Since the puzzle turned out to be way hairier than expected, my earlier answer got quite messy and hard to follow, so here's a complete rewrite. It's still very clunky, but I hope it's much clearer than the earlier one. Also, there's an executive summary at the end, if you are not interested in all the gory details.
First, we'll need to make some general ...
Now that we have three increasingly complex proofs (two deleted, one of them mine) that it's impossible, it's pretty clear that it must be
This is, without doubt, the most refreshing chess problem I've ever tried to solve. Thanks, OP!
If white can ever stay out of check for one turn, then it can promote its pawn and put black in checkmate. So in order to play perfectly, black must attempt to put white in check with every move. In turn, white should make sure that black has only one option for a check in the next move, or else the rook will "break free" and have much more influence over ...
This 40 move solution on lichess works, and while it may not be the most orderly solution, it is impossible to create a solution in fewer moves since I'm using optimal pawn movement at every step and no other pieces (6 fields to move, 2 in the initial move, so 5 moves for each pawn to promote).
PGN of the game:
1. e4 f5 2. exf5 Kf7 3. f6 Nh6 4. fxe7 Kg8 5. ...
I give you the following:
This solution can arise in a regular chess game, where the black king is superfluous. The conditions would still be met if the black king were removed. White needs to make 25 moves (Queenside castle, thanks to h34 in comments) to get into this position, but due to the need of capturing all opposing pieces, a game leading to this ...
Yes. The minimum number of pieces required is 5.
5 queens can be places such that they cover every space on the board, as in the following example:
There are 12 such arrangements, along with rotation and reflection of each of them.
Edit: The above proves that 5 queens is enough, but it doesn't prove that 4 queens isn't enough. According to this ...
Initially, there are an even number of knights on white squares (namely, there are two of them, at b1 and g8).
Every time a knight moves, the number of knights on white squares either increases by one (if a knight on a black square moves) or decreases by one (if a knight on a white square moves).
Either way, the parity changes each turn. Thus, if ...
I wrote a computer program and it showed that $18$ moves is the optimum.
Here is one such solution:
Oddly enough, even if you relax the condition of alternating white and black moves, it cannot be done in fewer moves.
For $3\times3$ the optimal number of moves is $16$.
Without the need to alternate moves the optimum is $14$ moves, for example just by ...
I'm not trying to solve the puzzle, I'm just interested in how many solutions there are, since the OP claims he doesn't know. I brute forced it with a program.
First of all, there are 32432400 configurations, not taking rotations and reflections into account. Since the board has 16 squares, if we were to place the two kings anywhere, we'd be ...
Unless I'm mistaken, the result is
Reading through the wall of text, the rules seemed a bit too complicated for it to be "just some random game", so figuring out the magic seemed interesting.
To figure out if the bishop can capture all the lolcats on his first move, we need a "hitbox" for the lolcat; that is, the set of all those squares ...
(Edit: so this is wrong..)
Brilliant puzzle! The answer is:
[Spoiler alert! Scroll down at your own risk]
We proceed by contradiction. Assume that indeed, White can castle. We have the following undeniable facts:
Neither the White King nor the White h-Rook have moved.
Since neither of them have moved, the only way the Black Rook could have gotten to ...
This has something to do with the way computers evaluate positions. They will first count the value of each side's pieces (usually: pawn = 1, knight/bishop = 3, rook = 5, queen = 9) and then some other things, like pawn structure and mobility (number of possible moves). Based on that, Black is much better in this position. Computers don't have a way to 'see' ...
Give these names to all the squares:
Each number can only be accessed by way of the numbers before and after it (where 8 wraps around to 1). That means they form a loop. Since they can never pass each other up on the loop, their relative ordering cannot change. Therefore it is impossible.