# Tag Info

29

If you are trying to lose the game as quickly as possible then you can do it in 1 turn and 2 rolls of the dice. This is using the UK version as I'm not sure about the US one. Roll 1: Roll 2 (Because of the doubles): The chance of this is:

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Okay. THAT. WAS. INCREDIBLE! It took me two solid hours of work to solve and even longer to write and draw it all up here - hopefully it'll be worth it! To start us off, here is the final maze layout and routes: In the following explanation all colours have been abbreviated to their initials as follows: G=Green, O=Orange, P=Purple, Y=Yellow. An ...

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So building on kristinalustig's answer they are indeed playing The cards are assigned as follows Alice's first turn Bob's first turn In particular Alice's next move produces Bob's next move produces "you could have got yourself 7 points more if you’d played the five of clubs first." Alice's last move is Alice's score For those unfamiliar with the ...

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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.

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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 ...

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The game is and the result is

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

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

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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:

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

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

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These board-gamers are playing: The board at the end of the game looks like this: Who wins the game? A final few Easter Eggs:

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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,...

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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 \$...

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I spent quite a while working on this yesterday and today. I only have a partial answer. I started with the assumption that I chose this assumption because: Given this assumption, I tried: However: I might proceed with:

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The clue answers: Sorted by group (and given in text form): And when you take the given pair of letters, and read those pairs of letters in clue order, it spells out the message: The full journey on a map (locations approximate), for reference (journey begins bottom-right): (Various clues solved by bobble, Chris Cudmore, Deusovi, North, Stiv, and ...

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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 ...

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

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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 ...

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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.

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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 ...

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From the facts about the first grid My first guess of the genre was Naturally, the next step would be It is not very hard, if we focus on the number 5: Therefore, Now to the second grid. With transcribed numbers, the grid to solve is this: Starting with R4 and focusing on fives, Then it gets a little harder... Finally... Reading the sums of each ...

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This should do it: White is in zugzwang, so black has

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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:

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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 ...

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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 ...

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

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