My friend René_Careenium and I just thought about something I recently asked here in L19 about a hypothetical ko situation that would admit no thermographic analysis and I believe we just found it. This beast will be hopefully intersting for you. Let's see if you agree with the strange conclusions that we reached.
The positionLet us assume Chinese rules, no ko threats and no komi for now.
So if you agree with me so far, the two big groups are in seki, and the ko fight enclosed by them, no player has good follow up moves after winning it. One could say it is like a bidirectional 10000-year ko. The ko will stay open until the scoring begins, the question is who will capture last?
Note that this effect happens with most common rulesets, with Chinese rules the swing is 4 points, with stone scoring the swing is 2, and with territory scoring the swing would be 1 point.
Observe also that the assumption of no ko threats is necessary: if one player has enough ko threats they can kill the whole thing. Therefore this ko is hyperactive.
We looked through various popular go bestiaries including SL and found no ko like this. If you know of anything similar, please let us know.
In the context of the whole boardAgain, with Chinese rules, if the position reaches the late endgame, and both players need the ko in their favour to win, then the player that has the ko turned against their favour would rather not pass (White in the example above). Instead, White would rather play in the ko. Black cannot pass or White will pass too then White wins but Black can
play in their own territory to lift the ko bans. White then passes, black recaptures. Then we have this strange ko fight where the environment plays and the subsequent pass act as ko threats. But these are not virtual tertiary threats, they are primary (!). Already a strange situation, as a move that is not sente at t=0 can act as a primary threat.
The situation is that then we have the following molasses-like sequence: w ko, b environment, w passes, b ko, w environment, b passes. The player that can play the most moves at temperature 0 will win the ko. Of course most deviations from the molasses loop will be punished: If Black 2 passes after White recaptures, White passes too and wins. If White 3 plays in the environment to say, fill an eye, then Black recaptures and now White has one move less, hence they are closer to losing. If Black 4 neglects to recapture then White just passes again, and Black lost a move and hence they are closer to losing. Note that alternatives for many of these might exist, but this would be the canonical sequence if all territory for both players is eyes.
So it might look like the player with the most territories (and captures don't count!) wins the ko however...
Two unsettling examplesThe following two examples have the same number of stones for each colour (75), Chinese rules, no komi, no captures, no threats, the position starts in the same configuration, and the territories are the same. However I believe White wins in the first one and Black wins in the second one. Note that the position is in the other orientation than in the examples above.
- Click Here To Show Diagram Code
[go]$$ Black to play (White wins)
$$ ---------------------------
$$ | . X X X X X X . . X X X X |
$$ | X . X X X X X . . X X X X |
$$ | X X X X X X X X X X X O O |
$$ | X X X X X X X X X O O . O |
$$ | X X X X X X X X O O . O . |
$$ | X X X X X X X O O . O . O |
$$ | X X O O O O X O O O O O O |
$$ | X X O . . X O O O O O O O |
$$ | O O O X X X O O O O O O O |
$$ | O X X O X O O O O O O O O |
$$ | . X . O X O O O O O O O O |
$$ | X X O O X X O O O O O O O |
$$ | X O . O . X O O O . O . O |
$$ ---------------------------[/go]
- Click Here To Show Diagram Code
[go]$$ Black to play (Black wins)
$$ ---------------------------
$$ | . X X X X X . X . X . X . |
$$ | X . X X X X X X X X X X X |
$$ | X X X X X X X X X X X O O |
$$ | X X X X X X X X X O O O O |
$$ | X X X X X X X X O O O . . |
$$ | X X X X X X X O O . O . . |
$$ | X X O O O O X O O O O O O |
$$ | X X O . . X O O O O O O O |
$$ | O O O X X X O O O O O O O |
$$ | O X X O X O O O O O O O O |
$$ | . X . O X O O O O O O O O |
$$ | X X O O X X O O O O O O O |
$$ | X O . O . X O O O . O . O |
$$ ---------------------------[/go]
The difference is that one player can play in the opponent's territory instead of their own, winning moves. This of course does not affect the score, but it makes the opponent's area more full, hence less available moves. I deliberately do not give sequences for these two claims, because the lines are long and boring and often it boils down to reading a bit then counting available moves. The hint is that in the first one white doesn't have to be fancy, they can just fill in their own eyes to win. In the second, if White threatens to split the square into two eyes Black must prevent this by playing in the midpoint of the resulting three-point shape. If you're still unconvinced we can discuss it or even play it.
