# Beginner's Query

In the attached diagram does white have two eyes that are solidly connected or can black play where the question mark is ?

Black can play there if white didnât capture a black stone at the question mark at his last move.
Thats called a ko.
http://senseis.xmp.net/?Ko

It doesnât really matter in this position though. Eyes are not that important when the stones in question can still run away or make a base. Here it is more likely that white captures black than the other way around.

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there is no eye where the ? is.

a stone on the edge of the board only has 3 liberties, the stone on the left has only 1 liberty left and therefore is in atari (can be captured with the next move). to prevent capture white has to play at the ? herself, filling up the âeyeâ. â> no eye in both cases.

black also doesnt commit suicide by playing there, because capturing takes priority over that.

EDIT: contact me if you need more of an introduction. ill be glad to help!

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What you see at the â?â point is called a false eye. A false eye is a point that looks like an eye, but really isnât one. This is because Black can play at the â?â point and capture a stone in the process. (White canât capture back because of the rule against repetition.)

You should know that the other âeyeâ isnât a true eye either. Black could make it false by playing on the same points as on the left part. The point isnât a true eye until White plays something to prevent Blackâs upper move.

For further discussion, see http://senseis.xmp.net/?RecognizingAnEye

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I think youâre giving him far more information that he asked for. I think the question regards the fact that if black plays at the question mark, both players will be surrounded, and would die.

Black can play there because the action of the active player takes precedence, i.e., even if both players surround the other, the black player removes the white stone first, creating a liberty, and therefore lives.

I also think identifying true eyes is complicated, but to me, an easier way to think of it is:
In order to live, each of your groups (stones that are connected) need to be adjacent to at least two areas that are completely surrounded by your stones.
I think that way of thinking even makes two headed dragons easy to understand (http://senseis.xmp.net/?TwoHeadedDragon).

If you look at your example it has five white groups (because none of the stones are connected). The stone to the right of your question mark is adjacent to two areas that are completely surrounded by white (to the left and right). However, none of the other white stones are. They are only adjacent to one such area. Therefore, the white stones are not safe.

I would re-word that, considering this context is ambiguous when we talk about groups âlivingâ in the sense of having two eyes, when that black âgroupâ would not gain an eye by capturing.

Thanks for the replies, but I am still puzzled. Some of you seem to recognise that White has two false eyes and others say that there are no eyes at all. My tentative thoughts on the matter were that white had two eyes which were solidly connected by a string on one piece, but I wasnât sure, so asked the question.
If there are two eyes solidly connected by a string of one piece than black would not be able to play into the eye at the ?
So I am afraid that I still really do not understand what you are trying to tell me.

Your stones are not connected. Only the orthogonals connect stones.

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Hello @Innes_Ewen,

have you gone through The **I**nteractive **W**ay **T**o **G**o already? Itâs available in many languages, I strongly recommend it âŚ IWTG should be a required course for all Go players, IMO, and I think youâll be able to answer this question (and many more) after finishing it. Plus, itâs fun (IMHO)

Good luck and lots of fun on your journey into Go!

A false eye is not an eye. Itâs just a clever name for something that looks like one.

This is explained in the first link I posted: http://senseis.xmp.net/?FalseEye

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False eyes and no eyes are basically the same thing. Black can play at the ? and kill the white stone, because the white stone only has 3 liberties, 2 of which are occupied by black stones. Stones that are diagonal are not solidly connected.

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