Title: New numbers set. Post by: FFDiaz on December 28, 2011, 09:35:40 AM Yesterday, in the Spanish section at this forum, I was speaking with Kali about this theory of numbers and I thought that perhaps someone more is interested in take a look.
The part that might be more useful for the generation of 3D Fractals probably is the numeric cycle-basis, an extension of complex numbers using its cyclical properties with the powers,so the first cyclo basis is g=1, g2=g=1 and the period is 1, the second is h=-1, h3=h=-1 and the period is 2,the third one is i, i5=i and the period is 4, the fourth one is j, j9=j and the period is 8 and so on. For a more detailed description see the attached article (isn't full 256 kb only). Greetings Paco Fdez. Title: Re: New numbers set. Post by: DarkBeam on December 28, 2011, 10:45:08 PM can you do a preliminar render of something? would be appreciated
:) Luca Title: Re: New numbers set. Post by: FFDiaz on December 28, 2011, 11:38:47 PM Sorry Luca, I do not have the necessary software to represent it because I have, only works for conventional complex numbers, it would have to develop the code to do this and since the 1990s not code in C.
I will try to do something in Basic but it is a bit crude. Precisely I joined this forum because I know that people very skilled in this type of software development there are and my take me much time to get up-to-date and the short time available to me I need it for my theoretical research. Thanks Paco Fdez. Title: Re: New numbers set. Post by: David Makin on December 29, 2011, 12:25:45 AM Basic or basic-style pseudo-code would I guess be fine for most of the programmers on the forum ;)
Certainly better that the "raw" math for me anyway ;) Title: Re: New numbers set. Post by: DarkBeam on December 29, 2011, 06:25:46 AM Ok I can try with Fragmentarium ... If it will look good will be implemented somewhere :)
Title: Re: New numbers set. Post by: fractower on December 29, 2011, 06:58:58 AM I think you can get there using complex numbers. Some periods have multiple solutions.
g^2 = g; g = 1 Special case I wish would go away. h^3 = h; h = -1 i^5 = i; i = sqrt(-1) j^7 = j; j = [(-1)^(1/3), (-1)^(2/3)] First multi-solution. k^9 = k; k = [(-1)^(1/4), (-1)^(3/4)] Exclude sub harmonic. x^n = x; x = [(-1)^(m*/((n-1)/2))] where m is a positive int less than (n-1)/2 and shares no common factors with (n-1)/2. Title: Re: New numbers set. Post by: DarkBeam on December 29, 2011, 09:24:29 AM Reduced to the bone your paper proposes this law;
float xx = real(zri) float yy = imag(zri) float zz = real(zjk) float ww = imag(zjk) zri = xx*xx + yy*yy + flip(2*xx*yy + zz*zz) + cri zjk = 2*zz*(xx-yy) + 2*ww*flip(xx-yy+zz) + cjk // I may have forgot some ww terms but they stay = 0 in 3D slices so... :) You should already know that all the quaternion-style formulas have already been explored (well, almost... or in any case the result is known) The result is not so satisfying? Or yes? :alien: It depends on what you wished in the end. ;D ^-^ Also tried more recipes; // pacobrot B zri = xx*xx + yy*yy + flip(-2*xx*yy - zz*zz) + cri zjk = 2*zz*(xx-yy) + 2*ww*flip(xx+yy-zz) + cjk // pacobrot B zri = xx*xx + yy*yy + flip(-2*xx*yy - zz*zz) + cri zjk = 2*zz*(xx-yy) + 2*ww*flip(xx+yy-zz) + cjk // pacobrot C zri = xx*xx - yy*yy + flip(2*xx*yy + zz*zz) + cri zjk = -2*zz*(xx+yy) + 2*ww*flip(xx+yy-zz) + cjk The quality is horrible because I used UF with brute force... :dink: Even the holy grail would look awful with that method. Anyway it is realistic at 70% ... So don't hope in miracles ;) Hi!!! Title: Re: New numbers set. Post by: FFDiaz on December 29, 2011, 06:35:39 PM I will respond in order, first to David Makin.
