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

For example,

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

Thanks for your kind attention.