***Request**: please, leave my notation alone.*
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Everything below is VERY SIMPLE while writing it down was still a tough challenge for me (*the introduced terminology may be used in a follow up, under much more general circumstances*).
**The board:**
I'll represent the GO (weiqi) board as
$$ \{(x\ y)\in\mathbb Z^2:\ \max(|x|\,\ |y|)\ \le\ 9\}
\ \subseteq\ \mathbb Z^2 $$
It's smoother to define the game for the entire $\ \mathbb Z^2.\ $ Then
one can show that only a small part of it -- actually, a small part of the GO board -- is sufficient for catching the sole black stone.
We assume the Manhattan metrics in $\ \mathbb Z^2$:
$$ \forall_{(u\ v)\ (x\ y)\in\mathbb Z^2}\quad
d((u\ v)\,\ (x\ y))\ :=\ |x-u| + |y-v|\qquad $$
There are the obvious exactly eight **linear** isometries of $\ \mathbb Z^2.$
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**GAME**
A **position** is an arbitrary ordered pair
$\ P\ :=\ (b\ X)\ $ such that
$\ X\subseteq\mathbb Z^2\ $ is finite, and
$\ b\in\mathbb Z^2\setminus X.$
Position $\ \mathbf I\ :=\ ((0\ 0)\ \ \emptyset)\ $ is, by definition, the ***initial position***.
An ***odd-move***, also called a ***black move***, in position $\ P:=(b\ X)\ $ is an arbitrary
$\ c\in\mathbb R^2\setminus X\ $ such that
$\ d(b\ c)=1.\ $ The resulting position is defined as
$\ Q:=(c\,\ X).$ An odd-move $\ c\ $ is said to be
***original*** $\ \Leftarrow:\Rightarrow\ c\ $ is different from all previous moves.
An ***even-move***, also called a ***white move***, in
position $\ P:=(b\ X)\ $ is an arbitrary
$\ y\in\mathbb Z^2\setminus\{b\}.\ $ The resulting position is defined as $\ Q:=(b\ Y),\ $ where
$\ Y:=X\cup\{y\}.\ $ (Even-move $\ y\in X\ $ would be silly but legal).
A ***game-score*** is an arbitrary finite or infinite
maximal (non-extensible) sequence of consecutive positions $\ G\ :=\ (P_0\ P_1\ \ldots),\ $ where
three conditions are satisfied:
- $ P_0 := \mathbf I; $
- each position $\ P_{2\cdot n+1}\ $ is a result of an odd-move (done by the sole black stone);
- each position $\ P_{2\cdot n}\ (n>0)\ $ is a result of an even-move (done by adding a white stone, if
any).
***Remark 1** every finite game-score ends in an odd-indexed position.*
The set $\ T(G):=\{(0\ 0)\}\cup\{b_{2\cdot n}:\ n=0\ 1\ldots\}\ $ of
all odd-indexed positions of $\ G,\ $ plus the origin, is called the ***trace*** of game $\ G.$
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**Winning:**
Let $\ W\subseteq\mathbb Z^2\ $ be an arbitrary finite set. Game $\ G\ $ is $W$-won (by white stones)
$\ \Leftarrow:\Rightarrow\ T(G)\subseteq W\quad $ (white stones can be played outside of $W\ $ (!))
Set $\ W\ $ is a winning domain
$\ \Leftarrow:\Rightarrow\ $ the player of the white stones has a strategy, call it $W$-strategy, under
which every game is $W$-won (regardless of the choice
of the black stone moves).
If finite $\ W\ $ is not a winning domain then we say that the black stone defeats $\ W.$
**Theorem 1** If the black stone defeats finite $\ W\ $ then it can defeat $\ W\ $ by playing original moves only.
Actually, a stronger theorem holds. We say that odd move
$\ c\ $ that leads to position
$\ P_{2\cdot n+1}:=(c\ Y)\ $ ***stumbles***
$\ \Leftarrow:\Rightarrow c=(0\ 0)\ $ or there exists an earlier position $\ P_{2\cdot k+1}:=(b\ X)\ \ (k<n),\ $ and
a linear isometry $\ S:\mathbb Z^2\to\mathbb Z^2\ $ such that
- $\ S(W)=W,\ \ $ and
- $\ X\subseteq S(Y).$
**Theorem 2** If the black stone defeats finite $\ W\ $ then it can defeat $\ W\ $ without ever stumbling.
**Remark 2** Consider black stone's move $\ c\ $ such that there is only one nearest neighbor point that is not occupied by white stones. Then the white stone player wins immediately by setting a new white stone on the point that was left by the black stone just a moment ago.
