(First published in The Ringing World, 20 February 1953)

A Note upon Some Aspects of Stedman Triples

By Q. E. D.

One of the oldest problems in ringing is whether or no it is possible to obtain peals of Triples by the use of common bobs only.

For such methods as Grandsire, which contain a bell in the hunt, the work of W. H. Thomson (and more recently that of Prof. Rankine) has effectively answered the question in the negative.

Whilst for methods, such as New Bob, containing no hunt bell, the possibility has definitely been proved by the peals of such men as John Holt.

However, in the case of principles, such as Stedman, no answer has, as yet, been obtained. Therefore, it is to the latter that I am going to turn my attention in the hope of throwing some light upon the matter.

Argument

The basis of Stedman is, as is well known, the ringing of a Slow Six followed by the ringing of a Quick Six. Thus any study of Stedman must start with the consideration of the group:

       1234567
       2135476
 Slow  2314567
  Six  3215476
       3124567
       1325476
       3152746
       3517264
 Quick 5312746
  Six  5137264
       1532746
       1357264
 Slow  3175624

This is merely a Slow Six followed by a Quick Six with rounds put as the first change of the Slow Six instead of, as is normal, its being the 4th change of a Quick Six. This is, in fact, the unit which may be regarded as a lead of the method.

For a start we will denote any group of this type, i.e., a Slow Six followed by a Quick Six, by the symbol (PsPq), thus showing that both the Slow and Quick Sixes are 'plained'. Obviously if we regard this group as a lead of the method we may call a Bob at the end of the group.

This will give us a group almost identical to with (PsPq), but having, instead of the change 3175624, the change 3175246. This modified group we will denote by (PsBq), showing that the Slow Six is plained, whilst the Quick Six is bobbed.

The great difference between Stedman and Treble dominated methods lies in the fact that there is another 'bobbing' position other than at the unit end (we are not considering 'Real Double Methods'). This is at the Slow Six end, which, of course may be regarded as just as much the unit end as is the Quick Six end. Thus we may obtain the modified group (BsPq) in which the Slow Six is 'bobbed' and then followed by a Plain Quick Six starting from 3152467 instead of 3152746.

Finally, of course we may call bobs at both 'unit ends' obtaining the group (BsBq), i.e., having a Bobbed Slow Six followed by a Bobbed Quick Six.

Upon exmination it will be seen that it is impossible to obtain any further types of group by merely bobbing the unit in the recognised manner. Thus if a peal of Stedman Triples is possible by the use of common bobs only it will consist of the above four groups and of those only - in future we shall drop the 'suffix-bracket' notation and shall refer to the groups as P2; PB; BP; B2 respectively.

Let us now suppose that it is possible to obtain the 5,040 changes which make up a peal of triples in 60 'P-blocks', i.e., in 60 sets of groups of the form (P2)7, where this notation denotes the 'set' consists of 7 units P2, following each other - i.e., a 'P-block' is of the same nature as a Plain Course. Then these 60 blocks will be mutually exclusive, i.e., no change will occur in more than one such block. (We do not know that such a set of 60 mutually exclusive P-blocks may be obtained. However there is no loss of validity in the assumption and the conclusions reached will still hold.)

In a similar way let us suppose that we have 60 mutually exclusive blocks (PB)7, and again 60 blocks (BP)7 containing the 5,040 changes. These 5,040 changes are also contained in 84 'B-Blocks' of the form (B2)5. It now remains to try to obtain peals using these blocks as a basis. Obviously though blocks of the same type are mutually exclusive, some relationship must exist between blocks of different types - these relationships give, in fact, the modified Q-set laws for the method.

To examine these relationships let us once more look at the Quick and Slow Sixes. It will be seen that each of these sixes (no matter whether Quick or Slow) contains all the combinations possible on the front three bells, and therefore, any repetition between Sixes may be observed by examining the position of the back four bells: more precisely, any six exhausts the rows of like nature with the same bells dodging in the same order in 4.5 and 6.7.

For example, if 4567 are the dodging bells at backstroke, then the particular six, whether Quick or Slow, will contain all the possible rows of like nature with 4567 as the four hindmost bells, no matter in which order the three front bells fall. These four hindmost bells are called the characteristic, or K, of the six.

Now consider the K's of the unit block P2. These are three in number: (1) 5476; (2) 7264; (3) 6542. The K 5476 remains unaltered throughout the modified blocks - hence we do not need to consider it in the following argument. And we merely consider the the two K's 7264, 6524. These we will call the foundation, F, of the group. Thus we have:-

F(P2) = 7264; 6524 - (1)

(where F(P2) is read as 'the foundation of the group P2 is'.)

Similarly we get:-

F(PB) = 7264; 2564 - (2)
F(BP) = 4276; 7562 - (3)
F(B2) = 4276; 6542 - (4)

A comparison of these F's will give us the Q-set laws which apply.

Firstly, let us compare (1) and (2). Then we see that the Slow Six of the P2 group may repeat with the Bobbed Quick Six of the PB block.

