Chess Sudoku

You posted that while I was writing.

The original problem is impossible. I proved that, look at #5.

Ah yes, sorry. Comes with the bad habit of posting without reading everything through.

Well, I believe I've done it. Feel free to examine this to see if I missed something, but I believe the below diagram proves that this is not impossible. I followed all the rules. As in a chess game, White starts from the bottom, so white's pawns threaten upward, and black's downward.

Now I feel dumb. It never occurred to me that you could move the pieces out so as to use a corner twice.

Great job ijgeoffrey.

Thank you! :) 

Wow, very nice!

Very Terrible

Good job!

How's this? (I couldn't remove the kings, but the knight concept is demonstrated.)

I was trying to include the kings. White's could go in a corner, but there's no place for black, I don't think it's possible unless you ignore the kings.

That's still a great idea though. I'm supposed to be good at this sudoku thing, but everytime you post I feel like an idiot 

Lol! Well for what it's worth, I like it when you post these questions. I enjoy finding the answers. :)

This one should stump you: Try to do it with queens 

You can get rid of the kings by placing another piece on top of them.

8 queens sudoku: Not touch another:

Solution:

(It is not letting me post without kings, I will let someone else)

a2, b4, c6, d8, e3, f1, g7, h5

There are 82 solutions. I did not make this up. for detailed info use this site:

(well I did the chess sudoku but not the queen puzzle)

http://en.wikipedia.org/wiki/Eight_queens_puzzle

Formulas got off site:

The examples above can be obtained with the following formulas. Let (i, j) be the square in column i and row j on the n × n chessboard, k an integer.

1. If n is even and n ≠ 6k + 2, then place queens at (i, 2i) and (n/2 + i, 2i - 1) for i = 1,2,...,n/2.
2. If n is even and n ≠ 6k, then place queens at (i, 1 + (2i + n/2 - 3 (mod n))) and (n + 1 - i, n - (2i + n/2 - 3 (mod n))) for i = 1,2,...,n/2.
3. If n is odd, then use one of the patterns above for (n - 1) and add a queen at (n, n).

Another approach is

1. If the remainder from dividing N by 6 is not 2 or 3 then the list is simply all even numbers followed by all odd numbers ≤ N
2. Otherwise, write separate lists of even and odd numbers (i.e. 2,4,6,8 - 1,3,5,7)
3. If the remainder is 2, swap 1 and 3 in odd list and move 5 to the end (i.e. 3,1,7,5)
4. If the remainder is 3, move 2 to the end of even list and 1,3 to the end of odd list (i.e. 4,6,8,2 - 5,7,1,3)
5. Append odd list to the even list and place queens in the rows given by these numbers, from left to right (i.e. a2, b4, c6, d8, e3, f1, g7, h5)

For N = 8 this results in fundamental solution 1 above. A few more examples follow.

• 14 queens (remainder 2): 2, 4, 6, 8, 10, 12, 14, 3, 1, 7, 9, 11, 13, 5.
• 15 queens (remainder 3): 4, 6, 8, 10, 12, 14, 2, 5, 7, 9, 11, 13, 15, 1, 3.
• 20 queens (remainder 2): 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 3, 1, 7, 9, 11, 13, 15, 17, 19, 5.

Brilliant! We should look for more solutions.

Haha! Nice one. :D I was thinking about your knights idea though, and I'm certain it's possible to include the kings and still have no threats. :)

Interesting, and very cool. :)

Well, I just came up with another one. :) What do you guys think: should I just post the solution, or should I try to make this one a puzzle? Just like real sudoku, I can post a diagram with just a few of the pieces already in place, and others can try to find the solution.

Edit: Compared to my last one, I would rate this new one as easy.

Give us a puzzle.

Ok. Here goes: