On the Mappings of Primes to Degrees of Equal Tunings

Hi,

the expression "n-equal" used to refer to the division of pure octave
into n equal parts. So it was completely self-evident that standard
mapping of primes was defined as round(n*log(p)/log(2)) where n is
the number of steps to the octave and p is the mapped prime. This
simply gave the best approximations to the primes.

Since the discovery of TOP tuning paradigm prime number 2 has lost
some of its specialness. The octave is no longer automatically
assumed to be just. Because TOP equal tunings are defined by their
mapping of primes there are many flavours of n-equal tunings. But I
think some of these mappings are somehow more natural than others.
Let me show what I mean.

The 7-limit standard mapping of primes to 12-tone equal division of
2:1 is [12 19 28 34] (or <12 19 28 34] if you want). Now let's see if
the division of 3:1 into 19 equal parts gives the same mapping. Yes
it does. The same holds for 28-tone equal division of 5:1 and 34-tone
equal division of 7:1. The primes 2, 3, 5 and 7 kind of "agree" about
the mapping.

Def. mapping m from a set P of primes to integers (or degrees of an
equal tuning) is natural for P if m(p)=round(m(q)*log(p)/log(q)) for
all p and q belonging to P.

Another example: there is no natural mapping of P={2, 3, 5, 7} to
integers where m(2)=16 because round(16*log(3)/log(2))=25 and round
(16*log(7)/log(2))=45 but round(25*log(7)/log(3))=44.

The name "natural" for these mappings is just a suggestion and if you
think it sucks please come up with a better one. Maybe something
derived from the verb "agree" would be better.

Some natural mappings:

P={2, 3, 5}

3 5 7
4 6 9
5 8 12
7 11 16
8 13 19
9 14 21
10 16 23
12 19 28
15 24 35
16 25 37
18 29 42
19 30 44

P={2, 3, 7}

4 6 11
5 8 14
9 14 25
10 16 28 (contorsion)
12 19 34
14 22 39
17 27 48
18 29 51
19 30 53
22 35 62
24 38 67
26 41 73
27 43 76

P={2, 3, 5, 7}

4 6 9 11
9 14 21 25
10 16 23 28
12 19 28 34
18 29 42 51
19 30 44 53
22 35 51 62
26 41 60 73
27 43 63 76
29 46 67 81
31 49 72 87
41 65 95 115

Notice that 5, 15 and 16 are missing.

P={2, 3, 5, 7, 11}

9 14 21 25 31
22 35 51 62 76
29 46 67 81 100
31 49 72 87 107
41 65 95 115 142
46 73 107 129 159
49 78 114 138 170
58 92 135 163 201
63 100 146 177 218
72 114 167 202 249
80 127 186 225 277
87 138 202 244 301

Kalle Aho

Hi Kalle,

No one ever replied to this? This looks something like consistency
to me.

-Carl

At 03:43 AM 7/17/2004, you wrote:
>Hi,
>
>the expression "n-equal" used to refer to the division of pure octave
>into n equal parts. So it was completely self-evident that standard
>mapping of primes was defined as round(n*log(p)/log(2)) where n is
>the number of steps to the octave and p is the mapped prime. This
>simply gave the best approximations to the primes.
>
>Since the discovery of TOP tuning paradigm prime number 2 has lost
>some of its specialness. The octave is no longer automatically
>assumed to be just. Because TOP equal tunings are defined by their
>mapping of primes there are many flavours of n-equal tunings. But I
>think some of these mappings are somehow more natural than others.
>Let me show what I mean.
>
>The 7-limit standard mapping of primes to 12-tone equal division of
>2:1 is [12 19 28 34] (or <12 19 28 34] if you want). Now let's see if
>the division of 3:1 into 19 equal parts gives the same mapping. Yes
>it does. The same holds for 28-tone equal division of 5:1 and 34-tone
>equal division of 7:1. The primes 2, 3, 5 and 7 kind of "agree" about
>the mapping.
>
>Def. mapping m from a set P of primes to integers (or degrees of an
>equal tuning) is natural for P if m(p)=round(m(q)*log(p)/log(q)) for
>all p and q belonging to P.
>
>Another example: there is no natural mapping of P={2, 3, 5, 7} to
>integers where m(2)=16 because round(16*log(3)/log(2))=25 and round
>(16*log(7)/log(2))=45 but round(25*log(7)/log(3))=44.
>
>The name "natural" for these mappings is just a suggestion and if you
>think it sucks please come up with a better one. Maybe something
>derived from the verb "agree" would be better.
>
>Some natural mappings:
>
>P={2, 3, 5}
>
>3 5 7
>4 6 9
>5 8 12
>7 11 16
>8 13 19
>9 14 21
>10 16 23
>12 19 28
>15 24 35
>16 25 37
>18 29 42
>19 30 44
>
>P={2, 3, 7}
>
>4 6 11
>5 8 14
>9 14 25
>10 16 28 (contorsion)
>12 19 34
>14 22 39
>17 27 48
>18 29 51
>19 30 53
>22 35 62
>24 38 67
>26 41 73
>27 43 76
>
>P={2, 3, 5, 7}
>
>4 6 9 11
>9 14 21 25
>10 16 23 28
>12 19 28 34
>18 29 42 51
>19 30 44 53
>22 35 51 62
>26 41 60 73
>27 43 63 76
>29 46 67 81
>31 49 72 87
>41 65 95 115
>
>Notice that 5, 15 and 16 are missing.
>
>P={2, 3, 5, 7, 11}
>
>9 14 21 25 31
>22 35 51 62 76
>29 46 67 81 100
>31 49 72 87 107
>41 65 95 115 142
>46 73 107 129 159
>49 78 114 138 170
>58 92 135 163 201
>63 100 146 177 218
>72 114 167 202 249
>80 127 186 225 277
>87 138 202 244 301
>
>
>Kalle Aho
>
>
>
>
>
>
>Yahoo! Groups Links
>
>
>
>

