►

Re: Pi as a siteswap.
11 Apr 2005 13:42:42 GMT
t...@jugglethisBUTNOTTHISBIT.net.nospam (Schwolop)

f(5) is 1 for a start. 31416 is jugglable... tis fun too! (Of course, this way of writing whatever it is you’re considering writing seems to assume you’ve ignored the decimal point... If you want to juggle fractional objects, it’s easiest to use rings, cut out the required proportion (for which purpose, knowing pi is quite useful!) and then juggle with the decimal number rounded down worth of extra objects. Of course, you almost certainly didn’t wish to know that, and I can’t for the life of me think why I bothered to write it...)

Tom - definitely in procrastination mode!—Schwolop

i’ve examined the first 10000 digits of pi. the longest siteswap in it is 13 digits long—strach

Did you just keep looking for longer siteswaps until you couldn’t find a 14 digit one or did you examine the whole thing for longer siteswaps? Are you certain there isn’t a 15 digit one, for example?—Rory

i checked the existence of siteswaps up to 50 length. i doubt if there are any longer ones in first 1000 digits. i wonder how far is the first 100 digits long siteswap in Pi.

f(n) looks as O(some very high exponent)—strach

Since you can’t have siteswaps with two digits n places apart where the first digit is the second digit plus n[1], it should be easy to find a lot of such pairs in whatever section of pi you’re looking at, so limiting the maximum length of a siteswap to the largest distance between these pairs. Actually it would be from the digit after the first digit in the first pair to the digit before the second digit in the second pair.

For example in 3.1415926535897932384 (twenty digits) there can be no ^^ ^^ siteswap of period > 8 because it would have to include 65 or 32, which aren’t allowed.—Rory

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

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

grrr you are right. I neglected identical SSs with a shifted starting throw [1] because JL doesn’t print them out.

-andy

[1] such as 51 and 15—adremeaux

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

f(14) = 293144 This is a 4 ball siteswap 84018257113637 f(15) = 394541 This is a 5 ball siteswap 362867229569307 f(16) = 756836 This is a 5 ball siteswap 4707528448389317—Guy

I’d call it 0, but then I’m prone to zero-based indexing. I’d also call it wrong because 6 only appears there if you round up. That digit should be a 5.

f(1) = 0 (3; 3 props) f(2) = 0 (31; 2 props) f(3) = 1 (141; 2 props) f(4) = 12 (9793; 7 props) f(5) = 8 (53589; 6 props) f(6) > 50, I can’t be bothered to find it. I might explore this programmatically in the future.—Rory

Well, one of the main conjectures with Pi is that it is a random number. This means that every sequence of digits that you can imagine repeats infinitely often in the digits of Pi. If this conjecture is true, then any proper siteswap of any length appears in Pi infinitely often. Now, that doesn’t answer your question regarding the position of the first n-length siteswap, but I hope it helps otherwise.—RPN