BOSE One-stone condensate

BOSE One-stone condensate is a method for guesstimating the value of the remaining endgame by using the simplification rules of BOSE. The name arises (besides the punny) from the property of BOSE that you only need to know the point value of the biggest endgame move in order to know how much (on average) you can gain by actually playing it.

Table of contents Value of tenuki The BOSE One-stone Condensate Derivation Examples Value of a 1 point move Value of a 2 point move Value of a three point move Proof Induction Hypothesis Endgame value of an N+1 point move (induction step) Base case (required to get the induction running) Properties of the result Points lost by passing Comparison with proper endgame analysis Applications

Value of tenuki

The value of remaining endgame (also called the "value of sente" or the "value of tenuki" or even "ambient temperature") is an important concept when considering any endgame positions. With a whole board full of various kinds of endgame, that value is almost impossible to calculate accurately. In an isolated endgame situation, just drop the "almost".

Using a lot of simplifications, it is possible to condense all that information into a single sentence.

The BOSE One-stone Condensate

A tenuki is worth half the size of the biggest available gote play.

Derivation

Taking BOSE rules into account, the result can be achieved by counting out all the possible outcomes and then calculating the weighted average of them.

Another way to get the result is to start from the simplest cases and then work up from there.

Examples

Value of a 1 point move

As per BOSE rule 2, we do not know how many one point moves there are. Since every two moves are miai, they cancel each other, as black gets one and white gets the other. Therefore there are only two cases:

• The number of 1 point moves is odd. (1, 3, 5, etc.)

In this case the player who plays first gains one more point than the other player.

• The number of 1 point moves is even.

In this case, both players get equally many of them, and the value of playing first is zero.

• Average

As we do not know the exact number of available 1 point moves, we have to average to get a result. Taking both cases to be equally likely, we get that the value of a 1 point move is (1+0)/2 = 0.5 points.

Value of a 2 point move

Again, there are only two cases to consider

• The number of 2 point moves is odd

In this case, the first player will get 2 points more than the second player, but the second player gets to play a 1 point move first. Using the value gained above, we get that the first player gains 2 - 0.5 = 1.5 points.

• The number of 2 point moves is even

In this case both the players get equally many 2 point moves, but the first player get to play a 1 point move first. The value of that, as described above, is 0.5.

• Average

Again, taking both cases to be equally likely, we get that the value of getting to play a 2 point move is (1.5 + 0.5) / 2 = 1 point.

Value of a three point move

• Odd

First player gets 3 more points, second player gets to play 2-pointers first. Value of first play is 3 - 1 = 2 points

• Even

Value equals the value of playing a two point move first, and is therefore 1 point.

• Average

(2 + 1) / 2 = 1.5

Proof

Induction Hypothesis

The endgame value of an N point move is N/2.

Endgame value of an N+1 point move (induction step)

Assume the hypothesis is true for some move size N.

Then, the endgame value of a N+1 point move is the average of the two cases:

1. there being an odd number of N+1 sized moves (value N+1 - N/2) and
2. there being an even number of N+1 sized moves (value N/2)

Taking both cases as equally likely, the average becomes ((N+1 - N/2) + N/2) / 2. This simplifies to (N+1)/2.

Therefore, if the hypothesis is true for some move size N, it follows that the hypothesis is also true for move size N+1.

Base case (required to get the induction running)

If the size of the biggest endgame is 0, there are no points left to gain. The value of playing first is 0, which equals 0/2 = N/2.

So, by induction, for every integer move size >= 0, the hypothesis is true.

By BOSE rule 1, this covers every possible case.

Properties of the result

Points lost by passing

Since playing first is worth half of the biggest gote, then passing loses that many points and the opponent gains the same amount. This means that if a player passes when the biggest available play is 4 points, the game result is shifted by 4 points on average.

Comparison with proper endgame analysis

A proper endgame analysis takes into account the whole board. In particular, the number of the various sized moves are known, and the values of follow-up moves can be calculated exactly. Techniques like reverse sente and giving up some points to turn a gote sequence into a sente one can be considered. Of course, kos can also be taken into account.

On the other hand "tenuki is worth half of the biggest move" is somewhat more concise and slightly easier to remember than the dozens of books written on the subjects of endgame theory and ko.

Applications

• Reverse sente is worth half of the second biggest gote