We can turn G&G into a math problem, which may be easier (or harder) to solve.
What we know:
1. Whenever we press a hex, it and the ones around it toggle. This means that the best solution doesn't press any hex twice.
2. Light and Shadow are indistinguishable, they both behave the same way. When labeling them 0 and one 1 it doesn't matter which is which
3. ##More to be added
We can label every hex with a number, like so:
In this case, there are 61 squares, so we label each with a number 1 - 61.
Whenever we press a number, some group of other numbers will toggle.
With this we can form a
matrix that represents interactions.
With
m spots, let's form a
m x m matrix. The entry a(
i,j) is 1 if
i changes when
j is pressed.
In our example of 61 spots, the matrix ends up looking like this (purple = 0, yellow = 1):
(Very neat!)
Now if we multiply this matrix by a vector of integers (whole numbers) (x1, x2, ..., xm), the result (
mod 2) is how the board would change if we pressed the corresponding hex x times.
Shadow/Light can be labeled by 1 and 0, light being 1 and shadow 0 (or vise versa)
Now to solve G&G, we want the change to be the same as the current board state or it's opposite. That way it becomes all the same.
Then the problem becomes this:
We need to find a vector of integers so that when multiplied with the interaction matrix equals the board state (mod 2).
That's what I spent the last few hours trying to solve, but math is hard.
Other notes: You may recognize the last equation as a way to represent linear systems of equations. That is this problem can also be seen as one.
