[r-t] Emergency peal composition

[Philip Saddleton]

Alexander Holroyd said  on 17/12/2012 21:59:
> Indeed.  Mark's proof seems right, but assumes a plain hunt lead end if
> I understand correctly.  Can all 13 leads be got by mixing plain lead
> ends (and lead end orders), and/or using half-lead calls (in methods
> with regular half leads)?

To get all 13 leads we must have all 13 lead ends and all 13 lead heads 
- certain of them happen at calls and the others cannot be paired together.

With half lead calls there is not the same problem, since we can get the 
same pairs to cross at the half lead in many different ways (this is the 
basis of Pipe's classic Cyclic 6 Maximus).

So

  234567890ETAB 8
  ABET907856342 3
  BTA0E89674523 A
  EA9B7T5038264 5
  T0B8A6E492735 Th
  8604T3B527A9E 7
  795E3A426B8T0 9
  08T6B423A5E79 9
  9E7A523B4T608 7
  T0B826A4E3957 Th
  8604T5B7392EA 5
  57496E8A02T3B A
  648507T9BE3A2 3
  4567890ETAB23
13-part
All regular 14th's place methods, with the pivot bell as shown. Th 
denotes T pivoting in first place with a PB half-lead, and a half-lead 
7ths place bob.

Similarly,

5192 Spliced Maximus
  234567890ET Zanussi S
  ET907856342 Maypole LA
  T0E89674523 Maypole A (h)
  648305T729E Maypole A
  57394E628T0 Ariel S
  8604T325E79 Zanussi S
  795E324T608 Zanussi S
  08T624E3957 Ariel S
  9E725T30486 Maypole A
  T028E694735 Maypole A (h)
  648507T93E2 Maypole LA
  4567890ET23 Zanussi S
11-part
(h) is a 16ET half-lead bob

PABS