If you understand BES v.s SES, you feel easy.

And, another question, white must make eye?

## 2009년 9월 24일 목요일

## 2009년 9월 22일 화요일

### coffee break

Maybe allmost think this black 1 sequence, isn't it?

I think this sequences are not good for black in some situation.

Because,

3-4 point is strong for triangle marks side, so white is usually challenge A, B, C, D point.

But, in second Dia, white takes triangle side very easily.

Then, which point is strong?

to be continued.

I think this sequences are not good for black in some situation.

Because,

3-4 point is strong for triangle marks side, so white is usually challenge A, B, C, D point.

But, in second Dia, white takes triangle side very easily.

Then, which point is strong?

to be continued.

## 2009년 9월 16일 수요일

### CR step 7. big ES v.s small ES /baduk/

Now, we study big eye-space v.s small eye-space capturing races.

At first, you already know "eye-space-number."

At this Dia, ES 1 group v.s ES 3, it is same to ES v.s ES, because in case of ES 1 ~ 3, eye-space is equal to eye-space-number.

This is 4 eye-space v.s 3 eye-space CR Dia.

In case of 4 eye-space, eye-space-number is "5", eye-space-number is bigger than eye-space.

So we call that black group "big-eye-space".

It is relativity of course.

So, in-spaces are the burden of small-eye-space.

You can imagine x-marks are filled with any stones, and you can count it easily.

At this Dia, there is 4 eye-space, but it is small eye-space because there is 5 eye-space.

So, in-space is the burden of black.

You can imagine like this.

White is 6 eye-space and black is 5 eye-space.

So white is big eye-space, black is small one.

You can count it like below Dia.

It's not end of BES v.s SES.

And, my English ability is poor, so I can't explain full of my knowledge.

I want to find another WRITER of this blog in English.

Please mail to me : hanayeol@naver.com

At first, you already know "eye-space-number."

At this Dia, ES 1 group v.s ES 3, it is same to ES v.s ES, because in case of ES 1 ~ 3, eye-space is equal to eye-space-number.

This is 4 eye-space v.s 3 eye-space CR Dia.

In case of 4 eye-space, eye-space-number is "5", eye-space-number is bigger than eye-space.

So we call that black group "big-eye-space".

It is relativity of course.

So, in-spaces are the burden of small-eye-space.

You can imagine x-marks are filled with any stones, and you can count it easily.

At this Dia, there is 4 eye-space, but it is small eye-space because there is 5 eye-space.

So, in-space is the burden of black.

You can imagine like this.

White is 6 eye-space and black is 5 eye-space.

So white is big eye-space, black is small one.

You can count it like below Dia.

It's not end of BES v.s SES.

And, my English ability is poor, so I can't explain full of my knowledge.

I want to find another WRITER of this blog in English.

Please mail to me : hanayeol@naver.com

## 2009년 9월 9일 수요일

### CR step 6. ES v.s ES /baduk/

If there is not in-space, it's simple.

Black out-spaces are 2, and white out-spaces are 4.

( Black out-spaces 1 and eye-space-number 1 )

( White out-spaces 3 and eye-space-number 1 )

So, white wins the race by two sequences gap.

There is an in-space.

Now, which side have out-space-advantage?

Black has 2, and white has 3.

In this case, white has "ALL" in-space burden.

Non-eye-space v.s non-eye-space with in-space, the side having more out-spaces has "in-space minus one " burden, do you remember?

However, eye-space v.s eye-space, all in-space burden demended.

So, you can imagine that x-marks are filled with any stone like below.

It has each two out-spaces.

But, there is important causion.

If the imagining-result is even, out-space-disadvantage-side has burden of goto-seki ( draw ).

You must not forget this.

White can go other side.

Because each side has burden of in-space, and expects disadvantage.

Black out-spaces are 2, and white out-spaces are 4.

( Black out-spaces 1 and eye-space-number 1 )

( White out-spaces 3 and eye-space-number 1 )

So, white wins the race by two sequences gap.

There is an in-space.

Now, which side have out-space-advantage?

Black has 2, and white has 3.

In this case, white has "ALL" in-space burden.

Non-eye-space v.s non-eye-space with in-space, the side having more out-spaces has "in-space minus one " burden, do you remember?

However, eye-space v.s eye-space, all in-space burden demended.

So, you can imagine that x-marks are filled with any stone like below.

It has each two out-spaces.

But, there is important causion.

If the imagining-result is even, out-space-disadvantage-side has burden of goto-seki ( draw ).

You must not forget this.

White can go other side.

Because each side has burden of in-space, and expects disadvantage.

### PJ step 7. 1 n one third (4) /baduk/

White sequence, there is nothing.

Black sequence, there is 1 ( A ) and one third ( B ) black cross.

Two sequences made 1 and one third, so each one sequence earns two thirds.

At the second Dia, white earns two thirds, but there is nothing, so there already exists two thirds black cross.

Each side has the right, so the value is 1 and one third.

Conclusion : exist two thirds / value is 1 n one third.

to be continued.

Black sequence, there is 1 ( A ) and one third ( B ) black cross.

Two sequences made 1 and one third, so each one sequence earns two thirds.

At the second Dia, white earns two thirds, but there is nothing, so there already exists two thirds black cross.

Each side has the right, so the value is 1 and one third.

Conclusion : exist two thirds / value is 1 n one third.

to be continued.

## 2009년 9월 3일 목요일

### PJ step 6. 1 n one third (3) /baduk/

White 1 sequence A, there is nothing.

Black 1 sequence A, there is 1 and one third black cross.

Two sequences make 1 and one third, the value of each 1 sequence is as well, and 1 sequence makes two thirds.

White 1 sequence A, white earns two thirds, but there is nothing, so there is already two thirds black at first.

