In article <[EMAIL PROTECTED]>,
"Riley" <[EMAIL PROTECTED]> writes:
> Hello All!
>
> I thought I'd open the following up to discussion.
>
> 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?
>
In problems in ternary it's often convenient to use a symmetric,
notation, so in this case name the columns -1, 0 and 1, and
number the cards in the original pile as
-n, -n+1....-1, 0, 1, ...n-1, n
Then the columns indices in reverse order are the ternary digits of the
position. In the example:
2,2,3 becomes (subtract 2, reverse) 1 0 0 ternary which is 9 decimal.
and indeed, QH is 9 right from the middle.
So since the number is n=9*a0+3*a1+1*a2, adding 1 to each of a0,a1,a2
corresponds to adding 9+3+1=13 to n. So the subtraction of 2 that's needed
corresponds to subtracting 2*13=26. And the middle of an array @F is at
(@F-1)/2, so generally the offset you will see is -2*13-1 = -27 (or -28
if you start from an array that's 2 longer).
in general, if you have a=1+3+...+3**(m-1), then 3a=3+9+...3**m, and
subtracting 2a=3**m-1, so a=(3**m-1)/2 (in our case m=3, so a=13).
And the (negative) offset was twice that+1, so 3**m (27 in our case).
So yes, it will always be a power of three, for the m-stage cardtrick it
will be 3**m.
