[r-t] Bobs only Stedman Triples

[edward martin]

On 23 June 2012 10:56, Robert Bennett <rbennett at woosh.co.nz> 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 close
> 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 composition
> 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..
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 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