# [r-t] An easy challenge

matthew at frye.org.uk matthew at frye.org.uk
Mon Jul 8 16:48:22 BST 2019

```Quoting Graham John <graham at changeringing.co.uk>:

> On Mon, 8 Jul 2019 at 10:22, Roddy Horton <roddy at horton.karoo.co.uk> wrote:
>> 6 bell composition is not my field and I have been asked to come up
>> with a 720 or 1440 of the following treble bob but non palindromic
>> method.
>> x36x16x12x16x12x16x12x16x12x16x34x16  = 142563
>> Can anyone help please?
>
> This is far from an easy challenge! It is straightforward to get a
> 2880 with 4 extents - any calling that has all 120 possible leadheads
> will do it, such as this:
> https://complib.org/composition/40288?substitutedmethodid=36085&methodaccesskey=15d3e1bf5be8039e4caeaa417910d5a946c0d3cd
>
> But a shorter length that is true will require a hand-crafted solution.

It's not too bad, actually. Due to the pattern of odd/even rows, you
only need to get all 60 in-course leadheads, which will give you a
true 1440. Unfortunately, Q-set rules mean you can't join these 12
courses with just normal bobs.

Using singles actually makes things harder to figure out, as it messes
up the parity of the rows. However, if you add an extreme call (12)
you can get around the Q-set issue while keeping parity.

What I came up with is:
F  I  O  W  H
-  -  3
-  3
-  -  e
2
e
-     -  e
2
e
-=14 e=12

This could be trivially rotated to put the block with extremes near
the start if you prefer.

(Health warning: this will have lots of 65's a back, but that's
inevitable when you ask for this kind of thing.)

(Also, I worked this out with pen and paper, and haven't run it
through any kind of computer checker. I'm relatively confident on this
sort of thing, but mistakes happen, so it's worth checking yourself.)

MF

```