52
votes
How can 3 queens control the white squares?
I think this arrangement works for the bonus question:
- 1,703
47
votes
Accepted
36
votes
Accepted
Mr. Hilbert and the Problem of the Erroneous Equation
I don't know how to do the formatting (thanks McMagister for the edit) but the answer is
- 636
28
votes
Accepted
Ernie and the Lock-down Puzzle
Ernie's jigsaw puzzle isn't as straightforward as it seems, as it's actually:
One way of assembling the pieces legally is:
How will you know when you have succeeded?
PS Ernie definitely has a sense ...
- 121k
28
votes
A pentagon that can measure the first 7 integer distances
A possibly more elegant solution for 1..7 if we don't insist on a convex pentagon.
- 25k
27
votes
Accepted
How did four chessmen disappear?
This seems to work:
And the position looks like this:
Apart from the symmetrical solution, this might very well be unique:
- 72.3k
25
votes
Accepted
24
votes
Mr. Hilbert and the Problem of the Erroneous Equation
Simply rearranging the symbols used in the intended solution.
- 13.4k
24
votes
Accepted
Consecutive captures on the same square
Since we are talking about a standard game of chess (although with both players co-operating), we know that there are four pieces that cannot possibly make a capture in the series:
the two bishops on ...
- 72.3k
24
votes
23
votes
Accepted
22
votes
Accepted
The 5040 Square
A possible solution is:
10 8 3 21
12 15 14 2
1 7 30 24
42 6 4 5
Strategy
$$5040=2^4 \times 3^2 \times 5 \times 7$$
First I decided ...
- 4,010
21
votes
Chess Construction Challenge #6: The One Move Royale
Vepir has helped twice in this answer, first in spotting a mistake and then with an improvement in the number of moves. Please got upvote their answer too if you like this one.
Here is a position with
...
- 131k
20
votes
Accepted
Create a Word Search
Here's my answer:
From this point on, there is no spoiler text, because it makes it easier to format, at least for me.
...
- 2,325
20
votes
Accepted
Your Task Is to Create the World's Hardest Irregular Sudoku!
The 'hardest' possible Irregular Sudoku has
and it looks like this:
- 144k
19
votes
What if I told you that guessing in Sudoku is very bad and might give you a bad karma?
Glorfindel's answer is sufficient for the main question.
To answer the bonus question:
Here is an example:
To construct this example,
As for a starting position,
- 894
19
votes
Accepted
17
votes
Occupy a field with tetrominos
Here is yet another solution with 9 pieces. This one is nice and symmetrical.
I have been trying to think of a way to show that 8 will not work by arguing in terms of the number of edge squares that ...
- 415
17
votes
Mr. Hilbert and the Problem of the Erroneous Equation
Probably fails the no letter criterion.
Or using Lagrange notation as a workaround (thanks to McMagister) we can also write
- 4,010
17
votes
Accepted
16
votes
Accepted
The Game of Golden Squares
I've achieved
tiles, and can prove that this is the optimal solution.
Reasoning:
Golly 4.0+ pastable RLE of this solution:
...
- 4,864
15
votes
Create a 1 meter measure
I can do it in just:
Initial configuration:
First:
We have:
Now:
We get the mark:
Finally:
You get:
And the required distance is:
Why this works:
- 21.9k
14
votes
14
votes
Accepted
No moves at all, not even to put yourself in check
Here are my first idea (both sides are essentially the same answer, so the hint fits too):
Both positions seem to be independently reachable by a legal game. It might be possible to find a legal game ...
- 72.3k
14
votes
Accepted
Is there a configuration of the 8-puzzle where locking a tile makes it harder?
As far as I know, the only way to figure this out is by letting a computer run through all the possibilities. It is a small puzzle, so this does not take long.
First I will assume that you want the ...
- 47.6k
14
votes
14
votes
Accepted
What's the most distant chess position?
This question was asked on Chess Stack Exchange a couple of years ago: Which chess position requires the most moves to reach?
Just like @loopywalt here in the comments, I remembered Tim Krabbé's diary ...
- 27.4k
13
votes
Accepted
Geometry From Hell
Suppose you can set your pair of compasses to length 1. Then
Then we know that the length of the line that would join those two points is
The simplest construction would use:
- 22.4k
13
votes
Accepted
Mark Two Points Which Have a Distance of $\sqrt{3.6}$
Of course we need to use Pythagoras.
This leads to the following solution:
Here is another more compact solution.
- 47.6k
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