Stedman Triples without Singles!

The event on 22 January of the first peal of Stedman Triples being rung with bobs only marks the end of a long debate: Is it possible or can it be proved impossible? This question has intrigued ringers for 150 years.

In the Bell News of 1 June 1889, the well-known composer J. W. Washbrook produced a proof that a bobs-only peal is impossible. He starts "It has long been known to the Exercise of the impossibility of producing a peal of Stedman Triples...". However, his argument is applicable only to peals of the kind so far produced at that time, twin-bob peals using Hudson's courses.

Early in the last century the maximum number of P-blocks (different plain courses) possible was the subject of correspondence in the Bell News, culminating in W. H. Thompson's proof in the issue of 19 June 1909 that 40 (already achieved) is the maximum. He showed how to group the 360 possible courses into 40 sets of nine, every course in a set being false with all the others of the same set, so that not more than one could be chosen from each set.

But in the last century it was known that the extent may be obtained in 84 B-blocks (bobbed courses of 10 sixes), and while ringers generally dislike the long dodges produced by runs of consecutive bobs, a few peals have been composed and rung by the B-block method, for instance Slack's 2-part, which was produced to confound a conjecture by J. Armiger Trollope that an exact two-part with two singles could not be had. Slack's peal has no more than 6 consecutive calls. Trollope, who was editor of the Ringing World, refused to publish the peal as being of insufficient interest, and the composition had to be submitted as an advertisement.

The would-be composer might think it a relatively simple matter to link B-blocks by means of sets of three plains (i.e. plaining bobbed Q-sets); in fact it is not at all simple. Firstly, there is an apparently insurmountable difficulty: as there are an even number of B-blocks to start with, the Q-set parity law defeats any attempt to link them into one peal; the Q-sets are of odd order and do not alter the parity (oddness or evenness) of the number of round blocks at any stage in the linking attempt.

Supposing this difficulty resolved, another lies in the fact that the 84 B-blocks are not unique; there are 504 distinctly-different ones (as with the 360 P-blocks, these are all the possible ones with in-course six-ends; the others are their reverses). They fall into 42 systems of 12 B-blocks, and within each system there is a choice of one mutually true pair among 6 such pairs. Thus, only one-sixth of the possible B-blocks are actually used in making up a peal, so that if one is trying to find a Q-set of plains linking a particular sixend of a B-block to two other B-blocks, the chances of success are only 1 in 6 for each of the other two prongs of the Q-set, that is, 1 in 36 for viability of the proposed Q-set. There is a further limitation on linking, explained below.

Any B-block has the same pair of bells in positions 6, 7 for each in-course row (the alternate rows, as Stedman changes are pure triples). The twelve B-blocks of a system have the same pair of bells in the same order in 6-7. The front five bells happen to ring Stedman Doubles. If all twelve of the system of B-blocks with a given pair of bells in 6, 7 are examined, it is found that any particular one is false with ten of the others; but there is one B-block, its complement, with which it is true. If you want the extent in B-blocks, then it must be in such complementary pairs. The 12 B-blocks may be coded by the "leader circle" notion of the old composers such as J. W. Parker, i.e. the order in which the front 5 bells perform any particular work ( such as quick bell leading) and have all the 12 possible in-course leader circles of the front 5 bells. Complementary pairs have mutually reverse leader circles; the order of the Stedman Doubles rows is reversed.

One does not have to use whole B-blocks (dissected into bits or not) exclusively in a peal, the great majority of compositions do not. Within a system of 12 B-blocks, the rows of every six occur in six different forms, all in different B-blocks. What is crucial to our subject, the 18 rows of any three consecutive sixes of kind S-Q-S may be found in two other B-blocks of the system in three consecutive S-Q-S sixes. The two bells which dodge in 4-5 with the quick bell, dodge together in the quick six; if these three bells be rotated, the same 18 rows remain; three different forms of the same rows.

The 12 B-blocks of a system fit nicely onto Thompson's dodecahedron. Mutually true B-blocks are on opposite faces and can be disposed to get the mutually true S-Q-S trios around the vertices. It is interesting to note here that Thompson was expounding the relationship of the in-course rows on 5 bells to the symmetry group of the dodecahedron, before Burnside published his Theory of Groups (1897), the first treatise on groups in English. Other information on groups was in mathematical journals, mostly German. But then Thompson was 23rd wrangler in the Cambridge Mathematical Tripos in 1863.

