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## 581EDO

- Dave Keenan
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### Re: 581EDO

Here's George's proposal in a form that makes it easier to evaluate flag and shaft arithmetic, and apotome complements.

- George Secor
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**Posts:**31**Joined:**Tue Sep 01, 2015 11:36 pm**Location:**Godfrey, Illinois, US

### Re: 581EDO

Hmm, if one were to choose between good, bad, and ugly, I guess the verdict is ugly.Dave Keenan wrote:That's quite a tour de force, George. I don't doubt that it is valid, however I find it unsatisfying. Of course it's easy to criticise, and much harder to come up with an alternative. It's unsatisfying that there is no regularity that might reduce the amount of information that needs to be remembered or looked up -- or at least none that I can discern. In particular, there is no consistent assignment of numbers-of-steps to all flags, and the double-shaft symbols do not repeat a flag sequence of the single-shaft symbols. And there are so many different core symbols.

There's a way to fix this, however, if we want, in order of importance:

1) consistent flag arithmetic for the single-shaft symbols;

2) minimize number of accented symbols;

3) minimize number symbol cores.

Using the following flag arithmetic: = 1, = 2, = 2, = 6, = 7, = 11, = 13, = 15, = 17

and assuming that is defined as the 17:121 schismina (1088:1089), I propose the following symbol sequence:

581:

- Dave Keenan
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### Re: 581EDO

Thanks George. I think that's a major improvement. Here it is numbered:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56Even when the double-shaft symbols don't recapitulate the flags for a subsequence of the single-shaft symbols, it is nice if a consistent value can be assigned to the second shaft, usually the value of the symbol (30). Unfortunately, this breaks down with the symbols for 7 and 38, and , and those for 18 and 49, and . And with the symbols for 19 and 48, and , there is an unfortunate crossover, relative to 18 and 49. [Edit: And 8 and 37, and , have the same kind of crossover with 7 and 38.]

- Dave Keenan
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### Re: 581EDO

I believe there was a typo for 6°581, which should be . I have corrected this, and added mixed-sagittal below. I have also checked that your double-shaft symbols correctly match the mixed sagittal, as per figure 13 on page 24 of http://sagittal.org/sagittal.pdf.

The remaining problems with this notation are all caused by the fact that the symbols for 7 and 8 degrees, which are the 23-comma and the 17-comma , are out of size order, relative to their untempered values, while the symbols for 18 and 19 degrees, which have the same flag combinations as the apotome-complements of 7 and 8 respectively, are not out of size order. They are the 5:23-small-diesis and the 5:11-small-diesis .

The remaining problems with this notation are all caused by the fact that the symbols for 7 and 8 degrees, which are the 23-comma and the 17-comma , are out of size order, relative to their untempered values, while the symbols for 18 and 19 degrees, which have the same flag combinations as the apotome-complements of 7 and 8 respectively, are not out of size order. They are the 5:23-small-diesis and the 5:11-small-diesis .

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56

- Dave Keenan
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### Re: 581EDO

So it all comes down to finding sufficient unaccented symbols between 2 and 11 degrees, without any size-order reversals, and with consistent flag arithmetic. We can have gaps of at most 3 degrees between unaccented symbols. Here are all the options. Size order reversals are indicated by "<->".

19s,5:7k 7:11k 17k 143C 23C 17C 11:49C 19C,5C <-> <-> 2 3 4 5 6 7 8 9 10 11

- Dave Keenan
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### Re: 581EDO

Of course we should use the lowest prime symbols for degrees 2 and 11. I only included the 19s and 19C symbols so you'd know I hadn't forgotten them, and that there really are no unaccented symbols that map to 3 or 10 degrees. But the left scroll must still have the value 2 since we want = 26 and = 28.

Here I show the implied value of the left boathook for each symbol that contains it. None of them is compatible with any other.

(a) at most one of them contains a left boathook,

(b) there is at most one symbol from each size-reversed pair,

(c) there are no gaps wider than 3 degrees.

Here I show the implied value of the left boathook for each symbol that contains it. None of them is compatible with any other.

5:7k 7:11k 17k 143C 23C 17C 11:49C 5C <-> <-> 2 3 4 5 6 7 8 9 10 11 =5 =4 =6 =4.5The only

*other*symbols that contain the left boathook are the following, but we are not forced to use either of them.49S 23S 17 20 =4 =5So the puzzle is to choose a subset of the symbols on degrees 4 thru 9 above, such that

(a) at most one of them contains a left boathook,

(b) there is at most one symbol from each size-reversed pair,

(c) there are no gaps wider than 3 degrees.

- Dave Keenan
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### Re: 581EDO

Since there are 6 symbols, there are 2⁶ subsets, and we can enumerate them as a binary count. The right-hand column indicates at least one condition that the subset violates. ">3" = gap wider than 3 degrees. "~|" = incompatible left boathook values. "<->" size order reversal.

And we find that there is one, and only one, solution.

And we find that there is one, and only one, solution.

7:11k <-> 17k 143C 23C <-> 17C 11:49C violates 4 5 6 7 8 9 >3 >3 >3 >3 >3 >3 >3 >3 >3 >3 >3 >3 >3 >3 >3 >3 >3 ~| ~| ~| >3 ~| ~| ~| >3 ~| ~| ~| >3 ~| ~| ~| >3 >3 >3 >3 >3 Hoorah! <-> ~| >3 ~| ~| ~| >3 ~| ~| ~| >3 >3 ~| ~| >3 ~| ~| ~| >3 ~| ~| ~| >3 ~| ~| ~|

- Dave Keenan
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### Re: 581EDO

So, the only change required to George's last proposal is to eliminate 17C for 8 degrees in favour of 143C for 9 degrees. However, I note that we do not need the non-athenian symbol 5:23S for 18 degrees and suggest we eliminate it to reduce the number of cores that must be learned. And I note that we obtain a more regular, and hence more easily learned pattern of accents, if we change 3 degrees from to . Here's the result.