[r-t] Plain Bob leadheads

Philip Saddleton cantab at saddleton.org.uk
Tue Feb 23 15:01:25 GMT 2021

On Fri, 2021-02-19 at 09:14 +0000, Mark Davies wrote:
> Please can someone produce, or point me at, a concise proof of this
> lemma:
> PB1: In a palindromic treble-dominated method with Plain Bob
> leadheads, 
> on an even number of bells, the lead end place notations must be 12
> or 1n.

A shorter proof:

Suppose place P is made at the lead end, and the place bell order is A,
P, B, ... . Since the method is palindromic, the bell becoming Bth's
place bell at the lead head is in Ath's place at the lead end. If the
course has more than two leads, then A and B are adjacent positions. If
the method has Plain Bob lead heads, then P lies mid-way between A and
B in the Plain Bob lead head order. There are two possibilities:

{A, B} = {2m-1, 2m}: then P=2
{A, B} = {2m, 2m+1}: then P=n

More generally, for any palindromic method, if we label the vertices of
a polygon with the place bell order for any cycle that includes a bell
that pivots, and select any change that is an apex of the palindrome,
the lines joining the pairs that cross at that change are parallel. 

This allows us to create a pseudo coursing order for short course
methods: once we have fixed a cycle containing a pivot bell, select any
unlabelled vertex for a bell that has not yet been assigned, and there
is only one way to label the bells that it crosses with at each of the
apices that satisfies the above rule. 


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