33

There are an infinite number of squares whose sides pass through exactly four circles. In fact, there are an infinite number of squares passing through the top four circles alone. Each frame of the above gif represents one possible square you can draw through those circles. In addition to the 90 squares represented in these 90 frames, there are infinitely ...


24

Yes, X can win. To simplify things I'm going to take advantage of your rules that O cannot win, so that X doesn't need to worry about O getting 1000 in a row. Just consider a 1 dimensional game, chose an origin, and label the locations with integers in order. Define the "bin $K$" as the set of spaces $x$ such that $1000\ K \le x < 1000 (K+1)$. A ...


20

I've got 20


19

I've done it in 21:


15

Here is another simpler proof This generalises to all odd n. For even n, however, X. X. X..X X..X .... .XX. X..XX. X..... ..X..X ..X..X X..... X..XX. X..XX..X X......X ..X..X.. ..X..X.. X......X X..XX..X ........ .XX..XX. X..XX..XX. X......... ..X..XX..X ..X......X X...X..X.. X...X..X.. ..X......X ..X..XX..X X......... X..XX..XX.


15

I would suggest an alternate (simpler) strategy:


14

9 squares that are $1$x$1$ (trivial) 4 squares that are $\sqrt{2}$x$\sqrt{2}$ (Diamonds centered at the 4 central circles) 2 squares that are $\sqrt{5}$x$\sqrt{5}$ (Each starts at a yellow circle. Move like a knight: example 2 right 1 up, 2 up 1 left, 2 left 1 down, 2 down 1 right) 2 squares that are $\sqrt{13}$x$\sqrt{13}$ (Each starts at the orange ...


12

Thanks to @xnor presenting a solution, I was finally able to fix the bug in my Python 3 program. from collections import Counter import sys grid = [[7, 3, 1, 2, 9, 8, 5, 6, 4], [4, 9, 5, 1, 8, 6, 7, 2, 3], [9, 4, 6, 7, 5, 3, 1, 8, 2], [1, 5, 3, 8, 7, 4, 2, 9, 6], [8, 6, 2, 5, 3, 1, 4, 7, 9], [2, 8, 9, 4, 1, 5, 6, 3, ...


12

My solution: Solution method:


9

I reckon the answer is


9

This game is called Sprouts. There is a discussion in Winning ways for your mathematical plays, Vol. 3, by Berlekamp, Conway, and Guy, on p. 598. Sprouts also has a Wikipedia page.


9

Going first, you should be able to force a win. Using @kaine's terminology: box - an area on the board that can be made into a square square - a box of 4 lines three - a box of three lines which can be made into a square on the next turn safe - a move that doesn't result in a three Moves are enumerated as follows: * 1 * 2 * 3 4 5 * 6 * 7 * 8 9 ...


9

First player wins Example:


8

I found 21 9 variations of black square 2 variations of orange square 4 variations of brown square 4 variations of gray square 2 variations of red square


8

Maybe I am misunderstanding the puzzle, but it seemed straightforward to I literally plopped down circles as random as I could (but aligned to a grid, for ease of working) in a convenient drawing tool, in a 16x16 square. Then I The result is: As $a = 16$, I wonder if we are allowed to break that requirement for the bonus? Of course, circle placement will ...


8

This game is So in this case If we take "rational" to mean that when either player had a winning move they took it then Now So


7

... seems to work. (Brief) explanation of how I found it: Nice puzzle!


7

I solve it in two stages. First stage is The second stage is


6

My answer is 21. It break down as follows: 9: Square from dots right next to each other 4: Squares with 1 dot in the middle 2: Squares with central 4 dots in the middle 2: Squares with previous 2 squares in the middle (The above can be more easily visualized from watching Kevin's answer) 4: From the 4 squares with 1 dot in the middle, extend out 1 more ...


6

I wouldn't call this 'elegant', but here's an argument that there can be no solution starting with Grid 1. In the first diagram, edges 1-6 must all have the same label: edges 1 and 2 because of the square they share, then edges 3 and 4 because of the vertex they share with 1 and 2, and finally 5 and 6 because of the square they share with 2 and 4. ...


6

A 4 letter word that I found to be a dead-end word ladder island Why it's a dead-end word ladder island


6

Here's a way that should be faster than writing down letters. Let's say Steve was handed a piece a paper for him to draw on. Steve quickly draws an ichthus and adds some grill marks on it then rips the paper in half. He takes the bottom half, stands it up on the ground, and places some rocks around it so it stands up straight. If the answer is a destroyed ...


6

Alice wins. She starts by playing in the center square of the center board (which we'll call board C). Bob sends Alice over to another board, and she plays in the center of that one. Eventually, Bob fills up the rest of the middle board, and Alice then has the center of all but one of the outer boards. On that turn, Alice takes the "self-square" of the ...


5

Note: the contents of this answer are not actually correct; however, I've been asked to undelete this because it contains content that might be useful. I'm still trying to work through what I missed, but in essence, I think it comes down to the assumption that Red plays pieces that are adjacent to existing pieces, or do not otherwise interfere with them. I'...


5

Here's an easy "intuitive" explanation how X can win: If X can get a 998 in a row and it's her turn, she has won. So, if she can get TWO 998s in a row, and it's Os turn, she has won. {O will only be able to cover one - so it's over} So, if she can get FOUR 997s in a row, and it's O's turn, she has won. (O will only be able to cover ..two times.., so ...


5

Solution to all 3: Sorry it's a bit sloppy but conveys the idea adequately. I either got lucky or am good at these (or they were easy and had multiple solutions), I had to backtrack like twice total for all three.


5

Since there were exactly 32 empty spaces, I tried to push the ability of my computer and enumerate the best possible move starting from all $2^{32}$ states. For each state, the code finds the maximum number of squares you can get more than your opponent. In a few minutes, the code found that the best result is -9. That is, whatever you do, you cannot win ...


5

Here's a way to do it in 20 moves:


5

The task is Number the grid 1 to 25, from left to right top to bottom. Also, color the grid like a checkerboard so cell 1 is black. Pressing a white cell only changes black cells, and vice versa. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 There are an odd number of black cells which all must be changed. Pressing an interior ...


5

For $n = 1$, it's trivial that first player can't win since he can't make any move. Also for $n = 2$, both players can make the moves so the board will be full (by placing two $1 \times 2$ tiles), hence the first player can't win too. But for $n \geq 3$, For more details:


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