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fractint-digest        Friday, January 22 1999        Volume 01 : Number 354




----------------------------------------------------------------------

Date: Wed, 20 Jan 1999 00:24:04 -0500
From: Gedeon Peteri <[e-mail]>
Subject: (fractint) Ultra Fractal pages

Two pages at my Geocities website have been extensively revised. The
Chebyshev and the Zf_Ang pages are now entirely made up of multi-layered
true color images generated with Ultra Fractal. Some of the images are
reworked Fractint pars; others
are new. I hope you'll enjoy your visit.

Gedeon

- --
- --------------------------------------------------------------
Fractals: http://www.geocities.com/~gedeonp/index.html
Member Infinite Fractal Loop
Last updated: January 20, 1999 - two Ultra Fractal pages added
Photography: http://members.xoom.com/gedeonp/index.html
Last updated: November 8, 1998
- --------------------------------------------------------------



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------------------------------

Date: Tue, 19 Jan 1999 23:44:27 -0600
From: "Paul N. Lee" <[e-mail]>
Subject: (fractint) The FOTD website....

Greetings,

I got an email this evening from the folks at my ISP.  A portion of that
message reads as follows:

    Your AT&T WorldNet Personal Web Page at
    http://home.att.net/~Paul.N.Lee/FotD/FotD.html
    is terrific-- so good, in fact, that weÆve added it to
    our Member Showcase listing for Science & Technology.
    ItÆs where we feature the most interesting and active
    community sites in AT&T WorldNet Community Port. YouÆll
    definitely get more traffic and visibility for your site
    as a result of your hard work.

To see the Web site listed with their other top community Web sites,
please go to  http://community.att.net/community/showcase/index.html.

P.N.L.
- --------------------------------------------------------------
http://www.fractalus.com/cgi-bin/theway?ring=fractals&id=43&go

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------------------------------

Date: Wed, 20 Jan 1999 00:44:12 -0500
From: Gedeon Peteri <[e-mail]>
Subject: (fractint) True color Zf_Angs

Paul,

I have missed your postings lately. Having started working with Ultra
Fractal a short time ago, the first formula I worked with in it was my
old favorite, your Zf_Ang. I have replaced the images on my old Zf_Ang
page at my Geocities website with these true color images. I thought you
might like to see them. Thanks for your fine formula!

Gedeon
- --
- --------------------------------------------------------------
Fractals: http://www.geocities.com/~gedeonp/index.html
Member Infinite Fractal Loop
Last updated: January 20, 1999 - two Ultra Fractal pages added
Photography: http://members.xoom.com/gedeonp/index.html
Last updated: November 8, 1998
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------------------------------

Date: Wed, 20 Jan 1999 10:38:54 -0600
From: "Damien M. Jones" <[e-mail]>
Subject: Re: (fractint) The FOTD website....

Paul, Jim,

Congratulations!

Damien M. Jones   \\
[e-mail]  \\  Fractalus Galleries & Info:
                    \\  http://www.fractalus.com/

Please do not post my e-mail address on a web site or
in a newsgroup.  Thank you.

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------------------------------

Date: Wed, 20 Jan 1999 16:27:31 EST
From: [e-mail]
Subject: Re: (fractint) The FOTD website....

Congratulations Paul and Jim!

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------------------------------

Date: Wed, 20 Jan 1999 18:09:34 -0500
From: Paul DeCelle <[e-mail]>
Subject: Re: (fractint) Fabio

Fabio Fazzi wrote:
>
>
> I can change colors in Winfract but not in Fractint, can somebody explain me how i can
> change colors (as Volcano, Blue, etc...)?

Fabio--

With your image loaded (or even while it's still loading), press "C",
then press "L".  This will allow you to load any colormap in the
Fractint directory.  You can cycle the colors by pressing "C" again, and
stop cycling by pressing "C" yet again.  The "<" and ">" keys change the
cycling direction.  Once you have the look you want, press "ESC" to get
back to normal Fractint operations.  I hope this is clear; I change and
cycle colors without really hinking about the steps...

Regards, Paul DeCelle

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------------------------------

Date: Wed, 20 Jan 1999 18:21:53 -0500
From: Paul DeCelle <[e-mail]>
Subject: Re: (fractint) Changing Colors

Oops, looks like Fabio's question already got answered.  Guess I should
read ALL my mail before replying (66 messages waiting after a two day
absence!)

