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11.8 Eye Topology with Ko

This section extends the topological eye analysis to handle ko. We distinguish between a ko in favor of `O'' and one in favor of `X':

 
.?O?   good for O
OO.O
O.O?
XOX.
.X..

.?O?   good for X
OO.O
OXO?
X.X.
.X..

Preliminarily we give the former the symbolic diagonal value a and the latter the diagonal value b. We should clearly have 0 < a < 1 < b < 2. Letting e be the topological eye value (still the sum of the four diagonal values), we want to have the following properties:

 
e <= 2     - proper eye
2 < e < 3  - worse than proper eye, better than half eye
e = 3      - half eye
3 < e < 4  - worse than half eye, better than false eye
e >= 4     - false eye

In order to determine the appropriate values of a and b we analyze the typical cases of ko contingent topological eyes:

 
      .X..      (slightly) better than proper eye
(a)   ..OO          e < 2
      OO.O
      O.OO      e = 1 + a
      XOX.
      .X..


      .X..      better than half eye, worse than proper eye
(a')  ..OO      2 < e < 3
      OO.O
      OXOO      e = 1 + b
      X.X.
      .X..

      
      .X..      better than half eye, worse than proper eye
(b)   .XOO      2 < e < 3
      OO.O
      O.OO      e = 2 + a
      XOX.
      .X..

      
      .X..      better than false eye, worse than half eye
(b')  .XOO      3 < e < 4
      OO.O
      OXOO      e = 2 + b
      X.X.
      .X..

      
      .X..
      XOX.      (slightly) better than proper eye
(c)   O.OO          e < 2
      OO.O
      O.OO      e = 2a
      XOX.
      .X..

      
      .X..
      XOX.      proper eye, some aji
(c')  O.OO      e ~ 2
      OO.O
      OXOO      e = a + b
      X.X.
      .X..

      
      .X..
      X.X.      better than half eye, worse than proper eye
(c'') OXOO      2 < e < 3
      OO.O
      OXOO      e = 2b
      X.X.
      .X..

      
      .X...
      XOX..     better than half eye, worse than proper eye
(d)   O.O.X     2 < e < 3
      OO.O.
      O.OO.     e = 1 + 2a
      XOX..
      .X...

      
      .X...
      XOX..     half eye, some aji
(d')  O.O.X     e ~ 3
      OO.O.
      OXOO.     e = 1 + a + b
      X.X..
      .X...

      
      .X...
      X.X..     better than false eye, worse than half eye
(d'') OXO.X     3 < e < 4
      OO.O.
      OXOO.     e = 1 + 2b
      X.X..
      .X...

      
      .X...
      XOX..     better than false eye, worse than half eye
(e)   O.OXX     3 < e < 4
      OO.O.
      O.OO.     e =  2 + 2a
      XOX..
      .X...

      
      .X...
      XOX..     false eye, some aji
(e')  O.OXX     e ~ 4
      OO.O.
      OXOO.     e = 2 + a + b
      X.X..
      .X...

      
      .X...
      X.X..     (slightly) worse than false eye
(e'') OXOXX     4 < e
      OO.O.
      OXOO.     e = 2 + 2b
      X.X..
      .X...

It may seem obvious that we should use
 
(i)   a=1/2, b=3/2
but this turns out to have some drawbacks. These can be solved by using either of
 
(ii)  a=2/3, b=4/3
(iii) a=3/4, b=5/4
(iv)  a=4/5, b=6/5

Summarizing the analysis above we have the following table for the four different choices of a and b.

 
case    symbolic        a=1/2   a=2/3   a=3/4   a=4/5   desired
        value           b=3/2   b=4/3   b=5/4   b=6/5   interval
(a)     1+a             1.5     1.67    1.75    1.8         e < 2
(a')    1+b             2.5     2.33    2.25    2.2     2 < e < 3
(b)     2+a             2.5     2.67    2.75    2.8     2 < e < 3
(b')    2+b             3.5     3.33    3.25    3.2     3 < e < 4
(c)     2a              1       1.33    1.5     1.6         e < 2
(c')    a+b             2       2       2       2           e ~ 2
(c'')   2b              3       2.67    2.5     2.4     2 < e < 3
(d)     1+2a            2       2.33    2.5     2.6     2 < e < 3
(d')    1+a+b           3       3       3       3           e ~ 3
(d'')   1+2b            4       3.67    3.5     3.4     3 < e < 4
(e)     2+2a            3       3.33    3.5     3.6     3 < e < 4
(e')    2+a+b           4       4       4       4           e ~ 4
(e'')   2+2b            5       4.67    4.5     4.4     4 < e

We can notice that (i) fails for the cases (c"), (d), (d"), and (e). The other three choices get all values in the correct intervals. The main distinction between them is the relative ordering of (c") and (d) (or analogously (d") and (e)). If we do a more detailed analysis of these we can see that in both cases `O' can secure the eye unconditionally if he moves first while `X' can falsify it with ko if he moves first. The difference is that in (c"), `X' has to make the first ko threat, while in (d), O has to make the first ko threat. Thus (c") is better for O and ought to have a smaller topological eye value than (d). This gives an indication that (iv) is the better choice.

We can notice that any value of a, b satisfying a+b=2 and 3/4<a<1 would have the same qualities as choice (iv) according to the analysis above. One interesting choice is a=7/8, b=9/8 since these allow exact computations with floating point values having a binary mantissa. The latter property is shared by a=3/4 and a=1/2.

When there are three kos around the same eyespace, things become more complex. This case is, however, rare enough that we ignore it.


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