Nothing further from my intention that insinuate that Basic is a little capable programming language, I was referring to what I am able to do with it. I apologize in advance because I'm not speaking English and it is easy to not me express correctly, so my sincere apologies, David. Greetings Paco Fdez. Title: Re: New numbers set. Post by: FFDiaz on December 29, 2011, 08:13:33 PM Now to Luca or Darkbeam as you prefer. ;D
I I am excited to see the first results, particularly in option B. It has been something like to see a son for the first time. Many thanks in advance for your interest and work. I would like to tell you, since I do not know if this is the approach that you have given in your programming, which are my ideas on the graphical representation. If t is a number belonging to RC3 them t=x0·g+x1·h+y·i+z·j. -x0 is in the positive part of the x axis and x1 is in the negative part of x axis. -y is on the y-axis and z on the z-axis. All initial values x0, x1, y and z are positive. In other words, all the coefficients of c are positive in the iteration formula f(t)=t2+c. t2=(x02+x12)·g+(2·x0·x1+y2)·h+(2·y·(x0-x1)+z2)·i+2·z·(x0-x1+y)·j And would to check if diverges the iteration for every number of the space so defined we take as square of the module|t|2=(x0-x1)2+y2+z2 Again many thanks Luca. Greetings Paco Fdez. Title: Re: New numbers set. Post by: FFDiaz on December 29, 2011, 09:05:39 PM And now Fractower.
Is not convenient to think i like sqrt (- 1) or (-1)1/2 because this brings us to i2 could be 1 or -1 depending on the properties of the powers that we use. So if we use the property (a·b)n=an·bn let see: i2=i·i=(-1)1/2·(-1)1/2=((-1)·(-1))1/2=11/2=1 or if we use an·am=am+n:i2=i·i=(-1)1/2·(-1)1/2=(-1)1/2+1/2=(-1)1=-1 Similarly it is not convenient to think of j=(-1)1/4, k=(-1)1/8... I think that the correct interpretation of the successive powers of the representatives numbers of each cyclo-base is given in column 3 of the table that is on page 16 of my article. Many thanks for your interest. Greetings, Paco Fdez. Title: Re: New numbers set. Post by: DarkBeam on December 29, 2011, 10:22:25 PM Ahh! So you did not made a normal quaternion ;D
Please please express your formula in clear basic language using common funcs and symbols! I need that you call x like x, y like y... Use how many symbols as you please... I need a clear transcription to do the translation correctly :D No need to reply in a day but be clear ;) por favor :D Title: Re: New numbers set. Post by: David Makin on December 29, 2011, 11:05:01 PM Ahh! So you did not made a normal quaternion ;D Please please express your formula in clear basic language using common funcs and symbols! I need that you call x like x, y like y... Use how many symbols as you please... I need a clear transcription to do the translation correctly :D No need to reply in a day but be clear ;) por favor :D Actually it's most definitely not a quaternion. The key is that the form concerned has *4* terms for an "observable" 3 dimensional system as I understand it. In terms of 4 separate "dimensions" addition and subtraction are as you would expect and the system is commutative and distributive but not associative(, nor does it have a unique inverse???) Anyway here's the standard multiplication table for up to 5 "dimensions" -> == 4 "observable" dimensions, I give this in terms of r,i,j,k,l as that's how one would normally expect but under Paco's system r and i are not "separately observable" and in fact combine to a single numerical dimension (as in the magnitude calculation): Code: r i j k l For 3 observable dimensions then r,i,j,k are required, addition and subtraction are as you would expect, multiplication follows the above table and the magnitude |xr+yi+zj+wk| is sqrt((x-y)^2+z^2+w^2) if I've understood correctly. Unfortunately as far as I can see there is no definition given for division - though I guess working out the appropriate matrix and inverting it would probably give a form to use though my math is not quite up to that. @Paco - have you a definition of division (for two 4D variables x and y in terms of your g,h,i,j or in terms of the above r,i,j,k) for an "observable" 3d system ? Title: Re: New numbers