Apalendromic valid siteswaps?
I’ve just watched this video, which is about cyclic sequences which are different forwards from backwards:
I found it interesting, but it’s constrained to two character sequences, and it got me wondering about other cyclic sequences such as siteswaps.
Some siteswaps are valid forwards and backwards (eg 441 and 144 are both valid) but if you shift the throws left two places those patterns are the same.
Some sequences are different forwards and backwards (eg 531 and 135 are different sequences) but the reversed sequence is not a valid siteswap.
Siteswaps have two important properties, the number of throws and the number of balls... So, what’s the shortest sequence with this property, and what’s the minimum number of balls? I don’t think it can be done with 1 ball, can it be done with 2?
Bonus points for video of you juggling your answer!
I should probably say it’s fairly easy to find the shortest possible period, but the smallest number of balls eludes me at the moment
All period 2 siteswaps are identical under cycling and under reflection, so the lowest possible period is 3. By construction from 111 and from 000, adding distinct multiples of the period, we have 147, 741 for 4 balls and 036, 630 for 3 balls.
This construction technique cannot produce a 2 ball siteswap*.That doesn't mean they don't exist, but I strongly suspect no 2 ball siteswap exists because of the zero problem - reflecting 20x to x02 (x is a variable not a weight 33 throw...) means x must be at least 2. Based on this, Ithink we can prove any such 2 ball siteswap cannot contain the throw sequence 11 at all, and then prove by induction on x that no such 2-ball siteswap exists. Haven’t actually done the legwork for that though..
*[Adding distinct mutiples of the period is the same as adding distinct nonzero numbers of balls. So the lowest possible number of balls for this construction is 0+1+2, i.e. 3.j
No induction needed. First remark that the sequence “x, some number of zeroes, 1” does not reflect. Then for any 2 ball siteswap of finite length k which remains valid when reflected, we find the first occurrence of the highest throw and call it x. If x is 1 then we have the siteswap 1 which is not apalindromic. Assume x is at least 2.
The string of length x following the highest throw is then
x, some number of zeroes, y
By reflectivity, y is at least equal to x and by our initial choice x is at least equal to y, so y = x.
The same logic then goes through for y and the followng string of x characters. After a finite number, i.e. ceiling(k /x), rounds of this same argument we find that the reflective siteswap must be precisely x00...0. This is reflective, but is identical with its reflection, so it is not apalindromic.
All of which actually that no apalindromic 1 ball siteswap exists instead. You’d thik I could count to 2 by now!
That seemed like something they would have had to go out of their way to model, surely it would have been easier on the animators if his ears had rotated with his head... but perhaps that doesn’t match the mickey style guide somehow?
Ehm, why can't it be done with 1 ball?
210 is a 1 ball siteswap. 012 is not.
Am I missing something?
Or were we looking for valid apalindromic siteswaps?
1 ball shortest sequence:
I found it by picking up two balls and trying... And then checking on paper. Didn't realize at first that I have done the 3-ball versions before. (52413, 53142)
.. playing on this, it turns out, that it works with these 5b versions too (added 2 to each digit in 53142):
[brainstorming] Must there be some distinct sequence of digits or numbers (( that calculates and could be shown in a graph, as a curve? )) to mirror a siteswap on, for it to be apalindromic?
Or is that wanting to use too much mathematics on a phenomenon that rather genders from the given orbits? .. see: in 52413 , the 5 has an own orbit (00500), while the other two balls each (shifted a few beats) do 0201300400, each on their side in matters of symmetry in this case. In 53142, there's similarly an own 5-orbit, with 03002 and 00140 ow orbits changing sides thereby here also keeping the symmetry of the whole.
.. but, is this orbits-thing part of the cause, or else an effect of this apalindrome?
Of course, it will work adding any number of balls using that method, since adding one to each number always gives a new valid siteswap with one more object. :)
Actually... Adding an object by adding the length of the siteswap to one of the numbers would also work. Some siteswaps could end up being the same forwards and backwards using this method (the 036 mentioned earlier in the thread for example), but not this one.
Little Paul asked for shortest siteswaps and fewest objects though, and while subtracting could work as well as adding it doesn't work if one number goes below zero. As John R showed there is no apalindromic 1-object siteswap.
It seems like length 5 is the shortest apalindromic 2-object siteswap. It could probably be proven, too, but I just checked since there isn't really that many 2-object siteswaps of length 3 or 4.
no, because 02 is the same as 20 if you rotate it by one place - so they're the same sequence.
This is true for any 2 digit siteswap.
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