I have a question about a go problem:

Dong-Dong grouping > (1) ~10k_6350 > (3) 30k~25k > (5) > 2824

Found at GoChild:http://gochild2009.appspot.com/?locale=en_US

http://bayimg.com/pAlAJAAde

The two solutions given are:
1) B A9
2) B D9, W E5, B A9

Is this the most favorable way to play for both sides regardless of
whether black or white has more ko-threats?

I was thinking that if white has at least 1 ko-threat more than black,
I'd expect white to set up a ko by playing at A8 instead of E5 in case
black tries the 2nd solution in vain.

Is there a difference in the number of ko-threats needed by black to
kill white in the corner between the
given solutions?

It seems to me that black would need at least 2 ko-threats more than
white for the 1st solution to work and only at least 1 ko-threat more
than white for the 2nd solution to work.

Kind regards and thanks in advance for any ideas, Niek

It seems to be the most favorable way and if white tries to set up ko with A8 instead of E5 as you suggest then white needs to find the first ko threat which seems to be to be a waste. (White will win if he fills all the outside liberties so black has to start the ko, if white starts, the ko is easier for black since white uses 1 more threat than he otherwise would).

Just FYI this thread will most likely be moved since this is Game analysis. (Study room is appropriate I think).

W wouldn't want to throw in to start the ko in any case; he knows that B will have to start the ko eventually (W can win by filling outside liberties, so eventually B will have to throw in at A9 and W will get to make the first capture without making a ko threat).

The ko is always direct for W, since W can ignore one threat and capture at B7. On the other hand, if B throws in at A9 immediately he has an approach ko and has to ignore two threats; if he plays D9 first, he only has to ignore one threat. -- So I can't figure out why the first solution would ever be better for B.

Great question, Eagle Eye!

This is one of those rare semeai where the correct play (by Black) is a mutual liberty, even though Black can play an outside liberty.



As for the number of ko threats that Black needs to win the ko, with the correct solution Black needs only one extra (primary) threat. With the incorrect variation he needs one or two, depending upon how the play goes. (White may play elsewhere after in the incorrect variation.)

_________________
The Adkins Principle:
At some point, doesn't thinking have to go on?
— Winona Adkins

Visualize whirled peas.

Everything with love. Stay safe.

D9,E5,A9 is correct sequence because it will only need one more ko threat.
if you start with ko...you will need two ko threat

_________________
"The more we think we know about
The greater the unknown"
Words by neil peart, music by geddy lee and alex lifeson

I think my impression that white might want to play at A8 was motivated by the idea that if black plays at A8, he prevents white from making two eyes, while if white plays at A8 and wins the ko, he'd have two eyes. But perhaps that reasoning would only apply when black's A5 group would be unconditionally alive (or at least having sufficient liberties on the outside to kill white from the inside).

I wasn't sure where to post this thread, but it's fine with me if an admin or moderator decides to move it at some point.

But if A5 were alive or has superfluous liberties, then Black kills unconditionally with A8. No need to risk a ko at all.

I did some counting in order to reproduce Bill magic.
From his diagrams we get:
B gets 15 points if B starts and B wins the ko b = 15
W gets 28 points if B starts and W wins the ko w₁ = -28
W gets 28 points if W starts w₂ = -28
So, maybe accidentally, here applies w₁=w₂=w
Bill has count c = -13_2/3 and temperature t = 14_1/3.
We find: b = c + 2t and w = c - t or inversed:
t = (b-w)/3 and c = (b + 2w)/3 = t + w
I won't try to explain the concept temperature. Too hot for me! Nor what if the w's are unequal.
edit: I don't care to much but the study group is the best place for this topic IMHO.

Give Black one more outside liberty and White's play is normally to throw in and make ko.

_________________
The Adkins Principle:
At some point, doesn't thinking have to go on?
— Winona Adkins

Visualize whirled peas.

Everything with love. Stay safe.

If White is komaster and Black plays first, White can play elsewhere or pass. So when White is komaster the corner is worth 29 pts. for White.

_________________
The Adkins Principle:
At some point, doesn't thinking have to go on?
— Winona Adkins

Visualize whirled peas.

Everything with love. Stay safe.

Understood. But I am trying to decypher your magics and it seems you only used 15 and -28 to calculate count and temperature.
BTW Are count and miai value the same concept?

When White is komaster Black is effectively dead. But when Black is komaster White takes one net move to kill (3d variation above), for a score of -28. Or Black takes two net moves to make and win the ko (1st variation), for a score of +15.

Each move in the ko gains on average (15 - (-28))/(2 + 1) = 43/3 = 14 1/3. The local count is -28 + 14 1/3 = -13 2/3.

Es claro?

_________________
The Adkins Principle:
At some point, doesn't thinking have to go on?
— Winona Adkins

Visualize whirled peas.

Everything with love. Stay safe.

Si, claro! Muchas gracias y keep smiling