The Bunny is a new chess piece. It can move in 2 different ways: Diagonally, but only exactly one space (so like a bishop with the limitations of a king). It can also "Bunny-Hop" over another bunny. Here's an explanation of how this is done:
Let's say that, in the following image, the white bishop is a bunny named Henry, and the black bishop is a bunny named George. Henry can move onto any of the pawns. He can move onto the pawns diagonal to him by simply moving there normally (bunnies can move diagonally in any direction 1 space). It can capture the pawn 3 squares north of it by jumping over George. When a bunny jumps, it moves 3 spaces in the direction that's it's jumping. George is north of Henry, so Henry must be jumping north. So, he's jumping north 3 spaces, onto where that pawn is. Additionally, George must move south one space, because when a bunny jumps over another bunny, the bunny it jumps over (George) must move to the original position of the bunny that is jumping (Henry)
The board starts with 2 bunnies, one on a white square, one on a black square. You may choose the initial positions as long as the bunnies are on squares of different colors.
How can you alternate moves (move Bunny 1 during 1 turn, then Bunny 2 during the next) so that each bunny moves to each space on the board, exactly once? Note that each bunny must step on every one of the 64 squares, so they will each move 63 times. It will be at most 63 turns.
Answer
Bunny: a chess piece which moves like a bishop but only one square from its current position. It may also hop over another piece.
Hop: a bunny hops when another piece is in a square touching the current square (no diagonals). In its turn it occupies the square 3 squares from its current square in the direction of the hopped piece. The hopped piece then takes the original square of the bunny. This is not considered a turn/move for the hopped piece.
Say, there are 2 bunnies on a board. $\alpha$ and $\beta$. $\alpha$ and $\beta$ begin on opposite colour squares of your choosing.
By what means might they both tour the board? No turn may be taken to a previously occupied square (being hopped does not constitute occupation). They take turns, $\alpha$ moves first, $\beta$ second, and they will take 63 turns each.
The bunny thus has 2 distinct modes of motion: the "step" and the "hop". If a piece is hopped over its motion (though not counted as a move) is a "pivot". A sequence of "step"s is a "walk".
When $\alpha$ moves to a square he paints the square $green$. When $\beta$ moves to a square he paints it $blue$. After both bunnies have moved to a square it is painted $red$ - and hereafter no bunny may enter except by "pivot", and then only immediately as the last bunny has moved to this square.
$Lemma\ 2:$ As a corollary to the above, there are only a subset of total squares of a given colour reachable from any arbitrary square - where reachable means that it can be reached by only using some sequence of steps (no hopping). If all squares or a colour were reachable then $Lemma\ 1$ would be false.
I will now claim that this bunny's tour is impossible.
For $\alpha$, given any starting square $s$ on colour $c$, we know that $s$ must be a square on only one of the four walks needed to cover the square of colour $c$ - call this walk $w$. Any square visited from $s$ must be part of $w$. It is clear that his tour cannot be completed from $s$ ($Lemma\ 1$). This means that $\alpha$ must escape and continue by means of a hop. Now, $\beta$ occupies one of the squares in $w$, $s^{\prime}$. $\beta$ may now process on his walk $w^{\prime}$. We know that only the same squares that were reachable to $s$ are reachable to $s^{\prime}$. This means that they are part of the same walk. Eventually $\beta$ will get stuck or otherwise need to change to a different walk to have a chance at covering the board (as we know that 1 is not sufficient). His only means of escape is a hop. Now $\alpha$ takes a square on $w^{\prime}$, BUT $w^{\prime}$ has exactly the same reachable squares as $w$. So, $\alpha$ has gained no new reachable squares for colour $c$ ($Lemma\ 2$). And there is nothing he can do to remedy this situation.
Thus, there is no 2-bunny tour of a chessboard.
I'm quite positive that there is no solution.
Let's say we have 2 bunnies, a black one and a white one. The black bunny paints the board with green paint, and the white bunny paints the board with blue paint. A square that has been painted with both green and blue paint, is red.
- a white bunny can no longer visit a blue square
- a black bunny can no longer visit a green square
- a red square can no longer be accessed by any bunny.
The answer to this question suggests that there is no "walking-bishop" tour. The bunny is a walking-bishop with a hopping capability.
Each bunny must visit every square. This means that at least 8 hops would be required. 3 squares are "involved" in a hop. And one of them will be guaranteed to be red after a hop. This means that the 8 hops will involve cancelling out at least $\frac{1}{8}^{th}$ of the entire board.
- they must change from black to white squares 4 time in order to get out of where they are stuck and to finish painting the board.
The dead-locks occur under the following conditions: (the first 3 are because the bunny's counterpart cannot each a hopping square)
- a white bunny is anywhere where the 4 opposite colour squares surrounding him are coloured green
- a black bunny is anywhere where the 4 opposite colour squares surrounding him are coloured blue
- any bunny is anywhere where the 4 opposite colour squares surrounding him are coloured red
- a bunny is on a red square and cannot move to non-red square without making it red (now hopping is not possible - as the other bunny would need to occupy a red square - i.e. a square he has already occupied in the past)
I will turn this into something solid soon.
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