# Forum for Endgame

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# Some dicussion on the endgame [#2308]

210.6.119.230: Some dicussion on the endgame (2010-06-12 17:04) [#7762]

Hello, I am an amateur player of GO, Recently I am confused in how to find the best sequence of the endgame.

Although there exists some approximation methods already. For example, for a gote move as follow It has an approximation value of 1.5 (1 by breaking white 's territory and 0.5 for the right to get another move at a)

A move of value 1.5

However, there are some cases which is not easy to determine the next move by the dividing method

For example,

Example 1 , When to play?

It is hard to determine if it is the correct move by finding its approximation value because it can be either a sente or gote.

Therefore I am considering a simple method to determine the right time for a move and then figure out the best sequence.

(My elementary thought)

Firstly, gote moves without following moves can be grouped in A

where A={a(1),a(2),...,a(n)} that a(1)<a(2)<...<a(n)

That a is a one-step

And then the moves with 1 following move can be represented by b(n)

That b is a two-step, e.g. b(1)= (α,β) where α is the value of the first move and β is the value of the second move

b can be of four types 1.(α',β') , where ' means AS(absolute sente) 2.(α',β ) 3.(α ,β') 4.(α ,β )

Obviously, those b of type one is a AS and can be played at once

For types 2

2. Adding a move a(n+1) of value β in A

For types 3 and 4. 3. if a(n) > (α+β) then play a(n) until (α+β) >= a(n)

```  if a(n)  <= (α+β) then play b(1)
```

4. ( This part may be quite complicated, please be patient )

```  Sente value ( denoted by Δ ) is used,
Δ is the value of moving first
Δ  of group A = a(n)-a(n-1)? + a(n-2) - a(n-3)? + .. + a(1) for odd n
///... + a(2) - a(1) for even n
By grouping i elements in A > β in group A'
j-i elements in A < β in group 'A
n-j  elements in A > α  in group 'A'
Then, firstly
if (α  > a(n)) or (β> a(n)) than play b(1)
if β <= α < a(n) than
if (n-j is odd) and (j-i is odd) and (Δ of 'A' <= Δ of 'A + α)
than play b
```
```  by considering
{a(1),..a(i),β,a(i+1),..a(j),α,a(j+1)...,a(n)}
```
```  because playing b will lose Δ of 'A' first
but the opponent will have to play β ( but this is not regarded as the
gain of the opponent)
therefore the alternate playing sequence of elements in 'A is changed
Then our side can benefict (α + Δ of 'A)
This is why the inequality above comes.
```
```  However, in this time I cannot figure out the rules for determination
for the cases of
1. a(n) > β > α
2. n-j is even
3. j-1 is even
```

There may be some mistakes in the preliminary algorithm. It is welcome for your suggestion and correction. I hope that the reader of this page who are interested in my algorithm could help me to solve the problem. Thank you very much . (Possibly more harder for rules to determine 3-step)

For 4-step and higher steps, we will use approximation method.

Nevertheless, if the algorithm is successful, we can analyse the move in example 1 as

For black, it is b(1) = 1 For white, it is b(1) = (1,1,5/3'), which is a 3-step for white