set. Post by: David Makin on December 30, 2011, 04:19:42 AM Code: r i j k l So the square of (x, y, z, w) would be (x^2+y^2, 2*x*y+z^2, 2*x*z-2*y*z+w^2, 2*x*w-2*y*w+2*z*w) Title: Re: New numbers set. Post by: DarkBeam on December 30, 2011, 09:08:44 AM Now to Luca or Darkbeam as you prefer. ;D I I am excited to see the first results, particularly in option B. It has been something like to see a son for the first time. Many thanks in advance for your interest and work. I would like to tell you, since I do not know if this is the approach that you have given in your programming, which are my ideas on the graphical representation. If t is a number belonging to RC3 them t=x0·g+x1·h+y·i+z·j. -x0 is in the positive part of the x axis and x1 is in the negative part of x axis. -y is on the y-axis and z on the z-axis. All initial values x0, x1, y and z are positive. In other words, all the coefficients of c are positive in the iteration formula f(t)=t2+c. t2=(x02+x12)·g+(2·x0·x1+y2)·h+2·y·(x0-x1)·i+2·z·(x0-x1+y)·j And would to check if diverges the iteration for every number of the space so defined we take as square of the module|t|2=(x0-x1)2+y2+z2 Again many thanks Luca. Greetings Paco Fdez. hmmm, if I understand right x0 is -x when x is lt 0 but when gt 0? 0? for x1 the same problem ;) and y z should be normal? cx cy cz must be preprocessed with abs() Probably? :) Title: Re: New numbers set. Post by: FFDiaz on December 30, 2011, 11:40:29 PM Indeed, these numbers have nothing to do with quaternions.
i is the imaginary number conventional but j isn't the j of quaternions. j^2 is i not -1 as in the quaternions. The real part of the number is constructed with the cycle bases g and h. The coefficient which multiplies to g is the numerical value of an observable of a positive nature (for example, the charge of proton) and the coefficient of h for a negative observable. Therefore, usually the coefficient of g or h is zero depending on the type of observable (this is important when it comes to the programming of the fractal). The square of a number of RC3 is indeed as have written you David (I made a mistake and not wrote the last term of the coefficient of i, but I have now corrected). Indeed, as suspicions David, this is not probably a division algebra, but I have to study it in more detail the inverse of a number, I believe that it is not unique. But although this is very important to mathematicians, the division is not a necessary operation in physics that I am creating. Luca in a while I will try to translate this into relationships that you can include in a program. Greetings, Paco Fdez. Title: Re: New numbers set. Post by: FFDiaz on December 31, 2011, 02:04:24 AM Let's suppose that we begin for the case in which the coefficient of h is zero, that is to say, the coordinate x of every point of the studied space it will be positive. So :
c=cxo*g+cy*i+cz*j I think that it is clear that the points that we have to study belong to a three-dimensional space. We shall define the following variables: cxo //x coordinate of each point in R^3 cy //y coordinate of each point in R^3 cz //z coordinate of each point in R^3 txo //xo coefficient of number t belonging to RC3 (t=xo*g+x1*h+y*g+z*j) tx1 //x1 coefficient of number t belonging to RC3 ty // y coefficient of number t belonging to RC3 tz // z coefficient of number t belonging to RC3 and to calculate the iteration function f(t)=t^2+c txo=txo*txo+tx1*tx1+cxo tx1=2*txo*tx1+ty*ty ty=2*ty*(txo - tx1)+tz*tz+cy tz=2*tz*(txo-tx1+ty)+cz to find out if diverges the iterative process we check if (txo-tx1)^2+ty^2+tz^2< a Title: Re: New numbers set. Post by: DarkBeam on December 31, 2011, 10:50:02 AM Let's suppose that we begin for the case in which the coefficient of h is zero, that is to say, the coordinate x of every point of the studied space it will be positive. So : c=cxo*g+cy*i+cz*j I think that it is clear that the points that we have to study belong to a three-dimensional space. We shall define the following variables: cxo //x coordinate of each point in R^3 cy //y coordinate of each point in R^3 cz //z coordinate of each point in R^3 txo //xo coefficient of number t belonging to RC3 (t=xo*g+x1*h+y*g+z*j) tx1 //x1 coefficient of number