Stedman Turning Courses

Simon Linford posed the following question on the Change-Ringers' mailing list:

Which row is furthest from rounds using conventional Stedman Cinques as the method (principle) for getting to that row? Put a different way, if asked to compose a touch of Stedman Cinques that would contain a specific row, which such row would cause the touch to be the longest?

I and others had speculated about a related question before, i.e. what is the maximum length required to call round a touch of Stedman Caters or Cinques from a given, random point. The general consensus was that it ought to be between one and two courses.

After a bit of thought I decided that the solution ought to be within the scope of a reasonably powerful PC. This is what I came up with a week or so after Simon's challenge:

First construct a table giving the minimum number of changes required to reach each quick six-end. We can start this off by putting T(rounds) = 0, T(132547698E) = 1, etc. Fill in all the results up to 11 changes manually - a total of 6 + 6 x 3 = 24, since there are three possible calls at the slow six-end.
Next, for any entry already filled, calculate the seven possible rows that can be reached after a further two sixes (-- and ss give the same result, as do -s and s-). If this entry does not already contain a lower number, enter the previous value + 12. This process took about half an hour on a 233 MHz PII with 64 Mb RAM. I wouldn't recommend trying it with less memory. (It was 6 hours on my 32 Mb machine.)
The highest value in the completed table was 159, for 4321967E805. Working backwards, we see that starting from the fourth row of a quick six, we need 156 changes to go from 23491768E50 to rounds, or from rounds to 512307684E9
To solve Simon's problem, consider each row in turn, and how it can occur at the end of a touch, i.e. six positions in a quick six, and six each in the following six after each of the three calls. For each in turn work out what that quick six-end would be and look up the number of changes to reach it. Adjust for the difference between the six-end and the point where the row occurs and take the lowest in each case.

The solution:

Three pairs of rows require 130 changes:

E0714852639
48057936E21

E0971485236
59068E47321

E0798634512
0E789635421

The results for Triples and Caters are more memorable rows:

However, this is not just an academic exercise. By using a slightly different table it is possible to construct the shortest touch between any two six-ends, and so generate useful turning courses or short touches containing given rows. The difference is that rounds is the only seeding row, and the table then gives the number of rows to get from rounds at a quick six-end to any other quick six-end. Since this is always a multiple of 12, dividing through by 12 we can fit the length into four bits (I hope - I haven't actually checked that none needs more than 30 sixes), and so half the storage required by the table.

The required calling can be reconstructed by working backwards: look up each of the seven possible previous quick six-ends in the table and find which one has a value of one less. If the touch is required to start or finish at a slow six-end, work out the touches for the three possible following or preceding quick six-ends respectively, and see which is the shortest.

Here is the software. It runs under Windows 95 and works for Triples, Caters and Cinques. Once constructed the tables are stored on disk and so the program runs very quickly. However, to construct the Cinques table took nearly two hours on my 180MHz PPro, so be warned!

A variation of this approach can be used to find all touches of a given length producing a required course-end. This program uses the same tables as above. It uses various dlls that should be in the \windows\system directory. If necessary they can be downloaded and extracted from redist.exe.

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Last updated 27 November 2000