[Edu-sig] casino math (unit three) (Saturday night fever)

[Gregor Lingl]

Laura Creighton schrieb:

> I have to think about this more, but on a very quick first reading it
> seems that you have missed a strategy that I have needed in order to
> solve some tough PileOn games.
>
> And that is breaking up groups of 2 or 3 that you already have built, or 
> dealt.  Given these 3 piles to start with
>
> A B X X
> C D X E
> F G X B
> (and then some more)
>   
This pattern does not occur in my sample configuration:

    (   (11,  6,  2,  1),
        (11, 10,  4,  1),
        ( 9,  2,  6,  3),
        ( 9,  4,  8,  3),
        ( 7, 12, 10,  5),
        ( 7, 12,  2,  5),
        ( 5,  8,  4,  7),
        ( 5,  6, 12,  7),
        ( 3, 10,  8,  9),
        ( 3,  2,  6,  9),
        ( 1,  4, 10, 11),
        ( 1,  8, 12, 11),
        (13, 13, 13, 13),
        (),
        ()    )

it has only even ranks in the second and third column, occurring
pairwise in these columns, and odd ranks in the fourth one.
(The thirteenth row is out of the game) So initially there are no
three equal ranks in the third column and no X occurring in the
second or third column can appear in the fourth.
> the correct way to solve things may be to put E and B in the empty
> slots, take one X  from the first pile and put it on the second
> and take the other and put it on the third.  Then move the B
> from the empty slot back to pile one.  
This also does not occur in my initial configuration, as some B
in the fourth column can only reappear in the first one and not
in the second one. One (ond only one) move combining a seven from
the first column with two sevens from the fourth one appears in
my proof of non-solvability of this layout.
> The same thing has also
> been needed when piles are
>
> A X X B 
Also this pattern does not occur: each row has only two different
even ranks. Of course only an a bit closer investigation (as
discribed in my pevious posting) shows that these patterns also
do not occur during the game.
>
> as well.
>
> Does your algorithm handle that, or does it assume that things that
> are stuck together belong together?
No this is not assumed.

I do not have a special algorithm to solve Pile On, but only search
the game tree using backtracking. Normally that gets out of hand.
The point is that I believe to have found one *special* layout that
can be investigated completely by hand, because due to its
symmetries there occurs only a very limited set of possible moves.

I think, that still some labour is needed to retrace my arguments.
But if you are convinced (or only strongly believe) that I am wrong,
you have an easier way to proof it: simply solve the configuration
given above (and possibly show me, that resolving those patterns
you displayed above is needed to do so.)

Very thrilling!

Regards,
Gregor

>
> Laura
>
>