Construct a tetromino by attaching two 2x1 dominoes along their longer
sides usch that the midpoint of the longer side of one domino is a
corner of the other domino. This construction yields two kinds of
tetrominoes wit opposite orientations. Let us lacc them S and Z
tetrominoes respectively.

    . X X   X .        X X .   . X
    X X .   X X        . X X   X X
            . X                X .
   S-tetrominoes      Z-tetrominoes

Assume that a lattice polygon P can be tiled with S-tetrominoes. Prove
that no matter how we tile P using only S- and Z-tetrominoes, we
always use an even number of Z-tetrominoes.

-------------------------------------------

How could there be an ambiguous shape?
. . . . . . . .
. . . . A A . .
. . B A A C C .
. . B B C C . .
. . . B . . . .
. . . . . . . .
This is not ambiguous, I can always find a peak.
I need shapes without peaks.
First C, then A, then B, let's try more:
. . . . . . . .
. . . A A B . .
. . A A . B B .
. . D . . . B .
. . D D . C C .
. . . D C C . .
. . . . . . . .
Well, this can be tiled with Zs only because of the
symmetry but it also looks a lot coincidental.
. . . . . . . . . .
. . A A . . G G . .
. A A C C G G H . .
. B C C F F I H H .
. B B F F J I I H .
. . B E E J J I . .
. . E E K K J . . .
. . . K K . . . . .
. . . . . . . . . .
Hm... Can I use some Zs in it?
Oh! Realization about the first example!
The peak can be also of a Z:
. . . . . . . .
. . . . A A . .
. . B C C A A .
. . B B C C . .
. . . B . . . .
. . . . . . . .
Actually, any shape
. . . . . .
. . X X . .
. X X X X .
. . X X . .
. . . . . .
Can be tiled with two Ss, or two Zs.
Now let's go back to the original problem --
why cannot we exchange one S for one Z?
There is likely some sort of parity invariance...
Let's see how a shape with a single Z can look like
(start with Z, letter Z in the middle)
. . . . . . . . . . . .
. . . E E A A G . . . .
. F E E A A B G G J . .
. F F D Z Z B B G J J .
. . F D D Z Z B I I J .
. . . . D C C I I . . .
. . . . C C H H . . . .
. . . . . H H . . . . .
. . . . . . . . . . . .
Now, let's try different tiling, start with the shape
. . . . . . . . . . . .
. . . * * * * * . . . .
. * * * * * * * * * . .
. * * * * * * * * * * .
. . * * * * * * * * * .
. . . . * * * * * . . .
. . . . * * * * . . . .
. . . . . * * . . . . .
. . . . . . . . . . . .
And tile:
. . . . . . . . . . . .
. . . B B H X X . . . .
. A B B Z H H X X E . .
. A A Z Z Y H G G E E .
. . A Z Y Y G G F F E .
. . . . Y D D F F . . .
. . . . D D C C . . . .
. . . . . C C . . . . .
. . . . . . . . . . . .
Three Zs: X,Y,Z.
It is working.
But the fundamental difference between
odd Zs and even Zs is still unclear.
What if we allow multi-shapes (with more layers)
such as
.. .. .. .. ..
.. .. AB A. ..
.. AB AB .. ..
.. .B .. .. ..
.. .. .. .. ..
Tiles the shape
. . . . .
. . 2 1 .
. 2 2 . .
. 1 . . .
. . . . .
And also, we can try to take it modulo 2
(when the question is about parity).
Then, I can compose Ss to the shape
. . . . .
. . . X .
. . . . .
. X . . .
. . . . .
Which means that all the following squares are equivalent:
. . . . . . . . . . . . .
. . . X . . . X . . . X .
. . . . . . . . . . . . .
. X . . . X . . . X . . .
. . . . . . . . . . . . .
. . . X . . . X . . . X .
. . . . . . . . . . . . .
. X . . . X . . . X . . .
. . . . . . . . . . . . .
(and shifts)
So how many equivalence classes do we have?
A B C D A B C D A B C D A
E F G H E F G H E F G H E
C D A B C D A B C D A B C
G H E F G H E F G H E F G
A B C D A B C D A B C D A
E F G H E F G H E F G H E
C D A B C D A B C D A B C
G H E F G H E F G H E F G
A B C D A B C D A B C D A
Eight: A,...,H
If it works, I guess that gaussian elimination on Z2 could
solve it automatically. Let's continue as a human being.
(it is still possible that it does not work)
The different orientations
of S are already equivalent through this
equivalence. What are the available positions?
a B C d     a b C D     A b c D     A B c d
E F g h     e F G h     e f G H     E f g H

e F G h     e f G H     E f g H     E F g h
C D a b     c D A b     c d A B     C d a B
That is nice, the lower row is identical to the bottom one.
It is hopefull.
Can we find something simple that would distinguish it from
the Z shape? For example the following?
. . * .
. * . .
It intersects every S-shape in two or zero positions:
. a A .    . . A a    a . * a    a a * .
a A . .    . A a .    . * a a    a * . a
What about Z-shapes?
z z * .    . z Z .    . . Z z    z . * z
. Z z .    . * z z    z * . z    z Z . .
Good, it works. Problem solved!
So the oficial solution would use the following pattern:
. . * . . . * . . . * . .
. * . . . * . . . * . . .
* . . . * . . . * . . . *
. . . * . . . * . . . * .
. . * . . . * . . . * . .
. * . . . * . . . * . . .
* . . . * . . . * . . . *
. . . * . . . * . . . * .
. . * . . . * . . . * . .
It intersects every S-shape in two or zero squares,
every Z-shape in one square, so the number of intersections
between a given shape and this pattern determines the parity
of necessary Zs to tile it.