[r-t] in-course 120

Philip Saddleton pabs-ant at tiscali.co.uk
Thu Oct 5 17:05:09 UTC 2006

edward martin said  on 04/10/2006 19:51:
> On 10/4/06, Andrew Johnson <andrew_johnson at uk.ibm.com> wrote:
>> With a bit of experimentation I came up with
>> which has some nice symmetry.
> Very nicely done. I too had a look at Stedman Doubles but failed to
> spot the route that you found
> mew

I think this and its reverse are the only possible 5-parts. Here's the 
graph (view in a fixed font):

. a=======k
. |       |
. |  c-d  |
. | /| |\ |
. |/ f-e \|
. b  | |  l
.  \ g-h /
.   \| |/
.    j-i

We need to visit each vertex twice, with an odd number of edges between 
each visit. We cannot revisit any of c-j without passing through b or l, 
as all circuits in this subgraph are even. Clearly we cannot follow the 
same edge twice in succession. Vertices a and k must be between b and l, 
so the only options are

l-k-a-b...b-a-k-l... or l-k-a-b...l-k-a-b...

where ... are chosen from c-j. But as there are only two ...'s and 
neither has any repeats, each contains each of c-j once. This means we 
can only have the first option, or the l-k-a-b are revisited after an 
even number of edges. Now what possibilities are there for b...b. We 
start with c and finish with j, or vice versa - one will simply be the 
reflection of the other. The only possibilities that visit all of the 
vertices is

b-c-d-e-f-g-h-i-j (or its reverse)

and for l..l

l-d-c-f-e-h-g-j-i-l (or its reverse)

once we choose the direction of one there is only one possibility for 
the other, giving the path


This is Andrew's result. You can now go back and label all of the edges 
with their place notation. the double link between a and k can be 1 or 
5: one of each is needed, or it comes round after one part.


More information about the ringing-theory mailing list