SOMA Crystal
SOMA News 27 Nov 2000
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Impossibility Proof for the Five Square

For a long time, SOMA players have wanted a proof, that the double set figure "Five Squares" really was impossible. So here it is, by Jerry Solinas. jsolinas@erols.com
(Originated J. Solinas, 1974)

The Five Square consists of a two-story center block and four wings of nine cubes each.
By an "outer wing" is meant the six cubes of a wing that are not adjacent to the central block.

The three non-flat pieces (#5, #6, #7) cannot contribute to an outer wing, since the distance from the second-story cubes is too great.
Thus, in any solution to the Five Square, the outer wings are comprised solely of flat pieces (#1, #2, #3, #4).

Moreover, the outer wings are far enough apart that no piece can contribute to more than one outer wing. Each outer wing requires parts of at least two pieces, so that at least eight flat pieces are required in all. But there are only eight flat pieces available. Therefore, each outer wing is comprised of the cubes of precisely two flat pieces.

To summarize so far: each outer wing contains cubes from precisely two flat pieces, and no other piece.
By an "outer central" cube is meant the center cube on the edge of a wing that is farthest from the center. [RED]

By a "wing central" cube is meant the central cube of a wing (which is contained in the outer wing). [GREEN]

No single piece can occupy both the outer central cube and the wing central cube in a given wing. For if it did, it would cut the the outer wing in half in such a way that we could not fill the outer wing with just one other piece.

In particular, neither Piece #3 nor Piece #4 can occupy an outer central cube.
Therefore, each of the four outer central cubes must be filled by a Piece #1 or Piece #2. Thus each of the four wing central cubes must be filled by a Piece #3 or Piece #4.

The only ways to make the outer wings are:
      422   411   322   311
      442   441   332   331
      .42   .4.   3.2   3..

Now to prove further by checker boarding, we conclude, that since Piece #1 cannot fill a wing central cube, then an outer central cube cannot be filled by the middle cube of a Piece #1.
Moreover, a wing central cube cannot be filled by the middle cube of a Piece #3.

Now checkerboard the Five Square figure in Black and White, in all three directions, so that the outer central cubes are White.
By counting, we find that there are 29 Black cubes and 25 White ones.

Pieces #2, #4, #5 and #6 each fill two Black cubes and two White ones. (either way we color them.) Piece #7 will be either 3 Black - 1 White or 1 Black - 3 White.
Since two wing central cubes (which are Black) are filled by non-central cubes of Piece #3, it follows that each Piece #3 fills 3 Black cubes and 1 White cube.

Since two outer central cubes (which are White) are filled by non-central cubes of Piece #1, it follows that each Piece #1 fills 1 Black cube and 2 White cubes.

(NOTE:Piece #7 can be 1 black+3 white or 3 black+1 white)

Piece: 1 2 3 4 5 6 7     1 2 3 4 5 6 7
Black: 1 2 3 2 2 2 1or3   1 2 3 2 2 2 1or3 = 26 or 28 or 30
White: 2 2 1 2 2 2 3or1   2 2 1 2 2 2 3or1 = 28 or 26 or 24

Adding them all together, the pieces fill an EVEN number of Black and White cubes.!

We expected 29 Black + 25 White. !!!

Since this disagrees with the totals of the Five Square.
The figure is impossible to build with the fourteen pieces.!




Written by Jerry Solinas, edited by Thorleif Bundgaard.

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