[r-t] Ben Constant's Yorkshire Royal
andrew_johnson at uk.ibm.com
Fri Jan 16 10:01:58 UTC 2009
ringing-theory-bounces at bellringers.net wrote on 15/01/2009 18:20:31:
> [image removed]
> Re: [r-t] Ben Constant's Yorkshire Royal
> Philip Saddleton
> 15/01/2009 18:21
> Sent by:
> ringing-theory-bounces at bellringers.net
> Please respond to ringing-theory
> Mark Davies said on 15/01/2009 08:58:
> > Bobs-only Stedman Triples is a good example. It is, I suspect, much
> > large ever to provide an exhaustive enumeration - what is it,
> something like
> > 10^25 or something? Very big, anyway. But consider this thought
> > Divide the search space up by establishing a large number of call
> > below which the search tree is accessible. Now write a simple
> program which,
> > given the call prefix, will output all the compositions from it. (You
> > assume a rotational sort if you want only the distinct roundblocks).
> > Plug all this (the prefixes and the search program) into a server and
> > a web front end over it. The first page of the website says, "This
> > contains all possible bobs-only touches of Stedman Triples". Some
> > print advises that, because there are so many compositions, only apage
> > time will be shown. The user can click "Next" to get to the next page,
> > type in a page number. There are probably hundreds of billions of
> pages, but
> > for any one, the webserver will come back after a few millisecondsand
> > you the compositions.
> > Who would be convinced, given such a website to play with, that every
> > bobs-only composition of Stedman Triples already exists?
> Won't most of the pages be blank? If so I don't think anyone would.
There are a lot of magic block peals. For each basic composition the magic
blocks link 10 B-blocks into 5 blocks, leaving 84-5 = 79 blocks in total.
These blocks are then linked with 39 (or more) Q-sets of omits. Each of
these Q-sets can usually be done in 3 ways as there are two places where
the same front bells and same back 3 bells come up (omit 6 bobs, or omit
set of 3 or another set of 3). That gives roughly 3**39 arrangements.
3**39 is 4,052,555,153,018,976,267. You've then got 840
giving 3,404,146,328,535,940,064,280. At 3 hours per peal 85,000 peal
bands could have been ringing different bobs-only Stedman Triples peals
since the Big Bang, 13.7 thousand million years ago, without repeats!
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