[r-t] Bobs only Grandsire

[Philip Saddleton]

Try having x for B. This means the second block is rung backwards, and 
rules out x for A, C, D, F. We can't have x for E, but can for G. This 
gives three blocks:

1xxxxx78 AB AC DB DG DE EE CG CB CD AB AB AB (5 part) = 960
1xxx7xx8 FC AD FG FG FG (4 part) = 320
1x8xxxx7 EE EC FG FD (2 part) = 128

(leads ending B or G are plained, the others bobbed).

If we are to join this set of blocks together by calling bobs, precisely 
half of the G Q-sets will have to be bobbed (we need one of the two in 
the 128 to be bobbed, and one of the two in the third bob block to be 
plain). The obvious way to do this is do it according to the nature of 
the rows. Magically they all join up:

2016 Grandsire Major
   2345678
   -------
   8726543  2
   3485627  2
   2537468  1
   -------
   6728543  1
   4863752  1
p 4257368  3
   -------
   6748235  1
   3865724  1
   2534876  1
   7426583  1
   3875624  2
p 3426578  3
   -------
   8735624  2
   6483275  3
   7365428  1
   -------
   2578346  1
   4826537  1
   3647852  1
   5732684  1
   4856237  2
   3647825  1
   2735684  1
   8524763  1
   6483572  1
   7362458  1
   -------
   4876532  3
p 4235678  3
   -------
3-part

There are 20 of these blocks making up the extent. Can we join them by 
bobbing Bs?

PABS

edward martin said  on 14/06/2006 14:10:
> On 6/12/06, Philip Saddleton <pabs at cantab.net> wrote:
>> With an even number of bobs in a Q-set, there is no obvious reason why
>> bobs-only extents of Grandsire shouldn't exist at even stages. For
>> Minor, an exhaustive search rules it out in a matter of seconds, but is
>> there a simple proof that it is impossible? It seems that with each
>> Q-set forcing four B-blocks to be rung in a particular direction there
>> is not enough room for manoeuvre.
>>
> 
> That about sums it up I think
> I don't have a simple proof or any proof at all really, but if you
> have nothing much to do then read on:
> The block from 21354768 neg to 51728364 pos is plain hunt. The choice
> of linkage to treble at lead is either x (which will retain the flow
> of pos neg) or 3-8 (which will alter the flow)
> What I did was set out the three bobbed blocks that would allow 1,7 &
> 8 to occupy every possible positional relationship (ie from LHs
> 1*****78;  1**7***8; & 17*****8) noting only the rows with treble in
> 1-2. I then labeled each member of the 7 possible q-sets A-G inclusive
> for example the first bobbed block ran:
> 1*****78 = A
> *1***7*8
> ------------
> *17*8*** = B
> 1*78****
> ------------
> 17**8*** = C
> 71*8****
> ------------
> 81*7**** = D
> 18**7***
> ------------
> 1*87**** = A
> *18*7***
> ------------
> *1***8*7 = B
> 1*****87
> ------------
> 1****8*7 = A
> *1****87
> ----------------
> and so on
> The three bobbed blocks were thus represented:
>>From 1*****78 to 1****7*8 : A B  C D  A B  A B  A C  D B  A B
>>From 1**7***8 to 1***7**8 : G F  G D  B C  G F  G F  D A  C F
>>From 17*****8 to 1*7****8 : C G  F D  E E  E C  F G  D E  E E
> 
> Now (apart from the initial lead which starts AB the others could flow
> in either direction, However, because we are ringing Grandsire,
> whichever direction they flow, the lead qset would have to be 3-8 The
> lack of maneuverability becomes apparent when we start assigning PN to
> the various qsets
> Thus out of the choice of 3-8 or x because we start from rounds with
> 3-8, then all members of qset A have to be 3-8  Note that if two
> members of the same qset occur consecutively (as with E in the third
> bobbed block) then since we are ringing Grandsire the whole set has to
> be 3-8
> I found no solution without using singles
> I had tried the same technique with Grandsire Minor & again found no
> solution without using singles
> I suppose that all this does is perhaps demonstrate the difficulties
> of hoping for a comp such as that by CK Lewis which got 40320 Bob
> Major by calling 3bobs at fifths
> 
> mew
>