[Retros] die Schwalbe 13237 - Gerald Irsigler

andrew buchanan andrew at anselan.com
Tue Jan 30 02:21:02 EST 2007


(SPOILER - my suggested solution.)

I agree with Roberto. I think the problem calls for a 3 stage process:

(1) Find an n-1 point independent set - not connected orthogonally,
diagonally or knightily :-).
(2) See if it's possible to add an nth point which will only be connected
to one other chosen vertex. if it's connected to two, then one must be
knight and the other non-knight. also it can't be adjacent to any other
chosen vertex.
(3) Verify that there are no strange retro issues.

Off the top of my head, the following n-1 set seems to work:

.x......
....x...
.......x
........
x.......
...x....
......x.
........

Then a1 is only connected to Xa4.
Or c1 is only connected to Xd3 & Xh6, but if a king there is checked, then
it can be a double check.

So either of these would work as a 7th point. Can anyone do better? I don't
think it's possible.

Cheers.
Andrew.

--- raosorio at fibertel.com.ar wrote:


>

>

>

> 13237 - Gerald Irsigler

>

> How many squares can you mark at most on a chess board so that every

> possible position with pieces on those squares (including wK and bK) is

> legal?

>

> I understand that the author assumes: a) no pawns on the 1st/8th rank

> b) one white king and one black king on the board.

>

> The first obvious restriction is that squares in contact are not posible

> to

> avoid kings in contact.

>

> But the hardest restrictions appear to avoid double illegal checks,

>

> 1) Double check over the same king: no double geometrical relation

> between one square and the others (orthogonal, diagonal and 2,1 knight

> type,

> unless some magic discovered check position appears)

>

> 2) Both kings under check (I'm having polemics enough in this respect due

> to

> our Die Schwalbe problem 13071): just one couple of squares in checking

> position (orthogonal, diagonal or 2,1 knight type). Otherway, both kings

> could

> be placed under check.

>

> Based on this, apparently 6 squares is the maximum (this is not a

> strict conclussion,

> it's what I found just "moving the hands").

>

>

>

>

>

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