View source for Easily rearranging arguments without using FA

{{See_also|rearranging arguments without using FA}}

==September 16 2015==
{{irci|Zearen|It's common knowledge that the cluster {{vlapoi|se|te|se}} swaps the second and third position. How do you construct a minimal {{selmaho|SE}} cluster to create an arbitrary place structure.}} 
{{irci|selpahi|I think any method is still too difficult to work for real-time human processing}} 
{{irci|durka42|... permutation groups}}
{{irci|Zearen|durka42: I tried that, but a lot of it's built on orbits and transpositions of the form (k (k+1))<br/>Our building blocks are different.  We use orbits of the form (1 k)<br/>I've shown three identities which are really all you need:<br/>Let {{selmaho|SE}} be the set of {{selmaho|SE}}.<br/>We consider strings of SE and call this set "SE*".<br/>Let Ø be the empty string.<br/>∀ a ∈ SE : aa {{=}} Ø<br/>We can observe that each {{selmaho|SE}} is a function and a permutation, and that a string can be interpreted as a composition in the normal order.<br/>I will now show that for r_k ∈ SE, r_1 r_2 ... r_k r_1 {{=}} r_2 ... r_k r_1 r_2<br/>i ≠ j <{{=}}> r_i ≠ r_j<br/>We call this operation a rotation.<br/>As a concrete example {{vlapoi|se|te|ve|se|broda}} {{=}} {{vlapoi|te|ve|se|te|broda}}.<br/>You can verify yourself, if you wish<br/>Proof-ish thing:<br/>The basic idea is that we're using the 1st position as a sort of swap space to build an orbit.<br/>Whenever we do the first swap we store the first position that spot then we store each successive position in the spot that follows.<br/>Finally, we restore the 1 position and swap the final value from the chain into the starting position, completing the orbit.<br/>Let's say we want the place structure 1 4 2 3 5.<br/>That is 4 -> 2 2 -> 3 and 3 -> 4<br/>In a permutation group, we call such a permutation an "orbit"<br/>We would write it this way: (4 2 3) or equivalently (2 3 4)<br/>We reverse it, and translate it into conversions: «ve te se», then tack on a «se» or a «ve» to restore the position to get {{vlapoi|se|ve|te|se}} or {{vlapoi|ve|te|se|ve}}.}} 
{{irci|selpahi|I'm assuming we'll be using a buffer position to break into new "orbits"?}} 
{{irci|Zearen|Correct.<br/>The reason we reverse is because the order of application is reversed for orbits and {{selmaho|SE}}.}} 
{{irci|selpahi|What about (3 4 2) ?}} 
{{irci|Zearen|Also equivalent.  It's the fact that we can rotate orbits like that that we use to prove (2) :)<br/>That would be «te se ve te», of course.}} 
{{irci|selpahi|This seems exactly the same as rubik's cube BLD cycles, except having to go backwards because of {{selmaho|SE}}.}} 
{{irci|Zearen|So we start with:<br/>1 2 3 4 5<br/>2 1 3 4 5 (se)<br/>3 1 2 4 5 (te)<br/>4 1 2 3 5 (ve)<br/>1 4 2 3 5 (se)<br/>This technique allows us to arbitrarily re-arrange the last 4 places.}} 
{{irci|selpahi|Well this is way easier than doing each step in one's mind.<br/>Only problem is that backwards SE is confusing}} 
{{irci|Zearen|It still takes a bit of mental gymnastics.<br/>At a certain point, just use {{selmaho|FA}} or some {{jvs|ce'u}} binding thing.<br/>We still need to be able to swap into x1.<br/>This is easy though:  There are two cases.<br/>(3.1) The x1 conversion is a place not being re-arranged in the last four<br/>We call a sequence "1-preserving" if c1 {{=}} x1 after the conversion is complete.<br/>We note that for 1-preserving sequences r_1 ... r_n, r_i ∈ SE we can tack a q ∈ SE on either side for free.<br/>Further, all conversions of the form in (2) are 1-preserving.  We already showed this.<br/>So that case is easy.<br/>(3.2)  The c1 is part of the cycle.<br/>This means that there is a position r {{=}} x1 by defintion.<br/>If we tack two conversions for position r onto a side of the conversion, we get an identical conversion (1) with a 1-preserving sub-conversion.<br/>We can then rotate the sub-conversion via (2)<br/>We then apply (1) again to eliminate it.<br/>Let's consider:<br/>2 4 1 3 5<br/>First we change it into a one preserving conversion:<br/>1 4 2 3 5 (te C where C ∈ SE*)<br/>We know how to solve this kind.  It's the orbit (4 2 3).  Therefore C {{=}} ve te se ve<br/>So our conversion is:<br/>te ve te se ve<br/>But wait, we can reduce this.  Let's rotate te (ve te se ve)<br/>te te se ve te<br/>se ve te (1)<br/>Now let's see if it works ^^<br/>1 2 3 4 5<br/>3 2 1 4 5 (te)<br/>4 2 1 3 5 (ve)<br/>2 4 1 3 5 (se)<br/>It works !}} 
{{irci|selpahi|Rotations work on any length of SE right?}} 
{{irci|Zearen|I think this is more of a for-fun thing.  I'm not sure it's practical.}} 
{{irci|selpahi|It's not practical, in that using multiple SE in a row is ugly and rare, but it's more practical than memorizing all the SE* combinations and their resulting place structures}} 
{{irci|Zearen|Kind of.  They have to be of form r_1 ... r_n r_1 and ∀ 1 ≤ i, j ≤ n; i ≠ j <{{=}}> r_i ≠ r_j<br/>I.e. you can't rotate {{vlapoi|te|se|ve}} nor {{vlapoi|se|te|ve|te|se}}.}} 
{{irci|selpahi|Can I rotate '''te se te''' or '''se te se'''? What would be a five-tuple that I can rotate?}} 
{{irci|Zearen|se te ve te se -> se te ve se se te se -> (te ve se te) se te se ->  te ve se te (te se te) -> te ve se te te se te -> te ve se se te -> te ve te<br/>I see an easier way to compute it.<br/>{{jvs|se}} is its own inverse and te ve te is 1-preserving :)<br/>But it's fun to do the arithmetic, too.<br/>Yeah, it's not clear.<br/>te se te can be rotated to se te se can be rotated to te se te etc.<br/>Actually, te se te {{=}} se te se is a fun corollary.<br/>More generally, a,b ∈ SE; a b a {{=}} b a b.<br/>Case a {{=}} b is trivial (a a a {{=}} a a a {{=}} a).  Case a ≠ b is a rotation.}}