[r-t] Bobs only Stedman Triples
Philip Saddleton
pabs at cantab.net
Thu Jun 28 14:11:56 UTC 2012
edward martin said on 27/06/2012 23:07:
> On 26 June 2012 12:51, Robert Bennett<rbennett at woosh.co.nz> wrote:
>>
>
>> > But I think that the bobs must still be used in threes.
>>
>> _I suspect that this is the case. These things are hard to prove. It might
>> be interesting to count up the bobs for Colin Wyld's peal and assign the
>> bobs and omits into complete, incomplete and pseudo q-sets..
>
> I really hope that you will spend time doing that
> As I recall there are 705 bobs (which number IS divisible by 3!)
> I'd be interested if you find any "pseudo q-sets" whatever you mean by that
> As for being hard to prove... I've told you :
A bobs only extent (or any decomposition of the extent into bobs only
blocks) must have precisely one six with each possible permutation of
the back four bells. Take the following sets of six ends:
A B C D
xxx4567 xxx4756 xxx4675 xxx5467
xxx3567 xxx3756 xxx3675 xxx5367
xxx2567 xxx2756 xxx2675 xxx5267
xxx1567 xxx1756 xxx1675 xxx5167
Each member of D must be preceded by either a bobbed member of A or a
plain member of B. Hence #(bobbed A) + #(plain B) = #(D) = 4, hence
#(bobbed A) = #(bobbed B) [=#(bobbed C)]. So of the twelve sixes from A,
B and C, the number bobbed is a multiple of three, but that does not
mean that there is necessarily a logical way to associate them into
"pseudo q-sets".
For instance, in a decomposition into twin bob blocks, in each of the 40
sets of 12 sixends where the seventh is unaffected, six are bobbed and
six plained, but not a q-set in sight.
PABS
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