[r-t] Crambo - mystery solved(?)

Richard Smith richard at ex-parrot.com
Fri Sep 8 13:42:24 UTC 2006

Alexander Holroyd wrote:

> It _is_ possible on 6 (and I would guess on all higher
> numbers).

Yes, it is possible on all higher numbers.  The proof for
n>6 is below, which together with your example for n=6 and
my earlier negative proof for n<6 covers everything.

> Here is an example (which I think it is safe to
> say no-one in their right mind would ring).

Cool.  Unfortunately it's not clear to me that the obvious
generalisation works; however, this has given me enough of
an insight into the problem to constuct a similar 7 bell
method that I *can* generalise to n bells.

> p = 234.345.

This is Erin Minimus with 5,6 covering that has been
'Orpheused' by inserting 1234 between each change meaning
that the plain course contains all 48 rows with 5,6 in 5-6
in either order.

(Tying in with the earlier part of the thread, it is also a
crambo based on Erin Minimus; but instead of Erin with 5,6
covering, it is a crambo of Erin with 5,6 dodging behind.)

Clearly there are 5x6/2 = 15 mutually-true courses that
between them include the whole extent.

This can be readily generalised.  Take any (n-2)-bell
extent, add two covering bells, and 'Orpheus' it by
inserting the change 123..(n-3) between each pair of actual
changes.  The plain course of this method has the property
that any one row can be removed leaving a legal touch.

We now have n(n-1)/2 mutually-true courses including the
whole extent.  Unfortunately it is not obvious if and how
these can be joined.  On five bells, for example, we know
they cannot be joined whilst retaining the property that any
row can be removed leaving a legal touch (because no touches
with this property exist).

> - = 234.345.

Ander's touch works having a 1234->1236 bob.  In order that
both the plain and the bob leads produce legal touches when
an arbitrary row is omitted, the bob must occur between two
changes with place made in 4-5-6 -- in Ander's example,
between a 3456 and a 1456.  This could easily be generalise
on higher numbers to a 1...(n-3)(n-2) -> 1...(n-3)n bob
inserted between two changes with places in (n-2)-(n-1)-n.
(And we can easily prove that for n>5, there exists an
(n-2)-bell extent with three or more blows behind.)

Unfortunately, that's as far as I can take a bob-based
solution as I cannot prove that in general we can find
suitable Q-sets to join the courses.  (I imagine this is
why Ander has just asked the circumstances under which an
odd number of courses can be joined by bobs, though on 4n
and 4n+1 bells, we have an even number of courses.)

Let's instead start with a 7-bell method based on Erin


The front four bells ring a whole-pull extent of Erin whilst
the back three ring whole-pull plain hunt.

If we introduce a 12345->12367 single between changes with
places in 4-5-6-7.  E.g.

  p = 2347.3456.2345.456.2347.3456.2345.456.2347.3456.2345.56
  s = 2347.3456.2367.456.2347.3456.2345.456.2347.3456.2345.56

As this Q-set has order 2, we can definitely join all the
courses together.

This generalises to n bells.  Start with any (n-3)-part
extent (that has at least three blows behind) and ring this
whole-pull on the front (n-3) bells.  Add whole-pull Plain
Hunt on the back three.  As there are suitable starting
extents on four or more bells (no three bell extents have
three blows behind), this proves the existence of extents on
any n > 6 that are still legal touches if an arbitrary row
is omitted.


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