Date: February 24, 1995
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The first peal of Stedman Triples was rung in 1731. Until now, all peals of Stedman Triples have had a least two singles. It has been a long standing question as to whether a peal could be obtained with just common bobs alone. This article examines the nature of the problem, the difficulties to be overcome and the steps taken to arrive at a solution.
The are $5040/6 = 840$ possible calling positions in a peal of Stedman Triples. Each of these calling positions can possibly have a bob, so a simple search would require examining $2^{840}=7.3×10^{252}$ possible touches. This is a huge number, far in excess of the number of atoms in the universe, so a direct search is out of the question.
An extent of Stedman Triples is 5040 changes, which is 60 whole courses, or 84 bobbed courses. A sequence of 3 bobs (for plain courses) or 3 omits (for bobbed courses) on the same three bells links 3 courses or blocks of courses into 1 block.
A Q-set is a set of calls affecting the same group of bells so forming a round block, such as 3 bobs at home for Plain Bob Major.
Consider the following six ends, which are followed with either a bob or an omit.
2314567 2314675 2314756
------- P ------- P ------- P
3241657 3241765 3241576
2314567 2314675 2314756
------- Bob ------- Bob ------- Bob
3241576 3241657 3241765
If an omit is rung after 2314567, then the next row is 3241657. This means that
a bob can not be rung after row 2314675, because otherwise the same row would be
rung again. Therefore, if row 2314675 is rung at a six end it must be followed
by an omit, so row 3241765 must also be rung. This means that six end 2314765
can not be rung with a bob, so must be rung with an omit.
If a bob is rung after 2314567, then the next row is 3241576. This means that an
omit can not be rung after row 2314756, because otherwise the same row would be
rung again. Therefore, if row 2314756 is rung at a six end it must be followed
by a bob, so row 3241765 must also be rung. This means that six end 2314675 can
not be rung with an omit, so must be rung with a bob.
The set of six ends 2314567, 2314675, 2314756, must therefore be rung either all
with a plain following, or all with a bob following, or at most two members of
the set can be present. This set is an example of a Q-set, and is why bobs come
in threes, affecting the same three bells, in compositions containing Q-sets of
six ends. Similarly for compositions based on bobbed courses, the omits come in
threes, affecting the same three bells.
These Q-sets contain an odd number of elements. If there was only one way of ringing a row then the Q-set rule would imply that courses have to be linked using Q-sets, joining an odd number of courses into a smaller odd number. As there are an even number of courses to start with, and an extent is all rows in one block (an odd number), this gives the basis of the proof that an extent of Grandsire Triples requires singles.
Fortunately for Stedman Triples, a row can be obtained in six ways, three in a quick six and three in a slow six. The Q-set rule does not disprove the existence of a bobs only peal, because the third element of a Q-set could be obtained elsewhere in a six.
The greatest number of mutually true plain courses of Stedman Triples is 40. The fact that you can not find 60 plain courses containing all the rows in an extent was shown by W.H.Thompson. This means that you can not take plain courses, link them by 3 bobs and hope to get a peal.
These use pairs of consecutive bobs, at positions 3,4 (=S, in slow); 5,6 (=H, first half turn); 7,8 (=L, last whole turn) and 12,13 (=Q, in and out quick). Peals based on Thurstans’ block are the most common example of this type of peal.
I investigated this type of peal, and found using computer searches several nice half peals which could be linked with singles to give a true peal. Linking the halves with twin bobs failed, with 12 changes repeated.
A row can be categorised as either odd or even depending on whether it takes an odd or even number of swaps of adjacent bells to return the row to rounds.
With Stedman Doubles, exactly two pairs of bells are swapped in each change in a plain course. This means that as a plain course starts with rounds, an even row, every row must also be even. A plain course of Stedman Doubles has 60 even rows. An extent of doubles contains 120 rows, of which 60 are odd and 60 even. The other 60 rows are odd, and the only way of getting to them is to swap an odd number of bells over for one change. This change with an odd number of bells being swapped is a ‘single’, because only a single pair of bells are swapped. The odd rows can then be rung. A further single is required to switch back to even rows, so that rounds can then be arrived at.
