Hopefully in the spirit of the Fortnightly Challenge 3rd Sept 2018 - Reusing Information
If we are given a binary string, say:
1100010101100011
and we wish to transmit it safely, we might use 'double reading'. This reads the string twice, to produce, using the example above, say:
(11)(10)(00)(00)(01)(10)(01)(10)(01)(11)(10)(00)(00)(01)
This decreases the chance of data corruption when the data is transmitted.
This puzzle involves a two dimensional version, but we hit a snag when we try to encode.
If we wish to transmit a $3\times3$ block, we encode by placing a $4\times4$ grid on top of it:
. . . . x x x . . . . x x x . . . . x x x . . . .
The x's are the data to transmit, and the dots are the encoding.
Unfortunately, encodings aren't unique! And nor are encoding methods!
One encoding method is to say if an 'x' is $0$, then it has an even number of its diagonally neighbouring '.''s set to $0$, otherwise set an odd number of $0$'s. Another is to use number of neighbours.
So if the block to encode is:
. . . . 1 1 0 . . . . 0 1 0 . . . . 0 1 1 . . . .
then:
1 0 0 1 1 1 0 0 0 1 0 0 1 0 0 0 0 1 0 1 1 1 1 0 0
is a solution, and we would transmit:
100001100001100000000111111000
Your job is to find the number of solutions to:
. . . . . . . 0 1 0 0 1 1 . . . . . . . 0 1 1 0 0 1 . . . . . . . 0 1 0 1 1 0 . . . . . . . 1 0 1 1 0 1 . . . . . . . 1 1 1 0 0 1 . . . . . . . 1 1 1 0 0 0 . . . . . . .