**edit**

As used Wlod AA points out, the optimization of the _number_ of stones feels more like a computer science problem than a math problem, so maybe I should point out some simplifications if we just want an upper bound on the Manhattan distance we can travel.

We can use an idea that is classical for this game. Namely, I claim we may assume the Angel never repeats a position (in the sense that it never lands on a cell it has already visited).

Consider the maximal Manhattan distance the Angel can travel, and take a strategy $S$ that takes it the furthest from the initial point against any Devil strategy. 

Make a new strategy for the Angel, where at each position, it looks at all possible ways the game can continue, depending on the Devil's strategy played against $S$. Among all these continuations, and among those the set of all turns where the current node is (re)exited, there is one after which the Angel gets the furthest before reentry (or before being trapped). Now just move directly in that direction, and pretend there are imaginary stones where the Devil would've put them.

Now the cage argument is trivial, the Angel with the new strategy will run in a circle and after a few moves it reaches a sticky situation. The pivot running is also simplified, you can just run around until you reach the end.

**original**

After the clarifications, this seems to be a version of Conway's Angel(s) and Devil(s) game, where the Angel is restricted to only cardinal movement. Usually it is allowed to move also diagonally (so like a chess king). Even if allowed diagonal movement, it is a result of Berlekamp that the Devil has a winning strategy if the game is played on the infinite plane.

I didn't look at Berlekamp's proof, but Conway writes in [1] that the strategy works on a $32 \times 33$ board. I looked also at the strategy given by Kutz and Pór [2], and if you follow it literally, I am not sure 19x19 suffices. However, with our more restricted speed 1/2, I think it is easy to give a winning strategy for the Devil directly.

I will call the players _Angel_ and _Devil_ following Conway. The black stone is also called the Angel, while I'll call the white stones _stones_, following the post.

I'll describe a naive strategy that gives an upper bound of 19 stones, where the Angel never moves more than $4$ steps away from the original position in Manhattan distance. In particular I claim that the Devil wins on a $11 \times 11$ board already. Furthermore, we track the movement of the Angel, and the Devil only puts stones directly in front of the Angel (after it moves) or diagonally in front, i.e. if the Angel moves east (I'll use cardinal directions), the devil will put the next stone stone east, southeast or northeast.

**The Devil's strategy**


The idea is as follows: The Devil wants to force the Angel into a situation where it is the Angels's turn, and the Angel has a stone (called the _pivot_) to the south, to the northwest, and northeast (or some rototranslation of this situation). Let's call this the _sticky situation_ (because we'll show that the Angel is stuck to the pivot stone).

In picture form, where $S$ and $P$ are stones ($P$ the pivot stone) and $A$ is the Angel:
$$\begin{array}{ccc}
S & . & S \\
. & A & . \\
. & P & . \end{array}$$.

I claim that once in a sticky situation, the Angel is captured
after at most 11 more stones are added, and the Angel will never be able to travel more than Manhattan distance 2 away from the pivot.

To achieve this, we'll indeed force the Angel to run around the pivot. If in the sticky situation, no matter where the Angel moves next, we will introduce a stone in front of it. If it moved north, it will be forced to come back south and return to the sticky situation. If it moved east, it is forced to come back or move south. If it moves south, we introduce a stone southeast of it, and it is back to another sticky situation.

In picture form, the aim is to keep going towards the following constellation, where $P$ is the pivot stone, $s$ are stones that are added on a per need basis, and $S$ are the stones ensuring the Angel rotates around the pivot:
$$\begin{array}{ccccccc}
. & . & . & s & . & . & . \\
. & . & S & . & S & . & . \\
. & S & . & . & . & S & . \\
s & . & . & P & . & . & s \\
. & S & . & . & . & S & . \\
. & . & S & . & S & . & . \\
. & . & . & s & . & . & . \\
\end{array}$$

At some point, the Angel repeats a position, and we can stop its movement around the pivot and trap it with at most 4 extra stones. So you need at most the $S$ stones, one of the $s$ stones, and 4 extra trapping stones. We assumed two $S$s were already in place so that's at total of 11.

To get the Angel into a sticky situation is easy. Namely, as long as the Angel avoids a sticky situation, it can be confined into a $2 \times 2$ area by just always putting stones in front of it when it moves.

I.e. the idea is to keep building this cage:
$$\begin{array}{cccc}
. & S & S & . \\
S & . & . & S \\
S & . & . & S \\
. & S & S & . \\
\end{array}$$
If the Angel never escapes from here, eventually we catch it with at most 11 stones in total.

If it escapes, a naive upper bound on the number of stones is the bounds of that box minus the escape corner plus the stone making the situation sticky (8 stones), after which the situation would be (up to rototranslation):
$$\begin{array}{ccccc}
. & S & S & . & S \\
S & . & . & A & . \\
S & . & . & P & . \\
. & S & S & . & . \\
\end{array}$$
and then the 11 stones from the previous strategy for a total of 19.
Of course these overlap, i.e. if the box is in place we could shave several stones from the previous strategy.

Finally, to get the upper bound on Manhattan movement, observe that when we escape the cage, the pivot is at most two steps away from the original position (two of the cage stones are at distance $3$, but cannot actually become pivots as far as I can tell). And we don't get more than distance $2$ way from the pivot.

-----

There are admittedly some points where I would either need to introduce some more concepts, or do the case analysis for a precise proof. I played this enough in my head to be completely sure I didn't miss anything, but games can be tricky... I'm not an expert but maybe one can get the upper bound by computer with just minimax+$\alpha\beta$+memoization, that might be something to try.

Also, if someone is interested in turning the arrays above into something more artistic, feel free of course.

**References**

[1] <cite authors="Conway, John H.">_Conway, John H._, [**The Angel problem**](http://www.msri.org/communications/books/Book29/), Nowakowski, Richard J. (ed.), Games of no chance. Combinatorial games at MSRI. Workshop, July 11–21, 1994 in Berkeley, CA, USA. Cambridge: Cambridge Univ. Press. Math. Sci. Res. Inst. Publ. 29, 3-12 (1997). [ZBL0872.90133](https://zbmath.org/?q=an:0872.90133).</cite>

[2] <cite authors="Kutz, Martin; Pór, Attila">_Kutz, Martin; Pór, Attila_, [**Angel, devil, and king**](http://dx.doi.org/10.1007/11533719), Wang, Lusheng (ed.), Computing and combinatorics. 11th annual international conference, COCOON 2005, Kunming, China, August 16–29, 2005. Proceedings. Berlin: Springer (ISBN 3-540-28061-8/pbk). Lecture Notes in Computer Science 3595, 925-934 (2005). [ZBL1128.91312](https://zbmath.org/?q=an:1128.91312).</cite>