Paul

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------------------------------

Date: Thu, 21 Jan 1999 00:35:57 +0100
From: "Ognen Ivanovski" <[e-mail]>
Subject: (fractint) Fw: Fractint & it's color decomposition

Hi everyone.
I found the address of Lee Crocker, the author of the Decomposition coloring
method on Fractint's cerdits screen.
I've sent the message below to that address but the author could not be
reached thatways.
So I am hoping that he'll read this message.
Also, if anyone else knows the answer, it would be nice to read it.
Once my project is built, be sure it will be a good source of unusual
"geneticaly engeineered" formulas.

Thanks in advance.
Ognen

- -----Original Message-----
From: Ognen Ivanovski <[e-mail]>
To: Lee Crocker <[e-mail]>
Date: Monday, January 18, 1999 6:56 PM
Subject: Fractint & it's color decomposition


>Hello,
>
>I'm a computer science student who has his love for fractals since early
>highschool. At last I got the oportunity to do something with it, i.e. to
>learn about fractals better.
>
>Now, I am working on a project that I call The Genetic Artist. Using
Genetic
>Algorithm I'm supposed to make a population of fractals (roughly speaking)
>to mate between themselves and produce "better" ones. The more the fractal
>is "fit" the better are the chances to spread it's genes. The fitness is
>based on votes from the visitors of the project's page.
>
>I borrowed the idea from a project that appears at  gatech.edu (called
DNA).
>They are using fractint to generate their images. But I want to make my own
>generator (so I can learn the process throughly) and some of the images
that
>I intend to use as generation one are that beautiful due to using
>decomposition.
>
>In Fractint's Help there is a rough explanation of how it is done. Could
you
>explain the process to me in short terms. All I need to know is:
>
>When the iterations stop at a value of (x, z) and the decomposition factor
>is 64 (I like that value best) how do I determine the color of the pixel
>(ranging 0..255)?
>
>I hope I'm not asking too much of your time. I'd love to send a .par file
in
>exchange for the information. Also when I finish the project I'd like you
to
>come to the web page and see how the generations are going.
>
>Needless to say that my first mean of relaxation at home is of course
>spending hours in front of fractint. :)
>
>Thanks in advance,
>Ognen Ivanovski
>student at the Institute of Informatics
>University of "St. Cyril and Methodius"
>Skopje, Macedonia (former Yougoslav Republic of)
>
>_____
>http://eon.pmf.ukim.edu.mk/~milkman/ [coming soon]
>


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------------------------------

Date: Wed, 20 Jan 1999 19:43:00 -0500
From: Barry N Merenoff <[e-mail]>
Subject: (fractint) Re: Decomposition

I would assume that the way this is done is by repeatedly squaring the
value from the final iteration. A binary number would be built up from le=
ft
to right. The first (leftmost) bit would be one when the final iteration
has a negative real part or zero when the final iteration has a positive
real part. The second bit would be determined the same way from the squar=
e
of the final iteration; the third, from the fourth power of the final
iteration; etc.

If any of these values has a zero real part, it would be grouped with
either the positive or negative real parts; it's up to you. However, you
would have to be consistent.

Note: I haven't actually checked the source code on this. It just seems t=
he
sensible way decomposition would be done. If someone knows a mistake in
this algorithm, please let both of us know.

Collin Merenoff

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------------------------------

Date: Wed, 20 Jan 1999 19:35:58 -0600
From: "Damien M. Jones" <[e-mail]>
Subject: Re: (fractint) Re: Decomposition

Collin,

Actually I'd expect the decomposition to simply be read off from the angle
directly. You could keep squaring the number, but if you don't add c each
time this has precisely the same effect, but it also increases the
magnitude of the number substantially each time, and if your bailout is set
too high you will overflow.

Damien M. Jones   \\
[e-mail]  \\  Fractalus Galleries & Info:
                    \\  http://www.fractalus.com/

Please do not post my e-mail address on a web site or
in a newsgroup.  Thank you.

