40

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.


20

Here is the optimal answer with Shown as knights;


18

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 ...


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


14

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


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 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 ...


13

Here is my attempt at the solution. I don't have any mathematical proof that this is the minimum no. of knights, but the steps I followed suggest that. In the figures below, the yellow cells denote the knight's location, and the corresponding numbers denote their covering cells.


11

The cheeky answer is: The reasoning is: For a not so cheeky answer there is an upper bound of We arrive at this by I implemented this algorithm here and golfed it here. The pieces I came up with are: There is a lower bound of A proof of this:


11

People have given some good upper bounds, how about a lower bound. However we can improve this ... However we can improve this ...


11

Here's another proof of the lower bound in Sriotchilism O'Zaic's answer.


10

I think this is different enough to warrant its own answer for the bonus question. Starting with Matthew Jensen's configuration, swap the last cube to get


10

I think the minimum number of knights is: the positions:


10

Partial answer, Tasks 2-4 Finished Task 2 Finished Task 3 (what a doozy) Finished Task 4 (easier than 3)


9

UPDATE (after a pretty sturdy hint from OP): (Again, one more number can be constructed, if flipping the dice is allowed.) Original answer: I got all the way up to with these dice: Here's how to construct the numbers If I can flip the dice, there's also


9

EDIT EDIT 2 Made a computer program to look for a solution with fewer moves. It managed to improve my previous solution so it now only takes Here they are: I had to make some assumptions to get a solution within a reasonable time, so I'm not absolutely sure this is the minimum. I'd love to see the minimum if this is not it!


9

Here is a solution using knights. I believe this is minimal, but I do not yet have a proof. The obvious lower bound is 12, as 11 knights can cover at most 88 squares, so there is at least $100-88-11=1$ square uncovered and empty. This lower bound can be improved slightly if you take into account that a knight covering a corner can cover at most 6 squares....


8

I think I figured it out after a bit of trial and error (if there is a method of calculating it please comment) *I have horrible formatting feel free to edit the spoilerblock to make it better/more readable


8

I came up with: Method:


8

Since the largest unpainted rectangle has an area of 6, So, Time for some case bashing! Continuing from there,


8

My one solutions:


8

Another solution, with bonus: Method: Edit: Had to fix broken spoiler tags.


8

I have an arrangement with 9 7? Lot of overlap, but not sure how to improve. Image of solution:


7

The answer is: Explanation: Further:


7

It seems to be possible to do this with nine painted cells.


7

One very efficient way to separate areas by squares is to put them in diagonal lines. This utilises the diagonal, which is the longest dimension of the square. Also, you can divide the plane into separate X-pentominoes (which have the maximum allowed connected area) by using diagonals of squares only. That pentomino pattern has the annoying property that ...


7

This is a great puzzle that had me really hooked! Partial answer. I can almost do it with 6:


7

Partial Answer Finished Task 1


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