[r-t] 8-spliced touch
Alexander Holroyd
holroyd at math.ubc.ca
Thu Aug 7 18:04:22 UTC 2008
I agree this would be nice to know. The number of possibilities (without
pruning the tree for truth) is 8!*8^3 = 260,000,000, which seems a bit on
the large size. Maybe just about doable with a fast program and computer.
But perhaps some more constraints would be desirable. E.g. every lead
should contain at least one run (this would allow for a much faster
search), and perhaps one should insist on a variety in types of runs (it
would be pretty sad if they were all 5678s) - but how best to formulate
this?
I would like to know similar things for normal length touches of e.g.
Grandsire triples, Stedman triples. First one must decide what one
wants...
> On a different point, I have a problem that should be fairly simple to
> solve. Much as I dislike them, I'm interested in the most musical short
> touch of the so-called "standard" 8 surprise major methods. The
> constraints are:
>
> - The touch should contain all 8 methods, with a change of method only at
> the leadhead.
>
> - Bobs (14) and Singles (1234) at the leadhead are fine.
>
> - The length of the touch should be between 253 changes (ie comes round at
> handstroke snap just before 8 leads) and 258 changes (comes round at
> backstroke snap).
>
> - Music is defined by occurance of <4-runs> - each appearance of 1234,
> 2345, 3456, 4567 or 5678 (or their reverses) in a row scores 1. (So rounds
> scores 5, the row 56784321 scores 2 etc).
>
> - Truth is not essential, but should ideally be recorded so a sub-list of
> true touches can be compared.
>
> - As a supplement, I'd also be interested in the results of touches up to
> both 9 leads (+ 2 changes), and 10 leads (+ 2 changes).
>
> The problem is so constrained that I would have thought it should be
> possible to solve exhuastively without too much trouble. Can anybody do
> this and supply the results? Indeed, has anyone done this before?
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