y [mailto:[EMAIL PROTECTED]
> Sent: Wednesday, March 03, 2004 6:32 PM
> To: [EMAIL PROTECTED]
> Subject: Re: A new golf "Cardtrick" has started
>
>
> Hello All!
>
> I thought I'd open the following up to discussion.
>
> In most of the solutions, we ha
In article <[EMAIL PROTECTED]>,
Stephen Turner <[EMAIL PROTECTED]> writes:
> Maybe someone else can shed some more intuition here.
>
Why ? You already basically answered it :-)
An ofset of 2 on all digits is ternary 222, which is just 1
below ternary 1000, 3**m. And the one extra comes fro
I think it's a bit more obvious than Juho's making it.
I think it's reasonably straightforward that the formula is (a+3b+9c). To
see this, notice that groups of three cards end up in the same place in the
second deal, and groups of nine cards in the third deal. Or just try out
some values:
n = 15
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,
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
Hello All!
I thought I'd open the following up to discussion.
In most of the solutions, we have something similar to:
-lap $_=$F[s/.. / /g/2+$_+3*$&+9*$'-27]
(The above is Stephen Turner's)
All of the solutions contain an inner expression of the form:
9a + 3b + 1c, (or 3**2 a + 3**1 b + 3**0 c
