[r-t] Plain Bob leadheads
Alan Reading
alan.reading at googlemail.com
Sat Feb 20 21:10:15 GMT 2021
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
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