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@...> wrote:
>
> Hi Kalle,
>
> No one ever replied to this? This looks something like consistency
> to me.
>
> -Carl

Yep, no one ever replied. :) It certainly gives similar results.

Kalle

--- In tuning-math@yahoogroups.com, "Kalle Aho" <kalleaho@...> wrote:
>
> --- In tuning-math@yahoogroups.com, Carl Lumma <ekin@> wrote:
> >
> > Hi Kalle,
> >
> > No one ever replied to this? This looks something like consistency
> > to me.
> >
> > -Carl
>
> Yep, no one ever replied. :) It certainly gives similar results.

But there are also differences i.e. 15-equal is 7-limit consistent but
doesn't show up when the primes are {2,3,5,7}.

I had already forgotten this idea of mine as it seems to generate no
discussion at all. :)

Kalle

--- In tuning-math@yahoogroups.com, "Kalle Aho" <kalleaho@...> wrote:

> But there are also differences i.e. 15-equal is 7-limit consistent but
> doesn't show up when the primes are {2,3,5,7}.
>
> I had already forgotten this idea of mine as it seems to generate no
> discussion at all. :)

I can't discuss it without knowing what it is. What do you mean by
saying 15-et "doesn't show up"?

--- In tuning-math@yahoogroups.com, "Gene Ward Smith"
<genewardsmith@...> wrote:
>
> --- In tuning-math@yahoogroups.com, "Kalle Aho" <kalleaho@> wrote:
>
> > But there are also differences i.e. 15-equal is 7-limit consistent but
> > doesn't show up when the primes are {2,3,5,7}.
> >
> > I had already forgotten this idea of mine as it seems to generate no
> > discussion at all. :)
>
> I can't discuss it without knowing what it is. What do you mean by
> saying 15-et "doesn't show up"?

Hi Gene,

I defined it here:

/tuning-math/message/10884

Kalle

--- In tuning-math@yahoogroups.com, "Kalle Aho" <kalleaho@...> wrote:
> Def. mapping m from a set P of primes to integers (or degrees of an
> equal tuning) is natural for P if m(p)=round(m(q)*log(p)/log(q)) for
> all p and q belonging to P.

A patemt val which maps everything in the p-limit diamond to the
nearest rounded value defines n-consistency. This is weaker, since it
only requires the same for primes, but is similar in being a property
of n. AI guess the main question is why primes deserve special
treatment when defining consistency. It might be interesting to look
for p-inconsistent n which are p-natural, I suppose.

This is another posting I overlooked completely recently, probably due
to Yahoo changing things on me.

>This is another posting I overlooked completely recently, probably due
>to Yahoo changing things on me.

The 'links not marked followed by clicking next' bug seems to be
fixed these last few days. Have you noticed?

-Carl

--- In tuning-math@yahoogroups.com, "Gene Ward Smith"
<genewardsmith@...> wrote:

> A patemt val which maps everything in the p-limit diamond to the
> nearest rounded value defines n-consistency. This is weaker, since it
> only requires the same for primes, but is similar in being a property
> of n.

Gene, do you mean that naturalness is weaker than consistency?

How can 5, 15 and 16 then be 7-limit consistent if there are no
natural mappings for them for primes 2, 3, 5 and 7?

