[r-t] Bobs-only Stedman Triples - 78 complete B-block peals

[Andrew Johnson]

78 complete B-block peals 

There are some more peals based on combining all the rows of several B-blocks into larger round blocks. If all the sixes were separated then for these peals it would be possible to arrange them into 78 whole B-blocks with some sixes left over, so in the peal all the pieces of all 78 B-blocks are present and complete. This is for the convenience of classifying the peals as the choice of which pair of B-blocks with the same bells behind does not affect this statistic, neither does adding or removing 3 bobs in a Q-set [where the same 3 bells are affected by the call and the other bells are in the same position at each call].

73 round blocks
Here is the first block.
2314567QS---------P--------P-P--------P---------P-P--------PP-------P---------PP-------PP---------PPP--------PP------P---------PP*1(1)

Rotated version of that block: https://complib.org/composition/90582
Signature: 12:1+21
Links the contents of 12 whole B-blocks into 1 big block using 21 plains, giving 84-12+1=73 round blocks including the untouched B-blocks.
This leaves 12 more B-blocks as the pairs of those, and 60 paired B-blocks, where each pair can be chosen from one of six, making assembly of the peal quite easy. There are then 73 round blocks to link. By choosing the B-block pairs carefully extra Q-sets can be found which can be plained to reduce the total number of bobs.
https://complib.org/composition/90595 576 bobs
https://complib.org/composition/85439 576 bobs

There are 73 round blocks formed using 21 plains, so there is the chance of finding 72/2=36 Q-sets to link them into one block, the peal, which would add another 36*3=108 plains, so 129 plains and hence 711 bobs, but I have not found a peal using this block with that many bobs.

75 round blocks
2314567QS---------P------PP---------PPP------P--P--------P--------P---------PP---------PP*1(1)
2341576QS---------P------P--P*1(1)
2761534QS--------P-------P-PP*1(1)

Signature: 12:3+21
This takes the contents of 12 whole B-blocks and puts them into 3 round blocks using 21 plains.
There are then 12 B-blocks which are pairs of those original blocks before they were reconstituted, and 60 paired remaining B-blocks.
https://complib.org/composition/93211 579 bobs

An upper limit of bobs in a peal on this plan is as follows: 21 plains, 75 blocks, so a possible maximum of 840 - 21 - 3 * (75 - 1) / 2 = 708 bobs.

77 round blocks (a)
2314567QS---------P--------P---------PP---------P-P------PP*1(1)
3671245QS---------P------P--P*1(1)
3517246QS--------P-------P-PP*1(1)
2314675QS--------P--P------P-*1(1)

Signature: 11:4+18
Upper limit of possible bobs: 840 - 18 - 3 * (77 - 1) / 2 = 708 bobs.

https://complib.org/composition/93112 555 bobs
https://complib.org/composition/85001 552 bobs (blocks reversed)
https://complib.org/composition/88208 549 bobs

Instead of minimising the number of bobs, we can choose to reduce the number of sets of odd numbers of bobs in a run. This helps maintain the alternation of quick and slow work for the bells. Doing that gives this peal, where the only runs of an odd number of bobs are single bobs, and there are only 18 of them. The 4-bob and 6-bob sets come at the start of the slow for some bell, whereas the  98 of the runs of 2 bobs are in quick, 23 are in slow. The treble does not have any odd bobs behind, just 0,2 or 6 bobs.
https://complib.org/composition/84174
594 bobs
18 isolated bobs
121 runs of 2 bobs
7 runs of 4 bobs
51 runs of 6 bobs

I think this peal would not feel like a Thurstans twin-bob peal as there are so many more bobs, but might be slightly easier on the band than other bobs-only peals.

77 round blocks (b)
2314567QS---------P-------P---------P-P------PP---------P-P*1(1)
2736514QS---------P------P--P*1(1)
2174563QS--------P-------P-PP*1(1)
5263417QS--------P--P------P-*1(1)

Signature: 11:4+18

https://complib.org/composition/82690 582 bobs
https://complib.org/composition/93059 546 bobs
This peal has 30 fewer bobs than any magic block peal.


TBC

Andrew Johnson