27

The answer:


22

The "best" move for this puzzle is the one most likely to hit the ship in one shot, as the puzzle doesn't allow for further follow up shots. There are 23 possible positions for that ship, and 68 spaces where a shot could be placed. Due to the size of the ship and previously placed shots, we can find both where shots cannot possibly hit the ship, and ...


21

It can be done in jumps. Here is the solution: If you start with the hole elsewhere, you need more moves to reach an end position. Here are all the optimal end positions for each case, but without the move sequence leading up to them.


20

The word was... And the reason you couldn't play the more common variant was...


19

D5 covers 4 locations D8 covers 3 F2 covers 2 H2 covers 1 G6 covers 0 E5 is the best; it covers 5 locations.


19

It is currently: because All that means it is now the


18



17

First rook is placed at any of Second rook is placed The third rook is placed And then the fourth rook A crappy example game below:


15

On my puzzle website there is an old puzzle called Hoo-Doo which is essentially this same puzzle. Some other board sizes:


14

The queen can always win. If the queen goes first, obviously it will win (just take the knight). If the knight goes first, there are only 2 places to go (mirrored, so effectively the same): - - - - - + . . . . . | . . K . . | . . . . . | . . . . . | All the queen has to do is move: - - - - - + . . . . . | . . K . . | . . Q . . | . . . . . | All the ...


14

Here is one way you could begin to prove that the board can be cleared for all values of $m$ and $n$. Proof by induction. [incomplete] Case 1: $k = 1$ Here we take k = 1 to mean the smallest possible board with the given conditions $(m \ge n \ge 2)$ which would be a $m = n = 2$ or $2m \times 2n$ = $4\times4$ board. Theorem: For an Othello board of size $...


13

For the updated version, I'll assume that after the placement of the pieces, the queen moves first. (Otherwise the knight could arrange things so that the queen is captured on the first move.) In this situation, the queen will still always win. Wherever the two pieces start, the queen can get to a space adjacent to the knight in no more than 3 moves. First,...


13

Yes, I found this video on YouTube that has this "perfect game". Terrible music by the way https://www.youtube.com/watch?v=prWG1OFgVqg


13

Using a computer to attempt to beat my original solution I found some Here is one: I reduced the search space by not allowing stones to moves to (although they can jump via) the spots marked with an * below: * * * * · · · · * * * · · · · · * * · · · · · · * · · · · * * * · · · · * * * * 1 2 3 * * * * * 4 5 6 * * * * * 7 8 9 * * * * * I then This ...


12

This kind of puzzle is commonly referred to as "lights out", after a board game by that name. I googled that phrase and came upon a solver at http://www.ueda.info.waseda.ac.jp/~n-kato/lightsout/. So I can't take credit for the answer but it's pretty simple:


12

I initially gave a slightly rambling stream-of-consciousness answer, which I have preserved below in case anyone prefers that. Here is a slightly slicker one. (It is the same argument, just possibly clearer.) It remains to show I have the suspicion that there is a way to express this idea that packs everything into one sentence, but I haven't quite found ...


12

I got a quick answer but not sure how good it is.


11

OK, I barely made it with 10 moves, but it seems pretty hard to prove this is optimal (and I suspect the answer may be lower). The last move you make is Rh8, which will be the mate.


11

The trick is to place them Full list:


11

This solution is no longer valid. The original posting of the problem had a typo (and lacked the restriction of "no auctions"); with the typo, at least one auction was required to solve, and this answer provided such a solution. For the corrected problem, @dcfyj's post provides the correct answer. It is Player 1: Battleship's turn Here's how the game plays ...


10

This is known as the "18-Queens Problem". I found this well-known solution by Friedrich Burchard & Friedrich Hariuc (1976) in 96 Half-moves. I can't make a claim to its optimality, but by looking at it and seeing that no better can be found, I'd say it may well be optimal. 1.e4 f5 2.e5 Nf6 3.exf6 e5 4.g4 e4 5.Ne2 e3 6.Ng3 e2 7.h4 f4 8.h5 fxg3 9.h6 g5 ...


10

Preamble Since you are forced to come up with a number of turns you will play, you can't simply walk out after winning the first bet. In order words, your wins will need to be big enough to cover your subsequent losses. If you bet on green (higher percentage than 0 or 00), then you have a 1/19 chance of winning a bet, and a 18/19 chance of losing. The ...


10

The upper bound limit is likely to be quite a bit lower than 60. For the purposes of this I will assume that you are WHITE. In order for you to have a valid move in Othello, you need a line formed by the empty space and at least one of each colour. The largest number of valid moves made possible by the presence of a single WHITE counter is 8 - one for each ...


10

Why?


10

The answer is indeed as stack reader discovered but here is a visual of the moves:


10

  A “piece” of the puzzle, having only inconclusively considered the second part...   If you think you know the answer for the above, don't be too hasty. Before writing your answer, consider this:   . . . $\require{begingroup} \begingroup \let \BS = \boldsymbol \let \S = \small \let \T = \...


10

It is: Reasoning:


10

The other guys already got the answer. (Maybe not with completely airtight arguments, but nevertheless.) The explanations would be a lot easier to follow if they had pictures, so here's one: Legend: Red square: after turn 1, you are in one of these. Green square: after turn 2, you are in one of these, or in some red square (except square 2). Red circle ...


9

Let's assume that opponent's Knight stands on A1, then my queen will be on {last letter}{last number} For both cases: For the updated question:


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