Life In 19x19

Posted: **Mon Dec 07, 2020 3:48 am**

I'd like to publish this go problem, which in the past would have been too hard for me to solve but now is surprisingly easy. I think I've become better at approaching L&D problems but it might be a kind of selective bias. Anyway, here's the problem

Click Here To Show Diagram Code: [go]$$B Advanced problem $$-------------------- $$| . . . . . . . . . . $$| . X X . X . . . . . $$| . O O X X . . . . . $$| X . O . O X X . X . $$| O . . . O O O X . . $$| . O O . . . X . . . $$| . X . O O . X . . . $$| . . X X X X . . . . $$| . . . . . . . . . . $$| . . X . . . . . . . $$| . . . . . . . . . .[/go]

Can I ask you: how many starting moves do you consider? How many moves do you have to read into the variation?

Here's my solving procedure:

The fundamental approach to tsumego, not only to solve it but to direct your reading, is "reduce space first, then play (find) the vital point". In this problem there are 3 places where we can reduce the eye space:

Click Here To Show Diagram Code: [go]$$B Fundamentals $$-------------------- $$| . . . . . . . . . . $$| . X X . X . . . . . $$| 1 O O X X . . . . . $$| X 2 O 3 O X X . X . $$| O . . 4 O O O X . . $$| . O O . 6 5 X . . . $$| . X . O O . X . . . $$| . . X X X X . . . . $$| . . . . . . . . . . $$| . . X . . . . . . . $$| . . . . . . . . . .[/go]

After all three exchanges (in any order) White is alive. Which of these exchanges is then problematic? Not 5-6: in this order it's irrelevant. Not 1-2 which is more difficult to appreciate: if White captures two stones, Black can throw in, so there's no eye at

. Without the exchange, White can play there and create an extra eye. But 3-4 looks problematic, because 3 forces Black to play what seems to be the vital point, dividing the eye space into two eyes.

So what if we postpone the problematic exchange?

Click Here To Show Diagram Code: [go]$$B Solution $$-------------------- $$| . . . . . . . . . . $$| . X X . X . . . . . $$| 1 O O X X . . . . . $$| X 2 O 6 O X X . X . $$| O . 7 5 O O O X . . $$| . O O . 4 3 X . . . $$| . X . O O . X . . . $$| . . X X X X . . . . $$| . . . . . . . . . . $$| . . X . . . . . . . $$| . . . . . . . . . .[/go]

Then we have found a solution! Now all that remains is checking what happens if White doesn't answer at 2 or 4

Posted: **Mon Dec 07, 2020 4:39 am**

The above was the "solving procedure", finding a solution. The remainder is "checking procedure", confirming the solution.

Click Here To Show Diagram Code: [go]$$B Resist 1 $$-------------------- $$| . . . . . . . . . . $$| 7 X X . X . . . . . $$| 1 O O X X . . . . . $$| X 5 O 8 O X X . X . $$| O 3 6 . O O O X . . $$| 4 O O 9 . 2 X . . . $$| . X . O O . X . . . $$| . . X X X X . . . . $$| . . . . . . . . . . $$| . . X . . . . . . . $$| . . . . . . . . . .[/go]

Click Here To Show Diagram Code: [go]$$B Equivalent $$-------------------- $$| . . . . . . . . . . $$| 7 X X . X . . . . . $$| 1 O O X X . . . . . $$| X 5 O 2 O X X . X . $$| O 3 6 . O O O X . . $$| 4 O O 9 . 8 X . . . $$| . X . O O . X . . . $$| . . X X X X . . . . $$| . . . . . . . . . . $$| . . X . . . . . . . $$| . . . . . . . . . .[/go]

and so on ... Any way White doesn't answer

will follow a similar pattern.

Click Here To Show Diagram Code: [go]$$B Advanced problem $$-------------------- $$| . . . . . . . . . . $$| . X X . X . . . . . $$| 1 O O X X . . . . . $$| X 2 O 4 O X X . X . $$| O . 7 . O O O X . . $$| . O O 6 5 3 X . . . $$| . X . O O . X . . . $$| . . X X X X . . . . $$| . . . . . . . . . . $$| . . X . . . . . . . $$| . . . . . . . . . .[/go]

Likewise, resisting here allows White to reduce the eyespace to 3.

Click Here To Show Diagram Code: [go]$$B The crux $$-------------------- $$| . . . . . . . . . . $$| 3 X X . X . . . . . $$| 1 O O X X . . . . . $$| X 2 O 4 O X X . X . $$| O . . . O O O X . . $$| . O O . 6 5 X . . . $$| . X . O O . X . . . $$| . . X X X X . . . . $$| . . . . . . . . . . $$| . . X . . . . . . . $$| . . . . . . . . . .[/go]

The crux of the problem is understanding that

is sente. Black doesn't need to defend at

and if he does, White lives.

