Juggling Pi as a siteswap.—strach
Re: Pi as a siteswap.
11 Apr 2005 15:28:53 GMT
kuba.straszew...@poczta.fm.nospam (strach)

f(5) is not 1 unfortunately   pi is 3.14159...   f(1) = 1. This is a 3 ball siteswap 3.  f(2) = 1. This is a 2 ball siteswap 31.  f(3) = 2. This is a 2 ball siteswap 141.  f(4) = 13. This is a 7 ball siteswap 9793.  f(5) = 9. This is a 6 ball siteswap 53589.  f(6) = 247. This is a 4 ball siteswap 190914.  f(7) = 120. This is a 5 ball siteswap 4709384.  f(8) = 461. This is a 6 ball siteswap 96274956.  f(9) = 3890. This is a 4 ball siteswap 783123552.  f(10) = 3889. This is a 4 ball siteswap 4783123552.  f(11) = 6048. This is a 5 ball siteswap 53779914037.  f(12) = 2599. This is a 5 ball siteswap 858900971490.  f(13) = 119. This is a 5 ball siteswap 6470938446095.

and of course I ignore the decimal point. i don’t what the 3.14 as real number has to do with siteswap? I will download more digits of pi, and try to find possibly longest siteswap there. Anyone else to try?—strach

Yay!, nice thread :-)

How then are siteswaps distributed in a long series of random digits ?—Peter

Similar, and a question i havent seen answered here before:

If one takes a random sequence of digits m, what is the probability that a subsequence of length n from a random starting point is a valid siteswap?

I guess the answer to that would be as simple as finding the total amount of siteswaps of a given length and dividing it by 10^n.

So then the question is more like how many valid siteswaps are there is a given length n?

-andy—adremeaux

i can only answer easier question:

given a random sequence of n numbers(not digits) what is the ppb, that this sequence is a valid siteswap. as every number mod n is equally propable. than we chose the numbers randomly. the first number can be chosen any way. the second can’t interfere with the first so it is (n-1)/n ppb , that it won’t be the reason that the sequence is not a siteswap. the third cant interfere with first, and second, so (n-2)/n

the ppb, that the given sequence of n random numbers is a siteswap is then n! / n^n.

but when you have digits only, than whole thing become much more compicated.

on the other side, when we consider siteswaps containing only digits there are only 2 possible siteswaps (zeros and ones)

so the longer alphabet(more digits), the bigger chance of finding siteswap?—strach

Sorry, that is not correct. There are 30 valid siteswaps with period two (I juggling lab’d it), out of a possible 100. The formula gives 50% (2/4).

The reason it is not correct is because it doesn’t take into account the need of only whole values for balls. That is, it contains patterns which are valid for x.y balls. I’m not quite sure how to fix it, though.

-andy—adremeaux

i don’t say it is always correct. this is the case when every number is equally possible - not every digit. This is the same as the case, when you can only pick up numbers beetween 0..n-1, where n is the length of a siteswap. and every such zumber is equally probable.

however There are 50 valid siteswaps with period two (I juggling lab’d it), out of a possible 100. (in a case i was talking about there are 4 possible siteswaps(00, 01, 10, 11)and only tow of them are correct)

xy - where x and y are both odd or both even is a siteswap. exactly 50% of two digit strings are siteswaps.

  00 02 04 06 08 - valid siteswap 01 03 05 07 09 invalid  11 13 15 17 19 - valid siteswap 10 12 14 16 18  20 22 24 26 28 - valid siteswap 21 23 25 27 29 invalid  ............—strach

The longest in the first 1000,000 digits is period 17 - there’s only 1.

  f(17) = 343721. This is a 4 ball siteswap 05268611514962570.

If anyone can get hold of more than 1000,000 digits then you can put them in a text file and use my program to find more: <atlas.walagata.com>—Peter

I’m currently in the process of doing just that. I’ve found a period 20 ss somewhere around the 8000000 mark, but since I’ve got 100,000,000 digits to play with, I would expect to find more. When I’ve got all of the results I’ll post them up here, along with the location fo the source file (which I can’t currently remember since I had to reboot into Windows to use the program). It has the first 4.2 billion digits if anyone has a really fast computer, fast connection and lots of time on their hands... Guy—Guy

I’ve written my own program now and it’s currently searching for all of the siteswaps between periods 14 and 30 in the first 100 million digits of pi. When it’s done, it’ll spit out a file with every ss between these periods, in order of where it occurs, with the number of balls. I can’t imagine it’ll take less than a few hours to process it all though.   Stay tuned. Guy—Guy

f(18) = 1928549 This is a 5 ball siteswap 538936820638552953.

I don’t know why I’m waisting my time on this really. I’ve updated my program to search a range of periods and it’s now much faster. <atlas.walagata.com>

I still can’t find more than 4 million digits though. I found the site with 4.2 billion but I think my firewall doesn’t like the ftp or something. It’s also stored as 2 digits per byte, which means I’d have to write another program to convert it to 1 digit per byte.—Peter

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

I haven’t done the others yet, but I might get round to it. An interesting one is the number of n period siteswaps in a given number of places for pi. "What would that look like?" I hear you ask. Well, in the first 100,000 digits of pi, here are the frequencies of the different siteswap periods:

  n frequency  1 100000  2 48910  3 22496  4 9328  5 3842  6 1673  7 642  8 260  9 95  10 37  11 6  12 4  13 5

A graph of which can be found here: <ntlworld.com> You’ll notice that the log graph is very linear (except at one end, but that’s because the frequencies get smaller - this would be fixed by examining more digits).—Guy

Yes, another power law. And with a gradient apparently close to -8/9 ... Does the gradient converge towards some value as we do more digits ?

I wonder how this depends on the 0..9 range of our digits ? Pi is presumably working as just a good source of random digits here.—Peter