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FRACTAL FORMULAS
Hey folks!
I am hoping to experiment with some simple fractal formulas in VB. I have some running ok, but some
I seem to be having trouble with. I am using VB 4.0, and am hoping someone out there in cybertopia
might be able to share insights or code snippets? I don't need anything fancy like complex graphic
rendering of the fractals, although if you are excited about this stuff and feel ambitious I'm happy to
check out any graphic stuff you might want to share.
Simply, all I'm looking for is simple number in and number out routines, i.e. you stick a number(s)
into the formula, set the iteration ceiling and then watch what ever numbers get pumped out, thats it really.
So we might have say two text boxes for inputs,
textbox1 = any number value, i.e. 129
textbox2 = any number value, i.e. 234
Then we feed those values into the fractals formula loop and for every interation the newly generated
values get dumped to some other textbox.
=================
Barry Martin - Hopalong
This one is probably one of the most interesting ones, it seems Barry Martin released a really old dos version called hopalong
and there are various exe takes on this floating about, but most of them don't work anymore because they are so old.
I had trouble with this one because I don't really understand the "sign" function, and the VB 4 sign function doesn't seem
to work in a way thats compatible with the formula, but then I'm just as likely ignorant of how to properly apply it.
2 plots resemble organic structures like cells
xnew=y-sqrt (abs(b*x-c))sign(x)
ynew = a-x
different values for constants a,b,c give different patterns and periodicities
========
For the Lorentz function, I'd like to get a really simple rudimentary one working, but I'm also curious about what would happen
if we had two parallel Lorentz functions working together, one accepting outputs from the other and then feeding back into the first one,
i.e.
Lorentz function (1) outputs cycle iteration values to Lorentz (2)
And then Lorentz(2) inputs to Lorent(1) and so on.....
By the way, does anyone know if a single Lorentz function can operate on more than the default # of inputs
Lorentz Function
xnew = x+d*a*(y-x)
ynew = y+d*(x*(c-2)-y)
znew = z+d*(x*y-b*z)
where a,b,c,d are contants, i.e. values of 10,8/3,28,0.003
========
I tried coding this one also, but instead of getting a rich set of chaotic values all I got were two values
that repeated exactly the same over and over again????
Mandlebrot
z ; iterate via ; z^2 + c
==========
For this one I tried implementing this but for some reason, when I coded this to VB, it only generated
the same values over and over again.... no chaotically generated values......
(folds and remaps oval onto itself
)
Michael Henons;
Germaine to stellar orbits
xnew = y + 1 - 1.4 * x^2
ynew = 0.3 * x
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cyndy_northrup wrote: I tried coding this one also, but instead of getting a rich set of chaotic values all I got were two values
that repeated exactly the same over and over again????
cyndy_northrup wrote: For this one I tried implementing this but for some reason, when I coded this to VB, it only generated
the same values over and over again.... no chaotically generated values......
Your problem seems to be a problem to update the old values with the new ones found:
cyndy_northrup wrote: xnew = y + 1 - 1.4 * x^2
ynew = 0.3 * x
are you updating x and y?
'step 1:
xnew = y + 1 - 1.4 * x^2
ynew = 0.3 * x
'store xnew,ynew somewhere
x=xnew;
y=ynew;
'goto step 1
Russell
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yes!
unfortunatelty, i cant seem to get vb to do this properly it just repeats the same numbers?
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ok, but it could be acceptable that the numbers repeats itself, it is only a sequence.
If you want to draw the images related to the fractals I'm thinking than you are not on the right way:
take the mandelbrot formula z(n+1)=z(n)^2+c
you do not have to prepare the sequence of z to draw the image, you have to study its convergence!
Fixing c, start from z=0 then you have to see if z(n) will escapes to infinity. You have to do this for many values of c (complex).
The image is only the real and imag part of c and the color is the first value of n that gives |z(n)|>2.
All this is true if my memory is well working
Russell
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I don't know what your problem is, and I am not going to cover all those fractals.
Mandelbrot needs complex numbers (z = z*z + c needs a complex z and a complex c);
and the Mandelbrot set requires you iterate this formula with some constant c, a
variable initial value z0 of z (that's the point on the drawing), and then you calculate
how many iterations it takes to satisfy abs(z)>=2 to determine the "height" or color
of point z0.
There are several articles on Mandelbrot here on CodeProject; maybe on the others too.
The basic info probably is available on Wikipedia, if not just Google.
BTW: fractals make most sense when you plot them, just showing some numbers is
hiding their beauty for sure.
Luc Pattyn [Forum Guidelines] [My Articles]
this weeks tips:
- make Visual display line numbers: Tools/Options/TextEditor/...
- show exceptions with ToString() to see all information
- before you ask a question here, search CodeProject, then Google
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i have a 16 bit variable(unsigned short).
now i want a best algorithm (function) which will do the following.