**Remark 3** Consider black stone's move $\ c\ $ such that there are exactly two nearest neighbor points that are not occupied by white stones. Then the white stone player can force the black stone into a repetition (hence stumbling) of its previous move by playing a new white stone onto the unoccupied nearest neighbor of $\ c\ $ different from the previous black stone move (one before $\ c$).
**Theorem 3** If the black stone defeats finite $\ W\ $ then it can defeat $\ W\ $ without ever stumbling, and by playing moves $\ c\ $ such that there are (at the time) at least three nearest neighbors of $\ c\ $ that are not occupied by white stones.
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**THEOREM 4** Set
$$ V\ :=\ \{(x\ y)\in\mathbb Z^2:\ |x|+|y|\le 3\} $$
is a $25$-point winning domain.
(*See the proof below*).
*I'll present a (natural) $V$-strategy against which the black stone has essentially **only one non-nonsense defense***.
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**Proof**
We have $\ P_0=\mathbf I.\ $ Due to the symmetry of $\ V,\ $
we may assume that
$$ P_1\ :=\ (\,(1\ 0)\ \ \emptyset\,) $$
Let me play white stones, I am playing $\ (2\ 0)\ $ hence
$$ P_2\ :=\ (\,(1\ 0)\ \ \{(2\ 0)\}\,) $$
Black move back to $\ (0\ 0)\ $ would create a position inferior to $\ \mathbb I.\ $ Up to a symmetry, only one black
move $\ (1\ 1)\ $ is left:
$$ P_3\ :=\ (\,(1\ 1)\,\ \{(2\ 0)\}\,) $$
Now, let me play move $\ (2\ 2)\ $ hence
$$ P_4\ :=\ (\,(1\ 1)\ \ \{(2\ 0)\,\ (2\ 2)\}\,) $$
Then the black stone's moves $ (1\ 0)\ $ and $\ (1\ 0)\ $
would stumble, and $\ (2\ 1)\ $ would have only two unoccupied
nearest neighbors. Thus, only one move $\ (2\ 1)\ $ is left:
$$ P_5\ :=\ (\,(1\ 2)\ \ \{(2\ 0)\,\ (2\ 2)\}\,) $$
Now, I am forced to play $\ (1\ 3),$
$$ P_6\ :=\ (\,(1\ 2)\ \ \{(2\ 0)\,\ (2\ 2)
\,\ (1\ 3)\}\,) $$
Here, the black stone has only one non-stumbling move
$\ (0\ 2),$
$$ P_7\ :=\ (\,(0\ 2)\ \ \{(2\ 0)\,\ (2\ 2)
\,\ (1\ 3)\}\,) $$
Thus, let me play $\ (0\ 3),$
$$ P_8\ :=\ (\,(0\ 2)\ \ \{(2\ 0)\,\ (2\ 2)
\,\ (1\ 3)\,\ (-\!1\ 3)\}\,) $$
In the view of the earlier remarks, there is only one non-nonsense black stone's move $\ (-\!1\ 2),$
$$ P_9\ :=\ (\,(-\!1\ 2)\ \ \{(2\ 0)\,\ (2\ 2)
\,\ (1\ 3)\,\ (-\!1\ 3)\}\,) $$
Now, move $\ (-\!2\ 2)\ $ is forced,
$$ P_{10}\ :=\ (\,(-\!1\ 2)\ \ \{(2\ 0)\,\ (2\ 2)
\,\ (1\ 3)\,\ (-\!1\ 3)\,\ (-\!2\ 2)\}\,) $$
Then the only black stone's non-stumbling move is
$\ (-\!1\ 1),$
$$ P_{11}\ :=\ (\,(-\!1\ 1)\ \ \{(2\ 0)\,\ (2\ 2)
\,\ (1\ 3)\,\ (-\!1\ 3)\,\ (-\!2\ 2)\}\,) $$
Finally, let me play $\ (-\!2\ 0),
$$ P_{11}\ :=\ (\,(-\!1\ 1)\ \ \{(2\ 0)\,\ (2\ 2)
\,\ (1\ 3)\,\ (-\!1\ 3)\,\ (-\!2\ 2)\,\ (-\!2\ 0)\}\,) $$
Here the black stone doesn't have any non-nonsensical moves.
**END of Proof**