Recreational maths: a very funny number
lazarus / 8-Other

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MarkMLl · 2021-06-27 20:21

I don't know whether this is coincidence, or a quirk of number theory.
If it's the latter it's well beyond me, and I'd be interested in any
comments.

I've got a test program which is sending binary messages to a
graphical output device, the configuration messages can include a
palette.

When I was defining palettes in the test program I decided that rather
than entering in lots of numbers I'd allow the user to specify either
a linear, logarithmic or antilogarithmic (exponential) transfer
function.

The palette can be of various sizes: 2^1, 2^2 ... 2^6. The smallest
value in the palette should be 0 and the largest 255, so in the case
of 2^1 it's (0, 255) and so on.

Of these, the most demanding palette is the exponential palette of
size 2^2, since the ideal is to have the smallest value zero and the
next non-zero... getting this right took some experimentation.
Assuming that entries indexed i = 0..3 are i^z, I get this for various
values of z:

Code: Pascal  [Select][+][-]window.onload = function(){var x1 =
document.getElementById("main_content_section"); if (x1) { var x =
document.getElementsByClassName("geshi");for (var i = 0; i (* 1.0001 
    249 251 253 255                                                
*)(* 1.001        198 215 234 255                                    
           *)(* 1.01          20  47 109 255                         
                      *)(* 1.1            0   0   0 255              
                                 *)(* 1.05           0   0   4 255   
                                            *)(* 1.025          0   4
 31 255                                                 *)(* 1.024   
     1   5  34 255                                                
*)(* 1.026          0   3  29 255                                    
           *) 
I'm assuming that the best z for a palette sized 2^2 is also best for
larger palettes, and I'm slightly spooked that the digit sequence
radix 10 is so familiar.

Is there any rational reason why the best z = (2^10+1)/10^3 or is it a
dumb coincidence? :-)

MarkMLl

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MathMan · 2021-06-28 06:29

Quote from: MarkMLl on June 27, 2021, 10:21:08 pmI don't know whether
this is coincidence, or a quirk of number theory. If it's the latter
it's well beyond me, and I'd be interested in any comments.

I've got a test program which is sending binary messages to a
graphical output device, the configuration messages can include a
palette.

When I was defining palettes in the test program I decided that rather
than entering in lots of numbers I'd allow the user to specify either
a linear, logarithmic or antilogarithmic (exponential) transfer
function.

The palette can be of various sizes: 2^1, 2^2 ... 2^6. The smallest
value in the palette should be 0 and the largest 255, so in the case
of 2^1 it's (0, 255) and so on.

Of these, the most demanding palette is the exponential palette of
size 2^2, since the ideal is to have the smallest value zero and the
next non-zero... getting this right took some experimentation.
Assuming that entries indexed i = 0..3 are i^z, I get this for various
values of z:

Code: Pascal  [Select][+][-]window.onload = function(){var x1 =
document.getElementById("main_content_section"); if (x1) { var x =
document.getElementsByClassName("geshi");for (var i = 0; i (* 1.0001 
    249 251 253 255                                                
*)(* 1.001        198 215 234 255                                    
           *)(* 1.01          20  47 109 255                         
                      *)(* 1.1            0   0   0 255              
                                 *)(* 1.05           0   0   4 255   
                                            *)(* 1.025          0   4
 31 255                                                 *)(* 1.024   
     1   5  34 255                                                
*)(* 1.026          0   3  29 255                                    
           *) 
I'm assuming that the best z for a palette sized 2^2 is also best for
larger palettes, and I'm slightly spooked that the digit sequence
radix 10 is so familiar.

Is there any rational reason why the best z = (2^10+1)/10^3 or is it a
dumb coincidence? :-)

MarkMLl

Hello MarkMLI,

I think I can follow the above up to, but not including, the list. Can
you pls show how you calculate the linear, logarithmic & eponential
transfer.

Depending on how you calculate the exponential transfer it is probably
no coincidence that the "best fit" is the one you are seeing. If you
are looking for an exponential fit that gives 0 (for i=0) and 255 (for
i=2^k-1) then the natural approach would be to map i equidistant on
the range 0..1 (in the reals) and calculate 256^(p(i))-1. You seem to
have opted for keeping i as is and adapt the exponent z. Depending on
the way you calculate it is reasonable that z is correlated to 256
(max range) times 4 (i variants) = 1024.

Cheers,
MathMan

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MarkMLl · 2021-06-28 07:31

Quote from: MathMan on June 28, 2021, 08:29:35 amI think I can follow
the above up to, but not including, the list. Can you pls show how you
calculate the linear, logarithmic & eponential transfer.

Here's the relevant bits in a test program:

Code: Pascal  [Select][+][-]window.onload = function(){var x1 =
document.getElementById("main_content_section"); if (x1) { var x =
document.getElementsByClassName("geshi");for (var i = 0; i program
precomputePalettes; uses  Math; var  i: integer;  p2, l2: array[0..3]
of double; begin // Working values for linear palettes (or various
sizes).   for i := 0 to 3 do    p2[i] := i * (255 / 3.0); // Working
values for alog (exponential) palettes.   for i := 0 to 3 do    l2[i]
:= 1.025 ** p2[i]; // Round alog palette to integers.   for i := 0 to
3 do    WriteLn(Round(255.0 * (l2[i] / l2[3])))end. 
QuoteAlso, looking at the list above, I would have choosen z = 1.025
or z = 1.026 as "best fit" by your explanation. As you have obviously
choosen 1.001 I must have misunderstood something - care to explain in
more detail?

No, that list was purely in the order I worked through to get the
results: the value I chose- and am puzzled over- is 1.025 which gives
the sequence (0, 4, 31, 255).

MarkMLl

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MathMan · 2021-06-28 08:32

Quote from: MarkMLl on June 28, 2021, 09:31:52 am
QuoteAlso, looking at the list above, I would have choosen z = 1.025
or z = 1.026 as "best fit" by your explanation. As you have obviously
choosen 1.001 I must have misunderstood something - care to explain in
more detail?

No, that list was purely in the order I worked through to get the
results: the value I chose- and am puzzled over- is 1.025 which gives
the sequence (0, 4, 31, 255).

MarkMLl

Thanks for sharing. Must have been a x-post - I misread one statement
from your original and modified my response later on.

Now that I saw your calculation I am convinced it is a coincidence
that you found z= 1.025 / 1.026 as best fit. The way you are
calculating requires a z for which holds z^255 >= 2*255 (to get the
lower rounding to 0 correct). If you solve that for z you get
1.0247499.... (for equality). If you had i.e. a LUT maximum of 1,023
(instead of 255) the equivalent calculation would spit out z =
1.007473... As described above (in my edited post) I would have done
the exponential fit slightly different.

Cheers,
MathMan

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MarkMLl · 2021-06-28 09:40

Quote from: MathMan on June 28, 2021, 10:32:09 amNow that I saw your
calculation I am convinced it is a coincidence that you found z= 1.025
/ 1.026 as best fit. The way you are calculating requires a z for
which holds z^255 >= 2*255 (to get the lower rounding to 0 correct).
If you solve that for z you get 1.0247499.... (for equality). If you
had i.e. a LUT maximum of 1,023 (instead of 255) the equivalent
calculation would spit out z = 1.007473... As described above (in my
edited post) I would have done the exponential fit slightly different.

Yes, I was wondering later what the rounding was contributing. Just
dumb luck then :-)

MarkMLl