CR TIP of faster way /baduk/

Now, does it come under your notice?

It can be counted 5-eye-space-number. 8
Also you can count that 12 - 4 = 8.

It can be counted 4-eye-space-number. 5
Also you can count that 8 - 3 = 5.

It can be counted 3-eye-space-number. 3

Also you can count that 5 - 2 = 3.

If there are 2-in-crosses ( triangle marks ), it's same of 1 lower phase.

So, it can be possible below.

one way : 12 - 3 + 4 = 13
another way : 8 + 1 + 4 = 13

It may be a particular thing, and also it may save your time.

CR Review test /baduk/

How many sequences do white take to pick out black stones?

PJ Review test /baduk/

Is there anyone else who solve these problems approximatly?
There is no exact answer.
It is intended to breed your positional judgement sensation.
I'll post a comment to first PJ Review test problem for instance.

Convinience below.

CR step 4. eye-space-number /baduk/

There is 3-eye-space-group.
White can remove it just 3 sequences "include A."
There calls "3" "eye-space-NUMBER" in this blog.

Because it doesn't count white B and black C.

What about 4-eye-space-group?

White A and black B doesn't count. 3-eye-space-group left. So, 2 + 3 = 5.
Eye-space are 4, but eye-space-number is 5.
It's very important. Let me explain last below.

Now, 5-eye-space-group.

At the same reason, white A and black B doesn't count. 4-eye-space-group left. So, 3 + 5 = 8.

Last, 6-eye-space-group.

Yes, eye-space-number is 4 + 8 = 12.

You must remember this result for speed.

three three, four five, five eight, six twelve.
three three, four five, five eight, six twelve.
three three, four five, five eight, six twelve.
three three, four five, five eight, six twelve.
three three, four five, five eight, six twelve.

1-eye-space-group, 2-eye-space-group, 3-eye-space-group : eye-space is equal to eye-space-number.
But,
4-eye-space-group, 5-eye-space-group, 6-eye-space-group :eye-space is NOT equal to eye-space-number.



These differences are related to below.

1-eye-space v.s 2-eye-space : small-eye-space v.s small-eye-space
1-eye-space v.s 3-eye-space : small-eye-space v.s small-eye-space

3-eye-space v.s 4-eye-space : small-eye-space v.s big-eye-space
3-eye-space v.s 5-eye-space : small-eye-space v.s big-eye-space
3-eye-space v.s 6-eye-space : small-eye-space v.s big-eye-space
4-eye-space v.s 5-eye-space : small-eye-space v.s big-eye-space
4-eye-space v.s 6-eye-space : small-eye-space v.s big-eye-space
5-eye-space v.s 6-eye-space : small-eye-space v.s big-eye-space

p.s If there is not in-space, it is meaningless.

It's not end of it yet.

PJ step 4. value one and one third (1) /baduk/

Two groups are same.
If black places A, there is nothing.
If white places A and B, there is 2 whte territories because of dead stone.
Total 3 sequences make 2 terrotories, so, one sequence makes two thirds.
Now, black places A, makes two thirds, but there is nothing!
Therefore, there exists already two thirds white territory.
And each side has chance to make two thirds, multiply 2 by two thirds, it's value is one and one third.

Suppose that white places A, and rightly black places B.
First white has two thirds, and white earns two thirds, and black earns two thirds.
Then, white must have two thirds, right?

Remember one black dead stone.
There is one third black cross on the board, and one black dead stone off the board.
So, one minus one third is two thirds.
I proved it.

Conclusion : Exists two thirds. / Value is one and one third.

It's not end of it yet.

CR TIP of easier way /baduk/

Out-space-advantage-side is white.
So, imagine x marks are filled with any stones.
And, calculate only out-spaces, and you can read it faster.
In this case, white-out-spaces are 5, and black-out-spaces are 6.
So, black wins the race.
And, white has 9 ko threats. ( out-spaces 5 + in-spaces 4 = 9 )

Imagine like this.

CR Review test /baduk/

There are two quizes.
Show your ability.

PJ Review test /baduk/

It's the quiz of positional judgement.
Which is advantage now?
Convinience below.

Is it difficult?

CR step 3. NES v.s NES (2) /baduk/

It's non-eye-space v.s non-eye-space and there are three in-spaces.

I will seperate two sides. White has more out-spaces. Black has less out-spaces.
First, more-out-space-side always shoulders in-spaces minus one.
Why minus one?
You can answer yourself if you have seen "CR step 2". ( hint : I said very important. )
Now, I provide you with convinience of that.
Triangle marks show burdens of black and square marks show that of white.

It is 4 to 3 for white. So, white can play elsewhere.

Second, if the burdens are SAME, more-out-space-side can win, but less-out-space-side can only make draw - it's the best plan.

