RE: triad plots

John,

I'm really confused.

Can you tell me the formula used in your triangular plots? I need to be able
to calculate x and y values for any point (to place it on the plot), using
the intervals as inputs. I've managed to get nearly symmetrical plots but
something is skewing the symmetry. Help!

-Paul

I think I figured it out -- it's
x=i1
y=i1/sqrt(3)+2*i2/sqrt(3)

right???

So the axes _are_ at 60-degree angles, after all!

Or rather, the axes are at 120-degree angles but the normals to the axes are
at 60-degree angles.

?

This is so confusing . . .

This is what it looks like, with integer limit 24:

http://www.egroups.com/files/harmonic_entropy/chalmers.jpg

In-Reply-To: <CE80F17667E4D211AE530090274662729C54D6@...>
> Or rather, the axes are at 120-degree angles but the normals to the
> axes are
> at 60-degree angles.

I seem to be only getting half of Paul's conversations. Can replies
please be directed to the list.

> ?
>
> This is so confusing . . .

Sure is. How can the normals be at different angles to each other than
the axes are?

60 degrees makes sense to me. Where the third interval is the sum of the
first two, you'd have like

(2,0,2)

(1,0,1) (1,1,2)

(0,0,0) (0,1,1) (0,2,2)

Call each point (x,y,z), and plot it as (x',y').

(x',y') = (x + y*sin(p), y*cos(p))

where p is the angle between the axes, usually 90 degrees, now 60 degrees.

For 60 degrees:

x' = x + sqrt(3)/2*y
y' = y/2

right????????????????????????????????????????????????????

Converting the other way, to find out what intervals a point on the graph
represents, you have

x = x' - sqrt(3)*y'
y = 2y'

That's almost, but not completely different to

>x=i1
>y=i1/sqrt(3)+2*i2/sqrt(3)

Confussing's the word!

Graham

Graham -- I take it you're straight now on all the geometry, etc.?

How easy would it be for you to calculate the triads for a bunch of product
limits? The Matlab interpreter seems uniquely slow for this, and the
compiler costs like $1000.

In-Reply-To: <CE80F17667E4D211AE530090274662729C54F5@...>
> Graham -- I take it you're straight now on all the geometry, etc.?

Geometry, I think so. I found a book with a chapter on information theory
in the market while wandering aimlessly around. I haven't had a chance to
read it yet.

> How easy would it be for you to calculate the triads for a bunch of
> product
> limits? The Matlab interpreter seems uniquely slow for this, and the
> compiler costs like $1000.

Product limits of up to 10^7 take a few minutes in Python. Java should be
an order of magnitude faster. The problem would be transmitting that
amount of data, or even holding it in memory. So what were you planning
to do with it? It's nearly 7e4 triads within an octave for a product
limit of a million. For those kind of numbers, the number of triads is
roughly proportional to the product limit.

Graham wrote,

>Product limits of up to 10^7 take a few minutes in Python. Java should be
>an order of magnitude faster. The problem would be transmitting that
>amount of data, or even holding it in memory. So what were you planning
>to do with it?

Why, calculate harmonic entropy surfaces, of course!

Perhaps we should start with a simple case and see if we can get our results
to agree. But first I have to get back to the geometry -- I screwed it up!

Graham Breed wrote,

>Product limits of up to 10^7 take a few minutes in Python. Java should be
>an order of magnitude faster. The problem would be transmitting that
>amount of data, or even holding it in memory. So what were you planning
>to do with it? It's nearly 7e4 triads within an octave for a product
>limit of a million. For those kind of numbers, the number of triads is
>roughly proportional to the product limit.

Graham,

Can you give me a zip file with a text file of the numbers making up all the
triads with a product limit of a million where the largest interval is less
than 13:1? Also, give me the same thing with a product limit of 343 so I can
check your program.

Thanks, this would be very helpful to begin calculating "true" triadic
harmonic entropy surfaces (which I don't think will have ridges!).

-Paul