No Pass Go problem
- Click Here To Show Diagram Code
[go]$$B No pass go. Who wins?
$$ ---------------
$$ | . . X . . X . |
$$ | X X X X X X X |
$$ | O O O O O O O |
$$ | . O . O . . . |
$$ ---------------[/go]
Who wins with perfect play?
Note: Tryss has solved the problem with Black to play.

Unhidden:
White.
Why?
Ahem.

If we all played no pass go, SDKS would see that White wins, no matter who plays first. But since we do not all play no pass go, let me explain.
Black has 2 pts. of territory. I.e., 1.5 + 1.5 + 1 - 2 = 2. To avoid confusion, let me refer to an empty point on the board as a space, so we don't mistake it for a point of territory. The one space eye in the top right is 1 pt. of territory. Each two space eye is worth 1.5 pts. of territory. Adding them together we get 4 pts. and then subtracting 2pts. for the group tax we get 2 pts. of territory. If you are familiar with chilled go (
https://senseis.xmp.net/?Chilling ) you know that some positions have fractional scores (
https://senseis.xmp.net/?Numbers ). The same is true in straight no pass go.
Now let's evaluate White's position. White scores are negative, so we have -1 -1 -2 + v + 2 = -2 + v. (v is an infinitesimal called
down. See
https://senseis.xmp.net/?DOWN ) Adding the Black and White results together we get the value of the whole board: 2 - 2 + v = v. v may be an infinitesimal, but it is negative, which means that White wins, no matter who plays first. No pass go SDKs would be familiar with those values and would agree that White wins.

Let me show that and explain the values as I go.
- Click Here To Show Diagram Code
[go]$$W No pass go, White first
$$ ---------------
$$ | 6 2 X 4 3 X . |
$$ | X X X X X X X |
$$ | O O O O O O O |
$$ | 5 O 7 O . 1 . |
$$ ---------------[/go]
After

each player has two 1 space eyes. Black to play must fill an eye and loses.

eliminates the v, leaving White with 2 pts. of territory and a net score of 0 pts. on the whole board. If it's your turn and the score is 0, you lose. (In fact, the definition of a 0 game is that the second player wins.) It is Black's turn, so White wins.

plays in one of the 1.5 pt. eyes, leaving a 1 pt. eye. That is,

loses 0.5 pt. Knowledgeable players, like our SDKs, would not bother to make plays that lose points, but would simply score the game.

plays in the other 1.5 pt. eye, leaving a 2 pt. eye. OC, in regular go that is worth 3 pts., but the value of the dead stone is different in no pass go, and differs according to the eye. We do not bother to evaluate it by itself.

captures the

stone, leaving a 1 pt. eye. That means that Black has 2 moves in that eye and White has none. 1 move = 1 pt. of territory, so that eye is worth 2 pts.
Now each player fills a 1 pt. eye until Black would have to play self atari. White to play wins the game.
- Click Here To Show Diagram Code
[go]$$B No pass go, Black first
$$ ---------------
$$ | . . X . . X . |
$$ | X X X X X X X |
$$ | O O O O O O O |
$$ | . O . O 2 1 . |
$$ ---------------[/go]

plays in the 3 space eye, leaving its value as -2 + *. * (pronounced STAR) is an infinitesimal that we have in regular go, namely a dame. Either player can play on a dame, reducing its territory to 0.

plays in the 3 space eye, reducing its value to -2, as we have already seen.
At this point we can see that each player has 2 pts., for a whole board score of 0, with Black to play. As knowledgeable players we do not bother to play the game out, but declare White the winner.

Note: v is a game in itself. White can play from v to 0, winning that game. Black can play from v to *, but then White can reply to 0, also winning. Since White wins, no matter who plays first, v is negative.
Edit: I have not shown that after Black plays in the 3 space White eye it is worth -2 + *. We have seen that White plays to -2 from there. Now to show that Black to play also moves to -2.
- Click Here To Show Diagram Code
[go]$$B No pass go, Black first
$$ ---------------
$$ | . X . X . X . |
$$ | X X X X X X X |
$$ | O O O O O O O |
$$ | . O . O 1 X . |
$$ ---------------[/go]
I have altered the board to make it simpler. Black now has four 1 pt. eyes, for the same 2 pts. of territory.
After

the eye is worth -5 in regular go, but only -2 in no pass go. Now to show that.
- Click Here To Show Diagram Code
[go]$$B No pass go, Zero game?
$$ ---------------
$$ | . X . X . X . |
$$ | X X X X X X X |
$$ | O O O O O O O |
$$ | . O . O X X . |
$$ ---------------[/go]
If the bottom right eye is worth -2, then the whole board is worth 0 and the second player wins.
- Click Here To Show Diagram Code
[go]$$W No pass go, White first
$$ ---------------
$$ | . X . X . X . |
$$ | X X X X X X X |
$$ | O O O O O O O |
$$ | . O . O X X 1 |
$$ ---------------[/go]

captures the two Black stones, leaving an eye worth -1.5. Knowledgeable players could now stop and score the board. The whole board has a net score of 0.5, so Black wins by ½ pt. So far, so good.
- Click Here To Show Diagram Code
[go]$$B No pass go, Black first.
$$ ---------------
$$ | . X . X . X 1 |
$$ | X X X X X X X |
$$ | O O O O O O O |
$$ | . O . O X X 2 |
$$ ---------------[/go]

must fill a 1 pt. eye. Now

captures the two Black stones, leaving a -1.5 pt. eye. The whole board has a net score of -0.5, so White wins by ½ pt.
The second player wins, as advertised.

So the bottom right eye is equivalent to two 1 pt. eyes. (Edit: As long as White is alive, OC.)