# What is the domino position corresponding to game $\{0|\{0|-2\}\}$? [ONAG]

On page 101 of the second edition of Conway's ONAG, there is a brief aside about a game $G=\{0|\{0|-2\}\}$, which Conway says is equivalent to the position in dominoes that looks like this:

o o
o o
o o
o o

Where each "o" is an open square on a 2x4 grid. (In this particular domino game, Left can place two-square, non-overlapping dominoes vertically, while Right can only play them horizontally. The winner is the player that makes the last legal move.) My understanding (which is, presumably, incorrect) is that the game $\{0|\{0|-2\}\}$ is a game wherein Left can make a move that creates a position equivalent to $0\equiv\{\varnothing|\varnothing\}$, meaning there are no further legal moves for either side, while Right can move to the position $\{0|-2\}$. No matter who moves first, L wins, because the only route to a winning position for R would require two R moves in a row. (I think)

The problem is that I can't for the life of me imagine how the 2x4 grid could possibly be equivalent to this game. There is no L move I can see that leaves both sides without legal moves--it seems like they could only move to the game $\{3|-2\}$, or, less effectively, to $\{2|-2\}$. I assume that Conway is just leaving out some steps that should be obvious, can anyone fill them in for me?

• In my (I think first) edition it is volume 1, page 119 under the heading "The Tiniest Games". Aug 14 '17 at 14:47
• @RossMillikan That seems likely. In my edition the heading is "The All Small Games", but that sounds roughly equivalent. Aug 14 '17 at 14:50

Left has two options from the initial $2 \times 4$ position. He can make a move that uses a corner or make a move that takes two middle squares out of one side. The position where he takes a corner is $\{2|0\}$because if he moves again he takes the opposite corner leaving himself two moves, while right can move to leave them each one which is equivalent to $0$. It is taking the middle that is claimed to have value $0$. If he takes a middle and moves again he can leave $*$ or $-2$. If he takes a middle square and Right moves next we have a vertical L tetromino, which is shown to have value $\frac 12$ a few pages earlier. The middle square position is therefore $\{*,-2|\frac 12\}$. We can see this is zero because whoever moves loses. If Right moves he moves to $\frac 12$ and Left wins because this is greater than $0$. If Left moves he goes to $*=\{0|0\}$ and Right moves and wins. This shows the Left options from your position are $0,\{2|0\}$ The second is reversible in that if Left moves to $\{2|0\}$ Right can immediately move back to $0$, so the whole Left option is equivalent to $0$.
• OK, thank you, this is very helpful. The only step I'm a little murky on is the $\{\ast,-2|\frac{1}{2}\}\equiv0$ step. I definitely understand intuitively that whoever moves first loses and thus it's a $0$ game, but doesn't $\{\ast,-2|\frac{1}{2}\}\equiv0$ imply that $\ast\in0^L$ which would mean $0\gt\ast$? My understanding is that $(0\ngeq\ast)\wedge(\ast\ngeq0)$, so this shouldn't be possible... Aug 14 '17 at 18:47
• You can't go from $* \in 0^L$ to $0 \gt *$. That only works when the value of the game is a number. When the value is a number, moving always decreases the value, so the value of the game must be greater than any Left option. This is the point of the number avoidance theorem, which is presented near here. When the value is not a number, it can be advantageous to move in the game and that inequality does not apply. Aug 14 '17 at 19:22
• Ah, OK, I knew that non-number Games no longer had the restriction that $x^L<x^R$, but somehow still though $x^L<x<x^R$ was true. Upon re-reading I see I was misunderstanding. Thank you! Aug 14 '17 at 19:39