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You have 2 bishops, 2 rooks and 2 knights (of no colour). Place them in as small a rectangle as possible such that each piece can be captured by atleast 2 other pieces in a single move. Bishops have to be on opposing coloured squares.

I managed it in a 3x4 rectangle. Lets see if a better solution exists.

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The smallest solution by area is the $4\times 2$ rectangle. The smallest solution with respect to longest side is the $3\times 3$ rectangle.

(1) The $3\times 3$ rectangle works:

N e N
R B R
e B e

(where N=kNight, R=Rook, B=Bishop, e=empty)

(2) The $4\times 2$ rectangle works:

N R R N
e B B e

(3) No $1\times n$ rectangle can work

Proof: Bishops and knights do not attack any other squares on such a rectangle; rooks only attack two squares. But altogether we need $2\cdot6=12$ attacks.

(4) The $3\times2$ rectangle cannot work

Proof: Consider the two middle squares. No knight can attack them; every bishop and every rook can attack at most one one of them. For having each of them attacked twice, the rooks and bishops must occupy the outer four squares, and the knights must be on the middle squares. Impossible.

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Got a 3x3

K R B
  R B
  K

The hardest part was keeping the bishops on opposite colors.

Thinking of a 2x4:

K R R K
  B B

Here's a thought: what if we had a moebius board?

B R B
K R K
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    $\begingroup$ On a moebius board, you cannot consistently define white and black squares. $\endgroup$ – Gamow Jan 30 '15 at 15:44
  • $\begingroup$ True. But it was just a stray thought. $\endgroup$ – JonTheMon Jan 30 '15 at 15:46
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    $\begingroup$ @Gerhard: If a Moebius board has an odd number of squares in one dimension and an even number in the other, why couldn't one define white and black? All squares would be horizontally and vertically adjacent only to squares of the opposite color. $\endgroup$ – supercat Jan 30 '15 at 18:43
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    $\begingroup$ @Tibos the question specified "at least 2 other pieces in a single move". $\endgroup$ – Thebluefish Jan 30 '15 at 19:52

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