[r-t] Plain Bob leadheads

[Alan Reading]

On Saturday, 20 February 2021, Ian McCulloch <ianmcc at physics.uq.edu.au>
wrote:

>
> On Sat, 20 Feb 2021, Martin Bright wrote:
>
>
>>
>> On Sat, 20 Feb 2021 at 17:18, Philip Saddleton <cantab at saddleton.org.uk>
>> wrote:
>>       On Fri, 2021-02-19 at 15:33 +0100, Martin Bright wrote:
>>       > Proof:  It's enough to prove it in the case h = (2 4 6 .. n n-1
>> ... 5
>>       > 3) in cycle notation.
>>
>>       Is it? It depends whether 'Plain Bob Lead Heads' means all of them.
>>       It's not true for methods with a 1-lead course, but what if n-1 is
>> not
>>       prime?
>>
>>
>> I was assuming “Plain Bob lead heads” meant a course of n-1 leads.
>> You’re right that this is the point at which it doesn’t work in other cases.
>>
>
> But what actaully happens in Royal with 3-lead courses?  A method with
> plain-bob lead ends, 3-lead plain course, and not 12 or 10 lead end change
> would be quite interesting - a novice band can learn 3 leads and ring a
> plain course, and it has some theoretical interest too.  Is this possible?
>
> Cheers,
> Ian
>


My suspicion is that if you take h to be either of the two none rounds
plain bob lead heads you actually have to have for a 3 lead course of Royal
(namely 1795038264, 1860492735) the rest of Martin's argument can still be
made to go through but I've not checked this...

It is possible to get a two lead course of Triples where the lead head
notation is not 1 or 127 (the equivalent thing for odd stages): eg
7.1.7.123.7.1.347,145
So care does need to be taken with courses shorter than n-1 leads for sure.

Cheers,
Alan
-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://bellringers.org/pipermail/ringing-theory/attachments/20210220/3040629e/attachment.html>