Solving a Mathematical Puzzle: Ways to Fill an Odiosis Board

[TheFallen018]

Hi,

First of all, I'm not sure what section to put this in, so if this is in the wrong section, please accept my most sincere apologies.

I'm trying to solve this puzzle I've come against, and I'm not quite sure what the best way to go about it is. Here's the question.

View attachment 7980

So, it seems to me that the first thing you'd do is take the maximum number of combinations, which I'm guessing should be 2^4036, due to there being 4036 boxes that each have a possible 2 combinations. Let's take the example with A and B, since the top row is 1 in both boxes, A cannot equal B. Therefore, if A = 0, B has to be equal to 1, or vice versa. By that logic, that takes two possible combinations out of a 4x4 block, making the block instead of having 2^4 possibilities, it now has 2^3 possible combinations. By that logic, there should be 2^(3027) possible numbers.

I feel though that I've missed a good number of constraints. What do you guys think?

Thanks.

 

[castor28]

[sp]
An equivalent way of stating the rules would be that no two adjacent columns can be the same. That gives you 4 ways to fill the first column, and 3 ways to fill each of the remaining columns, for a total of $4\times3^{2017}$.
[/sp]