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------------------------------

Date: Wed, 20 Jan 1999 18:47:28 -0700 (MST)
From: Kerry Mitchell <[e-mail]>
Subject: Re: (fractint) Re: Decomposition

That's my understanding, too.  Setting "decomp=n" will cause Fractint to
color the image by breaking the final polar angle into n segments.  Binary
decomposition would come from setting decomp=2.  Then, all the angles from
0 to pi would be rendered in one color, and pi to 2*pi in another.

Kerry

- -------------------------------------------------------------------------------
Kerry Mitchell
[e-mail]    www.primenet.com/~lkmitch/
- -------------------------------------------------------------------------------

On Wed, 20 Jan 1999, Damien M. Jones wrote:

> Collin,
>
> Actually I'd expect the decomposition to simply be read off from the angle
> directly. You could keep squaring the number, but if you don't add c each
> time this has precisely the same effect, but it also increases the
> magnitude of the number substantially each time, and if your bailout is set
> too high you will overflow.
>
> Damien M. Jones   \\


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------------------------------

Date: Wed, 20 Jan 1999 21:12:53 -0500
From: JoWeber <[e-mail]>
Subject: (fractint) web-site has been moved

Hi All,

I moved my site to

http://www.joweber.de

and added a new page. Cheers --Jo--

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------------------------------

Date: Thu, 21 Jan 1999 10:36:45 GMT
From: "Andrew Coppin" <[e-mail]>
Subject: (fractint) Algorithm for Decomposition

To draw a fractal with binary decomposition, e.i. decomp="n":

  1. Do the usual calculation for the pixel colour exactly as normal.
  2. Instead of using the iteration count for the colour value...
    a) Work out the imaginary part of the final value of Z (or
whatever) divided by the real part.
    b) Take the arc-tangent of this value.
    c) You now have an angle (radians!) between +Pi and -Pi.
    d) Split that range into "n" intervals, and see which interval
contains your angle. The number of this interval is than the pixel
colour.

Simple.
Nam et ipsa scientia potestus est!
Andrew Orphi Coppin
DMU MK.

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------------------------------

Date: Thu, 21 Jan 1999 11:23:37 GMT
From: "Andrew Coppin" <[e-mail]>
Subject: (fractint) Set Union

I'm working with the formula (Z^3) - 3(A^2)Z + B, which is
(apparently) a generalization of all possible 3rd degree polynomials.
I've got Julia sets working fine, but I'm having trouble with the M
set. You see, with the good old quadratic M you must iterate one
point, zero, and see where that goes. However, for this Cubic M,
there are two points to worry about, +A and -A. So there are two M
sets, M+ and M-. I've got formulas for these working just fine, but M
itself is the Union of M+ and M-. How do I do that in FractInt?

P.S. For anyone interested, you can do these Julias with

CubicJulia = {
  A = p1,
  B = p2,
  T = 3*A*A,
  Z = Pixel:
    Z = Z*Z*Z - T*Z + B,
  |Z| < 4
}

For anyone wanting to see the M set (or M+ or M-), there's an
additional problem: they're 4-dimensional!!! I can do slices parallel
to the "A" plane of the "B" plane (B looks best). For example

CubicMandelbrotBP = {
  A = p1,
  B = Pixel,
  T = 3*A*A,
  Z = A:
    Z = Z*Z*Z - T*Z + B,
  |Z| < 4
}

That gives you M+ in the B plane. To change it to the M- set, change
"Z=A" to "Z=-A". To change to the "A" plane, replace "A=p1" with
"A=Pixel" and replace "B=Pixel" with "B=p1".

P.P.S Yes, I *KNOW* that M+ and M- are symmetrical.

- ---------------------------------
Nam et ipsa scientia potestus est!
Andrew Orphi Coppin
DMU MK.

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------------------------------

Date: Thu, 21 Jan 1999 11:50:24 GMT
From: "Andrew Coppin" <[e-mail]>
Subject: Re: (fractint) Set Union

O.K., I've solved it. I've managed to find a quick work arround. If
you use two orbit variables, X and Y, and initalize them to +A and -A
respectivly, then run them both though the normal iteration formula,
then at the end of each iteration assign the one with the highest
magnitude to be the value of Z, then use |Z| < 4 as your bailout.
Behold:

CubicMandelbrotB = {
  A = p1,
  B = Pixel,
  T = 3*A*A,
  X = A,
  Y =-A,
    X = X*X*X - T*X + B,
    Y = Y*Y*Y - T*Y + B,
    if(|X|>|Y|)
      Z=X
    else
      Z=Y
    endif
  |Z| < 4
}

By reversing the ">" in the if() to an "<" you can do intersection
instead. So this method works, but it's sssssllooooowwww. If anyone
has a faster idea I'd just love to hear it.

