[r-t] Bobs only Stedman Triples

[Philip Saddleton]

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