SOMA Crystal
SOMA News 15 Feb 1999


The W-Wall has long been a pain for SOMA players. Why can't we solve it. It's such a simple figure, and all our sences tell us it should be simple, and yet we don't manage - WHY?

Let us first divide the Soma pieces into two groups - the planar pieces 1,2,3,4 and the non-planar pieces 5, 6 and 7.

Consider piece 7 first, Solutions have to have piece 7 in one of the positions of A, B, C, D. We do not have to test mirror projections around X and Y, so these 4 positions of piece 7 cover ALL possibilities.

Due to their shapes, 5 and 6 cannot occupy the same edge X, Y or Z therefore they must be at 2 different edges.

By 2019-12-30, while thinking upon another SOMA problem of "two colored faces". I realised that 5 + 6 actually 'can' occupy the same corner. If they are placed like this:
But - Alas, it does not help us, the figure is still unsolvable.

Consider position A:
The side P cannot hold piece 5 or 6. Piece 4 will leave a spot of 1 cube anyhow it is placed so 4 wont do. This leaves pieces 1,2,3.
Piece 2 can only be in position AA leaving a fixed spot for piece 1.
Piece 3 can only be placed as A1. A1 and AA give the same remaining shape, so we continue from A1.
As pieces 5 and 6 MUST be at differing edges at Y and Z we place piece 5 in A2.
This forces piece 6 into A3 or A4
There is NO WAY pieces 2 and 4 can fill out the remaining spaces.

If we continued A1 with piece 6 we get A5 with an impossible hole. or A6 and A7 which also leave impossible holes

Now consider the A8
Only Piece 6 will fit below piece 2, (piece 5 will leave a void) forcing piece 5 into positions A9 or A10, in neither case will pieces 4 and 1 be able to fill the void.
So the A method DONT WORK

Consider position B:
As piece 3,4,5,6 cannot be at side P this leaves 1 and 2 for placements B1 or B2
As B1 and B2 look identical, we continue using B2. Only piece 6 may be placed at the top, as piece 5 leaves a void that can only be filled by 1 or 2, and they are used already. Now this leaves pieces 3, 4, 5.
Piece 5 have to be at the edge Z, and the only place left is seen in B4, leaving a small spot under 6 for a single cube.

Consider position C:
Piece 5 have only 3 positions at left side.
In C1 NO pieces can fill the P shape at the left side.
In C2 have a small void under 5.
At C3 piece 2 and 1 are the only possibilities for continuing, they have only one possible arrangement as C4.
We then try to place piece 6, finding two possible positions, both displaying a small void, so this method DONT WORK.

Piece 6 have 3 possible positions C7, C8, C9, but only in C9 do P have a shape that can be filled, We do it by using piece 1 and 2. We then try to place piece 5 but the only two positions at C11 and C12 lead to impossible voids at the top

Consider position D:
We try to place piece 5 and find only 1 position at left side
(Remember that placing 5 in right side is a mirror of placing 6 at left, so we dont have to consider this way.)
using this we are forced to place 2 and 1 at the remaining left space. D2
Now we try to place piece 6 and find only one position possible, D3. but this would require piece 2 at right, and 2 is already in use.

Now we try to place 6 at left side, and find only one position, D4.
This position force us to use Piece 2 and piece 1 and piece 5 have only one possible place now. D5.
This leave a 1 cube hole under 7 so:

As you see, the W-WALL IS IMPOSSIBLE.

Proved and submitted by Thorleif Bundgård Feb. 1999.

BACK to news index