[r-t] Bobs only Stedman Triples

Robert Bennett rbennett at woosh.co.nz
Mon Jun 25 11:32:26 UTC 2012


 On Mon 25/06/12 8:46 PM , edward martin edward.w.martin at gmail.com sent:

On 23 June 2012 10:56, Robert Bennett  wrote:
 > These peals are an amazing achievement.

 They certainly are an amazing discovery;but the achievement was
 surely figuring out how to programme a computer to look for them and
 then having the expertise in knowing how to make use of what the
 computer had found...no perhaps you are right an amazing achievement

 > To reduce the number of consecutive bobs down to 5 makes the peals
 > to the ideal.

 In my very limited experience, strings of 9 consecutive bobs are
 inevitable if looking for 5 or 10 part comps. It is possible to plain
 these out in the basic part but the result will be a round block thus
 to link all parts this string has to be included

 > Are there any other bobs-only plans waiting to be discovered?
 > What about Erin Triples? Is it possible to have a bob-s only
 > in that method?
 >  R.Bennett.

 The problem as I see it is that in Stedman it is possible to thwart
 the q-set rule because initially the building material can be set out
 in bobbed blocks (= 60 changes of Stedman Doubles with the pair in 6-7
 continually dodging) There are 21 different pairs to come into 6-7;
 each pair has a complementary bobbed block as with bobs on quick six
 end 1234567 and on quick six end 2135467 In addition, each pair has as
 many positive b-blocks as with say from quick six end 2134576 and from
 quick six end 1235476. Thus the initial building material is 84
 mutually exclusive bobbed blocks.. 

_It is interesting that in the special blocks used to make the bobs-only
peal, the 3 omits are on the same Q-set. The situation is analogous to
bob minor, with the calling WHW x3, where the Q-sets are all bobbed or
plained, but the three bobbed leads are two at W and one at H. In PB
minor, the leads are reversible and in stedman, the sixes can be rung
different ways._ 

_The first block starts from rounds: omit on 567, two bobs, then another
omit on 567, 5 bobs, omit on 567, and 9 bobs._ 

_Since Erin can also have irregular sixes, the law of Q sets does not
exclude a peal with bobs. The question is how. The number of possibilities
is a lot less, because each six can only be rung 3 ways, but it is still

 Before the solution had been discovered by Colin Wylde, I had the
 theory that if you write out every bobbed block which contains rounds
 at the 2nd, 4th or 6th row of either a quick or a slow six, then note
 that there are three instances where the sequence of slow, quick slow
 contain exactly the same rows but in different order, therefore plain
 out from the B-Block before one of these sequences, enter the great
 unknown eventually enter to include ether one of the other two
 sequences, plain out again into who knows what, eventually to return
 to the original b-block without having repeated anything; ie
 essentially there are four important omits which thwart the initial
 q-set requirements. Unfortunately my brain was not up to the task of
 considering all the if this then thats, but eventually this theory was
 shown to be the essential part of the solution.
 I've not related all that with any intent to boast but to demonstrate
 that I do understand how singles can be avoided in Stedman Triples.
 In Erin Triples you cannot set out the 5040 in mutually exclusive
 bobbed blocks; therefore the hope is to find mutually exclusive,
 equally structured blocks of plain and bobbed sixes.
 If this is achieved then is it possible, within these blocks to
 duplicate Stedman's phenomenon by having a relatively short sequence
 of plain and bobbed six-emds which can contain the same rows but in a
 different sequence? I don't know and don't see how to find out. 

_I think one way would be to start with complete Q-sets. That is to decide
which q-sets are to be omitted and then work out which way round the sixes
between the omits have to be rung, so that they link up, and don't run


I have discovered numerous bobs only 5040s of spliced Erin & Stedman
 Triples (all 21 part comps with no consecutive quick sices and no
 consecutive bobs) but the publication of these would be of very
 limited interest I suppose. Incidentally, I believe that it is not
 possible to obtain a 21-part bobs only of Stedman Triples (This belief
 was instrumental to my investigating why not? and the idea of splicing
 in Erin sixes)

 Eddie Martin

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