It's not my scheme, it's Conway, Berlekamp, and Guy's scheme.RobertJasiek wrote:But where in your scheme do P ≥ Q and P ≤ Q fit?
I think you meanBill Spight wrote: Let Black play first in G, with correct play. The results are the opposite if White plays first.
1) If Black wins, then G |> 0.
2) If White wins, then G ≤ 0.
RobertJasiek wrote:I think you meanBill Spight wrote: Let Black play first in G, with correct play. The results are the opposite if White plays first.
1) If Black wins, then G |> 0.
2) If White wins, then G ≤ 0.
2) If White wins or ties (that is, Black cannot win), then G ≤ 0.
In in the great majority of cases when S(B) = S(W) = 0 the optimal sequences are gote for who plays first.Bill Spight wrote: What you should be interested in is scores, not counts. If you do not play when the position has a score, then in {-1 | 1} White does not play to 1, but stops at 0; i.e., does not play at all. That solves that problem.
The particular scores you are interested in are stops. A stop is the first score reached with optimal minimax play. For game, G, let S(B) be the stop when Black plays first and S(W) be the stop when White plays first. OC, S(B) ≥ S(W).
1) If S(B) < 0 then G < 0.
2) If S(W) > 0 then G > 0.
3) If S(B) > 0 > S(W) then G <> 0.
4) If S(B) = 0 and S(W) < 0 then G <| 0.
5) If S(B) > 0 and S(W) = 0 then G |> 0.
6) If S(B) = S(W) = 0, then G = 0 or G is an infinitesimal.
Now, we may not always realize when we have reached a score without playing the game out. For example, {3|3||0|||0||-5|-7} = 0.
Also, for difference games who gets the last play in a jigo may well matter, because sente usually matters.
Not IMX. The positive and negative infinitesimals are sente for one side, gote for the other. Then you have infinitesimals that are confused with 0, which are gote for both sides, and the cases where G = 0, which are sente for both sides, unless a play by either side is non-existent or a loss.Gérard TAILLE wrote:In in the great majority of cases when S(B) = S(W) = 0 the optimal sequences are gote for who plays first.
Well, Robert's example, where G = 0, is one.Gérard TAILLE wrote:Do you have an example in which one of the optimal sequence (black plays first or white plays first) is sente with S(B) = S(W) = 0 ?
Oops I wrote the contrary of what I had in mind. Sorry for that.Bill Spight wrote:Not IMX. The positive and negative infinitesimals are sente for one side, gote for the other. Then you have infinitesimals that are confused with 0, which are gote for both sides, and the cases where G = 0, which are sente for both sides, unless a play by either side is non-existent or a loss.Gérard TAILLE wrote:In in the great majority of cases when S(B) = S(W) = 0 the optimal sequences are gote for who plays first.
Well, Robert's example, where G = 0, is one.Gérard TAILLE wrote:Do you have an example in which one of the optimal sequence (black plays first or white plays first) is sente with S(B) = S(W) = 0 ?
P - Q has the same number of Black stones and White stones, and an even number of empty points. A jigo result is thus very likely to be sente for both sides.
How silly of me! Consider {0,*|0}. If we were to use a difference game to choose between the Black options, we could get DG = * - 0 = *, which is confused with 0.Gérard TAILLE wrote:In the majority of cases, when S(B) = S(W) = 0 in a difference game, the two areas P and -Q are miai and the sequence is sente for who plays first (=> in CGT context it is a loss for the player who plays first)
Do you know situations where one of the sequence is gote and S(B) = S(W) = 0?
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IIRC, {2F0 | 0 || -M0} + C0 in CGT notation is equivalent to M0|F0 in your notation, where C0 is a number, but F0 and M0 need not be. If F0 and M0 are numbers, then we can use a difference game to prove things about M0|F0 and M1|F1. In that case, whether we call the positions sente or gote is irrelevant to the proof.RobertJasiek wrote:Remark
We have studied the following theorems.
Definitions
The following local endgames M0|F0 and M1|F1 each with one simple follow-up have the sente(?) move values M0, M1 and follow-up move values F0, F1.
Presuppositions
Suppose the local endgames {2*F0 | 0 || -M0} and {2*F1 | 0 || -M1} with M0 ≥ M1 and F0 ≥ F1 and a constant number C.
Theorem 1
Playing to {2*F0 | 0} + {2*F1 | 0 || -M1} is at least as good as playing to {2*F1 | 0} + {2*F0 | 0 || -M0}.
Definitions
If M0 ≥ M1 and F0 ≥ F1, we write M0|F0 for the larger local endgame and M1|F1 for the smaller local endgame.
Presuppositions
Suppose the local endgames with simple follow-ups M0|F0 and M1|F1 of the creator with M0 ≥ M1 and F0 ≥ F1 in a (possibly empty) environment E without kos now or later.
Theorem 2
The creator's start in M0|F0 is at least as good as in M1|F1.
Presuppositions
Suppose the local endgames each with one follow-up M0|F0 and M1|F1 of the creator with M0 ≥ M1 and M0 + 2*F0 ≥ M1 + 2*F1 in a (possibly empty) environment E without kos now or later.
Theorem 3
The preventer's start in M0|F0 is at least as good as in M1|F1.
Presuppositions
Suppose the local endgames each with one follow-up A|B and C|D of the creator with A ≥ C and A + 2B ≥ C + 2D.
Theorem 4
A correct start of the creator is playing at A.
Remarks and Questions
- For theorems 1 + 2, you have said that the local endgames can be local gotes, ambiguous or local sentes.
- For theorem 1, it is obvious that M0 and M1 are sente move values.
- For theorem 2, we have seen in the proof that {2*F0 | 0 || -M0} and {2*F1 | 0 || -M1} appear so we also know that M0 and M1 are sente move values.
- Is it correct for theorem 3 that M0 and M1 are sente move values even if one or both local endgames are local gotes?
- Is it correct for theorem 4 that A and C are sente move values even if one or both local endgames are local gotes?
I follow the convention of using lower case letters for numbers and upper case letters for games (which might also be a number). In what follows below I will take all of your letters to indicate numbers, to keep things simple.RobertJasiek wrote:I need to understand some basics of adding a number K (call it the komi, if you like) to a game.
If F, M ≥ 0, yes.RobertJasiek wrote:Is {2F | 0 || -M} + K = {2F + K | 0 + K || -M + K}?
To compare moves in an environment without a ko caveat, we may set up a difference game.RobertJasiek wrote:Presuppositions
Suppose the local endgames {2*F0 | 0 || -M0} and {2*F1 | 0 || -M1} with M0 ≥ M1 and F0 ≥ F1 and a constant number C.
Theorem 1
Playing to {2*F0 | 0} + {2*F1 | 0 || -M1} is at least as good as playing to {2*F1 | 0} + {2*F0 | 0 || -M0}.
Hence, may we also write the following?
Presuppositions
Suppose the local endgames {2*F0 + K | 0 + K || -M0 + K} and {2*F1 + K | 0 + K || -M1 + K} with M0 ≥ M1 and F0 ≥ F1 and a constant number (C - K).
Theorem 1 + K
Playing to {2*F0 + K | 0 + K} + {2*F1 + K | 0 + K || -M1 + K} is at least as good as playing to {2*F1 + K | 0 + K} + {2*F0 + K | 0 + K || -M0 + K}.