On Wed, Mar 03, 2004 at 11:32:24AM -0600, Riley wrote:
> What strikes me as particularly odd is the fact that we all
> needed to add a certain amount to the above expression,
> namely @F/2 - 27, where the y-intercept is 27 (or 3**3).
Actually no. Those who used @F/2 needed to add 28 (since @F/2 is too
big...)
> I guess my question is this: Is this just a coincidence, the
> fact that our y-intercept is also a power of 3, and not
> just any power of the three, but the next in sequence?
It's not a coincidence, but I don't think there's any deep meaning
there either. It's just the way that the maths work out [obviously
everyone did the maths, or I missed some really easy shortcut :-)].
Blatantly reused from my blog [1]:
[quote]
The way I thought of the formula was this: Given an index I into the
set of cards (for example with the 21 cards below, 0-20) and the pick
P (1-3) we need a formula for determining from I and P the index which
would get translated to I.
0 1 2
3 4 5
6 7 8
9 10 11
12 13 14
15 16 17
18 19 20
Let's determine this by hand for for the interesting elements (we're
only interested in cards that are rearranged into the middle third):
7 8 9 10 11 12 13
1 0 3 6 9 12 15 18
2 1 4 7 10 13 16 19
3 2 5 8 11 14 17 20
Obviously ( "so the professor said 'I thought about this for three
days, and you were right; it is obvious'"... ) our formula looks like
I' = P + 3I - 22. To generalize this we just note that 22 is the
number of cards + 1 (which in turn happens to be @F-2). Now, to solve
the problem we just need to remember that the card that we're
interested in ends up in the middlemost position (i.e. for 21 cards in
index 10, @F/2-2), and we can just repeat the formula three times to
find out the original position:
I_0 = @F/2-2
I_1 = P_3 + 3*I_0 - (@F-2)
= P_3 + 3*(@F/2-2) - @F + 2
= P_3 + @F/2 - 4
I_2 = P_2 + 3*I_1 - (@F-2)
= P_2 + 3*(P_3 + @F/2 - 4) - @F + 2
= P_2 + 3*P_3 + @F/2 - 10
I_3 = P_1 + 3*I_2 - (@F-2)
= P_1 + 3*P_2 + 9*P_3 + @F/2 - 28
[/quote]
The constant I_x is always -(3^i*x + 1), since the operations we do on
it (multiply by 3, add 2) always transform it back to the same
form (just incrementing "i" by 1). If instead of @F/2 we use something
that happens to be smaller (for example s/ ..(.)//g), we can drop the
"+1" from that, and just get 3^i.
Or were you asking about something else? :-)
[1] <http://www.iki.fi/jsnell/blog/>, currently has mainly
explanations on some recent minigolf solutions, so I don't
feel too bad about mentioning it here... ;-)
--
Juho Snellman
