You may recall that, a while ago (some months ago), I asked for a proof (not by demonstration) that the knight chess piece can reach any square on the board, starting from anywhere.
While it is true that this is so, I got no proof which presents this in a generalized way. So now I can provide my own:
We will call the knight K, and say that K is a set of four positions a,b,c,d. The three first we will set as being in the same horizontal line, and write them as a1,a2,a3, and d is in the same vertical line with a3, so we will call it b3. We'll also allow that this set of four can alter its elements by pivoting from any ending element, so from a1 and b3, and define the pivot move as couples of mirror-symmetric relocations, at most 8, checked also by the board.
This means that K is also defined by having its one end act as the center of a circle, and the element next to b3 acting as a point in the circumference of that circle, thus making the circle be of radius=distance of a3 from a1= 2.
Given the chessboard has 2^6 squares (64), the number of distinct full circles (by which is meant circles that have a different square as the center, and full means they include all four possible circumference elements) is given by calculating how many circles of circumference=4 you can fit. Naturally you have to be at least 2 squares away from any edge, and to count all you have to include all those that differ exactly by 2 as well. You can think of this circumference4 circle as a point, also altering the chessboard to 1/4 of its size (cause your circle is a 4 circle). So 64/4=16= 2^4, which is how many of those full circles you can have (easy to check, eg starting from bottom left, the first circle covers the area inscribed in the square from a1,a4 to d1,d4, and you then add the analogous starting from b1,b4 to e1,e4 etc; there are in total 4 circles horizontally and 4 vertical groups formed by them, thus the total is 4(4)=16 )
img1
Now that we know the number of full circles formed (ie the number of positions on the chessboard from where the Knight has all its 8 moves available), we can check for part-circles. Those are of the same number as squares from which the knight has 2-7 moves available on an empty board. But instead of coming up with a construction of those multiple cases, we can calculate their numerical relation to that of the full circle, and we do that by examining the possible shifts of the chessboard when the full circles are on display: moving the chessboard (erasing one line) to the left would cancel all of your vertical circles and one of the horizontal ones, so from 2^4 you'd go to 3^2. But we also notice that adding the removed line to the other side, the right side, we are able to once again construct a new circle there, returning to 2^4. And the same is true for removing a vertical line and adding it to the other end. From this follows that regardless of how close the Knight is to the edge of the altering chessboard, it will retain the ability to form 16 full circles, and from this follows that the "alterable" full circles are 8(8) (this is the number of possible line relocations) the original full circle number, therefore they are 2^6. But 2^6 is also the number of squares on the board, so you have as many centers, thus as many positions the Knight can take.
The above needs another addition to be a proof of the movement ability of the Knight (cause up to know we started by hypothesizing the knight having as the center of the circle each of the squares in a given line differing by two at the very least from the edge of the board, and showed that if this is possible then all squares can host the knight. So we only have to show that the Knight can indeed position itself in the 4(4)
central tiles of the board. Here is a proof of that:
The square is 4(4), and we want to show that from any starting position on it, the Knight will reach (given enough moves) all of the tiles:
Figure 2
Obviously no full circle can occur here (we have a circumference of 3 here, while 4 is needed; moreover, these full circles would need an even integer as circumference, cause the center occupies a full square and thus isn't counted). But we can instead examine this as distinct cases of distributions of the 4 elements of "Knight", and their possible progressive relocations (factoring their own new starting distributions). Say the Knight begins at the southwestern edge of the 4(4) board shown in figure 2. It has two possible relocations, one with the unique row element ending at e4, the other ending at d5. Noticing that this all occurs within a 3(3) square, we say that in 3(3) the Knight can occupy 3 squares (the two mentioned and the starting one, which was c3). We also notice that it could only occupy one square in a 2(2) square or smaller, and that there is a possibility to occupy just one in a 3(3) if and only if it starts at its center. But given we only have to show the Knight can reach any other square of the 4(4) if it starts on one, we'll say that it occupies an x and two ys, the ys being symmetric and the x on their axis of symmetry. This means that our triangle of starting position+two new positions currently has a distance a from its center of symmetry, allowing for the end points to differ by two squares at each direction. This isn't enough by itself to show evidence of covering a 3(3) square, cause 3 is not perfectly divisible by 2. However we can a second triangle (green triangles in figure 3) and then show that by connecting the end points we finally get a set with the center of symmetry allowing for a distance of one square from two end points, so allowing for all 3(3) squares to be reached by rotations and new positioning.
Figure 3
Meh, this is a rather tedious proof, of mixed methods, and I'd rather have a more elegant version. But it's a start. Ideally I'd like to express everything as progressive distributions of a set's elements within a region.