t belonging to RC3 ty // y coefficient of number t belonging to RC3 tz // z coefficient of number t belonging to RC3 and to calculate the iteration function f(t)=t^2+c txo=txo*txo+tx1*tx1+cxo tx1=2*txo*tx1+ty*ty ty=2*ty*(txo - tx1)+tz*tz+cy tz=2*tz*(txo-tx1+ty)+cz to find out if diverges the iterative process we check if (txo-tx1)^2+ty^2+tz^2< a Mumble, I can implement a version of this in MB3D using w coordinate instead of tx1, but MB3D checks divergence in the standard way only; x^2+y^2+z^2+w^2< a and not (x-w)^2+y^2+z^2< a the difference is; -2*w*x :-\ anyway... I try to implement it in MMFwip3d at first to see the correct effect. :) If I am able to understand correctly everything that is :police: Title: Re: New numbers set. Post by: weavers on December 31, 2011, 12:44:41 PM Must, Trust, in the force Darkbeam! Do it before the year of the 12! Title: Re: New numbers set. Post by: David Makin on December 31, 2011, 04:40:52 PM <snip> the difference is; -2*w*x :-\ anyway... I try to implement it in MMFwip3d at first to see the correct effect. :) If I am able to understand correctly everything that is :police: Actually even I don't use that usually when testing an idea quickly - I normally use a copy of Ron Barnett's "3D Fractal Raytrace" formula from reb.ufm and simply replace either the QuatJulia or QuatMandelbrot sections with the test code (with DE parameter set to Vepstas) - in this case it'll be much easier to adjust any magnitude calculations in that than in my wip formula ;) Title: Re: New numbers set. Post by: DarkBeam on December 31, 2011, 05:03:41 PM Hey you! I just implemented your Makin mandelbrot sets 1,2,3,4! ;)
The new number set is harder to do I think ^-^ Happy new year :banana: :banana: :banana: Title: Re: New numbers set. Post by: David Makin on December 31, 2011, 10:06:47 PM Hey you! I just implemented your Makin mandelbrot sets 1,2,3,4! ;) Yey ! Though I much prefer the non-standard Mandelbulb types - "Sim.Rot. ZX" etc. - I'm sure some of which haven't been transferred to other 3D software yet. I really must get around to documenting that formula and releasing it properly as a non-WIP one to the formula database !! Title: Re: New numbers set. Post by: DarkBeam on December 31, 2011, 10:55:19 PM I have seen! Some formulas are kimda weird or similar to standard ones some others look good. Maybe I will implement some (what are the best ones?). And there is also Mandelview available in Mandelbulb3d ... found in a forgotten place :D
Title: Re: New numbers set. Post by: David Makin on January 01, 2012, 06:07:11 PM I have seen! Some formulas are kimda weird or similar to standard ones some others look good. Maybe I will implement some (what are the best ones?). And there is also Mandelview available in Mandelbulb3d ... found in a forgotten place :D The best thing I can do there is probably let you make up your own mind: Rucker: http://makinmagic.deviantart.com/art/Out-on-the-Rim-142136488 (http://makinmagic.deviantart.com/art/Out-on-the-Rim-142136488) (old version of code, renders better now) http://makinmagic.deviantart.com/art/Sporocarp-quot-fruiting-body-quot-142383128 (http://makinmagic.deviantart.com/art/Sporocarp-quot-fruiting-body-quot-142383128) Thornton 2: http://makinmagic.deviantart.com/art/Woodfest-142210135 (http://makinmagic.deviantart.com/art/Woodfest-142210135) http://makinmagic.deviantart.com/art/Woodhome-143575363 (http://makinmagic.deviantart.com/art/Woodhome-143575363) Sim. Rot. XYZ (Simultaneous Rotation in 3 angles) http://makinmagic.deviantart.com/art/Hairy-Ball-144652915 (http://makinmagic.deviantart.com/art/Hairy-Ball-144652915) Sim.Rot. ZX http://makinmagic.deviantart.com/art/Terraformed-Mars-144722355 (http://makinmagic.deviantart.com/art/Terraformed-Mars-144722355) Norm.Sim.Rot. XYZ (Norm.