Chasing game on the Go board In Go (Weiqi), two players take turns placing stones on the vacant points of a board. Once placed, stones can only be removed from the board if a stone or a group of stones are surrounded by their opponents on all orthogonally adjacent points, in which case the stone or group is captured.


Now we consider a variant of the game. A black stone is put on the center of an empty 19*19 Go board, and then the white player puts its stones on the vacant points.
Different from traditional rules, the black will not put any other stones on the chessboard. Instead, he can move his unique stone to all orthogonally adjacent points on the board at will, as long as there is no white piece on the target point.


When the black moves one step, his opponent will take turns placing a white stone on the vacant points to intercept the black, until it is surrounded by white stones on all orthogonally adjacent points, in which case the black stone is captured and the white wins. If the black stone makes it to flee to the edge of the chessboard without being captured, the black will win.
Then my questions are:

*

*How many white stones at least are needed to capture the only black stone? Or the black player is destined to win the game?

*What if the board is infinite without the edge? Would the black be destined to win in that case?

 A: Request:   please, leave my notation alone.


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 $\ \mathbb Z^2.$

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 $\ G.$

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.

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.

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 $\ (1\ 2)\ $ 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 $\ (-\!1\ 3),$
$$ P_8\ :=\ (\,(0\ 2)\ \ \{(2\ 0)\,\ (2\ 2)
 \,\ (1\ 3)\,\ (-\!1\ 3)\}\,) $$
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 $\ (-\!2\ 0),$
$$ P_{12}\ :=\ (\,(-\!1\ 1)\ \ \{(2\ 0)\,\ (2\ 2)
   \,\ (1\ 3)\,\ (-\!1\ 3)\,\ (-\!2\ 2)\,\ (-\!2\ 0)\}\,) $$
Here the black stone doesn't have any non-nonsensical moves.
END of Proof