This is why Stedman Doubles requires pair(s) of singles to ring the extent.
With Stedman Triples, exactly three pairs of bells are swapped each change. This means that the rows generated are alternately odd and even, and there is no requirement for singles to generate odd rows.
Singles are useful however to join a pair of blocks together, rather than three blocks. This is why conventional compositions uses singles; to link an even number of blocks into an odd number of blocks, so then three bobs or three omits can be used link three blocks into one, reducing the number of blocks by two each time until finally there is one block, the actual peal.
Peals based on bobbed courses (B-blocks) are another major class of Stedman compositions, for example Slack’s two-part. The front work of a bobbed course is that of Stedman Doubles, while the back bells dodge for 60 changes as 10 bobs are called, with 30 changes one way round, and 30 changes the other.
Henceforth, course will mean bobbed course. Bobbed courses are easier (for the composer, not the conductor) to work with, because they are true against all courses with other bells behind. As odd and even rows alternate if no singles are used, they are true against all courses with the back bells swapped. My investigations of B-blocks revealed several ideas for bobs only peals, most of which proved unsuccessful, but did give some ideas for conventional peals with singles.
For a given pair of bells at the back with a bobbed course, there are 5 front bells. This does not mean that there are $5!=120$ possible different courses. If there are no singles, then the end of a course must be an even row, so that restricts number of different courses to 60. Each course can be started in 5 different places, by moving on two sixes. This gives 12 different courses.
By ringing the front work backwards, the same bells on the front will come up back and hand instead of hand and back, and so will come up with the back bells the other way around. These two possible courses are mutually true, and can be termed partners of one another.
For a composition, once a course is decided, there is no choice for the partner course, so there are only six possibilities for a pair of courses that can be chosen for each permutation of bells at the back.
If there was some way of combining all the rows in an even number of courses into an odd number of blocks, then this would be a significant step in obtaining a bobs only peal of Stedman Triples. This is because then together with the unused courses there would be an odd number of blocks containing all the rows in an extent. These blocks could then in theory be linked using Q-sets so as to form just one block, containing all the rows, which would be a peal. Similarly, it would be useful if there was a way of combining an odd number of courses into an even number of blocks.
If the front work is considered as Stedman doubles, and the changes on 23145 as
a six end are written out, and the changes on 34125 are written out, then it can
be seen that the rows in sixes 5,6,7 of the first course are also in the other
course in sixes 7,5,6.
This was discovered just using pencil and paper, whilst investigating how falseness between courses is distributed, by choosing a course which is clearly false against the other in the first six. Falseness incidence between courses is important because the Q-set rule suggests that a peal can not be obtained with whole courses alone.
Full details of sixes 5,6,7
41253 B 21354 B
----- 5 ----- 5
14523 H 12534 H Same rows as
41532 B 21543 B in 7th six of
45123 H 25134 H course 23145
54132 B 52143 B
51423 H 51234 H
15432 B 15243 B
----- 6 ----- 6
51342 H 51423 H Same rows as
53124 B 54132 B in 5th six of
35142 H 45123 H course 23145
31524 B 41532 B
13542 H 14523 H
15324 B 15432 B
----- 7 ----- 7
51234 H 51342 H Same rows as
15243 B 15324 B in 6th six of
12534 H 13542 H course 23145
21543 B 31524 B
25134 H 35142 H
52143 B 53124 B
----- 8 ----- 8
25413 H 35214 H
The rows occur at handstroke in both courses or at backstroke in both courses.
This means that all the rows in the course 21345 can be rung by ringing sixes
1,2,3,4,8,9,10 of course 23145 and sixes 5,6,7 of course 34125. By symmetry,
sixes 5,6,7 of course 42135 could be used instead of sixes 5,6,7 of course
34125.
The common rows occur at handstroke in both courses or at backstroke in both courses. This means that if instead of a course of Stedman Doubles, a bobbed course of Stedman Triples is rung, then the same arrangement of front bells in the two courses will occur with the back bells the same way around, so the whole row will be the same.