==normalised) http://makinmagic.deviantart.com/art/A-Martian-Delicacy-144901891 (http://makinmagic.deviantart.com/art/A-Martian-Delicacy-144901891) http://makinmagic.deviantart.com/art/Under-Red-Skies-145191644 (http://makinmagic.deviantart.com/art/Under-Red-Skies-145191644) http://makinmagic.deviantart.com/art/Centauri-Prime-145239769 (http://makinmagic.deviantart.com/art/Centauri-Prime-145239769) "Nylander" (from an old formula version - a suggestion by Paul Nylander that I tried and can be interesting) My implementation here: Code: magn = (z1=sqrt(magn))^@mpwr Cons.Rot. ZX (consequetive rotations about 2 axes).. http://makinmagic.deviantart.com/art/The-Lost-Isle-145246194 (http://makinmagic.deviantart.com/art/The-Lost-Isle-145246194) Also there are more from about here: http://makinmagic.deviantart.com/gallery/?offset=24 (http://makinmagic.deviantart.com/gallery/?offset=24) In particular Alien Archeology, On the reef (Norm.Sim.Rot.ZX), Limestone Pavement, Shropshire Crags, (Norm.Sim.Rot.XYZ), Assorted Toffee Wrappers (Sim 3D angles), Summer on the Tundra (Sim.Rot.ZX), Temple of the Anointed (Cons.Rot.ZXY) Title: Re: New numbers set. Post by: DarkBeam on January 01, 2012, 06:38:04 PM Well, I arrived at this conclusion;
I would exclude the formulas that give strong "whipped cream" effects, strange stuff in general, or with visual outcome nearly identical to existing ones. :) Thornton formulas are really similar to existing ones, except for thornton2 that is interesting but imo has a strange look. (In fact it uses strange angles...) ;) "cons.rot." fmlas can be obtained with my "General" formula, probably? (It handles all formulas obtainable with two angles) Norm.Sim.Rot. XYZ; I like a lot the visual result of this one so I included it! :beer: Sim.Rot. XYZ - It makes less sense ... and in fact "not Normalized" formulas always look less interesting than the "corrected ones" :D "Nylander" ... i don't know how it looks, the detail you chosen shows strong deformations... :) Sorry but I can't include just everything ;) time constraints and so on :) Title: Re: New numbers set. Post by: FFDiaz on January 01, 2012, 10:50:32 PM I am annihilated before so much deployment of technical resources. I have the sensation of having woken up the attention of the big gurus of Mandelbrotland.
Again, thank you very much for the effort that you are doing for giving form to my ideas. Happy 2012 to all. Paco Fdez. Title: Re: New numbers set. Post by: DarkBeam on January 01, 2012, 11:26:15 PM David is a guru for sure, I am a kiddy ;)
Sorry for the hijackment, wait wait... One day I will reach that brot ;) Title: Re: New numbers set. Post by: David Makin on January 02, 2012, 12:24:10 AM I am annihilated before so much deployment of technical resources. I have the sensation of having woken up the attention of the big gurus of Mandelbrotland. Again, thank you very much for the effort that you are doing for giving form to my ideas. Happy 2012 to all. Paco Fdez. Apologies from me too for going OT in your thread - I must remember that it's possible to use messaging as well as threads ;) At *some* point I will get back to both my Wip3D formula for UF (and try implementing your new numbers and similar ideas) *and* to my semi-updated 3D IFS formula which is halfway updating to being able to render primitives in a similar way to Jesse's dIFS but handles any (affine) IFS "correctly". Then on to a full 3D IFS tree formula that uses DE instead of Hart's method so that any non-linear divergent transforms can be used as well as affine ones. And then......I may be dead ! Title: Re: New numbers set. Post by: David Makin on January 02, 2012, 12:27:44 AM <snip> I would exclude the formulas that give strong "whipped cream" effects, strange stuff in general, or with visual outcome nearly identical to existing ones. :) :) Tempting idea - but some whipped cream can look really good - for instance "Theme Park Ride" (from Paul Nylander's suggestion) is all about whipped cream, but I think it's still a great image - plus of course once converted to glass objects in Bryce or whatever even some Julia quaternions look pretty cool :) Title: Re: New numbers set. Post by: Kali on January 11, 2012, 02:42:29 AM Must, Trust, in the force Darkbeam! Do it before the year of the 12! Yeah... Luca, the force you must use. Do Paco's brot and my mandelbox surface suggestion for tomorrow, or to Yoda I will tell you are lazy Fractal Jedi. ;D ;D ;D Title: Re: New numbers set. Post by: FFDiaz on January 24, 2012, 12:52:13 AM Luca,I have a doubt, when you calculated the attached image now you used (tx0-tx1)^2+ty^2+tz^2 for to find the end of iteration?