1.if contiguous bits at LSB are high then the function should return 1.
for example, consider the following bit patterns
0000000001111111
or
0000111111111111
or
0000000000000001
2.if continuous bits at MSB are high then the function should return 2.
for example,
1110000000000000 and so on.
3.other wise it should return 0 for the examoples such as
1010000000000011
0000011101010011 and so on
thanks in advance.
any more clarifications required, please get back.
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Test the int with this constant variables if you want to not use a for loop with SHIFTs and bit-comparing.
To return 1:
1111111111111110
1111111111111100
1111111111111000
...
1100000000000000
1000000000000000
to return 2:
0000000000000001
0000000000000011
...
0111111111111111
else return 0
Russell
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There might be some nifty bit processing, but this should work with no more than about 6 comparisons (32 bits - 62 options)
8-bit version:
int fn(int x)
{
switch (x)
case 0x01:
case 0x03:
case 0x07:
case 0x0F:
case 0x1F:
case 0x3F:
case 0x7F:
return 1;
case 0x80:
case 0xC0:
case 0xE0:
case 0xF0:
case 0xF8:
case 0xFC:
case 0xFE:
return 1;
default:
return 0;
}
I think compilers are quite good at optimizing this sort of code, but I have never tested it so I'd be interested in comments.
Peter
"Until the invention of the computer, the machine gun was the device that enabled humans to make the most mistakes in the smallest amount of time."
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thanks,
the above logic is simple and works as of now.
but the ultimate goal is to design a function in the following prototype.
int GetWordStatus(int word,int word length)
{
//code here
}
if the word length is certainly 8 or 16 or 32, then teh above logic is appropriate.
but the checking has to be done on the first specified number of bits. i.e. wordlength.
hope i dont need to give any example,
if required, i can.
thanku.
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You should be able to extend this - do something like:
const int hi1 = {0xFF, 0xFE, 0xFC, 0xF8, 0xF0, 0xE0, 0xC0, 0x80, 0};
int GetWordStatus(int word, int wordlength)
{
ASSERT(wordLength <= 8 && wordLength >= 0);
int x = word & ~hi1[wordLength];
switch (x)
{
case 0x01:
case 0x03:
case 0x07:
case 0x0F:
case 0x1F:
case 0x3F:
case 0x7F:
return 1;
}
x = word | hi1[wordLength];
switch (x)
{
case 0x80:
case 0xC0:
case 0xE0:
case 0xF0:
case 0xF8:
case 0xFC:
case 0xFE:
return 2;
}
return 0;
}
It may be a bit inefficient for small wordlengths, if it is a problem you should spend some time and make it better.
-- modified at 6:57 Monday 3rd September, 2007
The reason I am suggesting a switch statement rather than a loop is that I thought the reason that in C/C++ the case values had to be constants was so that the compiler could order them and implement the switch as a simple binary search, so for N case statements there would be only log2(N) comparisons. I can't easily find a reference for this now - does anyone know if this happens?
Peter
"Until the invention of the computer, the machine gun was the device that enabled humans to make the most mistakes in the smallest amount of time."
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cp9876 wrote: I can't easily find a reference for this now - does anyone know if this happens?
Try to look the disassembly window opening it to the code you suggested ...
I think that this shortcut is applyed from the compiler only if the number of case is greather then a specified number ... and only if you turn on that optimizations that is probally related with faster applications, but not applyed in small exe file dimension.
Russell
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In this case I think that the better solution is a general solution...so a loop like
tmpword=word;
for(i=0;i<wordlength;i++){
if(tmpword == 1) return 1;
if(tmpword & 1 ) tmpword >>=1;
else break;
}
tmpword=word;
for(i=0;i<wordlength;i++){
if(tmpword == 0x80) return 2;
if(tmpword & 0x80 ) tmpword <<=1;
else break;
}
return 0;
or somethig similar to this
(I haven't tested the code...it is only an idea of what I'm meaning above... )
Russell
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try this vb code:
Private Function GetWordStatus(ByVal word As Integer) As Integer
Dim significantLength As Integer = (word & "").Length
'Calculate significant Length of number
Dim temp As Integer
temp = (2 ^ significantLength) - 1
If Convert.ToInt64(word, 2) = temp Then
Return 1
Else
Dim cntr As Integer
temp = 2 ^ (significantLength - 1)
cntr = (significantLength - 1)
While cntr > 1
If Convert.ToInt64(word, 2) = temp Then
Return 2
Else
temp += 2 ^ (cntr - 1)
End If
cntr = cntr - 1
End While
End If
' Return 0
End Function
Its not the best but an alternative
Any systematic work reflects its significance for a long time. So let's discuss the best...
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Hi,
it is much simpler than that.
one spec is missing: what if input is zero ?
