Infinite chess exists, now, infinite go?
A go position is a (partial) function \(f: \mathbb{N} \to \{0, 1, 2\}\).
Define \(c: \mathbb{N}^2 \to \mathbb{N}\) as \(c(a, b) = 2^{a+1} + 3^{b+1}\).
In fact, the set of values for which \(f\) is defined is precisely the image of \(c\).
Define \(l(a, b) = a\) and \(r(a, b) = b\).
Define the following:
- \(\uparrow: \mathbb{N} \to \mathbb{N}\) as \(\uparrow(n) = c(l(c^{-1}(n)), r(c^{-1}(n))+1)\).
- \(\downarrow: \mathbb{N} \to \mathbb{N}\) as \(\downarrow(n) = c(l(c^{-1}(n)), r(c^{-1}(n))-1)\).
- \(\rightarrow: \mathbb{N} \to \mathbb{N}\) as \(\rightarrow(n) = c(l(c^{-1}(n))+1, r(c^{-1}(n))\).
- \(\leftarrow: \mathbb{N} \to \mathbb{N}\) as \(\leftarrow(n) = c(l(c^{-1}(n))-1, r(c^{-1}(n))\).
The white score in a go position \(f\) is the cardinality of the set \(\{a: f(a) = 1 \land f(\uparrow(a)) = 2 \land f(\downarrow(a)) = 2 \land f(\rightarrow(a)) = 2 \land f(\leftarrow(a)) = 2\}\).
The black score in a go position \(f\) is the cardinality of the set \(\{a: f(a) = 2 \land f(\uparrow(a)) = 1 \land f(\downarrow(a)) = 1 \land f(\rightarrow(a)) = 1 \land f(\leftarrow(a)) = 1\}\).
A go position \(g\) results from a position \(f\) if \(f(x) = g(x)\) for all \(x\) so that \(f(x) \neq 0\) and the cardinality of the set \(\{x: f(x) = 0 \land g(x) \neq 0\}\) is exactly 1.
White has won in position \(f\) if the white score in \(f\) is \(\geq 5\) and the black score in \(f\) is \(< 5\).
Black has won in position \(f\) if the black score in \(f\) is \(\geq 5\) and the white score in \(f\) is \(< 5\).