What Hackenbush game represents oof?

Question

I have been reading about Hackenbush recently and have learned that the surreal numbers can be represented using RGB Hackenbush. I am having a hard time understanding On, Off, and Oof. What Hackenbush games represent these numbers?

Answer 1

$\textbf{on}$, $\textbf{off}$ and $\textbf{oof}$ are loopy game values in combinatorial game theory, meaning they occur in games in which a position can be repeated.

$\textbf{on}$ is the value of a game in which Left (Blue in Hackenbush) can move to $\textbf{on}$ (the same position, so this is just like skipping a turn) and Right (Red in Hackenbush) has no legal moves. We usually write it as $\textbf{on} = \{\textbf{on}|\}$.

$\textbf{off}$ is similar to $\textbf{on}$, just with the players switched. We write $\textbf{off} = \{|\textbf{off}\}$. $\textbf{on}$ is the highest possible game and $\textbf{off}$ is the lowest: if you add $\textbf{on}$ plus pretty much every game you get $\textbf{on}$ back, a win for Left, except if you add $\textbf{on} + \textbf{off}$, because you would get $\textbf{dud} = \{\textbf{dud}|\textbf{dud}\}$ (Dead Universal Draw).

$\textbf{oof}$ is a game in which Left can move to 0 (the game in which the second player wins) and Right can move to $\textbf{off}$: $\textbf{oof} = \{0|\textbf{off}\}$. It's a hot game in the sense that the players are very incentivized to move there, but it's not the hottest game out there (it's $\textbf{hot} = \{\textbf{on}|\textbf{off}\}$). We similarly also have $\textbf{ono} = \{\textbf{on}|0\}$.

$\textbf{on}$ and $\textbf{off}$ are possible to represent in Hackenbush as a position with infinitely many blue or red branches at ground level, independent of each other (unlike in ordinal-valued positions, where they are in top of each other and so you always leave less than before after chopping one). However, $\textbf{oof}$ and $\textbf{ono}$ are impossible to represent in standard Hackenbush, because we would have to have infinitely many branches of one color being able to be chopped by a single branch of the other color, but we can't have them be on top of the single branch of the other color because then we would have $\textbf{over} = \{0|\textbf{over}\}$ or $\textbf{under} = \{\textbf{under}|0\}$.

For more information, check out the second volume of Winning Ways or this wonderful video for SOME1 that summarizes Winning Ways with Hackenbush examples.