</math>
!<math>d - r</math>
!<math>{d \choose r}</math>
!<math>{d \choose {d - r}}</math>
!count
|-
|3
|2
|1
|<math>(2,3) (2,5) (3,5)</math>
|<math>(2) (3) (5)</math>
|3
|-
|4
|3
|1
|<math>(2,3,5) (2,3,7) (2,5,7) (3,5,7)</math>
|<math>(2) (3) (5) (7)</math>
|4
|-
|5
|3
|2
|<math>(2,3,5) (2,3,7) (2,3,11) (2,5,7) (2,5,11) (2,7,11) (3,5,7) (3,5,11) (3,7,11) (5,7,11)</math>
|<math>(2,3) (2,5) (2,7) (2,11) (3,5), (3,7) (3,11) (5,7) (5,11) (7,11)</math>
|10
|}
Each set of one side corresponds to a set in the other side which has the exact opposite elements.


Note that multicommas could just as well serve the purpose of a unique identifier for temperaments. There’s no particular reason to prefer the multimap to the multicomma. By convention, however, the multimap is the one that’s used.
Note that multicommas could just as well serve the purpose of a unique identifier for temperaments. There’s no particular reason to prefer the multimap to the multicomma. By convention, however, the multimap is the one that’s used.
If you need to do this process for a higher dimensionality than 6, then you'll need to understand how I found the symbols for each cell of Figure 6a. Here's how:
If you review the seven steps in the process for taking the complement, you may notice that a lot of it is busywork that will never change from one multimap to another. It all amounts to a specific sequence of 1’s and -1’s corresponding to a given rank and dimensionality. In consideration of this, I have gone ahead and prepared a table with the sequences you need to multiply the terms of your multimap by before flipping them:
# Take the rank, halved, rounded up. In our case, <span><math>\lceil \frac{r}{2} \rceil = \lceil \frac{2}{2} \rceil = \lceil 1 \rceil = 1</math></span>. Save that result for later. Let’s call it <span><math>x</math></span>.
# Find the lexicographic combinations of <span><math>r</math></span> primes again: (2,3), (2,5), (3,5). Except this time we don’t want the primes themselves, but their indices in the list of primes. So: <span><math>(1,2)</math></span>, <span><math>(1,3)</math></span>, <span><math>(2,3)</math></span>.
# Take the sums of these sets of indices, and to each sum, also add <span><math>x</math></span>. So <span><math>1+2+x</math></span>, <span><math>1+3+x</math></span>, <span><math>2+3+x</math></span> = <span><math>1+2+1</math></span>, <span><math>1+3+1</math></span>, <span><math>2+3+1</math></span> = <span><math>4</math></span>, <span><math>5</math></span>, <span><math>6</math></span>.
# Even terms become +'s and odd terms become -'s.
Yes, it's a lot of busywork. I could probably write a program to do it faster than I took explaining it. Maybe I will one day.


=== tempered lattice fractions generated by prime combinations ===
=== tempered lattice fractions generated by prime combinations ===
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