[r-t] Gangnam

[James Smith]

> Message: 9
> Date: Wed, 14 May 2014 18:37:48 +0100
> From: Robin Woolley <robin at robinw.org.uk>
> To: ringing-theory at bellringers.net
> Subject: [r-t] Gangnam, etc.
> Message-ID: <5373A9EC.6020301 at robinw.org.uk>
> Content-Type: text/plain; charset=ISO-8859-1; format=flowed
>
> ...
>


> 3. For those asym. methods which cannot have an extent, Bailey's 1440 in
> the diary works. I worked out, when lying in bed one afternoon, that if
> we had a 1440 in which each row occurs at both hand- and back-stroke, it
> should be true. The reason is as follows. Consider the method in its
> half-leads. This type of comp. has each half-lead rung as the first half
> of the lead and each half-lead rung backwards as the 2nd half. This is
> permissible as it is easy to show that any row in a *half-lead* can be
> in only one half-lead. Therefore, in the 1440, we have every row rung in
> the top half of a lead somewhere, and in the bottom half of some
> half-lead. All that remained was to search for a suitable comp. This was
> the only one in the on-line collection and works because it contains
> just two singles - 720 rows apart. (It is possible to construct those
> with more than two singles but care needs to be taken as to the actual
> positioning so every lead end group has a slightly different version.)
>
>
Two 'late greats' - Roger, and Eddie Martin, pointed me in the same
direction, though via a different route, when I was looking for
compositions of twin-hunt Minor methods a few years ago.  The first halves
of all of the 120 possible leads of a plain Minor method must together
contain the extent as they represent both half-leads of a group of 60
mutually true leads of a symmetrical method. That will also be true of
asymmetric plain methods as this logic is unaffected by anything that
happens in the second half-lead. Similarly, the second half-leads of the
120 possible leads must also contain the extent - since they are the first
half-leads of the different asymmetric method that one would obtain if one
wrote out the rows of the lead in reverse. Such a 1440 therefore contains
all the rows twice.  The examples I produced contain all the rows once at
hand and once at back, as in Robin's example, though whether a) this is a
universal quality of 1440s using all 120 leads or b) achieving rows once at
each stroke requires the use of all 120 leads I don't know.

James