To return to the original question, many attempts have been made to resolve the matter. In correspondence on the subject in the Ringing Worlds of January 1946, Rev. E. S. Powell declared it impossible. In the early 1950s the prototype computer at Manchester was programmed by R. A. Brooker using my data, in an attempt to produce a round block for a 21-part peal with no singles, without success (this must have been the first change-ringing program). Another "proof" of impossibility was bublished in the Ringing World of 20 February 1953, to be lauded by one ringer and demolished by another. The breakthrough came about five years ago when the five "magic blocks" were discovered by the patience and ingenuity of Colin Wylde. I confess that I had given up hope of such a thing; but when I heard from Australia the news that a no-singles peal had been composed (it was buzzing round the world on e-mail), like others I exercised the ringing equivalent of industrial espionage, picking up enough clues for me and my computer to find the blocks ourselves:

1.  3152467  2.  7253416  3.  6357421
   -1234567     -2374516     -3764521
   -1245367     -2345716     -3745621
   -2513467     -3527416     -7536421
   -2534167     -3574216     -7564321
    5426371      5431762      5472613
   -5463271     -5417362
   -4352671     -4753162        etc.
   -4326571     -4731562
   -3645271     -7145362     (5 blocks)
   -3652471     -7153462
   -6234571     -1374562
   -6245371     -1345762
   -2563471     -3517462
   -2534671     -3574162
    5427316      5436721
   -5473216     -5467321
   -4352716     -4753621
    4321567      4732516
   -3145267     -7245316
   -3152467     -7253416

The blocks are "magic" because 5 is odd and 10 is even. The parity barrier has been broken! Each block has a whole B-block of 10 sixes (9 consecutive bobs) plus a selection of 7 sixes (6 bobs) from another B-block and 3 (2 bobs) from a third. Each section of three sixes S-Q-S, for instance those ending 16 in block no.1, marries up with a set of 7, in this case the 16s in block no.2. But the 3+7 do not make up an actual B-block; each S-Q-S block is an alternative version of the three sixes S-Q-S missing from the B-block of the block of 7. Also, the plain sixend changes do not fall into whole Q-sets; thus is the Q-set parity law circumvented.

So we have the equivalent of 10 B-blocks in 5 magic blocks. Then there are the 10 complementary blocks (as explained above) to those 10, which will have to be used - have to be, because for the rest of the peal we are dealing in whole B-blocks only, no more dicing with parity; there is no choice over them. But ... there are still 32 systems of B-blocks to be deployed, and with a choice of one of 6 different true pairs from each, the total number of choices is 6 raised to the power 32, a number with 25 decimal digits. Nevertheless, the parity barrier has been stormed, we have 5 magic blocks with 74 B-blocks to join up with them, and 79 is an odd number. So there is a chance of plained Q-sets linking them up into one round block. Some of the bobs in the magic blocks will be plained for linkage, but the plains in them must on no account be bobbed, they are outside Q-set law and are vital to the Wyld magic spell.

Here, another obstacle appears. If three B-blocks can be joined by plaining a Q-set, then there is a further linking Q-set available for joining the same three blocks; the two arise where the same front bell makes its bobs. This has the beneficial effect of enabling more plains to be put into a peal, for if both Q-sets are plained, the three blocks still make one round block of 30 sixes. Slack was aware of this over fifty years ago, using it in his 2-part to attain the maximum of 6 consecutive bobs. But it has the unfortunate efect of reducing the chance of available links between different B-blocks. When one discovers a Q-set which will join up an extra two B-blocks, three sixend positions are ruled out for links elsewhere.

The task of assembling a bobs-only peal starting with the Wyld magic blocks is excessively long-winded and frustrating. Despite many attempts with computer assistance, from 1953 to 1995, no-one has succeeded in producing a bobs-only peal in parts. This now seems to be the impossibility - or is it? Philosophical questions are raised: Did a bobs-only peal exist in abstract form before the Wyld breakthrough? Was it created during the Big Bang?

Returning to recent events, what happened? Pandora's box had been opened. News of success got around pretty quickly. Composer-espionage got to work and soon there were rival peals being rustled up and hatched for diverse and urgent first performances. When I became within sniffing distance of composing one myself, not knowing (but guessing) what was going on among the High-Powered Peal-Ringing Fraternity, I rang up a friend to put out a feeler for interest in ringing one of mine, only to be told that the first no-singles peal had been rung the previous day! Alas, Colin Wyld was not in the successful band, but it is he who wins the laurels.

The old rivalries, such as between the ancient bands in Norwich and in London, still thrive, and long may they do so.

February 1995

Stedman Triples

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