Hi,

I'm working through 321go, and I'm finding it very good.

Now I have reached a chapter on reducing liberties, and I'm doing almost all exercises wrong.

Please have a look at the attached picture.

The task is to give the best move to reduce black's liberties.

I thought it was A, because A blocks black in the move towards the centre.
But the website wants C.
I don't understand this because to me it seems that pushing black to the side of the board is better than removing liberties there while leaving the way to the centre open.

Could someone please explain?
Thanks!

Playing at C connects your two white groups.

Some of these examples might be a little advanced, but take a look at:

https://senseis.xmp.net/?LameDuGo
https://senseis.xmp.net/?SqueezingOutTheToothpaste

Hitting someone on the "nose" at 1 in the diagram below is usually bad unless you think you can actually kill them or you have a specific tactical purpose you have read out. This is because they can just give you a ripped keima in response.

Click Here To Show Diagram Code

[go]$$Wc Ripped keima between marked stone and W1, white squeezes the black toothpaste.
$$ | . . . . . . . . .
$$ | . . . . . . . . .
$$ | . . . 1 . . . . .
$$ | . . 2 X . . . . .
$$ | . . W X . . . . .
$$ | . . O X O O . . .
$$ | . . O . . . . . .
$$ | . . . . . . . . .
$$ | . . . . . . . . .
$$ --------------------[/go]

Of course in this position, they could also just play here.

Click Here To Show Diagram Code

[go]$$Wc Super ripped-apart shape between the marked W stones.
$$ | . . . . . . . . .
$$ | . . . . . . . . .
$$ | . . . 1 . . . . .
$$ | . . . X . . . . .
$$ | . . W X . . . . .
$$ | . . W X W O . . .
$$ | . . W 2 . . . . .
$$ | . . . . . . . . .
$$ | . . . . . . . . .
$$ --------------------[/go]

This problem seems a little strange to me to be given the aim of "reduce liberties". "What is the best move" would surely be C or clealer aims would be "connect your groups" or "stop Black splitting your groups". Connection is the reason C is important, not liberties.

In terms of liberties A and C both leave black with 4 liberties. A is actually better in terms of reducing the amount of liberties Black can have after 1 more move: with white C black A next gives 6 liberties, with A nose tesuji-or-not the most liberties Black can give his chain is 5 with C. So in that respect A is the correct answer to the stated problem.

Actually, I wonder if it's not that bad. I mean, the natural continuation is this

Click Here To Show Diagram Code

[go]$$Wc Black stones are heavy
$$ | . . . . . . . . .
$$ | . . . . . . . . .
$$ | . . . 1 . . . . .
$$ | . . 3 X . . . . .
$$ | . . W X . . . . .
$$ | . . W X W O . . .
$$ | . . W 2 . . . . .
$$ | . . . . . . . . .
$$ | . . . . . . . . .
$$ --------------------[/go]

And here the black group is under severe attack. If white has support on the right, it may be playable (white two stones could be considered as light)