[r-t] Treble Dodging Royal (Universal Four Part)
J. J. Bissell
j.j.bissell at cantab.net
Tue Jul 12 18:16:30 UTC 2016
Dear All:
Strictly speaking I don’t think it is possible to produce a bobs-only four-part peal of Treble Dodging Royal, but the effect can be achieved on a `three-courses-per-part’ plan by inserting a course to bring up a suitable half-way part-end. For example, last week I found the following:
------------------------------------------------
5,040 Treble Dodging Royal (Universal Four Part)
23456 M W H
--------------
42356 -
25634 x -
43652 - - -
--------------
Four part : call x = MW in Parts I and III only.
True to all CPS methods in LHG b, c1, & f.
------------------------------------------------
which is based on a 24365 half-way part-end, and can be reversed for tenth place methods using eighth place bobs:
------------------------------------------------
5,040 Treble Dodging Royal (Universal Four Part)
23456 V O I
--------------
45236 - -
53462 - -
34562 x -
--------------
Four part : call x = IV in Parts II and IV only.
True to all CPS methods in LHG l, k1, & g.
------------------------------------------------
An attractive feature of these arrangements is that each contain eight LB courses in full (including all five LB5 courses); this means that the LB music content is fairly respectable regardless of chosen method, while coverage of incomplete LB courses can be maximised by selecting methods in LHG f and g. I’ve posted two examples, `5040 Bristol Surprise Royal (Universal Four Part)’ and `5040 Yorkshire Surprise Royal (Universal Four Part)’, at
http://bb.ringingworld.co.uk/comps.php?machine_comp&stage=10 <http://bb.ringingworld.co.uk/comps.php?machine_comp&stage=10>
I think that this result is interesting, and raises a few further questions. In particular, if restricted to the following plan:
(a) universal four-part structure (conventional bobs-only)
(b) half-way part-end 24365 or 65432
(c) all five LB5 courses in full
how many other compositions can be found? I’m sure that somebody good with machine proofs can answer this question quite quickly, as the search space should be small.
I would be especially interested to know of a composition for the lead-head groups not covered by my arrangements above.
Best wishes,
John
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