The weird situation is that for an arbitrary environment this position forces us to play this strange game of having more stones captured than the opponent (or bigger empty territories than the opponent), or alternatively, splitting your territory into eyes to make the opponent run out of non suicidal moves as fast as possible (a feeling known to those that have played against rude people online perhaps?). So even though the environment is at t=0 it's quite hard to compute who wins in a more open board. Just scoring this board is another game and it doesn't look like go anymore.
- Click Here To Show Diagram Code
[go]$$ Black to play. Who wins?
$$ ---------------------------
$$ | . . . . . . . . . . . . . |
$$ | . . X . . X . . X . X X X |
$$ | . X . . X . . X . X X O O |
$$ | . . . , . . , X X O O O . |
$$ | . . . . . . . X O O . O . |
$$ | . X X X X X X O . . O . . |
$$ | . X O O O O X O . O . . . |
$$ | X X O . . X O O . . . . . |
$$ | O O O X X X O . O . . O . |
$$ | O X X O X O . . . , . . . |
$$ | . X . O X O O . O . . O . |
$$ | X X O O X X O . . . . . . |
$$ | X O . O . X O . . . . . . |
$$ ---------------------------[/go]
Japanese rulesFor stone scoring the position behaves quite similarly, and in general for any area scoring rules with or without group tax, and any ko rule. For Japanese rules I believe it becomes a lot more boring because of the way ko works in the first encore. In order to recapture, players would have to pass and point at the ko, at which time they risk the other player passing and scoring the game.
Therefore, the strange situation only happens at 1/2>=t>=0 under Japanese rules. A similar game of wanting to not lose while maximizing the number of plays will happen in order to win. In particular, if only dame remain, then the outcome depends on the parity of the number of dame.
A hotter version of the same pathological ko, say, if the ko were a two-stage ko that nobody can fill in, would have the same effect in japanese rules as both players would rather lose a point and stay in the game than pass and give the ko up.
But I'm not an expert in the Japanese rules so please let me know what you think.
Environmental goFinally, the reason we're doing all of this. Consider the position in isolation plus a coupon stack, again with Chinese rules this time. Then the outcome of the ko depends on the parity of the number of coupons. Indeed, the player that begins with the ko against them can just play coupons and the opponent can only play coupons as well (no passing until all the coupons are depleted). Eventually if there is a coupon remaining then the first player can win.
Meaning this position has no thermography, as the limit as the granularity goes to zero is not defined. One could make two coupon stacks of the same temperature but with different parity of coupons and as delta->0 the score is 4 points larger in one than it is in the other.
However, Berlekamp et al "proved" (Actually the article does not provide a proof but Siegel's excellent book "Combinatorial game theory" does) that any simple game admits a thermography, even gives an algorithm for its computation. Simple is supposed to mean "all loops are kos" and "every subposition has at most one left ko option and one right ko option". The proof depends on the Forcing Pass Lemma which seems to assume that all subpositions are hot, which in this case they are not. I wonder, did I misunderstand their assumptions on the game? Or perhaps the definition of simple? Is this example wrong? Or is their proof just wrong?
A very interesting problem for me now is characterizing the positions that admit a thermograph. Molasses ko does so it must not be just not having a molasses sequence. Curiously enough I'm not sure even for this one the molasses sequence is the optimal play assuming the coupon stack as an environment.
Also imaginative suggestions for a name for this position are welcome.
What do you guys think?