[r-t] Grandsire Major
pabs at cantab.net
Sun Jun 27 20:58:21 UTC 2010
Having fixed the direction of two of the three types of B-block, there
is still another type of q-set we can plain. However, if we plain all of
these, we finish up with three types of block that we cannot join with
p 3624578 3
p 4327568 3
p 6478235 1
The problem is that singles have to occur in bobbed leads. It is easy
enough to join up these blocks with pairs of singles, one of which does
not have thirds made before the lead end. To get any further with
standard calls we have to start putting back bobs. I think it ought to
be possible to finish up with a block of 105 leads, with 1, 6, 7, 8 in
all possible positions by adding back three of the five types (i.e.
positions of 6 relative to 7 and 8) of q-set we have just plained. Three
is the minimum as there are four types of Type C block once we introduce
the 6 in different positions - two of these can be joined with a single
at the second lead, but the q-sets available to join the remaining three
blocks to the rest are different in each case.
Another approach would be to produce an irregular composition by
starting with these blocks and adding q-sets of bobs until the remaining
blocks can be joined. Keeping the original set of plains there are only
120 q-sets to consider.
I leave both of these approaches as an exercise for the reader.
Philip Saddleton said on 27/06/2010 21:26:
> Keeping the idea of a
> block with 1, 7 and 8 to ring in each position, there are effectively 42
> possible calling positions, giving seven types of q-set. Starting with a
> B-block with 78 at home, we can immediately rule out those that contain
> elements that come after the lead head, e.g. those with 7 or 8 before.
> Plains at the other five leads all belong to the same type. The sixth
> element is in a block of type 1xxx7xx8. If we have plains at all of
> these q-sets, we join both types of block, and what's more finish up
> with a 5-part block:
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