[Note. - By this terminology, which I am afraid is rather misleading, I mean that the first K of F(P2) may repeat with the last K of F(PB) for some blocks P2, PB.]

Also the second K of F(P2) may repeat with the second K of F(PB), and finally the second K of F(P2) may repeat with the first K of F(PB). Thus we see that repetition between the blocks P2 and PB may occur in three ways, i.e., there exists a Q-Set law, with three members, between them.

We will denote this by:-

C(1,2) = Q(3)

i.e., a comparison of 1 and 2 gives a three member Q-Set law. Similarly:-

C(1,4) = Q(3) }
C(3,2) = Q(3) } A*
C(3,4) = Q(3) }

Upon a further examination we find

C(1,3) = Q(4) } B*
C(2,4) = Q(4) }

For any two groups for which the relationship C(A,B) = Q(3) holds, i.e., for any of the pairs of groups in A*, it is obviously impossible to obtain a peal using only the two groups under consideration - for by arguments exactly similar to those in the case of Plain Bob Major, etc., we find that for a peal to be possible a relationship of the form

2n + 1 = 60
or 2n + 1 = 84

must hold, where n is an integer.

Now let us consider the combinations of groups contained in B*, i.e., groups between which a 4 member Q-Set law holds. We need only work this out in one case as mathematically the two cases are exactly similar.

If we start with the groups A, B such that

C(A,B) = Q(4)

we find that if in any A block we insert a Q-Set of the B blocks (note, do not confuse these with 'B-Blocks') we join on to the A block three further A blocks and so on for all further Q-Sets.

Thus for a peal to be possible under these conditions the relationships

1 + 3n = 60
or 1 + 3n = 84

must hold where n is an integer. Hence a peal may not be obtained by the use of two groups for which a Q(4) law holds.

Thus it is impossible to obtain a peal using only two of the above groups.

It may still be possible, however, to obtain a peal by using three different types of group or even by using all four. Firstly, let us consider the case of three different types of group A, B and C. By an examination of the relationships A* and B* it will be seen that the three relationships

C(A,B) = Q(3)
C(A,C) = Q(4)
C(B,C) = Q(3)

must hold, and thus if a peal be possible using three blocks a relationship of the type

1 + 3m + 2n = 60
or 1 + 3m + 2n = 84

must hold, or combining these two equations:- 1 + 3m + 2n = 0(6) where 0(6) means a multiple of 6.

Now, as n is governed by both a Q(3) relationship and a Q(4) relationship, it is a multiple of both 2 and 3 and is therefore a multiple of 6. Similarly for m. Hence the equation takes the form

1 + 0(6) + 0(6) = 0(6)
i.e., 1 + 0(6) = 0(6)

which is absurd. Hence we cannot obtain a peal using three of the groups. An exactly similar argument holds for the case of four groups. Therefore, we see that it is impossible to obtain a peal of Stedman Triples using only the four blocks P2, PB, BP, B2.

But, as we saw at the beginning, any peal of Stedman Triples obtained with common bobs only must contain these four groups only. Hence it is impossible to obtain a peal of Stedman Triples with common bobs only.


(RW 6 March 1953)

Congratulations to 'Q.E.D.'

Dear Sir,

Our contributor, 'Q.E.D.', is to be congratulated on having presented his argument (about the impossibility of obtaining peals of Stedman Triples with the Stedman standard bobs only) in a not too mathematical form for the average ringer interested in composition-theory. Not an easy task!

The late W. H. Thompson also employed ingenious reasonings of like nature in his treatment of common-bob peal composition in Grandsire Triples. But Dr. Rankin's similar dissertation resorted to the application of principles of Group Theory - a branch of mathematics which is, of course, not exactly elementary.

With this acknowledgment, perhaps Q.E.D. will permit a small measure of comment, intentionally helpful.

Some writers on the Stedman method suggest that his reason for placing rounds as the fourth row of a six is not apparent, and others, that perhaps it was to give a start like Grandsire. The reason, however, seems fairly evident. As the founder of the method-form which remains basic today, Stedman no doubt realised the value of symmetry in the construction of the divisions ('leads' in a hunt-controlled method) of a plain course. In other words, the pattern of the successive permutations throughout the change-rows is a mirror-reflection about the half-division.

The Stedman division consists of 12 change-rows, to include both forms of 'six' - Quick and Slow. But since the grid-patterns of these two sixes are different, symmetry of a Stedman division can only be obtained by sandwiching one type of six between the two halves of the other type - it does not matter, technically, which way. This means that rounds must be placed at the middle of a six. The 'snag' about starting from the middle of a Slow six is the retrograde hunting therein, which would cause the treble to behave in an unusual fashion at the go-off, by remaining on the lead for another blow.

So the start is made from the middle of a Quick six (after the third row) where the direct hunting allows No. 1 bell to run straight up.