Thanks. Paco Title: Re: New numbers set. Post by: DarkBeam on January 24, 2012, 09:08:06 AM No, because I can not modify bailout conditions in MB, as I said this makes your fractal impossible to be ported in Mandelbulb. (Actually)
It can be ported in Fragmentarium for example, but I don't have the time to do much investigation. Syntopia can help you I think. Good luck :) (As the image states, that was a Makin 'brot) Title: Re: New numbers set. Post by: FFDiaz on February 08, 2012, 07:52:53 PM I have changed the structure of RC3 to unify the cycle basis (g) and (h) in a single that would be the conventional real numbers, in such a way that the square of a number z=x0+x1i+x2j which belong to RC3 would be:
z2=(x02-x12)+(2x0x1+x22)i+2x2(x0+x1)j and the module of z would be: |z|=sqrt(x02+x12+x22) with these changes, I made a small program in visualbasic with which I have obtained a series of sections of the Mandelbrot3D set thus generated. As you can see in the attached images in the plane XY(z=0) gets the conventional Mandelbrot set. As we move away from it, the symmetry with respect to the X axis in the direction of the axis Y disappears, although there is a symmetry with respect to the XY-plane in the Z direction. I also include an image of the silhouette of the complete set. The code that I have generated the set is this if it is useful: Public Class mandel3D Dim cx, cy, cz As Double Dim zx, zy, zz, x, y, z As Double Dim iternum, iter As Short Dim bailout, modulo As Single Dim xp, yp As Single Dim col As New System.Drawing.Pen(System.Drawing.Color.ForestGreen, 1) Dim formGraphics As System.Drawing.Graphics Private Sub Comenzar_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Comenzar.Click bailout = 2 iternum = 1000 For cz = -2.2 To 2.2 Step 0.005 For cy = -1.2 To 1.2 Step 0.005 For cx = -2 To 1.2 Step 0.005 modulo = 0 zx = 0 zy = 0 zz = 0 iter = 0 While iter < iternum And modulo < bailout x = zx * zx - zy * zy + cx y = 2 * zy * zx + zz * zz + cy z = 2 * zz * zx + 2 * zy * zz + cz zx = x zy = y zz = z modulo = Math.Sqrt(zx * zx + zy * zy + zz * zz) iter = iter + 1 End While If modulo < bailout Then plotset() Next cx Next cy Next cz End Sub Private Sub plotset() Dim centrox, centroy As Integer centrox = 200 centroy = 200 xp = centrox - (0.8660254038 * cx - 0.7071067812 * cy) * 150 yp = centroy - (cz - 0.5 * cx - 0.7071067812 * cy) * 150 formGraphics.DrawLine(col, xp, yp, xp + 1, yp + 1) End Sub Private Sub mandel3D_Load(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles MyBase.Load formGraphics = Panelplot.CreateGraphics() End Sub End Class From this new perspective I will also explore RC4 and represent only the complex part, now I'll tell. Greetings Paco Fdez. Title: Re: New numbers set. Post by: Syntopia on February 08, 2012, 09:58:43 PM So these are standard complex numbers, with a third component, satisfying: j*j=i, and j*i=i*j=j, right? This will give images like the attached one.