>AI guess the main question is why primes deserve special
> treatment when defining consistency.

Sorry, what special treatment?

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" wrote:
>
> --- In tuning-math@yahoogroups.com, "Kalle Aho" wrote:
> > Def. mapping m from a set P of primes to integers (or degrees of
an equal tuning) is natural for P if m(p)=round(m(q)*log(p)/log(q))
for all p and q belonging to P.
>
> A patemt val which maps everything in the p-limit diamond to the
nearest rounded value defines n-consistency. This is weaker, since
it only requires the same for primes, but is similar in being a
property of n. AI guess the main question is why primes deserve
special treatment when defining consistency. It might be interesting
to look for p-inconsistent n which are p-natural, I suppose.

Gene, just to get this in perspective: how is the concept of n-
consistency useful?

Regards,
Yahya

--- In tuning-math@yahoogroups.com, "Kalle Aho" <kalleaho@...> wrote:

> How can 5, 15 and 16 then be 7-limit consistent if there are no
> natural mappings for them for primes 2, 3, 5 and 7?

Sorry, you are right. Naturalness is stronger than consistency, not
weaker.

--- In tuning-math@yahoogroups.com, "yahya_melb" <yahya@...> wrote:

> Gene, just to get this in perspective: how is the concept of n-
> consistency useful?

For one example, if you want to approximate a JI chord, you may want n
to be consistent relative to that chord, so that all of the intervals
in the chord take their best values.

--- In tuning-math@yahoogroups.com, "Kalle Aho" <kalleaho@...> wrote:
>Def. mapping m from a set P of primes to integers (or degrees of an >equal tuning) is natural for P if m(p)=round(m(q)*log(p)/log(q)) for >all p and q belonging to P. Apparently.

Anyway, I've been fiddling with this. You know for the rounded-to-nearest primes approximation (what some are calling the patent val) that the error in each prime must be less than half a scale step. So:

|d(p)| < 1/2m(2)

where d(p) is the deviation in interval p in octaves and m(2) is the number of steps to an octave. You can prove it if you like from |round(x)|<=0.5 for any x, m(p) = round(m(0)*log2(p)), d(p) = m(p)/m(2)-log2(p), and the fundamental theorem of arithmetic.

If the equal temperament is consistent in the tonality diamond formed by the prime harmonics, we also know that the deviation of any interval between primes must be less than half a scale step, so

|d(p) - d(q)| < 1/2m(2)

This is weaker than odd-limit consistency, because factors of 9 aren't included in the 11-limit. By setting q to 0 and making the octaves pure, so the error in an octave (d(2)) is zero, you can show that this implies the nearest-primes approximation. So if a mapping is consistent, it must be the nearest-primes mapping. Even for this weakened version of consistency that shouldn't be surprising.

I think naturalness as defined in the quote above implies that

|d(p) - d(q) log(p)/log(q)| < 1/2m(2)

Because 2 is one of the primes naturalness is defined for, it shouldn't be a surprise that setting d(2)=0 gives the nearest-primes formula. The primes p and q are independent, so you can also say

max[|d(p) - d(q) log(p)/log(q)|, |d(q) - d(p) log(q)/log(p)|] < 1/2m(2)

Play around with that and I think you can prove that naturalness implies consistency -- but only consistency for the diamond of prime harmonics, not the full odd-limit diamond. Set k=log(p)/log(q) (which has to be positive) and you get consistency provided

|d(p) - d(q)| <= max[|d(p) - d(q) k|, |d(q) - d(p)/k|]

That's the same as

|d(p) - d(q)| <= max[|d(p) - d(q) k|, |d(p) - d(q) k|/k]

if k>1 that means

|d(p) - d(q)| <= |d(p) - d(q) k|

Because p and q are arbitrary, we only need to look at one case, so let's say k > 1. If p and q have opposite signs, the errors compound to give

|d(p) - d(q)| = |d(p)| + |d(q)|

|d(p) - d(q) k| = |d(p)| + |d(q)|k

and the inequality holds. If the errors cancel,

|d(p) - d(q)| <= max(|d(p)|, |d(q)|)

and no nearest-primes approximation will break consistency for this p and q.

An interesting question might be what naturalness tells you about the Tenney-weighted prime errors.

Gene wrote:
> A patemt val which maps everything in the p-limit diamond to the > nearest rounded value defines n-consistency. This is weaker, since it > only requires the same for primes, but is similar in being a property > of n. AI guess the main question is why primes deserve special > treatment when defining consistency. It might be interesting to look > for p-inconsistent n which are p-natural, I suppose.

What's the p-limit diamond? What's n?

Graham

Graham, thank you for this excellent post!

