[r-t] 4-bell runs
Alan Reading
alan.reading at googlemail.com
Wed Feb 15 12:44:20 UTC 2012
>
> The 11-part doesn't stand up to this sort of analysis (or rather the one
> that resulted in the figure 4356). Each part generates a different number
> of 4-runs varying from 253 in the penultimate part up to 399 in the last
> part.
>
I agree there will be a different number of runs (now we are defining
things like ET23's not to count) in each part but there must be the same
number of each of 90ET, 890E, 7890, 6789, ..., 2345. There are 198 90ET's
anywhere within the change, so therefore there are 198x8 = 1584 forward
four bell runs, and because the composition is palindromic there are 1584
backward runs. So total=3168.
The 1234's and 4321's (if you count them) must be done separately for an
11-part and I've no idea what the total number is but if it's 42 we are in
agreement!
Cheers,
Alan
On 15 February 2012 12:33, Simon Bond <simon.bond.lists at googlemail.com>wrote:
> On 15/02/2012 12:22, Alan Reading wrote:
>
>> I presume that figure is based on removing 6x198 (to discount the
>> none-runs) from 4356 which gives 3168, I guess the extra 42 are 1234's and
>> 4321's?
>>
>
> The 11-part doesn't stand up to this sort of analysis (or rather the one
> that resulted in the figure 4356). Each part generates a different number
> of 4-runs varying from 253 in the penultimate part up to 399 in the last
> part. I'm afraid my method for counting these is just to put all the rows
> into a spreadsheet and do some string processing. It turns up the same
> figure of 3456 for the 12-part.
>
> SAB
>
>
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