[r-t] Bobs only Grandsire

[edward martin]

I had discovered these three blocks but didn't take them so far as you
did ie by bobbing half the Gs
mew

On 6/14/06, Philip Saddleton <pabs at cantab.net> wrote:
> 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
> >
>
>
>
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