A way of switching in and out of bobbed courses at the 5th or 8th six is to put an omit at those positions. The same bell makes 5ths at calling position 5 or 8, so if there are omits at 5 and 8, a third omit at 8 gives a round block.
Calling positions
Course end 1 2 3 4 5 6 7 8 9 0
2314567 AAAAAAAAAAAAAA-EEEEE
2614573 CCCCCCCC- BBBBB
2714536 DDDDD-
The course end is the six end that would result if the course was allowed to run until just before calling position 1. The order the sixes are rung is indicated by the letters.
This gives a block in which one course (2314567) is rung in its entirety, a
course (2614573) in which 7 sixes are rung and a course (2714536) in which 3
sixes are rung.
Having taken these three course, the gap in course 2614573 could be filled in
trivially by ringing sixes 5,6,7 from the same course. This is not very useful,
because the course would have to be entered from 2314567 at calling position 5,
which has already been rung.
However, there are other ways of ringing these rows. The missing 5,6,7 sixes in
course 2614573 could be rung by ringing sixes 5,6,7 from course 6412573 or
course 4216573. These two courses are the course 2614573 with the bells in the
same starting positions that bells 234 in course 2314567 transposed as 234 to
342 or 234 to 423.
If course 6412573 is chosen for example, these courses can be written out:
Calling positions
Course end 1 2 3 4 5 6 7 8 9 0
2314567 XXXXXXXXXXXXXX-XXXXX
2614573 XXXXXXXX- XXXXX
2714536 XXXXX-
6412573 XXXXX- fills in
course 2614573
Now the part of course 6412573 can be generated as part of another round block.
There is a gap in one of these generated courses, so this process can be
repeated.
Consider the following 5 sets of two-course blocks, with omits at 8,5,8.
X = part of course rung
# = part of course rung by rows
from another course elsewhere
% = part of course which fills in
a gap in a course rung
elsewhere
The corresponding course is on the right.
Calling positions Other
Course end 1 2 3 4 5 6 7 8 9 0 course
2314567 XXXXXXXXXXXXXX-XXXXX
2614573 XXXXXXXX-##### XXXXX (6412573)
2714536 %%%%%- (4217536)
6712534 XXXXXXXXXXXXXX-XXXXX
6312547 XXXXXXXX-##### XXXXX (3216547)
6412573 %%%%%- (2614573)
3416572 XXXXXXXXXXXXXX-XXXXX
3716524 XXXXXXXX-##### XXXXX (7613524)
3216547 %%%%%- (6312547)
7213546 XXXXXXXXXXXXXX-XXXXX
7413562 XXXXXXXX-##### XXXXX (4317562)
7613524 %%%%%- (3716524)
4617523 XXXXXXXXXXXXXX-XXXXX
4217536 XXXXXXXX-##### XXXXX (2714536)
4317562 %%%%%- (7413562)
This gives a set of 5 two-course round blocks, with a cyclic relationship between them, so each block has part of a course filled in by part of the next block, and the last block has part of a course filled in by the first block.
When this process has been completed, it is seen that there are five round blocks, which contain all the rows in the following ten courses, and no other rows.
2314567 6712534 3416572 7213546 4617523
2614573 6312547 3716524 7413562 4217536
We have therefore managed to reduce an even number of courses to an odd number of blocks. This is the essential step which makes possible a peal of Stedman Triples with bobs only. I termed these five two-course blocks the ‘magic blocks’ because of this. After writing the first draft I learnt that Colin Wyld discovered this step three years before me, and that Brian Price worked out these blocks some weeks after I discovered them. Colin then used this technique last autumn to produce the first bobs only extent of Stedman Triples.
If we also consider the 10 one-course blocks which are the partners of the special ten courses, then that gives 15 blocks containing all the rows in 20 B-blocks. To get a peal these 15 blocks need to be linked with the remaining 64 B-blocks using Q-sets of omits. This is a reasonably well known, though not always easy technique. The set of 15 B-blocks to be linked is as follows.
