I am confused by the following problem given in Graded Problems for Beginners (problem 219 on p45)...

Click Here To Show Diagram Code

[go]$$B
$$ --------------
$$ | . . . . . . .
$$ | . . . . . . .
$$ | . . X . X X X
$$ | . . . X 2 O O
$$ | . . X O 4 O .
$$ | . . X O 3 5 1
$$ | . . . X O O O
$$ | . . . X X X .
$$ | . . . . . . .[/go]

I can't work out a refutation for 1 at 5 instead, all the variations I can think of lead to the above diagram or to White being dead again... is there something easy I'm missing out (probably)? Thank you

Click Here To Show Diagram Code

[go]$$B
$$ --------------
$$ | . . . . . . .
$$ | . . . . . . .
$$ | . . X . X X X
$$ | . . . X . O O
$$ | . . X O 3 O .
$$ | . . X O 4 1 2
$$ | . . . X O O O
$$ | . . . X X X .
$$ | . . . . . . .[/go]

White is alive. (It sort of looks like there is a false eye, but since both white strings share the two liberties they can't be captured.)

Also, welcome!

Thanks! Either I missed 2, or it didn't register as life somehow

And thanks for the welcome

Click Here To Show Diagram Code

[go]$$B
$$ . . . . . . . |
$$ . . . . . . . |
$$ . . X . X X X |
$$ . . . X 3 O O |
$$ . . X O . O . |
$$ . . X O 2 1 4 |
$$ . . . X O O O |
$$ . . . X X X . |
$$ . . . . . . . |[/go]

Click Here To Show Diagram Code

[go]$$B
$$ . . . . . . . |
$$ . . . . . . . |
$$ . . X . X X X |
$$ . . . X 4 O O |
$$ . . X O . O 6 |
$$ . . X O 2 1 3 |
$$ . . . X O O O |
$$ . . . X X X 5 |
$$ . . . . . . . |[/go]

Here are two other variations to live.

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[go]$$B
$$ . . . . . . . |
$$ . . . . . . . |
$$ . . X . X X X |
$$ . . . X . O O |
$$ . . X O . O C |
$$ . . X O . 1 2 |
$$ . . . X O O O |
$$ . . . X X X . |
$$ . . . . . . . |[/go]

OC, here is best, forming two eyes right away, and leaving no sente for Black.

Very often when making two eyes after an inside play, the right play is next to the inside play. The reason is that, as here, you make one eye by capturing the inside stone and another eye on a diagonal from it.

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