Conclusion : Exist two thirds black. / Value is 1 n one third.

White crosses are 12.

Black crosses : decided area 10 + non-decided area two thirds = 10 and two thirds.

So, white has strong 1 advantage.

to be continued

### coffee break

Sorry, I was busy today.

My dream is global guidance of baduk.

So, I met some people concerned today.

My 6 boards were made by me and magnetic stones may sell anywhere because it's only magnetic plastics and I use merely stones just like using a book to pillow.

I think my posting speed is some fast. Isn't it?

It will be some difficult from now on.

See this below.

I often used this white 1 sequence, but, 99% people were taking in so far as high level-er.

What about your opinion of it for?

I will post about it after listening to your opinions.

My dream is global guidance of baduk.

So, I met some people concerned today.

My 6 boards were made by me and magnetic stones may sell anywhere because it's only magnetic plastics and I use merely stones just like using a book to pillow.

I think my posting speed is some fast. Isn't it?

It will be some difficult from now on.

See this below.

I often used this white 1 sequence, but, 99% people were taking in so far as high level-er.

What about your opinion of it for?

I will post about it after listening to your opinions.

## 2009년 9월 1일 화요일

### my baduk life

It's my baduk-investigating-room.

There are six 13-lined-boards made by hanayeol and many magnet-stones .

I am doing analysys in this room.

### PJ Review test /baduk/

Which side has an advantage?

It's not ending game quiz. So you need not think next progresses. Just judge at this instant on the board.

It applies to all PJ Review tests.

Convinience below.

It's not ending game quiz. So you need not think next progresses. Just judge at this instant on the board.

It applies to all PJ Review tests.

Convinience below.

### CR step 5. ES v.s NES /baduk/

There are eye-space-group and non-eye-space-group.

And, there is not in-space.

If there is not in-space, eye is meaningless. ( It's very important! )

So, at this dia, they have two out-spaces each other.

( In case of white, out-space 1 + eye-space-NUMBER 1 = 2 out-spaces for convenience' sake. )

But, there is an in-space.

It's defferent.

A is a burden of black. ( burden of NES )

So, imagine that x marks are filled with any stone.

( One in-space matches one NES out-space )

Like this.

It's easy way to count. How about you?

If you do not agree, see the next dia.

It has many many many spaces.

But, if you use my "offsetting method", it can be easy.

White has 8 out-spaces ( out-spaces 3 + eye-space-number 5 = 8. )

Black has 9 out-spaces.

So, black will wins the race 1 sequence gap.

And, there is not in-space.

If there is not in-space, eye is meaningless. ( It's very important! )

So, at this dia, they have two out-spaces each other.

( In case of white, out-space 1 + eye-space-NUMBER 1 = 2 out-spaces for convenience' sake. )

But, there is an in-space.

It's defferent.

A is a burden of black. ( burden of NES )

So, imagine that x marks are filled with any stone.

( One in-space matches one NES out-space )

Like this.

It's easy way to count. How about you?

If you do not agree, see the next dia.

It has many many many spaces.

But, if you use my "offsetting method", it can be easy.

White has 8 out-spaces ( out-spaces 3 + eye-space-number 5 = 8. )

Black has 9 out-spaces.

So, black will wins the race 1 sequence gap.

### PJ step 5. value 1 n one third (2) /baduk/

At this post, more detailed explanations, so more dia.

White 1 sequence, there is nothing. (W)

Black 1 sequence. If black places A, there are 3 black crosses.

So, if there is not ending game bigger than this*, white must respond.

( *2 n one third I think. It's almost unusual anyway. )

Now white responded.

Can black go other side?

If black goes other side, white will do ponuki and you ought to remember second dia - there is noting. But here, white do ponuki. - So, black can't go other side.

So, this is a set of black 1 sequence. (X)

If black 1 sequence more, there are 2 black crosses include of 1 dead stone. (Y)

3 sequences ( W, X, Y ) make 2 crosses, so each 1 sequence makes two thirds.

Now, black 2 sequences ( X, Y ) make 1 and one third, but there are 2 crosses, so at first, there already exists two thirds black cross.

Got it? It may be difficult.

Conclusion 1 : Exist two thirds black cross in first dia.

White 1 sequence, there is two thirds black cross because of 1 minus one third. ( remember dead stone ) (Z)

Because W, X, Y are the same, and there is two thirds black cross, so Z is as well.

And, as you know, black and white has same right. ( so, two thirds multiple 2 )

Conclusion 2 : W, X, Y, Z has value of 1 n one third.

Not yet, to be continued.

White 1 sequence, there is nothing. (W)

Black 1 sequence. If black places A, there are 3 black crosses.

So, if there is not ending game bigger than this*, white must respond.

( *2 n one third I think. It's almost unusual anyway. )

Now white responded.

Can black go other side?

If black goes other side, white will do ponuki and you ought to remember second dia - there is noting. But here, white do ponuki. - So, black can't go other side.

So, this is a set of black 1 sequence. (X)

If black 1 sequence more, there are 2 black crosses include of 1 dead stone. (Y)

3 sequences ( W, X, Y ) make 2 crosses, so each 1 sequence makes two thirds.

Now, black 2 sequences ( X, Y ) make 1 and one third, but there are 2 crosses, so at first, there already exists two thirds black cross.

Got it? It may be difficult.

Conclusion 1 : Exist two thirds black cross in first dia.

White 1 sequence, there is two thirds black cross because of 1 minus one third. ( remember dead stone ) (Z)

Because W, X, Y are the same, and there is two thirds black cross, so Z is as well.

And, as you know, black and white has same right. ( so, two thirds multiple 2 )

Conclusion 2 : W, X, Y, Z has value of 1 n one third.

Not yet, to be continued.

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