[r-t] Question about shortest link methods - does truth make a difference?

[Alexander E Holroyd]

Dear Ed,

This sounds like a promising approach, but I don't understand part of 
your argument.  Suppose we can establish your interesting claim:

For any two changes X and Y such that X.Y contains a long place (in 
other words such that X and Y share a place), there is a true 3-change 
block U.V.W that achieves the same permutation.
(By the way, I think you intend U.V.W to have no long places, and I 
don't see why it needs to be true).

Then how do you know you can substitute U.V.W for X....Y in the method 
that achieves Dp(R) without creating long places?  Perhaps U or W 
creates a long place in conjunction with the place notations before and 
after X and Y in the method.  Am I missing something?

However, I like your idea of considering a minimal counterexample.  Can 
we somehow argue that there is a shorter counterexample, by omitting 
irrelevant parts before and after the false section?

Ander

On 04-Apr-20 12:48 AM, Edmund Mottershead wrote:
> I believe you can expand your argument for Dt to Dp.
> 
> Suppose Dp(r)=Dpt(r) does not hold for all r.
> So there exist some rows r such that Dp(r) < Dpt(r).
> Let us take the row from this set such with the shortest link method.
> Call this row R and we have that Dp(R) < Dpt(R) and for all r such that 
> that is true, Dp(R) <= Dp(r)
> 
> If we prick out Dp(R) we will find some row k which appears twice.
> So in the place notation for Dp(R) we will have a piece of notation X 
> such that for a row l, X(l) = k
> In addition we will have another piece of place notation Y and another 
> row m such that Y(k) = m
> 
>  From the definition of Dp(R), the concatenation of X and Y must create 
> a long place, else we would just omit the intervening rows.
> However I claim that for any X.Y creating long places we can devise a 3 
> change, true method section accomplishing the same result.
> 
> This method section will be shorter than the false section leading from 
> k to k, since there must be at least two changes between the ks, eg 
> l,k,*,*,k,m
> as opposed to l,*,*m.
> 
> This is a contradiction, so Dp(r) = Dpt(r) for all r.
> 
> The tricky part is rigorously showing how to do your trick of getting 
> from 34.14 to 12X14 for all concatenations of place notation.
> If one of the changes has bells making places in adjacent pairs then we 
> can replace that change with the 'opposite' change, then the cross 
> change, eg 36.56 becomes 36.1234X.
> If both the place are split, then it seems that you can rewrite the 
> changes to be on a smaller sets of bells, eg 16.36 becomes 14X1256.
> I can't get my brain to think about the equivalent scenarios for odd 
> numbered bell methods at this time of night, but I think that it works.
> 
> Best,
> Ed
> 
> 
> On Fri, Apr 3, 2020 at 10:23 PM Alexander E Holroyd <holroyd at math.ubc.ca 
> <mailto:holroyd at math.ubc.ca>> wrote:
> 
>     Let r be a row, and consider possible sequences of non-jump changes
>     that
>     get from rounds to r.  Let D(r) be the minimum possible length of
>     such a
>     sequence.  So for example,
>     D(1234)=0
>     D(2314)=2  (the unique minimum length sequence is 34.14)
>     D(4321)=4  (one minimum sequence is plain hunt, -14-14).
> 
>     Now let Dp(r) be the minimum length if no bell is allowed to make 3 or
>     more consecutive blows in the same place.  So for example
>     Dp(2314)=3 (which is larger than D(2314)=2)
>     (34.14 is not allowed because 4 makes 3 blows in 4ths, but we can do
>     12-14 instead).
> 
>     Also let Dt(r) be the minimum length if the sequence of rows must be
>     true, i.e. no row is allowed to be repeated.
> 
>     Finally let Dpt(r) be the minimum length if it must be true AND have no
>     long places.
> 
>     Obviously for every row r,
>     D(r)<=Dp(r)<=Dpt(r) and
>     D(r)<=Dt(t)<=Dpt(r)
> 
>     In fact it is easy to prove that for every row r
>     D(r)=Dt(r):
>     if some minimum length sequence is not true, we can just omit the rows
>     between two identical rows to get a shorter sequence, a contradiction.
> 
>     We saw above that for some rows r (e.g. 2314)
>     D(r)<Dp(r)
>     and because both sequences are true, this same example also has
>     Dt(r)<Dpt(r).
> 
>     My question is:
>     Does Dp(r)=Dpt(r) hold for every r?
>     Or is there a row r for which Dp(r)<Dpt(r)?
>     In other words, if forbid long places, is there a row that you can get
>     to quicker if you allow falseness than if you don't?
> 
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