Sierpinski triangle revisited

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Forum Attachments

Summary:

Seeking help with a new algorithm to draw the figure

Sierpinski triangle revisited

runez 10 Aug 2008 11:45

Hi, so I might have found an alternative way to draw the Sierpinski triangle without using the chaos game.
As the chaos game is actually just an algorithm used to plot an IFS a different algorithm could also be used. Quote from wiki:

"An alternative algorithm is to generate each possible sequence of functions up to a given maximum length, and then to plot the results of applying each of these sequences of functions to an initial point or shape."

I know this could be done with recursive programming in an ordinary programming language. But wouldn't it also be possible to make a program on the TI that would do something similar?
I have a couple of ideas but im having a hard time structuring them. Since the functions in the IFS are all linear (simply distance function divided by two every time) it should be possible to calculate a point at a given level without doing all the iterations.

Any ideas are welcome and please ask if theres something cryptic in my text :p

Re: Sierpinski triangle revisited

DarkerLine 10 Aug 2008 15:09

If I'm reading this correctly, you want to write the algorithm that will, instead of making a random decision at every iteration, make all three possible decisions (creating a list of points)?

Re: Sierpinski triangle revisited

runez 10 Aug 2008 17:16

Exactly :)

Doesnt have to create a list of points though - I was thinking it could just plot them right away.

Re: Sierpinski triangle revisited

builderboy 10 Aug 2008 17:47

But then don't you choose all 3 option from all 3 new points?
And then choose all 3 new options from all 3 of the new points
and so on?

that would require you to store which points you will have plotted.

But i don't see why everyone seems to do it this way, when the xor way "Points down to next post" works just as well.

Re: Sierpinski triangle revisited

runez 10 Aug 2008 19:54

Simply to see if its possible..

Re: Sierpinski triangle revisited

builderboy 10 Aug 2008 19:58

thats true
:)

Re: Sierpinski triangle revisited

DarkerLine 10 Aug 2008 21:40

You might use plots to graph the points. Using lists seems to limit you to 6 iterations (3^6=729), though.

Re: Sierpinski triangle revisited

Zaphod Beeblebrox 11 Aug 2008 02:49

If you truely want to remove the randomness factor, try this:

:0→Xmin:1→ΔX
:0→Ymin:1→ΔY
:{0,1,0→L1:47→S
:For(A,0,64
:augment({0},seq(sum(L1,N,N+1),N,1,A+2
:2fPart(.5augment(Ans,{0→L1
:For(N,2,A+3
:If L1(N
:Pt-On(N+S,62-A
:End
:S-.5→S
:End

I just whipped this up real fast so if could probably use some optimization but the concept remains. Basically it creates each row of Pascal's triangle and takes it mod 2 and then outputs the values that are 1 (true).

Last edited on 11 Aug 2008 02:49 by Zaphod Beeblebrox Show more

Re: Sierpinski triangle revisited

builderboy 11 Aug 2008 03:06

that works really well!

Re: Sierpinski triangle revisited

DarkerLine 11 Aug 2008 03:20

That's not any different from the xor method, really.

Re: Sierpinski triangle revisited

builderboy 11 Aug 2008 03:32

its true, because of the way psacalls triangle works.
each 2 numbers on a row are added and the result is put below.
odd are black, even are white, you get serpinkis triangle.

adding 2 numbers is like xor because
even even = even
odd odd = even
even odd = odd
odd even = odd

just like

1 1 = 0
0 0 = 0
1 0 = 1
0 1 = 1

Re: Sierpinski triangle revisited

builderboy 10 Aug 2008 16:05

I once made a program that created a very nice Sierpinski triangle.
It started by putting a pxl in the top left corner.

Then you run a program that takes every singe pxl on the screen, and XORs it against the pxl to the right of it, and stores the result below the first pxl (1 is on).

Eventually, after every pxl has been tested, you will get a Sierpinski triangle.

Last edited on 10 Aug 2008 17:45 by builderboy Show more

Re: Sierpinski triangle revisited

runez 10 Aug 2008 17:19

Do you think you could get your hands on this program (post it etc.)? :P

Caus' from what youre writing I dont really see how it would function..

Re: Sierpinski triangle revisited

builderboy 10 Aug 2008 17:44

AxesOff
ClrDraw
PxlOn(0,1)

For(F,0,61)
For(L,0,F+1)
If PxlTest(F,L) xor PxlTest(F,L+1)
PxlOn(F+1,L+1)
End
End

That should do it.

Last edited on 10 Aug 2008 17:44 by builderboy Show more

Re: Sierpinski triangle revisited

runez 10 Aug 2008 19:54

Gives a DOMAIN error at the If test.

Re: Sierpinski triangle revisited

builderboy 10 Aug 2008 19:57

are you sure you set the for loops correctly, i just checked it and it works fine.