Many people have tried 3-component Mandelbrot system with Cayley table like-multiplication (especially extensions of the complex numbers). I have also tested the most reasonable combinations myself (link (http://"http://blog.hvidtfeldts.net/index.php/2011/09/distance-estimated-3d-fractals-iv-the-holy-grail/")). If you are interested, I used Fragmentarium with the following script - you can see the 3-vector multiplication in the code below, if you want to check it: Code: #include "DE-Raytracer.frag" Title: Re: New numbers set. Post by: Alef on February 10, 2012, 06:21:06 PM Coincidence. I tried this too. But using Chaos pro. It have pretty usefull compiler, some 3D engine, and I alredy have alike formula, so just needed to switch my equations with posted. Could be that there are some mistake in code or in formula as 3D looks at the time producing standart 2D mandelbrot. In 2D Chaos pro just cuts off unused quaternion parts, so write whatever in j and part, and it still will generate mandelbrot. So rewriten code to loop all variables and not just quaternionic z.
Bailout conditions included too. In 2D formula 1 generates mixed inside and outside in real positive side. The colouring aren't the most beautifull, but it nicely accents features. Both formulas produced similar results, what means, that at least there is no accidental mistakes. Much the same as above. All including julia versions have the same streching. Code: FfDiaz (quaternion) {Title: Re: New numbers set. Post by: Alef on February 10, 2012, 06:32:27 PM During implementation I made a mistake, and it generated very interesting 2D, requiring bailout value of 3. Very catholic, throught I would prefared something kosher and 6 sided;) Fractal made of main square set with antenna with crosses and smaller square minibrots around. Like angular mandelbrot. Looks promising, a bitt like kalisets. Bailout value 2 produces vicsek fractal on antenna (jerusalem cross).
Pictures are pretty basic, coz I didn't used any colouring. IMHO this shows more of fractal, even if don't looks so good. Inside are coloured by bof60. This thing needs good tuning, and you'll have new kaliset like angular pattern 2D formula. In attachments I included Chaos Pro fromula file. Just rename it to FFDiazBrot.cfm p.s. This forum part are very deep, probably too deep. Code:
Title: Re: New numbers set. Post by: Alef on February 13, 2012, 03:38:08 PM I meant, way how cyclic numbers are implemented could contain mistake so the streching along y=-x. Maybe conditionals if, then, else, or some other non ordinary aproach...
Don't want to steal the thread or distract from seeking what everyone here are seeking. Here I optimised and tuned formula error I made. Negative equation influences just bailout, and not final z. Allways the squares and crosses. Post function abs create some teuton orden fashioned burning ship. Features revealed by Kali suggested method using low iterations and exp smoothing. (http://www.ljplus.ru/img4/a/s/asdam/Celtic_and_templar_smll.jpg) Maybe the same thing deffined with higher powers would generate 5, 6, 7 sided figures, just that equations of it definetely aren't obvious. It looks that critical point is some x=0.7, which generates the most mandelbrot like thing. (http://www.ljplus.ru/img4/a/s/asdam/Traffic.jpg) Here dgoes Chaos pro: Code: Zuzubrot {Code: Zuzubrot {Title: Re: New numbers set. Post by: DarkBeam on February 13, 2012, 06:44:16 PM Looks like Kali's Mandelbrot on whatever numbers... ;)
Title: Re: New numbers set. Post by: FFDiaz on February 14, 2012, 10:01:07 PM In keeping with the simplification that I already commented of the ciclo-bases (g) and (h), I have represented the RC3 and set some RC4 to some fixed value x and drawing only the complex part. I include images and code in basic with which they have generated.
Code: Public Class mandelRC4 As you can see my ability as a programmer it is rather limited. Best regards Title: Re: New numbers set. Post by: FFDiaz on February 15, 2012, 01:38:38 AM Other also interesting results.
Title: Re: New numbers set. Post by: Alef on February 28, 2012, 05:35:06 PM Result looks somewhat stretched and irregular, indicating that it may have some problems..
Maybe using unused 4D (all 5 number parts) could generate more nice mandelbrot set. At least hypothesis seems to be pretty logical and completely fit with exponent function colouring observation. Title: Re: New numbers set. Post by: FFDiaz on March 18, 2012, 01:29:21 AM I have been exploring RC3 for the iteration function f (z) = z ^ 4 + c which shows the lack of associativity and the cyclical properties of j. I have also improved the subroutine of representation with which I get a stronger 3D effect images. Greetings Paco Fdez. |