[r-t] Grandsire Triples
Mark Davies
mark at snowtiger.net
Fri Oct 22 21:37:20 UTC 2004
OK, so the correct numbers are 10+10x10+10x10x10 = 1110. Easier to count
Glint's way, too!
Actually I think it IS possible to argue, without iterating all cases, that
the linkage from three calls is good enough for calling round in Grandsire
Triples. What a call does it basically rip one bell out of the working
coursing order and replace it with the hunt bell. That's exactly what a
single does, a bob is almost the same except that the hunt bell doesn't go
back into exactly the same spot that was vacated, but one bell later in the
coursing order.
So... let's consider quite a bad case, where the 2nd is in the hunt already
but the other bells are not in the plain course. We can write the coursing
order like so:
(2) 7abcd
Now there are 23 such cases (the 24th is the plain course). The only
operation we have is to swap a working bell with the hunt. At first glance
three swaps doesn't seem like enough - the "worst" case is when abcd are in
the reverse order of what they should be, i.e. 6435 instead of 5346. In this
case at least three bells have to be moved, and so three swaps through the
hunt are not enough. However actually this is a trivial one, because it is
the reverse of the plain course - just put the three in the hunt with a
single and rounds is at hand.
This reversal get-out clause helps solve any combination where three abcd
bells have to move, such as 5643 or 4356 - if we go for the handstroke
finish, by symmetry only one bell is out of place. We can easily get round
if only one bell is misplaced, as there are only two possibilities: in a
five-bell coursing order, the out-of-place bell can be one or two positions
away from home, no further. If it (call it x) is one position away from home
we just swap it with its neighbour (y) as follows (all calls singles):
2 swaps with x, x is in the hunt.
x swaps with y, y is in the hunt
y swaps with 2, 2 is in the hunt - plain course
(Actually you can do this with just two calls - a bob and a single.)
If the x bell is two positions away from home, we need two bobs. For
instance, in coursing order 75634 the 6th is two positions from home, 75346.
If you call the 6th (x bell) into the hunt, then the 2nd back into the hunt,
magically the coursing order comes right:
(2) 75634 start point, 6th two positions out.
(6) 75324 call 6th into the hunt
(2) 75346 call 2nd back into the hunt, 6th will be in right place.
I reckon there are nine coursing orders for abcd where only one bell is out
of place. With the reverses of those, we get 18 orders accounted for, or 20
if we add the plain course itself and its reverse. So, only four coursing
orders to consider... and these must be the ones where two bells are out of
place. By inspection these are:
3564
3654
4563
4653
These look more difficult, but there are lots of ways of tackling them. The
first and last can be done by swapping the 7th into the middle of the
coursing order, getting either the plain course or its reverse straightaway.
Alternatively all of them can be tackled by this system:
1. Choose two out-of-order bells, x & y.
2. Pick up x into the hunt.
3. Dump x out, pick up y.
4. Dump y out, pick up the 2nd again.
You have to be careful which order you do x and y in, and choose bobs and
singles carefully to leave the bell coming out of the hunt in the right
place. However, this system works for all.
So that I think shows that the linkage is good if either 2 or 3 is in the
hunt. The only other case is if one of the 4-7 is in the hunt. In some ways
this is easier, because then both 2 and 3 are in the working abcd coursing
order, and we can choose which to pull out; you've sort of only got three
working bells left to consider. I won't bother with the analysis, but it
isn't going to be any more difficult than the above. Percy can work it out
for himself.
MBD
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