[r-t] in-course 120

[edward martin]

On 10/5/06, Philip Saddleton <pabs-ant at tiscali.co.uk> wrote:
> 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
> >> 3.1.3.5.1.3.5.1.3.5.3.1.3.1.3.5.1.3.5.1.3.5.3.5
> >>
> >> 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
>
> a-b-c-d-e-f-g-h-i-j-b-a-k-l-d-c-f-e-h-g-j-i-l-k-a
>
> 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.
>
> pabs
>
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