Click Here To Show Diagram Code: [go]$$B Advanced problem $$-------------------- $$| . . . . . . . . . . $$| 6 X X . X . . . . . $$| 3 O O X X . . . . . $$| X 4 O 5 O X X . X . $$| O 1 . 7 O O O X . . $$| 2 O O . . . X . . . $$| . X . O O . X . . . $$| . . X X X X . . . . $$| . . . . . . . . . . $$| . . X . . . . . . . $$| . . . . . . . . . .[/go]

A good variation to check is this one:

seems to work too. However

Click Here To Show Diagram Code: [go]$$B Catch $$-------------------- $$| . . . . . . . . . . $$| . X X . X . . . . . $$| 3 O O X X . . . . . $$| X a O . O X X . X . $$| O 1 4 . O O O X . . $$| 2 O O C . . X . . . $$| . X . O O . X . . . $$| . . X X X X . . . . $$| . . . . . . . . . . $$| . . X . . . . . . . $$| . . . . . . . . . .[/go]

thanks to the 1-2 exchange, White can now play

and create an eye at the circled point. Without the exchange, Black can play A to falsify the other eye.

Click Here To Show Diagram Code: [go]$$B Check $$-------------------- $$| . . . . . . . . . . $$| . X X . X . . . . . $$| 2 O O X X . . . . . $$| X 4 O . O X X . X . $$| O . 6 . O O O X . . $$| . O O 5 3 1 X . . . $$| . X 7 O O . X . . . $$| . . X X X X . . . . $$| . . . . . . . . . . $$| . . X . . . . . . . $$| . . . . . . . . . .[/go]

FInal check: what happens if

here.

Posted: **Mon Dec 07, 2020 6:43 am**

Click Here To Show Diagram Code: [go]$$B Advanced problem $$-------------------- $$| . . . . . . . . . . $$| . X X . X . . . . . $$| 1 O O X X . . . . . $$| X . O . O X X . X . $$| O . . . O O O X . . $$| . O O . . . X . . . $$| . X . O O . X . . . $$| . . X X X X . . . . $$| . . . . . . . . . . $$| . . X . . . . . . . $$| . . . . . . . . . .[/go]

a looks obvious, as a big point for both Black and White.

But let's do a little analysis.

Click Here To Show Diagram Code: [go]$$Bc One point eye $$-------------------- $$| . . . . . . . . . . $$| . X X . X . . . . . $$| . O O X X . . . . . $$| B . O . O X X . X . $$| O . . W O O O X . . $$| . O O C . . X . . . $$| . X . O O . X . . . $$| . . X X X X . . . . $$| . . . . . . . . . . $$| . . X . . . . . . . $$| . . . . . . . . . .[/go]

Finding potential one point eyes is a good heuristic for life and death problems, since eyespace is limited. OC,

lives with two eyes, but is the key point to make a one point eye. It also looks like a key point for a second eye, as well.

So suppose we start with the less obvious play at

?

Click Here To Show Diagram Code: [go]$$Bc One eye $$-------------------- $$| . . . . . . . . . . $$| . X X . X . . . . . $$| 2 O O X X . . . . . $$| B 5 O 3 O X X . X . $$| O . 4 1 O O O X . . $$| . O O . . . X . . . $$| . X . O O . X . . . $$| . . X X X X . . . . $$| . . . . . . . . . . $$| . . X . . . . . . . $$| . . . . . . . . . .[/go]

If

captures the

stone,

connects, and White can make only one eye. So, as we thought from the start,

at 3 looks like it's necessary for White to live.

Click Here To Show Diagram Code: [go]$$Bc Seki $$-------------------- $$| . . . . . . . . . . $$| 0 X X . X . . . . . $$| 3 O O X X . . . . . $$| X 4 O 2 O X X . X . $$| O . 5 1 O O O X . . $$| . O O 7 . 6 X . . . $$| . X 8 O O . X . . . $$| 9 . X X X X . . . . $$| . . . . . . . . . . $$| . . X . . . . . . . $$| . . . . . . . . . .[/go]

throws in at

means that Black is able to prevent two eyes, but the result is seki.

Click Here To Show Diagram Code: [go]$$Bc Try again $$-------------------- $$| . . . . . . . . . . $$| . X X . X . . . . . $$| 1 O O X X . . . . . $$| X 2 O 5 O X X . X . $$| O . 4 . O O O X . . $$| . O O C . 3 X . . . $$| . X . O O . X . . . $$| . . X X X X . . . . $$| . . . . . . . . . . $$| . . X . . . . . . . $$| . . . . . . . . . .[/go]

prevents the seki, as well as the eye at

.

makes a one point eye, but

prevents a second eye.

Life In 19x19

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