The Bunny's Tour

Question

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)

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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 1

Reiterate problem

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.

Vocabulary

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.

$Lemma\ 1:$ It is a known fact that no tour purely constistuted of steps is possible. In fact, it is known that it takes 4 distinct walks (sharing no squares) to cover all the black (or white) squares on the board.

$Lemma\ 2:$ 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.

Answer

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.

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early observations for historical reasons only

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.

Answer 2

This isn't a complete answer, but I put it here to see if it can help in the generation of a complete answer.

Let us consider a necessary condition (though not a sufficient one) for a solution.

Lets say a bunny "diags" when it transitions to a diagonally adjacent square. This counts as a visit of the square onto which it arrives. Let us say a bunny "jumps" when it makes an allowed transition to a square three squares away in one of the side directions (which requires a second bunny in an appropriate location). This counts as a visit of the square onto which it arrives. Let us say a bunny "lands" when it is repositioned due to the other bunny's "jump".

When a bunny jumps another, the second bunny lands (by that jump) on the jumping bunny's initial position. This landing of the second bunny either does, or does not count as a visit. If it does count then the problem is insolvable as written as in a circuit of 63 moves (j jumps and 63-j diags) 63+j squares will have been visited. Thus j must be 0 which is impossible as the bunny could have visited squares of only a single color had there been no jumps. Thus we conclude that the landing of a bunny on a square caused by the other bunny jumping must not count as a visit.

Consider one of the two black corners Let us call it c. It has only a single diagonal neighbor which we call n, and two white squares along the side at a distance of three which we call w1 and w2.

The two lines tracing through that square must change colors either immediately before or after that corner as there must be a jump to allow the other bunny to exit that corner (as the entering bunny cannot also leave without visiting a square a second time).

For each bunny, c must either be entered by a diag from n, or by a jump from one of w1 or w2. Similarly c must either be exited by a diag to n or a jump to one of w1 or w2. The entrance and exit of a bunny cannot both be diags as this would revisit c. They cannot both be jumps as it would take the other bunny three diags to get in position for the second jump and the rules disallow either bunny skipping moves. Thus we know that the entrance and exit from c must consist of a diag and a jump in some order. This is true of each bunny. The second bunny lands on c when the first bunny jumps in or out. This second bunny cannot jump as the first bunny cannot be in position to be jumped thus it must diag out. Thus the second bunny's entrance to c must be by a jump.

We have tightly constrained what must happen at each of the four corners, but have neither discerned a possible move sequence, not shown it to be impossible.

Comments? Extensions?

Answer 3

Reiterate problem

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.

Vocabulary

The bunny thus has 2 distinct modes of motion: the "walk" and the "hop". If a piece is hopped over its motion (though not counted as a move) is a "pivot".

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.

$Lemma\ 1:$ It is a known fact that no tour purely constistuted of walks is possible. In fact, it is known that it takes 4 distinct walks (sharing no squares) to cover all the black (or white) squares on the board.

$Lemma\ 2:$ As a corollary to the above, there are only a subset of squares reachable from any arbitrary square. If all squares or a colour were reachable then $Lemma\ 1$ would be false.

Answer

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.

Answer 4

The bunnies can cover the entire board by following this path:

Black moves first, and the bunnies start in the lower left-hand corner (or, I suppose, red starting in the other corner would work just as well). I used red and black for the bunnies for ease of illustration. I took some liberties with the rules - I hope by "forward" you meant "whichever of the four compass points the bunny wants to face".

Answer 5

There is no solution.

If a bunny starts on a X-coloured square, without jumping another bunny that bunny will remain on that coloured square.

In order to move a bunny from a bunny to move from one coloured square it must hop over an adjacent bunny that is either north, east, south or west (diagonal bunny hops, if they are legal, will place the bunny on the same coloured square; diagonal moves also remain on the same colour). These adjacent squares are always a different colour than the square the bunny that is hopping is on.

The issue is this: In order for both bunnies to cover each square on the board each bunny must hop at least once (the black bunny needs to get to the white squares and the white bunny needs to get to the black squares) but, it is only for one bunny to hop over another during the course of a game.

For example, say the bunny on the white square hops over the bunny on the black square, at this point both bunnies are on black squares, and here they must remain the entire game.

Answer 6

Jim already made the observation about that a corner square requires either entering or exiting by means of a jump.

I have some more observations:

• Diagonal moves keep a bunny constrained to only white or only black squares; the only way to change square color is by means of a jump.
• A jump requires both bunnies to be on different square colors, and the jump causes them both to change square colors.
• Hence, both bunnies need to start off on different square colors (this was also stated as a requirement)
• Because both bunnies need to cover both all white and black squares, one bunny needs to jump at least once so that they can reach the other square color.
• A corner square can also be covered if a bunny starts of at that square.
• There are four corner squares. At most 2 squares can be covered by a bunny's starting position, so there are at least 2 jumps required to cover the other 2 corner squares.