- ---------------------------------
Nam et ipsa scientia potestus est!
Andrew Orphi Coppin
DMU MK.

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------------------------------

Date: Thu, 21 Jan 1999 10:25:21 -0300
From: "Alejandro Kainer" <[e-mail]>
Subject: RE: (fractint) Set Union

Hi Andrew,

I know that maybe it's nonsense, but have you tried putting A^2 and X^3 and
Y^3 instead of A*A and X*X*X and Y*Y*Y?. Maybe the formula parser process it
faster if it doesn't have to work with three or two variables at the same
time...

Ale

PS: I'm still trying to understand how the formulas work, so forgive me if I
ignore some facts that may result obvious to others.

________________________________
Alejandro Gustavo Kainer
e-mail: [e-mail]
________________________________

- -----Original Message-----
De: Andrew Coppin <[e-mail]>
Para: [e-mail] <[e-mail]>
Fecha: Jueves 21 de Enero de 1999 09:01
Asunto: Re: (fractint) Set Union


O.K., I've solved it. I've managed to find a quick work arround. If
you use two orbit variables, X and Y, and initalize them to +A and -A
respectivly, then run them both though the normal iteration formula,
then at the end of each iteration assign the one with the highest
magnitude to be the value of Z, then use |Z| < 4 as your bailout.
Behold:

CubicMandelbrotB = {
  A = p1,
  B = Pixel,
  T = 3*A*A,
  X = A,
  Y =-A,
    X = X*X*X - T*X + B,
    Y = Y*Y*Y - T*Y + B,
    if(|X|>|Y|)
      Z=X
    else
      Z=Y
    endif
  |Z| < 4
}

By reversing the ">" in the if() to an "<" you can do intersection
instead. So this method works, but it's sssssllooooowwww. If anyone
has a faster idea I'd just love to hear it.

- ---------------------------------
Nam et ipsa scientia potestus est!
Andrew Orphi Coppin
DMU MK.

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------------------------------

Date: Thu, 21 Jan 1999 10:24:14 -0500
From: Paul Derbyshire <[e-mail]>
Subject: Re: (fractint) Set Union

At 11:23 AM 1/21/99 GMT, you wrote:
>I'm working with the formula (Z^3) - 3(A^2)Z + B, which is
>(apparently) a generalization of all possible 3rd degree polynomials.
>I've got Julia sets working fine, but I'm having trouble with the M
>set. You see, with the good old quadratic M you must iterate one
>point, zero, and see where that goes. However, for this Cubic M,
>there are two points to worry about, +A and -A. So there are two M
>sets, M+ and M-. I've got formulas for these working just fine, but M
>itself is the Union of M+ and M-. How do I do that in FractInt?

I've created images for this type of fractal by superimposing images for
both critical points. I even have Fractint formulas to generate such
images; they use color ranges to color various areas...
- --
   .*.  "Clouds are not spheres, mountains are not cones, coastlines are not
- -()  <  circles, and bark is not smooth, nor does lightning travel in a
   `*'  straight line."    -------------------------------------------------
        -- B. Mandelbrot  |http://surf.to/pgd.net
_____________________ ____|________     Paul Derbyshire     [e-mail]
Programmer & Humanist|ICQ: 10423848|

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------------------------------

Date: Thu, 21 Jan 1999 11:22:14 -0500
From: George Martin <[e-mail]>
Subject: RE: (fractint) Set Union

Ale wrote:

>
I know that maybe it's nonsense, but have you tried putting A^2 and X^3 a=
nd
Y^3 instead of A*A and X*X*X and Y*Y*Y?. Maybe the formula parser process=

it  faster if it doesn't have to work with three or two variables at the
same time...
<

The "fast" parser used with floating point math calculates x*x with the
same function as x^2, so there is no speed up here.

George Martin

 =


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Date: Thu, 21 Jan 1999 10:03:47 -0600
From: "Damien M. Jones" <[e-mail]>
Subject: RE: (fractint) Set Union

Ale,

 - I know that maybe it's nonsense, but have you tried putting A^2 and
 - X^3 and Y^3 instead of A*A and X*X*X and Y*Y*Y?. Maybe the formula
 - parser process it faster if it doesn't have to work with three or
 - two variables at the same time...

Not at all--exponentiation is a very expensive process, but multiplication
is relatively cheap. In fact, sqr(A) is faster than A*A.

Damien M. Jones   \\
[e-mail]  \\  Fractalus Galleries & Info:
                    \\  http://www.fractalus.com/

Please do not post my e-mail address on a web site or
in a newsgroup.  Thank you.

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------------------------------

Date: Thu, 21 Jan 1999 14:11:50 -0300
From: "Alejandro Kainer" <[e-mail]>
Subject: RE: (fractint) Set Union

Damien,

OK, I understand that exponentiation is expensive. But you wrote "In fact,
sqr(A) is faster than A*A". How could that be? I mean, I think that the
function sqr() acts the same as A*A, but first the parser has to translate
it. Doesn't it mean an expensive way of doing the same thing too?

Ale

- -----Original Message-----
De: Damien M. Jones <[e-mail]>
Para: [e-mail] <[e-mail]>
Fecha: Jueves 21 de Enero de 1999 13:52
Asunto: RE: (fractint) Set Union


>Ale,
>
> - I know that maybe it's nonsense, but have you tried putting A^2 and
> - X^3 and Y^3 instead of A*A and X*X*X and Y*Y*Y?. Maybe the formula
> - parser process it faster if it doesn't have to work with three or
> - two variables at the same time...
>
>Not at all--exponentiation is a very expensive process, but multiplication
>is relatively cheap. In fact, sqr(A) is faster than A*A.
>
>Damien M. Jones   \\
>[e-mail]  \\  Fractalus Galleries & Info:
>                    \\  http://www.fractalus.com/
>
>Please do not post my e-mail address on a web site or
>in a newsgroup.  Thank you.
>
>--------------------------------------------------------------
>Thanks for using Fractint, The Fractals and Fractint Discussion List
>Post Message:   [e-mail]
>Get Commands:   [e-mail] "help"
>Administrator:  [e-mail]
>Unsubscribe:    [e-mail] "unsubscribe fractint"


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------------------------------

Date: Thu, 21 Jan 1999 12:25:10 -0500
From: Barry N Merenoff <[e-mail]>
Subject: (fractint) Re: Sqr function

The expansion of (a+bi)^2 is (a^2-b^2)+2abi. If this were done with regul=
ar
multiplication, you would multiply a by a, multiply b by b, and then
subtract. However, a faster way to do this, which Sqr probably takes
advantage of, is to factor the real part as (a+b)(a-b). This would give y=
ou
one addition, one subtraction, and one multiplication. Since addition is
cheaper (faster) than multiplication, this is probably how Sqr(z) can be
faster than z*z. Note that this shortcut works only when the two complex
factors are equal, so it couldn't be programmed into the regular
multiplication operation.

Collin Merenoff

PS: Yes, of course, repeated squaring would cause overflow; I should have=

realized that. However, an easy way to avoid this would be to halve each
square until it is within a certain "safety zone." I think this would be
faster than computing the arctangent.

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Date: Thu, 21 Jan 1999 11:59:25 -0600
From: "Damien M. Jones" <[e-mail]>
Subject: Re: (fractint) Re: Sqr function

Collin,

 - The expansion of (a+bi)^2 is (a^2-b^2)+2abi. If this were done with
 - regular multiplication, you would multiply a by a, multiply b by b,
 - and then subtract. However, a faster way to do this, which Sqr
 - probably takes advantage of, is to factor the real part as (a+b)(a-b).
 - This would give you one addition, one subtraction, and one
 - multiplication. Since addition is cheaper (faster) than
 - multiplication, this is probably how Sqr(z) can be faster than z*z.

