[r-t] Plain Bob leadheads

Philip Saddleton cantab at saddleton.org.uk
Sun Feb 21 09:27:41 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.
> Thank you!

Here's a counterexample:


But assuming you mean that the apices are at the half lead and lead
end, and there is more than one lead in a course, we have:

a) n-1 lead course
Suppose 2 is in 4th's place at the mth lead head. Draw a polygon with
2(n-1) vertices, label these alternately with the position of 2 at lead
end and lead head, stepping m leads between each pair. Thus the lead
head vertices are in PB coursing order. Since the method is
palindromic, the lead ends are in the reverse of the coursing order.
Each pair must be adjacent positions or the same, so 2 is paired with
either 2 or 3. The pairs are then either (2,2), (3,4), (5,6)... or
(3,2), (5,4), (7,6)... i.e. 2nd's or nth's place lead ends.

b) m lead course, 1 < m < n-1, (m odd)
Suppose bell p makes a place at the first half lead. Then as the method
is palindromic, the places that p occupies at the lead ends are the
same as those at the lead heads. There are m of these, equally spaced
around the (n-1 bell) PB coursing order, and since m is odd these
include the lead end place. This is not p, otherwise the course is one
lead long, so the set includes p-1 or p+1. Label the n-1 vertices of a
polygon with the PB coursing order and the vertices corresponding to
the m positions have symmetry about any line from the centre to the
mid-point between two of the vertices. If these two vertices are
adjacent positions the line of symmetry passes through either 2 or n.


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