Sweet, no more tech difficulties. I tried to respond to Supra's first post; I'll do so now (I saved what I wrote).
"2727716533, In this position, why is 3 a winning move for Red but not 2? I wanna hear your thoughts... I have my own thought I won't post yet. ^^"
The answer to your "why" is that 2 makes the B5 square immediately available (and by process of elimination you can tell that that's the only possible reason, without having to even know why B5 will matter). But I think you already knew that and are wanting to know how one might look at the position and see ahead to B5's importance and why 2 would fail.
First thing I glance for: would a horizontal chain of 2 on the left side be enough to eventually form a threat?
Normally, yes, because there are 3 odd squares A3, D3 and E3 that could form the threat. Odd number of odd squares means Red gets the first one.
But next thing I notice is Blue's potential on Row 2. At some point, Red will need to take E2 or F2 to prevent a Blue D2 threat. If Red goes 3 followed by 4, Blue can go 5, followed by Red 6 and Blue 5, so now Blue has the aforementioned E3 square. Blue's 2nd-row minor threat allows her to take the E3 square and reduce Red's potential threat to one with only 2 odd squares, meaning he won't be able to make a 3rd-row left-side threat in the absence of other factors.
Of course, Blue can still do all that if Red goes 3 instead of 2. But we've figured out already that 3rd-row left-side won't be enough for Red to win. He'll either need to combine it with something or he'll need the 5th row. So already, in the heat of making a decision during a game, I'd rule out 2 (given a minute-per-move clock at least). What's the point of 2 if the Row-3 threat won't be sufficient anyway, and going 2 might mess up my offense elsewhere (like on the 5th row which is where B5 would come in)? It seems there's nothing to gain by it and much to lose.
More analysis is needed to actually have the position figured out, but I think that addresses Supra's specific question.
