Knights $3\times 3$ riddle

The answer is:

Black can always win

First let's introduce some notation in order to be simpler to denote the moves:

We will call the squares $$a_3 \ \ \ \ b_3 \ \ \ \ c_3$$ $$a_2 \ \ \ \ b_2 \ \ \ \ c_2$$ $$a_1 \ \ \ \ b_1 \ \ \ \ c_1$$

We recognize that if the following position is reached and its black's turn, then he can win:

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Then, if white's first move is some knight to $c_2$, then black can win:

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We need then to find the winning continuation for black if white's first move is for example:

enter image description here

In this case, we would play N$b_1$ and force white to play N$c_2$ (otherwise we would get the first case):

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In this case we would play the knight right back to where the white one came from:

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Black has a winning strategy.

First let's number the squares, so one can talk about possible moves:

Numbered 3x3 chess board

Note that the middle red square doesn't get a number, as is can't be reached from either of the other squares by a Knight's move, so the Knights can't end up on it.

It's now easy to see that a from each white corner square a Knight can move to exactly the 2 'opposing' black squares, while a Knight on a black Square on the middle of a side of the chess board's can move to the 2 corners that don't belong to that side.

The 8 squares and the possible moves between them can easily be illustrated ny the following graph with 8 vertices (representing 8 squares) and 8 edges, representing how a Knight can move from one square to the other:

Knights moves as graph

It's a nice circle with 8 vertices and I've marked the initial positions of the 4 Knights. This representation has the advantage that it's much easier to see "what happens", the exact nature of how a Knight move works is already gone and this method can be used for many similar problems.

As we can see, if we follow the the circle starting at 1 going clockwise, we have first the 2 white Knights, then the 2 black Knigths. So to win, one player needs to force the 2 opposing Knights on neighboring (in this graph) squares, then put their own Knights on the 2 outer squares so they have no way to move.

In the initial position, there is 1 emmpty square between the Knights on 6 and 8 (one pair of opposing Knights) and one empty square between the Knights on 1 and 3( another pair of opposing Knights). I'l describe this situation as "The distances between opposing Knights are 1 and 1".

If White starts with moving either Knight to 5, increasing the "distance" between it and it's opposing Knight to 2, black can just move the opposing Knight towards the moved white Knight, making the distances between opposing Knight again 1 and 1. But now White has its Knights neighbouring each other, so can't "back away further".

If on the other hand White moves one Knight towards the opposing black Knight, black will do the same with it's other Knight, so now the distances between opposing Knights are 0 and 0. It should be clear now that Black wins in this case: White can not move a Knight towards the opposing Knight, so can only back away, which allows Black to follow and make the distances again 0 and 0. White can't back away forever, as there are only 8 vertices; at some time the 2 white Knight will be next to each other and white can't back away further.

If the first case, we already have the situation that the white Knigths are next to each other, and the opposing Knights are each 1 and 1 away. White's only move is to move one Knight towards an opposing Knight, then Black can answer by moving the other Black Knight towards the opposing Knight, making the distances again 0 and 0. So then Black wins as shown before.

While this method may not immediatly be easier to understand than Babado's answer, it allows a general insight into similar problems. For example, if the initial position has the Knights of the same color on opposing corners of the board, they become interleaved on the movement graph. Now to win a player has to simultaneously have their 2 Knights occupy the 2 possible movement opportunuties for both oppoenent's Knights, which is impossible, as this are 4 different squares. The method to win in the actual problem only works because the Knights are not interleaved, so can be forced into neighboring (in the movement graph) squares, so they reduce the others movement options.

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