Actually, the differences in cost between multiplication and addition are
almost negligible on a Pentium, and still minor on a Pentium-II. Unless you
are doing assembly programming, you might as well totally ignore the
difference and treat them as equal (at least for floating-point work;
integers are another matter). And when you DO get to the floating-point
level, other considerations come into play, such as pipelining, register
availability, etc. I actually found in several cases it's faster to skip
this "optimization" entirely and use a^2-b^2, especially if you need
a^2+b^2 for bailout (as you do for the M-set and J-sets).

For architectures where multiplication is still expensive (80x486 and
earlier; 680x0; others, no doubt) this optimization makes more sense.

 - PS: Yes, of course, repeated squaring would cause overflow; I should
 - have realized that. However, an easy way to avoid this would be to
 - halve each square until it is within a certain "safety zone." I think
 - this would be faster than computing the arctangent.

Not so much as you'd think. Figure at least ten cycles per bit for the
squaring method, versus 35 to 150 cycles for directly computing the
arctangent (on a Pentium). You can get eight bits in eighty cycles, which
is faster than arctangent in some cases. If you need more bits, you might
as well compute it directly. And don't forget, you only do this calculation
once per point, not once per iteration; this makes optimizing it
considerably less important.

Damien M. Jones   \\
[e-mail]  \\  Fractalus Galleries & Info:
                    \\  http://www.fractalus.com/

Please do not post my e-mail address on a web site or
in a newsgroup.  Thank you.

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Date: Thu, 21 Jan 1999 11:41:55 -0600
From: "Damien M. Jones" <[e-mail]>
Subject: RE: (fractint) Set Union

Ale,

 - OK, I understand that exponentiation is expensive. But you wrote "In
 - fact, sqr(A) is faster than A*A". How could that be? I mean, I think
 - that the function sqr() acts the same as A*A, but first the parser
 - has to translate it. Doesn't it mean an expensive way of doing the
 - same thing too?

sqr(A) returns the same result as A*A, of course. But as Collin points out,
it is slightly faster because the calculation can be optimized a bit. The
FractInt parser only parses the formula once, it doesn't parse it over and
over. So the cost of parsing sqr(A) versus A*A is only incurred once.

Damien M. Jones   \\
[e-mail]  \\  Fractalus Galleries & Info:
                    \\  http://www.fractalus.com/

Please do not post my e-mail address on a web site or
in a newsgroup.  Thank you.

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Date: Thu, 21 Jan 1999 13:16:06 -0500
From: Paul Derbyshire <[e-mail]>
Subject: RE: (fractint) Set Union

At 02:11 PM 1/21/99 -0300, you wrote:
>Damien,
>
>OK, I understand that exponentiation is expensive. But you wrote "In fact,
>sqr(A) is faster than A*A". How could that be? I mean, I think that the
>function sqr() acts the same as A*A, but first the parser has to translate
>it. Doesn't it mean an expensive way of doing the same thing too?

No, the parser doesn't change sqr(x) into x*x and reparse it...
it parses x*y where x=a+bi and y=c+di as (ac-bd) + (bc+ad)i, four fmuls and
an fadd and fsub. (It doesn't care whether y=x in particular so x*x is
still four fmuls...)
It parses sqr(x) into (a+b)(a-b) + 2abi, which is two fmuls and two fadds
and a fsub. (The 2ab is one fmul and a fadd since, assuming Tim Wegener
wasn't drunk or deranged on the day he wrote that particular line of code,
he has it actually work as ab + ab, since the add is less expensive than
another multiply.)

Since multiplies are the main contributors to cycle-eating in mathematical
iterations, we highlight this as follows: x*x causes four multiples and
sqr(x) only two. sqr(x) may be as much as twice as fast depending on how
much more expensive fmuls are than fadds. This might depend on your
processor of course...
- --
   .*.  "Clouds are not spheres, mountains are not cones, coastlines are not
- -()  <  circles, and bark is not smooth, nor does lightning travel in a
   `*'  straight line."    -------------------------------------------------
        -- B. Mandelbrot  |http://surf.to/pgd.net
_____________________ ____|________     Paul Derbyshire     [e-mail]
Programmer & Humanist|ICQ: 10423848|

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Date: Fri, 22 Jan 1999 01:16:03 -0600
From: "Paul N. Lee" <[e-mail]>
Subject: Re: (fractint) The FOTD website....

