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Entropic efficiency of a scale

🔗Gene Ward Smith <genewardsmith@coolgoose.com>

2/14/2006 12:46:18 AM

The efficency of an alphabet of n letters can be defined as the ratio
of the entropy of the alphabet over the maximum entropy for such an
alphabet. Here each letter ni appears with proabilbility pi, and the the
entropy of the alphabet is -sum_i pi log2(pi). If each pi is 1/n, we
get the maximum entropy, which is just log2(n). Hence the efficiency
can also be written -sum_i pi log_n(pi).

We may consider the scale steps si of an octave-repreating scale to be
an alphabet, if we assign a probability of log2(si) to each scale
step. While larger steps are not proportionally more likely to occur,
they do take up more room, so there's some idea of efficiency involved
here. Then the entropic efficency of a scale with scale steps si would be
-sum_i log2(si) log2(log2(si)) / log2(n). It is a measure of how
nearly equal the step sizes are, achieving the value of 1 in the case
of equal temperaments.

🔗Carl Lumma <clumma@yahoo.com>

2/14/2006 3:37:16 AM

> The efficency of an alphabet of n letters // Then the entropic
> efficency of a scale with scale steps si would be
> -sum_i log2(si) log2(log2(si)) / log2(n). It is a measure of
> how nearly equal the step sizes are, achieving the value of 1
> in the case of equal temperaments.

How is this different from Rothenberg efficiency?

-Carl

🔗Gene Ward Smith <genewardsmith@coolgoose.com>

2/14/2006 10:49:04 AM

--- In tuning@yahoogroups.com, "Gene Ward Smith" <genewardsmith@...>
wrote:
>
> The efficency of an alphabet of n letters can be defined as the ratio
> of the entropy of the alphabet over the maximum entropy for such an
> alphabet.

Sorry for posting this here by mistake. Dicsussion should be moved to
tuning-math, I expect.

🔗Gene Ward Smith <genewardsmith@coolgoose.com>

2/14/2006 10:47:41 AM

--- In tuning@yahoogroups.com, "Carl Lumma" <clumma@...> wrote:
>
> > The efficency of an alphabet of n letters // Then the entropic
> > efficency of a scale with scale steps si would be
> > -sum_i log2(si) log2(log2(si)) / log2(n). It is a measure of
> > how nearly equal the step sizes are, achieving the value of 1
> > in the case of equal temperaments.
>
> How is this different from Rothenberg efficiency?

Rothenberg efficiency is a totally different thing, isn't it?