# Tedomari / Solution 1

Sub-page of Tedomari
Tedomari problem 1

The value of the plays:

Move a has a miai value of 1 1/2. The local count is -1 1/2 (negative count for White). Whoever plays at a gains 1 1/2 points.

Move b has a miai value of 2. The local count is -2.

Move c has a miai value of 1 3/4. The local count is 1/4. Black can play to a local score of 2, while White can play to a local count of -1 1/2. (Mistake?) [1]

The current overall count:

Black has 3 points of solid territory in the top left plus 8 points in the bottom right. The area around c is worth 1/4 point, for a total of 11 1/4 points.

White has 8 points of solid territory plus 2 points around b and 1 1/2 points around a, for at total of 11 1/2 points.

The current overall count is -1/4.

Normally the correct play is b, the largest one, but Black should make a small sacrifice to get tedomari.

Black mistake

Black mistake:

takes the largest play. takes the largest remaining play. Now 3 and 4 are miai.

Each player has 11 points of territory, for jigo.

Black gets tedomari

Optimal play for Black:

The starting count is -1/4. gains 1 3/4 points, for a count of 1 1/2. gains 2 points, for a count of -1/2. And now Black gets tedomari. gains 1 1/2 points, for a final score of +1. Black has 13 points and White has 12.

The relation of infinitesimals to optimal play:

a and b are gote close to 1 3/4 in value. c has a value of 1 3/4. If White plays in c the next play is almost that big. So a and b are like * (STAR) and c is like ^ (UP). In the infinitesimal game ^ + * + * = ^, Black can win by playing in ^ (c).

White plays first

Optimal play when White plays first:

After White takes the largest play, Black's optimal play is again to make the second largest remaining play to get tedomari. Black has 11 points of territory and White has 12. White wins by 1 point.

Optimal play and infinitesimals:

We have seen above that the related infinitesimal game is ^, which is positive. That means that White cannot get tedomari, and should not even try.

After , the related game is ^* (UP STAR). Black wins that game by playing in * (a).

(That's not good enough to win the go game, but it is optimal play. :-))

[1]

Migeru I got stomped by c. Let's see if I can understand where white's -1 1/2 comes from.
It seems that the value of this board is {0|-3} + {0|-4} + {2|{0|-3}}. My mistake was to evaluate c to {2|0} assuming that the sequence gets played in full.
{0|-3}
This is worth 3 points for white.
{0|-4}
This is worth 4 points for white.
{2|{0|-3}}
2
This is worth 2 points for black. White fills at e and black at d, in either order.
{0|-3}
0
-3

aLegendWai: To me, this concept is an application of sente-gote relationship. In playing this endgame, sente/gote is counted.

For the endgame c, it is a White Sente endgame. If B plays first, this endgame completes and W has a choice to play the next endgame. If W plays first, B has to respond in order to complete this local endgame. So W is sente. [2]

So if you take the concept of sente-gote into account, you can still find the optimal order of play without knowing the concept of Tedomari.

(My ideas though. Any comments are welcome.)

[2] More advanced note: It is possible for B not to respond. The cost of playing elsewhere is 3-point worth. So B may not respond provided that:

• B can offset the loss by finding a bigger endgame move
• B decides to play mutual damage. If the damage made by B in another endgame is larger than the damage made by W in this endgame, B gains (and W is forced to respond another first), vice versa.

Tedomari / Solution 1 last edited by 88.107.3.58 on February 7, 2007 - 22:22