emailed my lecturer and he said 2 bytes is all u need 1 for each position 1 for each symbol, however 9 squares 9 potential symbols 9x9 = 18 thats 2 bytes and 2 bits?
Let your lecturer know that he is an idiot. If I've done the little mathematics needed and accounted for mirroring correctly, you only 15 bits.* Are you expected to save that last bit as well?

I have contributed all that I can to this goose chase; with about 10 lines of code he could take what I gave him and be done.
O_o

Yep. I read this stupid thread thinking I might find something interesting. I found a solved problem and a pedantic fool causing confusion. Lovely.

Soma

*Edit: I just realized that a much easier option exists. ;_;