# [r-t] Bobs only Grandsire

edward martin edward.w.martin at gmail.com
Wed Jun 14 13:10:17 UTC 2006

```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

```