2314567 or 2614573; 3215467 and 6215473
6712534 or 6312547; 7615234 and 3615247
3416572 or 3716524; 4315672 and 7315624
7213546 or 7413562; 2715346 and 4715362
4617523 or 4217536; 6415723 and 2415736
Where there is a choice of courses, either course can be linked into the touch with a set of three omits. The other course need not be included, because it will then be linked in as above, using omits at 8,5,8.
There is then a free choice for the order of the front bells for the other $84-20=64$ B-blocks, to give a set which can be linked together. In fact once one course has been chosen, its partner is fixed, so this gives a choice of $64/2=32$ pairs of courses. There are 6 possible pairs of courses with each combination of back bells, which gives $6^{32}=7.9×10^{24}$ possibilities. Although this is big, this is nothing like the $2^{840}$ possibilities considered earlier.
There are $15+64=79$ blocks to be linked together. Three blocks may be linked into one by three omits on members of a Q-set. This means that at least $(78/2)×3=117$ omits are required to link 79 blocks into one. There are also $3×5=15$ omits in the 5 magic blocks. This gives at least $117+15=132$ omits, so at most $840-132=708$ bobs in a peal on this plan.
There are two places in a bobbed course where the same three bells are affected by an omit. This can be used to have a round block using six omits and not three.
For example:
Calling positions
Course end 1 2 3 4 5 6 7 8 9 0
2314567 A-EEEEE-IIIIIIIIIIII
2314675 G-BBBBB-FFFFFFFFFFFF
2314756 D-HHHHH-CCCCCCCCCCCC
The letters show the order in which the parts of the course are rung.
This can be done for any block of three, providing an omit does not occur in a missing part of a partial course in a magic block, so allowing the bob count to be reduced in multiples of three.
This would allow up to another 117 omits, giving a total of 591 bobs. As the omits can be added independently for any set of 3 omits, the number of bobs can be varied in multiples of three from 591 to 708. There are therefore perhaps $2^{39}$ trivial variations on any possible peal. As a peal could be started in any quick six, this multiplies the variations by 480, and by another 2 if the peal is rung backwards.
This gives an estimate of $2×480×2^{39}=5.3×10^{14}$ variations of any peal generated. Ringing them all would take about 10 times the current age of the universe, surely enough for any peal tour.
When linking courses together, it is easier to write out the ends of courses. Transposing bells as the result of omits at certain positions is easily done.
Omit at 1 (or 4) 2314567 -> 2314675
Omit at 7 (or 10) 2314567 -> 2316574
Omit at 5 (or 8) 2314567 -> 2614573
Omit at 3 (or 6) 2314567 -> 6314572
Omit at 9 (or 2) 2314567 -> 2364571
The same course is given by the end of any quick six, so the following courses are all equivalent:
2314567 3451267 4125367 1532467 5243167
The partner of course 2314567 is course 3215467. These courses are all
equivalent:
3215467 2541367 5134267 1423567 4352167
In general, if a partner of an existing course in a block is linked into the
block, then several other partner courses can easily be linked in as well.
This is done by applying three omits on the same three bells on the partner
course as on the original course. The additional blocks linked will also be
partners of existing courses. This process can be repeated until all the
partners are obtained.
I succeeded in linking 11 of the 15 blocks in 2280 or 2400 changes, and 13 of the 15 blocks in 3000 changes by linking in two new courses using sets of 3 omits, but could not link all 15 blocks.
The linking of 11 of the blocks was done by hand, after writing out all the five equivalent six ends for each course, to make any similarity between courses more obvious. The five whole courses in the magic blocks were linked, then one of the partners of split course of the magic blocks was linked in. This meant that by the rule for linking partners of blocks, starting from the split course in a magic block I could apply the same omits to it as was applied to the partner of the split course which was already linked in. This linked all the partners of all the courses so far, apart from partners of the four split courses in the magic blocks. The four blocks to be linked were partners of courses in the magic blocks. The magic blocks were already connected into the main part of the composition.
The unused courses were not sufficient to link the partners into the main block, as one course was required to be used in different ways for several of the partners.
A variation of the original main block allowed one of the magic blocks to be linked in via one of the split courses. This meant that the whole course in the magic block was the course without a partner, and that as different unused courses were available linking the unlinked partners was not obviously impossible.