Michael Traynor wrote:
>
> Paul,
>
> No surprise, and well deserved.  Congrats.
>

Damien M. Jones wrote:
>
> Paul, Jim,
>
> Congratulations!
>

[e-mail] wrote:
>
> Congratulations Paul and Jim!
>


Thank you (one and all) very much !!!

The comments and praise are most welcome.   :-)

Later,
P.N.L.
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http://www.fractalus.com/cgi-bin/theway?ring=fractals&id=43&go

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Date: Fri, 22 Jan 1999 10:25:10 GMT
From: "Andrew Coppin" <[e-mail]>
Subject: (fractint) Re: Alejandro Kainer's Message

Yeh, it is nonsense. If you read the documents (I can't remember
exactly where it says so) you'll see that the FractInt authors
strongly warn *against* using, e.g. Q^5, and say that Q*Q*Q*Q*Q is
raster (about 4 times faster in my experience). The reason is that
the "^" operator is designed to work with complex number powers as
well, which means it's programmed with an X. X. Complicated Formula.
But never mind, and thanks for the message all the same.

- ---------------------------------
Nam et ipsa scientia potestus est!
Andrew Orphi Coppin
DMU MK.

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Date: Fri, 22 Jan 1999 10:41:48 GMT
From: "Andrew Coppin" <[e-mail]>
Subject: (fractint) MIDI

I had a look on the semi-offical FractInt Wish List, and at the
botton one of the Authors wrote "MIDI support requires someone who
understands MIDI". Well, I do, so what on earth is it going to be
used for in a graphics program?!?!

- ---------------------------------
Nam et ipsa scientia potestus est!
Andrew Orphi Coppin
DMU MK.

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Date: Fri, 22 Jan 1999 09:07:23 -0500
From: George Martin <[e-mail]>
Subject: (fractint) Re: Alejandro Kainer's Message

Ale wrote:

>
Yeh, it is nonsense. If you read the documents (I can't remember =

exactly where it says so) you'll see that the FractInt authors =

strongly warn *against* using, e.g. Q^5, and say that Q*Q*Q*Q*Q is =

faster (about 4 times faster in my experience). The reason is that =

the "^" operator is designed to work with complex number powers as =

well...
<

This is correct. If the power is a real integer constant -1, 0, 1, or 2,
the parser recognizes this and does not run the power function ("^"), but=

instead runs recip, one, ident, or sqr respectively. Any other power,
whether real, imaginary or complex, runs the power function, which is one=

of the slowest in the arsenal. Thus real postive integers greater than tw=
o
are likely to run much faster if expressed as multiplication rather than
power, e.g. X*X*X is better than x^3.


Paul wrote:

>
It parses sqr(x) into (a+b)(a-b) + 2abi, which is two fmuls and two fadds=

and a fsub. (The 2ab is one fmul and a fadd since, assuming Tim Wegener
wasn't drunk or deranged on the day he wrote that particular line of code=
,
he has it actually work as ab + ab, since the add is less expensive than
another multiply.)
<

Paul is right. Here is the ASM code for the sqr function:

      fld          st(0)               ; x x y
      fld          st(0)               ; x x x y
      fmul         st,st(3)            ; xy x x y
      fadd         st,st               ; 2xy x x y
      fxch         st(3)               ; y x x 2xy
      fadd         st(2),st            ; y x x+y 2xy
      fsubp        st(1),st            ; x-y x+y 2xy
      fmulp        st(1),st            ; xx-yy 2xy

Note line 4. This code was written by the eminently sober Chuck Ebbert. :=
)

George Martin

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Date: Fri, 22 Jan 1999 15:14 0000
From: [e-mail]
Subject: RE: (fractint) MIDI

Hi Andrew
> "MIDI support requires someone who
>understands MIDI". Well, I do, so what on earth is it going to be
>used for in a graphics program?!?!

It's OK, I'm dealing with it :-) there have been a number of people
interested in turning the chaotic orbits of  fractal caculation into music
of some sort, and since most modern sound cards include
comprehensive wavetable support I guess folks just want their favorite
program to use their new hardware :-)

Cheers,
        Robin.




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End of fractint-digest V1 #354
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