THEORY (assume unsigned integers)
1.
if all zeroes followed by all ones, then one more becomes a power of two
2.
if all ones followed by all zeroes, then bitwise complement reverts to first case
3.
a power of two has just one bit set; it can be tested in one line:
if (( x & (-x) )==x) then either x = zero or x = power of 2 !!!
CODE
if (x==0) return whatever you think is appropriate
if ((x&(-x))==x) return 1;
x=~x;
if ((x&(-x)==x) return 2;
return 3;
Luc Pattyn [Forum Guidelines] [My Articles]
this weeks tips:
- make Visual display line numbers: Tools/Options/TextEditor/...
- show exceptions with ToString() to see all information
- before you ask a question here, search CodeProject, then Google
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You won
Russell
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I knew there would be a clever bit twiddling answer to this - very neat!
Peter
"Until the invention of the computer, the machine gun was the device that enabled humans to make the most mistakes in the smallest amount of time."
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fn(WORD w)
{
WORD mark=0x8000;
if(w&mark)
{
mark >>= 1;
while( (w&mark) && mark )
mark >>= 1;
if(mark==0)//w must = 0xffff,return 1 or 2
return 2;//also can return 1
if(w&(mark-1))
return 0;
else
return 2;
}
else
{
mark >>= 1;
while( ((w&mark)==0) && mark )
mark >>= 1;
if( (w&(mark-1))==(mark-1) )
return 1;
else
return 0;
}
}
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Hi,
what about...
int GetWordStatus(int word,int word length)
{
if (word == 0)
return 0;
int nHigh = word / pow (2, word_length);
int nLow = word % pow (2, word_length);
if (nHigh > nLow)
return 1;
else if (nHigh < nLow)
return 2;
else
return 3;
}
Greetings.
--------
M.D.V.
If something has a solution... Why do we have to worry about?. If it has no solution... For what reason do we have to worry about?
Help me to understand what I'm saying, and I'll explain it better to you
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There are two sets
one set has ten thousands elements
other set has 30 element
Which loop would be the faster
for(i = 0; i < sizeoffirstset(10000); i++)
{
for (j=0; i < sizeofsecondset(20); j++)
{
if(set1[i] == set2[j])
//do some thing
}
}
for(i = 0; i < sizeofsecondset(20); i++)
{
for (j=0; i < sizeoffirstset(10000); j++)
{
if(set2[i] == set1[j])
//do some thing
}
}
Best Regards,
Mushq
Mushtaque Ahmed Nizamani
Software Engineer
Ultimus Pakistan
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Both are doing the same operations, just in a different order.
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Mushq wrote: sizeofsecondset(20)...sizeoffirstset(10000)
const UINT Sizeoffirstset =sizeoffirstset (10000);
const UINT Sizeofsecondset=sizeofsecondset(20);
for(i = 0; i < Sizeofsecondset; i++)
{
for (j=0; j < Sizeoffirstset; j++)
{
if(set2[i] == set1[j])
}
}
I'm looking for differences, I see only that the initialization j=0 is computed Sizeofsecondset times ... so let Sizeofsecondset be the smaller number.
Russell
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Mushq wrote: for(i = 0; i < sizeoffirstset(10000); i++)
{
for (j=0; i < sizeofsecondset(20); j++)
Ahahah
faster yes .... but to go where?
Russell
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First one is faster and because of internal smaller array.
run following code in VB:
Dim i, j As Integer
Dim strArr1(10000) As String
Dim strArr2(20) As String
For i = 0 To 10000
strArr1(i) = i & ""
Next
For i = 0 To 20
strArr2(i) = i * i * i & ""
Next
Response.Write(DateTime.Now.Millisecond.ToString())
For i = 0 To 10000
For j = 0 To 20
If (strArr1(i) = strArr2(j)) Then
Response.Write("
Hello" & strArr1(i) & strArr2(j))
End If
Next
Next
Response.Write(DateTime.Now.Millisecond.ToString())
For i = 0 To 20
For j = 0 To 10000
If (strArr2(i) = strArr1(j)) Then
Response.Write("
Hello " & strArr2(i) & strArr1(j))
End If
Next
Next
Response.Write(DateTime.Now.Millisecond.ToString())
Any systematic work reflects its significance for a long time. So let's discuss the best...
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Yes: finally the best way to find out who is the faster is use ... the clock
Russell
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Sure, measuring is good, smart measuring is even better.
if you have two alternatives that are likely to be almost equal, whichever you try first
will take longer, due to different starting conditions (maybe code needs to be JITted,
probably data needs to be loaded in memory and/or will be loaded in cache, etc).
Remedy: put everything in a for loop, and run a few passes; ignore the first pass,
take the average (or the best time!) of the remaining passes.
Luc Pattyn [Forum Guidelines] [My Articles]
this weeks tips:
- make Visual display line numbers: Tools/Options/TextEditor/...
- show exceptions with ToString() to see all information
- before you ask a question here, search CodeProject, then Google
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