Life In 19x19

Euphony wrote:

mitsun wrote:W clearly has 7 liberties. Counting B liberties is slightly trickier. In order to capture, B needs to play moves at N19 and N15, so imagine those stones are already on the board. Now how many liberties does B have? The answer is 7 liberties. It might look like only 6, but to play those 6 moves, W needs an extra approach move (K19 to approach at the top or L15 to approach at the side). Since it is B to play, he should be able to capture W, simply by filling liberties.

I am interested in this approach, as currently I pretty much use the brute force method for situations like this. You mentioned imagining N19 and N15 are already on the board. How exactly does this work? Can you always do this in a capturing race and still count the correct number of liberties?

Thanks so much,
Euphony

My understanding was:

Imagining those stones are already on the board when counting black's liberties is the same as not counting shared liberties. It makes it slightly easier to count because then you can more easily see how many liberties black has left, but since two liberties = two moves, it doesn't affect the calculation. We can see that white has 7 liberties including shared ones, and black has 7 excluding shared liberties, so if black goes first he wins.

That's interesting, Splatted. So one way of calculating a fight is to calculate the total liberties of the opponent's group, and the liberties of your group after all of the moves are made on the inside, and then compare.

That seems like a sound strategy. I'll have to try it out.

Thanks for the explanation.

Wait, that strategy doesn't work in general at all. Filling internal liberties not only takes a move, it also reduces the total liberties of the group. It works in this situation to assume those stones are there because there are two internal liberties, and filling those internal liberties actually adds a liberty to the group in both cases. One is a bamboo joint, so filling the internal liberty adds a liberty. The other adds a liberty because it is on the edge of the board, and the opponent must use an additional move after the hane.

I am going back to brute force.

I think you've misunderstood. You Should only count the internal moves as played when counting one side. In this game we counted white as having 2 liberties on the inside, but when counting black we didn't count any inner liberties. This means that it is irrelevant that playing a shared liberty takes one from black because we didn't count those anyway, and by counting them as white liberties we know that black has time to play those imaginary inner moves.

Also, white would only have to make one extra approach move because the final one would be a capture, and that's irrelevant to whether or not this works because if the inside plays didn't force an extra approach move we would have counted black as having one less liberty and seen that he couldn't capture white.

Edit: From re-reading your earlier post it seems you were doing it the way I described so maybe this post is useless...

Ok, I think I didn't apply it right in the game I played. I'll have to try again.