At the opening of his argument Q.E.D. cites a block of 12 Stedman rows, consisting of a slow six followed by a quick six. This is not, of course, a true Stedman 'division', and is not a symmetrical row-block. Consequently a little later (middle of second column) it has to be explained that the first of the three 'K's' arising from this exemplary block is not needed for consideration.

If, however, the true Stedman division be cited instead, and the head-rows of the sixes therein be signalled instead of the end-rows, there are only two 'K's' arising for consideration. The whole argument is in no way violated. And the effect of the calls may thus be theoretically considered where they are felt in practice, i.e., at the ensuing hand-stroke row after the call is made. (Calls being orthodox in Stedman, of course, at both the Link and the Junction, of its divisions.)

Later (bottom of second column of the article), when comparison of groups is begun, i.e., 'C(A,B)=Q(3)', and again in column three, where further group comparisons are considered, it is suggested that for 'A', 'B' and 'C', other letters, e.g., 'X', 'Y', 'Z', might be preferable to avoid possible confusion.

'Q.E.D.' will surely have earned the thanks of all students of composition theory.

Yours very truly,

A. YORK-BRAMBLE
Netley Abbey.


(RW 3 April 1953)

More About Q-Set Laws

Dear Sir,

Q.E.D.'s article on page 125 is quite inadequate.

The main flaw lies in imperfect knowledge of the Q-set law and its validity. The law is briefly: 'The alteration of calling of one Q-set off odd order does not alter the parity (oddness or evenness) of the number of round blocks. The alteration in calling of one Q-set of even order alters the parity of the number of round blocks.' But this law is subject to two conditions. Firstly, every member of the Q-set must be utilised, otherwise it becomes possible for some members to be left out and the others to be treated extraordinarily. Secondly, there must be one-way traffic through all Q-sets. It so happens that Grandsire Triples obeys both conditions.

To give examples, Thompson, by usoing 5ths place bobs in addition to 3rds place bobs, composed peals of Grandsire Triples, the Q-sets embracing the 5ths place bobs being treated extraordinarily, having both plain and 3rds place bob leads. Again, in the case of original minor 360 changes are reachable by bobs only and this number can be attained as leads can be used forwards or backwards and the round block uses some Q-sets in mixed directions.

These two conditions emerge from a rigorous proof of the Q-set law.

To return to Q.E.D.'s article, the main fault lies in the confusion of the K or characteristic of the six, which was proposed as a means of proof, with unique members of Q-sets. The six changes of a six can be arranged as a six in six different ways, for there is a choice of three odd changes for the first one, and the six can be quick or slow. These six ways are, of course, identical in regard to proof, having the same K, but for the purposes of composition they are quite different, and therein lies the snag - in effect, the composing of a peal uses only one sixth of all the possible different sixes, and while Q-sets do exist, one is by no means obliged to use every member. It is not possible to make aQ-set of Ks.

This is the main fallacy in Q.E.D.'s article, but not the only one. It is not possible to get 60 exclusive P-blocks, in fact it was proved in the 'Bell News' that 40 true courses is the maximum number, neither is it possible to get 60 (BP)7 or (PB)7 blocks (I have proved this). Despite his sweeping statement that ' there is no loss of validity in the assumption, and the conclusion will still hold', he is basing an argument about a number of blocks derived from others which do not even exist. This is indeed a novel development of the 'reductio ad absurdum' argument! However, this is not a real obstacle as the 84 B-blocks exist, and one could derive from those.

The most surprising error is the mistake in dealing with 4-member Q-sets, which in fact render the whole argument of parity of blocks invalid. Certainly the idea of 1 + 2n as the number of blocks joined in is inadequate and erroneous, and springs from a lack of rigorous thought. Q-set behaviour never predicts the number of blocks or number of courses united in the general case.

Again, near the end of his article, Q.E.D. assumes that different Q-sets do not interfere with one another. This is quite untrue, for all sixes can take part in any one of his Q-sets, and thus they interfere. The 5ths place bob peals of Grandsire illustrate this for either 5ths or 3rds place bobs alone cannot give a peal.

Q.E.D. should know his theory thoroughly before trying to use it on such tricky ground. Good luck to him, but he won't find it easy. I have been trying to resolve this matter since 1944, without success. During the last five years I have been trying to compose a peal with bobs only, and I think that this approach deserves as much attention as the negative one. Some years ago in Bristol I heard of a ringer who was certified as insane after trying to compose such a peal (perhaps someone can verify this).

Recently, I persuaded the staff of the Computing Machine Laboratory at Manchester University to try composing such a peal, on a 21-part plan I evolved, but I have not heard from them conclusively, and I hope to report later. They have kindly consented to run the problem at slack periods, and estimated that it would take 24 hours altogether. Considering that this 21-part plan is but a flea-bite of 1-part composition, and that the machine can perform 600 12-figure multiplications in a second, one realises the immensity of the task; last August the machine actually succeeded in compiling a block of 40 sixes, which, however, did not join up to the start, and so the extent is in 21 unjoinable blocks. This was no mean feat, although useless for our purposes.

Yours sincerely,

Brian D. Price
Brynmawr, Breconshire.


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