
SOMA News 
15 Feb 1999
EMail. 
The WWall 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 nonplanar 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 20191230, 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.
So this DONT WORK
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.
so B IS NOT POSSIBLE
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
C IS IMPOSSIBLE
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.
So D3 IS IMPOSSIBLE
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:
D IS IMPOSSIBLE