--- In tuning-math@yahoogroups.com, Graham Breed <gbreed@...> wrote:
>
> --- In tuning-math@yahoogroups.com, "Kalle Aho" <kalleaho@> wrote:
> >Def. mapping m from a set P of primes to integers (or degrees of an
> >equal tuning) is natural for P if m(p)=round(m(q)*log(p)/log(q)) for
> >all p and q belonging to P.
>
> Apparently.

:)

> Anyway, I've been fiddling with this. You know for the
> rounded-to-nearest primes approximation (what some are calling the
> patent val) that the error in each prime must be less than half a scale
> step. So:
>
> |d(p)| < 1/2m(2)
>
> where d(p) is the deviation in interval p in octaves and m(2) is the
> number of steps to an octave. You can prove it if you like from
> |round(x)|<=0.5 for any x, m(p) = round(m(0)*log2(p)), d(p) =
> m(p)/m(2)-log2(p), and the fundamental theorem of arithmetic.
>
> If the equal temperament is consistent in the tonality diamond
formed by
> the prime harmonics, we also know that the deviation of any interval
> between primes must be less than half a scale step, so
>
> |d(p) - d(q)| < 1/2m(2)
>
> This is weaker than odd-limit consistency, because factors of 9 aren't
> included in the 11-limit. By setting q to 0 and making the octaves
> pure, so the error in an octave (d(2)) is zero, you can show that this
> implies the nearest-primes approximation. So if a mapping is
> consistent, it must be the nearest-primes mapping. Even for this
> weakened version of consistency that shouldn't be surprising.
>
> I think naturalness as defined in the quote above implies that
>
> |d(p) - d(q) log(p)/log(q)| < 1/2m(2)
>
> Because 2 is one of the primes naturalness is defined for, it shouldn't
> be a surprise that setting d(2)=0 gives the nearest-primes formula.

Actually the set P doesn't have to contain 2. P could be for example
{3, 5, 7}.

The
> primes p and q are independent, so you can also say
>
> max[|d(p) - d(q) log(p)/log(q)|, |d(q) - d(p) log(q)/log(p)|] < 1/2m(2)
>
> Play around with that and I think you can prove that naturalness
implies
> consistency -- but only consistency for the diamond of prime harmonics,
> not the full odd-limit diamond. Set k=log(p)/log(q) (which has to be
> positive) and you get consistency provided
>
> |d(p) - d(q)| <= max[|d(p) - d(q) k|, |d(q) - d(p)/k|]
>
> That's the same as
>
> |d(p) - d(q)| <= max[|d(p) - d(q) k|, |d(p) - d(q) k|/k]
>
> if k>1 that means
>
> |d(p) - d(q)| <= |d(p) - d(q) k|
>
> Because p and q are arbitrary, we only need to look at one case, so
> let's say k > 1. If p and q have opposite signs, the errors
compound to
> give
>
> |d(p) - d(q)| = |d(p)| + |d(q)|
>
> |d(p) - d(q) k| = |d(p)| + |d(q)|k
>
> and the inequality holds. If the errors cancel,
>
> |d(p) - d(q)| <= max(|d(p)|, |d(q)|)
>
> and no nearest-primes approximation will break consistency for this
p and q.

So naturalness implies consistency in the diamond of primes in P, very
interesting!

> An interesting question might be what naturalness tells you about the
> Tenney-weighted prime errors.

Yes, any insights about that?

Kalle Aho wrote:
> Graham, thank you for this excellent post! Somebody read it then!

>>I think naturalness as defined in the quote above implies that
>>
>>|d(p) - d(q) log(p)/log(q)| < 1/2m(2)
>>
>>Because 2 is one of the primes naturalness is defined for, it shouldn't >>be a surprise that setting d(2)=0 gives the nearest-primes formula. > > Actually the set P doesn't have to contain 2. P could be for example
> {3, 5, 7}. In which case it won't say anything about octave-equivalent consistency.

> So naturalness implies consistency in the diamond of primes in P, very
> interesting! I think it means you keep consistency for any tuning where one of the primes is set to be pure. And I don't think there's anything special about prime numbers. You can do this for any octave-specific diamond. If it includes a 2, then you have consistency for the octave-equivalent diamond when octaves are pure.

My guess (and I haven't checked it) is that the consistent temperaments that fail to be natural will also fail to be consistent where one of the primes is set to be just. And this is why consistency doesn't imply naturalness.

>>An interesting question might be what naturalness tells you about the >>Tenney-weighted prime errors.
> > Yes, any insights about that?

It looks like the relationship is to consistency for tonality diamonds, and nothing to do with weighted primes.

Graham