> Carl Lumma wrote,

>

> >If we accept an odd-limit measure of dyadic consonance (and I do,

> more

> or

> >less), doesn't this lead to a prime-limit measure for chords?

>

> Not at all! Why would you think so? The consonance of a chord depends

> on

> (a) the consonance of the individual intervals and (b) the closeness

> of

> the chord to a harmonic series (due to combination tone and virtual

> pitch effects). In no way does this imply a prime-limit measure.

>

> Carl wrote me privately and suggested comparing these two chords:

>

> 1/1-5/4-3/2-15/8 vs. 1/1-5/4-3/2-11/6

>

> Here's an odd-limit analysis:

>

> The first chord has 2 3-limit intervals, 3 5-limit intervals, and 1

> 15-limit interval.

>

> The second chord has 1 3-limit interval, 2 5-limit intervals, 2

> 11-limit intervals, and 1 15-limit interval.

>

> Clearly, then, the first chord is more consonant based on constituent

> intervals. Based on other non-dyadic criteria, such as conformity to a

> harmonic series and equivalence of the lowest note to the fundamental,

> the first chord is again more consonant. I don't think the prime limit

> adds any information. In fact, you can construct arbitrarily dissonant

> chords in the 3-prime-limit, while many chords of higher prime limits

> will clearly be more consonant.

>Message: 19

> Date: Tue, 2 Mar 1999 15:29:26 -0500

> From: Joseph L Monzo <monz@juno.com>

>Subject: prime vs. odd limits

>

>Gary Morrison wrote:

>

>> The concept underlying prime limits is that we

>> can factor out stacks of any interval (i.e., powers

>> of any number) just as we can any arbitrary

>> number of octaves. That being the case, powers

>> of a given number become a minor consideration

>> compared the number being there at all, which in

>> turn means that the measuring stick has to be

>> primes. The concept then is that you can stack up

>> as many 3s (for example) as you'd like and the

>> high-level character doesn't change much from just

>> having one three. That just as a factor of 4 or 8

>> doesn't change the character of an interval much

>> from a factor of 2.

I think we must start to consider just how much of a consideration is it to "factor out" (I'd say cast out or ignore) the extra occurences of any given prime. How much does it "cost". Clearly the cost for casting out 2's is very low. The cost of casting out a single 3 is middling, but casting out 2 or more 3's has a high cost, while casting out even a single extra 5 has a marked effect. i.e. 25/16 is quite different from 5/4.

Clearly prime limits err on the side of assuming they all have zero cost (sort of), while odd limits err by ignoring the moderate cost of a single 3 and assuming the cost of 2's is zero. Integer limits err by ignoring even the low cost of 2's.

How about, when given a number, with prime factorisation

2^a * 3^b * 5^c * 7^d * ...

we define something like the "musical complexity" of that number as

2^(a*k_2) * 3^(b*k_3) * 5^(c*k_5) * 7^(d*k_7) * ...

Where the k_p's (read as k subscript p) are the "cost factors", for the various primes. Surely someone has proposed something like this before? Maybe even determined the factors by experiment?

Now we get integer limits by setting them all to 1, and odd limits by setting k_2 to zero and all the others to 1.

Prime limits seem rather more tricky at first. They don't seem to correspond to a static choice of cost factors, instead it seems that the highest prime to have a non-zero exponent has its exponent reduced to one, while all others go to zero. As such it seems rather hard to defend except for the fact that it almost treats 9 more realistically than do odd limits.

However, the prime limit turns out to be an approximation to a *logarithmic* complexity measure! The prime limit is astonishingly close to the natural *logarithm* of the "musical complexity" when the following values are used for the cost factors.

k_p = p/e

where p is the prime whose cost factor this is,

and e is the base of the natural logarithm ~= 2.71828...

i.e.

k_2 = 2/2.718 ~= 0.74

k_3 = 3/2.718 ~= 1.10

k_5 = 5/2.718 ~= 1.84

Now we can just argue about the relative values of the k_p's instead. ;-)

I suggest, very roughly:

k_2 ~= 0.3

k_3 ~= 0.8

k_5 ~= 0.9

k_p ~= 1, for p >= 7

You can download my Excel 97 spreadsheet (27kB) for calculating and plotting this "Musical Complexity", and find your own favourite values for the cost factors.

http://dkeenan.com/Music/MusicalComplexity.xls

Regards,

-- Dave Keenan

http://dkeenan.com

On Wed, 3 Mar 1999, Dave Keenan wrote:

> How about, when given a number, with prime factorisation

>

> 2^a * 3^b * 5^c * 7^d * ...

>

> we define something like the "musical complexity" of that number as

>

> 2^(a*k_2) * 3^(b*k_3) * 5^(c*k_5) * 7^(d*k_7) * ...

Wouldn't it then make sense to view the "musical complexity" of a ratio as

being the above formula applied to the product of it's numerator and

denominator?

-Bram

I (Dave Keenan) wrote:

>> How about, when given a number, with prime factorisation

>> 2^a * 3^b * 5^c * 7^d * ...

>> we define something like the "musical complexity" of that number as

>> 2^(a*k_2) * 3^(b*k_3) * 5^(c*k_5) * 7^(d*k_7) * ...

>> I suggest, very roughly:

>> k_2 ~= 0.3

>> k_3 ~= 0.8

>> k_5 ~= 0.9

>> k_p ~= 1, for p >= 7

>> http://dkeenan.com/Music/MusicalComplexity.xls

bram <bram@gawth.com> replied:

>Wouldn't it then make sense to view the "musical complexity" of a ratio as

>being the above formula applied to the product of it's numerator and

>denominator?

Certainly the next step is to get the complexity of a ratio from its

numerator and denominator, but I'm unclear whether to multiply them, or

take the maximum value (as is normally done with odd or prime limits). If

you multiply them you should then take the square-root (i.e. you should

find the geometric mean). This is to keep them commensurate with the

odd-limit or prime-limit where the max value is taken.

So if MC(x) is the Musical Complexity of x

Then for ratios n/d, either

MC(n/d) = Max( MC(n), MC(d) )

or

MC(n/d) = Sqrt( MC(n) * MC(d) )

What do others think?

Regards,

-- Dave Keenan

http://dkeenan.com

Dave Keenan wrote,

>Clearly prime limits err on the side of assuming they all have zero

cost (sort of), while odd limits err by ignoring >the moderate cost of a

single 3 and assuming the cost of 2's is zero. Integer limits err by

ignoring even the low >cost of 2's.

>How about, when given a number, with prime factorisation

>2^a * 3^b * 5^c * 7^d * ...

>we define something like the "musical complexity" of that number as

>2^(a*k_2) * 3^(b*k_3) * 5^(c*k_5) * 7^(d*k_7) * ...

>Where the k_p's (read as k subscript p) are the "cost factors", for the

various primes. Surely someone has >proposed something like this before?

You betcha! That's Barlow's Indigenstibility Function! You can read

about it in Georg Hajdu's paper, to which you provide a link from your

own web page (the latter seems inaccesible these days -- is it

offline?). Hajdu doesn't go into the justification for Barlow's

function, and I find odd-limit to be much more reliable.

>Maybe even determined the factors by experiment?

Why don't you propose one?

> I think we must start to consider just how much of a consideration is it to "factor out" (I'd say cast out or ignore) the extra occurences of any given prime. How much does it "cost".

Am I correct in inferring that, by how much they "cost", you mean how much taking it out affects the high-level harmonic character?