# Tag Info

47

I have a hunch that the answer is Explanation: Continuing this way, we see that

41

Assuming standard Minesweeper rules, here’s one solution (with $X$ = a mine): EDIT: In response to Euphoric in the comments, I solved this purely by logical deduction with a bit of educated guessing to make things easier on me. But if you really want to know how I did it, here’s a rigorous solution:

41

I was having a slow work day, so I fired up Blender and made this: In 13 hops, the block of 9 pegs can be moved two places down and to the right. By repeating the process two more times, the pegs can be moved to the bottom right corner.

28

Here is a simple strategy of how they could do it Proof

27

First of all, It will probably make a difference So let's concentrate on the other ones We can already stop here; so

24

My first attempt was done by hand. It used 28 commands: but this was not optimal. I have now done a computer search to find optimal solutions. It found 180 solutions of length No shorter solutions exist. I'll illustrate the following solution: To see how it works, start with a robot on every available square, and send the commands. The robots should all ...

22

I read the Wired article that was linked in the comments, and applied the ideas mentioned there to this problem, and my computer managed to find a solution that is moves long, and cannot find a solution of more moves. I do not know if this is an optimal solution. My solution is: I've split it by hand into chunks that return the robot to the centre spot, ...

21

This is, I'm sure, answered somewhere else. It is also related to Pascal's triangle. Simply fill out the grid as follows: In this grid, each number represents the number of ways of getting to that particular intersection. And that number is precisely the number of ways to get to the intersection below it added to the number of ways to get to the ...

21

4-button method

21

Here is the optimal answer with Shown as knights;

20

Perform tests of nine sheep on all but one sheep according to the illustrated patterns: The two important properties exhibited are The claim is that given a set of test results there is at most one possible group of five wolves. Suppose instead that some set of test results could have been produced by two different groups of five wolves A and B. Then both ...

20

The X-pentomino tiles the plane, so that tiling is a good way to start. There are two ways to cut an 8x8 region out of that tiling. If one of the 4 central squares of the 8x8 region has an X centred on it, you get this or else you get this The latter can be easily improved by replacing the ones at the edges to give this A different way to get the same ...

20

The answer is PROBABLY I don't have a proof yet, but experimenting with it will yield something like My strategy is Now my reasoning is that One idea to rigorously prove it is to use some result similar to and reason that EDIT: There was an error in my picture above: I put $18$ to the right of $15$. But this is very easy to correct. In the beginning ...

18

Although the puzzle is most likely to be solved without a computer, and we already have a winner, here are all 16 solutions, just for the record: There are some symmetries in there, of course. Whether rotations and flips should count as "different boards" is a matter of interpretation. Found with the following (quick and dirty) Java program that jus ...

17

Answer: Consider for instance You get the sequence of numbers In both cases,

17

It is a 10x10 map and you need to go east 9 times and south 9 times. There are total ${18\choose9}$ = 48620 paths which are impractical to be done with pen and paper in a reasonable amount of time. But if you observe This can be done pretty quickly with just 10x10=100 calculations. The answer is Proof of work Method

17

The 2 buttons can do the following With these settings we can reach any target in at most Proof that this is minimal

16

To specify a rectangle

16

15

15

I think the correct password is Reasoning

15

15

There are infinitely many solutions, some of them are in this form: To generate them: And to make them infinity:

15

I think the following would work Strategy

15

It is possible to solve this without a computer search. The proof of $\min|O|$ is below.

14

14

Absent the no-computers tag, I just wrote a quick Python script: The formulation of the problem was pretty easy too, just I did some more exploring, and

14

I really enjoy puzzles like these. Bonus question: Curiosity:

14

I think this works. Method: Oh, I don't think the bonus question was in there when I was solving this. I can take a look later. Here we go: Method:

14

I managed to find the 8 in So we are trying to find the 5th largest card in a bunch of 12, by measuring them in batches of four. Here's my strategy: Now we have identified, for certain, a couple of cards we can exclude: A1 and A2 (both have at least 8 cards smaller than them), and B4, C2, C3 and C4 (all have at least 5 cards bigger than them). We also have ...

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