What I would have done is started filling in letters randomly and every time a C or W ends up next to an O, choose the same letter or an O to put on the opposite side of the O.
Its hard to prove, but I’m pretty sure there isn’t a situation where a space can’t be filled in with this algorithm.
What I would have done is started filling in letters randomly and every time a C or W ends up next to an O, choose the same letter or an O to put on the opposite side of the O.
Its hard to prove, but I’m pretty sure there isn’t a situation where a space can’t be filled in with this algorithm.