It is 4 to 4.
white can win. because white has more out-spaces.

White can kill out with A.
But black can only make draw. because black has less out-spaces.

Conclusion : More-out-space-side always shoulders in-spaces minus one. / If the burdens are SAME, more-out-space-side can win, but less-out-space-side can only make draw.

cf. White can go elsewhere maximum ( never die ) the gap of out-spaces plus in-spaces minus one.

PJ step 3. value 1 /baduk/

There is the chance for white to make 1 cross.

But, if it's black turn, there is nothing.
So, there EXISTs a half white cross.
And if it's white turn, white can create 1 cross. If it's black turn, black can terminate a half white cross.
Therefore, its value is 1.

Conclusion : There exists a half white cross. / Its value is 1.

Then, I'll prove it.
There is two groups of a half white cross below.
So, total white cross is - a half plus a half - 1.

How about that?
Is it right?

p.s My English ability is some poor, so please point out my fault in using English.
I will rightly modify it. Thank you.

p.s I think that positional judgement is very very very important to fight to the finish in minuteness.
But its technique is nowhere! Even if Korea.
It seems some boring, complex, and so on... but, think about Lee Chang-Ho 9D in Korea.
His best arms is positional judgement ability!
And I think if one is strong of positional judgement, his other baduk skills follow.
Because baduk is the game of selection.
If I am of great advantage, I will concede and be safe to keep victory.
If I am of some disadvantage, I will shake ( "shake" means "make complex" ) all on the board to capture the chance of victory.
How about your opinion?

CR Review test /baduk/

I will not present black first or white first.
It's meaningless in my blog. You will know the truth of capturing race soon.
Show me your ability of capturing race.

PJ Review test /baduk/

Show your ability of positional judgement.
You can calculate decided-territory first, and add non-decoded-territory.


Convinience for you, show you below dia. ( decided-territory checked triangle marks. )

CR step 2. non-eye-space v.s non-eye-space (1) /baduk/

There are non-eye-space black and non-eye-space white.
And there is not in-space anywhere.
So, the result is decided by "out-space."
Black has 3 out-spaces, and white as well.
Therefore, the side going first will win.

There are non-eye-space black and non-eye-space white.
And there is one in-space.
Likewise, the result is decided by "out-space."
Because in case of non-eye-space v.s non-eye-space, "only one" in-space owns in common. ( It's very very important. )
Black has 3 out-spaces, and white as well.
Therefore, the side going first will win.

p.s I hope to communicate with visitors more and I will be very happy if you vote your country (nation).

PJ step 2. the most small value (2) /baduk/

There are triangle marks. It means decided-territory.

Black will do like this and created 1 black cross.

White will do like this and created one third black cross. You already knew if only you read "the most small value (1)."
So, there exists two thirds black cross.
If it's black turn, there created 1 black cross. The numeral difference is one third.
Therefore, each turns earn one third and its value is two thirds.

Conclusion : Exists two third black cross. / Value is two thirds.

There is a convenient way.
Black will want to play elseway instead of sequence 2.
Of course it can!

But anyhow white do 2 sequence, then, black must do 3 sequence.
Because 3 sequence is bigger than two thirds.
And because reasonal white player thought that there is not bigger one.

CR step 1. decide on terminology /baduk/

A, B, C = out-space
D = in-space
C = belongs out-space, but especially “eye-space”, and there is an "eye-space-number."
E = in-ko
F = out-ko
Please remember it for going on wheels.

PJ step 1. the most small value (1) /baduk/

There is a “ko”.

If it’s black turn, black will connect it by one sequence and there is nothing.
If it’s white turn, white will capture it and connect it later by two sequences and there is one captive.
All sequence is three, and quarrel a cross.
So one sequence earns one third.
Then, if black connect it, there is nothing.
If so, there was already one third white cross.
Bacause black earns one third.
And, each side earns one third, so its value is two thirds.

Conclusion : there EXISTs one third white cross. / its ending game(yose) value is two thirds.

Can you count it?
A exists one third white cross. B and C are as well.
So, there exists one white cross on the board.

prolog : real basic of Capturing Race /baduk/

Can you read the result?

Yes, white wins by 1 sequence gap.
It’s very easy.

Then, how about this?

Is it easy too?

It seems very difficult, but you can read it in a few seconds.
I will explain the method step by step.

prolog : real basic of Positional Judgement and ending game /baduk/

Can you do this positional judgement?

Yes, black is 18, white is 18, so even.

It’s very easy, because all territory was desided.

Then, how about this?

Is it easy too?
It seems very difficult, because of non-desided-territory.
However, I will explain the method step by step.

p.s It contains positional judgement and ending game, but main subject is positional judement.