[r-t] Ben Constant's Yorkshire Royal

[Andrew Johnson]

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 
> 
> to:
> 
> ringing-theory
> 
> 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 
too 
> > large ever to provide an exhaustive enumeration - what is it, 
> something like 
> > 10^25 or something? Very big, anyway. But consider this thought 
experiment. 
> > Divide the search space up by establishing a large number of call 
prefixes 
> > 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 
can 
> > assume a rotational sort if you want only the distinct roundblocks).
> > 
> > Plug all this (the prefixes and the search program) into a server and 
stick 
> > a web front end over it. The first page of the website says, "This 
site 
> > contains all possible bobs-only touches of Stedman Triples". Some 
small 
> > print advises that, because there are so many compositions, only apage 
at a 
> > time will be shown. The user can click "Next" to get to the next page, 
or 
> > 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 
show 
> > 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.
> 
> -- 
> Regards
> Philip
> http://myweb.tiscali.co.uk/saddleton/
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 
one
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 
rotations/reversals,
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!

Andrew Johnson






Unless stated otherwise above:
IBM United Kingdom Limited - Registered in England and Wales with number 
741598. 
Registered office: PO Box 41, North Harbour, Portsmouth, Hampshire PO6 3AU