Linking in single courses without a partner is difficult without leaving another course linked in without a partner. This occurs because courses must be linked in twos.
One way of solving this problem of linking the partner courses called say A,B,C,D was to link A to the partner of B, which was already in the main part of the touch. By symmetry B could then be linked to the partner of A, also in the main touch. This links in two of the courses. Having done this I could not then link the other two courses, because there were not enough useful B-blocks left.
Philip Saddleton took a more systematic approach to linking courses. He chose by hand which courses to link, but had computer assistance in listing which other courses would not be accessible if certain choices were made.
The aim was to find a set of 42 pairs of courses, including the courses making up the magic blocks, with as many Q-sets of omits as possible available to link the courses together. Philip maximised the connectivity between courses by where possible completing Q-sets where two members were already present, thus only adding one course. The partners were linked with complementary Q-sets. He also made sure there was always a potential exit point from each block. When this was finished all the courses has been chosen, and there were 22 pairs of Q-sets of omits. The courses were then connected by choosing all available Q-sets of omits and the 15 omits used in the magic blocks. Two Q-sets could not be used, since they contained an element in the substituted sixes of the magic block. Adding all possible omits gave seven blocks. These blocks were then joined by replacing four Q-sets of omits with bobs. After considerable work, Philip had managed to link the 15 blocks together with another 64 B-blocks to give a peal. The peal thus contains $3×5+22×(4×3)-2×3-4×3=261$ omits, hence 579 bobs.
The actual composition differs in the precise 20 courses chosen to make up the initial 15 blocks illustrated earlier, but the principle is the same. Philip’s composition is based on 5 blocks with calls at 1,4;4, rather than 8;5,8.
A Cambridge University Guild band, with Philip Agg conducting, rang this composition at St John the Evangelist, Waterloo Road on 22nd January 1995. This was the first peal rung of Stedman Triples with bobs only.
Andrew Johnson
5040 Stedman Triples
by A Johnson and P A B Saddleton
2314567 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0
________________________________________________
2416753 - - - - - - - - - - - - - - -
3415762 - - - - - - - - - - - - - - -
3425716 - - - - - - - - - - - - - - -
7214635 - - - - - - - - - - - - -
5247136 - - - - - - - - - - - - - -
5274631 - - - - - - - - - - - - -
3715642 - - - - - - - - - - - - - -
7451623 - - - - - - - - - - - -
4752613 - - - - - - - - - - - - - - -
3156247 - - - - - - - - - - - - -
4756231 - - - - - - - - - - - - - - -
3456271 - - - - - - - - - - - - - -
7356241 - - - - - - - - - - - - - -
1456237 - - - - - - - - - - - - - - -
4752361 - - - - - - - - - - - - -
3752641 - - - - - - - - - - - - - -
4752136 - - - - - - - - - - - - - - -
7453612 - - - - - - - - - - - -
1732645 - - - - - - - - - - - - - -
1532746 - - - - - - - - - - - - - - - -
1432657 - - - - - - - - - - - - -
1472635 - - - - - - - - - - -
6534721 - - - - - - - - - - - - -
1534762 - - - - - - - - - - - - - - -
2134765 - - - - - - - - - - - - - -
1734652 - - - - - - - - - - - - - -
1745263 - - - - - - - - - - - - -
1742653 - - - - - - - - - - - - - -
7152643 - - - - - - - - - - - - -
3417652 - - - - - - - - - - - - - - - -
3217564 - - - - - - - - - - - - - - -
5413672 - - - - - - - - - - - -
5247361 - - - - - - - - - - - - - -
4517632 - - - - - - - - - - - -
4516372 - - - - - - - - - - - - - -
4517326 - - - - - - - - - - - -
3215674 - - - - - - - - - - - -
3215746 - - - - - - - - - - - - - - - -
3465721 - - - - - - - - - - - - -
3456712 - - - - - - - - - - - - - - - -
7316542 - - - - - - - - - - - - -
2314567 - - - - - - - - - - - - -
________________________________________________
Contains 579 bobs.
Rung at St John's, Waterloo Road, on 22nd January 1995,
conducted by Philip J Agg.