[r-t] Calling Round Stedman - Minimum Sixes from Rounds

Andrew Johnson andrew_johnson at uk.ibm.com
Thu Nov 18 11:44:13 GMT 2021

> From: "Andrew Rawlinson"> 
> Does anyone happen to know of a proof of the minimum number of sixes
> to rounds from any given position in Stedman on each stage? I've 
> heard a few figures being rumoured and am curious if anyone knows of
> a conclusive proof.
> Best, 
> Andrew
It's been asked before, but no proof other than exhaustive search. Philip 
Saddleton answered it, the original web page has gone, and I can't find 
the original question in the Change-ringers archive, so I am adding it 

Stedman Turning Courses
Simon Linford posed the following question on the Change-Ringers' mailing 
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 
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:



The results for Triples and Caters are more memorable rows: 
53 to

84 to
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 

Andrew Johnson

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