[r-t] parity

Michael Schulte michaelfschulte at yahoo.com
Tue Aug 9 20:51:00 UTC 2005

--- Mark Davies <mark at snowtiger.net> wrote:
> Depends on what you're used to, but for most people it'll work best with
> 5-figure COs. For 6-figure COs like you quote it's still pretty immediate
> though - bring the 7 to the front with 3 shunts (negative) and the 5-figure
> CO that's left (26543) is obviously positive.

Here's my problem, although admittedly I may be missing something very obvious. I don't do it this

Say you have the row 16732845.

What is the coursing order to which you propose we convert it?

If this row occured at a handstroke in Plain Bob, I would give it a coursing order of 367245. If
it occured at a backstroke I should give it a coursing order of 542763. Which do I use?

The handstroke result, 367245, gives an in course (positive) result, as follows:

=> 736245 (2)
=> 753624 (4)
=> 753264 (1)
=> 753246 (1)
total = 8 moves.

The backstroke result, 542763, gives an out of course (negative) result:

=> 754263 (3)
=> 753426 (3)
=> 753246 (1)
total = 7 moves.

So you need to be more explicit about how you convert it to a coursing order, no? It would appear
at first glance that you need to convert it as a backstroke in Plain Bob.

This just seems like a lot of confusion to me. I like John David's suggestion better, since it is
a simple counting procedure that essentially amounts to figuring out how many calls it would take
to get back to rounds. (The call 'rounds' notwithstanding!)

This especially becomes evident on more than eight bells, at least to me. The coursing order of
the row


is not immediately obvious to me, but it seems really easy to calculate 7+8+7+0+6+4+1+1+1+2+0+0 =
37, an odd number. I suppose it does partly depend on taste, but I think I'd recommend this way to
someone rather than using a coursing order.


Mike Schulte
Sewanee, Tennessee, USA

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