[r-t] Hudson stuff
Richard Smith
richard at ex-parrot.com
Mon Sep 13 17:24:36 UTC 2004
Leigh Simpson wrote:
> Does anyone mind telling a muppet like me what Hudson
> methods/groups are, and how they work?
Hudson's Group is a group of order 60 that is generated by
the changes 12, 16, 34 and 56. (Actually, only one of 34
and 56 is needed to generate the group.)
The group is isomorphic to A_5, the group of even-parity
rows on five bells. In fact, it is the dual of A_5 under
the outer automorphism of S_6. This is described in some
detail towards the end of Brian Price's paper, "The
Composition of Peals in Parts", which is linked to from
Roger Bailey's web page.
Hudson's group, H, like any other group, can be used to
partition the full extent, E, into its right cosets.
Selecting exactly one element from each right coset produces
a right transversal, T. This has the property that
H T = E,
where multiplication of sets of rows is defined in the
obvious manner:
A B = { a b : a in A, b in B }.
What relevance has this got to ringing? If the transversal
can be expressed to a sequence of changes,
T = { 1, c1, c1.c2, c1.c2.c3, ..., c1.c2...cn },
then it produces a set of mutually true parts that can
(potentially) be joined to produce a multi-part composition
with H as the part-end group.
To see this, let's consider a simpler example on four bells.
Let E = S_4 be the extent and H = { 1234, 1342, 1423, 1243,
1432, 1324 } be the part-end group. The right cosets of H
are
1234 2134 2314 2341
1243 2143 2413 2431
1324 3124 3214 3241
1342 3142 3412 3421
1423 4123 4213 4231
1432 4132 4312 4321.
To find a transveral, we need to choose one element from
each coset, and for it to be of use in ringing, consecutive
elements need to be related by changes. An example would be
1234
x
2143
14
2413
x
4231.
Writing out the these three changes (four rows) starting
from each element of the group will produce all 24 rows on
four bells.
A good way to find such transversals is with a Schreier
graph. The vertices of a Schreier graph are the cosets,
which can be labeled by choosing an arbitrary element from
each one. There is an edge between two cosets A and B if
there is a change, c, such that a c = b for some a in A,
b in B.
For example, there is an edge between the cosets labeled
1234 and 2134 because they are related by the changes 34
and x (x because 2143 is in the same coset as 2134). The
complete Schreier graph is given below.
1234 -------- 2134 -------- 2314 -------- 2341
x,34 14 x,12
A suitable transversal corresponds to a Hamiltonian path on
this graph. Using this, it is trivial to enumerate the
possible ringing transversals: they are x14x, x14.12,
34.14x, 34.14.12.
The remaining question is how to join these parts together
to form an extent. Usually, this is done in one of two
ways. One is to make the transveral a part of the
composition. This means finding a change taking us from the
end of the transversal back to the part end group -- in this
case from the row 4231 to one of the rows of the form 1xxx.
Clearly not. (In terms of the Schreier diagram, this
corresponds to finding a Hamiltonian circuit which clearly
does not exist in this case.)
The other case is to make the composition palindromic and
have the transversal be half of a part. This requires
changes to be found at both ends of the transversal that
take the end row to a differnt row in the same coset. This
turns out to be possible: when the treble is leading, 12 or
14 can be used; when the treble is lying, 14 or 34 are
possible. Three of these four possibilities produce an
extent (of Plain Bob, Reverse Bob and Double Bob
respectively); the fourth possibility is Plain Hunt, and,
although the three leads of Plain Hunt are mutually true,
they do not join up.
Applying this idea to the other three blocks of place
notation gives rise to the eleven plain minimus methods.
Returning to Hudson's group, all of the elements of this
group are of even parity, which eliminates the possibility
of a principle. (A six-cycle has odd parity.) A
single-hunt minor method is the next possibility, and there
are two obvious ways of achieving this: an asymmetric (or,
at least, non-palindromic) plain method or a palindromic
treble-dodging method.
(Because the cosets have no natural interpretation in terms
of the treble's position, other paths, such as little
treble-dodging with double dodges in 1-2 and 3-4, are
possible.)
>From the possible palindromic treble-dodging minor methods,
one method stands, in my opinion, head and shoulders above
the rest. This is Hudson Delight Minor, place notation
&3-3.4-2-1.4-4.5,2. Although this does have three
consecutive blows at lead, all the possibilities have at
least three blows in one place, and most have more.
Phil's web site contains more examples of Hudson methods
with various treble paths.
Richard
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