Hot answers tagged

109

White must Indeed Note that


76

I see Evargalo has found the right answer, but the first thing that occurred to me was


66

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 Here's why: This is, without doubt, the most refreshing chess problem I've ever tried to solve. Thanks, OP!


55

Explanation:


49

The mystery The mate Additional note


49

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. ...


46

If my Python programming is to be believed, the minimum number of moves required is 41:


45

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 ...


43

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 ...


38



37

Proof: 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 ...


37

I can get the knight trapped in 15 moves: $$\begin{array} {ccccccccc} \cdot & \cdot & \cdot & \cdot & 5 & \cdot & \cdot & \cdot & \cdot\\ \cdot & 3 & \cdot & \cdot & \cdot & \cdot & \cdot & 7 & \cdot\\ \cdot & \cdot & \cdot & 4 & \cdot & 6 & \cdot & \cdot & \cdot\...


37

Here's a solution: Here's some of the logic for finding it: Here's a link to the moves: Clear the board!


37

I've found the actual answer: Then: Because: Picture:


35

Here's a solution that only takes 28 points:


34

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 ...


32

Because: According to Wolfram-Alpha, there are One possible solution is: A list (and images!) of all


32

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. There are 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 ...


31

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 ...


31

I think the answer is: Pawn: Rook: Knight: Bishop: Queen: King: Hidden word: Title:


31

It's simple: This gives the position: which is clearly mate. NB:


30

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 from which there ...


30

(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 ...


27

62 is the optimal result, I wrote an (admittedly somewhat hackneyed (and in retrospect not very good)) brute force program in Java which can be found here. It returns that 2 is the smallest number of uncovered squares possible using four queens. In short, it places four queens on a board, scans the board for each position, and outputs the case with the ...


27

The answer:


27

I hope I didn't make any mistakes: Edit : Replace queens by king


26

I found a solution that uses 16 moves. After exhaustively checking that there is no solution in 14 moves, I conclude that 16 moves is optimal, because after any odd number of moves the number of white and black squares occupied by knights cannot be equal.


26

Give these names to all the squares: 163 4 8 725 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.


26

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' ...


26

This seems to work: And the position looks like this: Apart from the symmetrical solution, this might very well be unique:


Only top voted, non community-wiki answers of a minimum length are eligible