## Chess Sudoku

For fans of Killer Sudoku, Samurai Sudoku and other variants

### Chess Sudoku

I have recently obtained a puzzle, which contains an 81-square board divided into 9 blocks of 3x3 squares. Additionaly, there are 81 chess pieces, 63 of them are queens, 9 each of 7 different colours (black, white, gold, silver, blue, green, red), and 18 of them are rooks, 9 each of 2 different colours (black & white).

The instructions are:
1. Each block must contain a set of 9 distinct pieces.
2. Each piece must not attack another piece identical to itself.
(In "attacking" there is no "parrying", i.e. a white queen will be seen as attacking another white queen 3 squares to its right even if a red queen and a black rook stand between them.)
3. Only black or white pieces can stand on the 4 corner squares.
4. Within each block, at least one rook must attack one or both pieces of the opposite colour.
(For example, a white rook attacking a black queen/rook, or both.)
That's it.

On the instruction sheet one example initial setup is shown as below:

And it seems to suggest there is only one unique answer:

Hopefully they have got it right?
udosuk

Posts: 2698
Joined: 17 July 2005

Well, at the very least there are two answers with the reflection along the main diagonal also working. There is no way to specify which is which and rule 3 points out it is one of two possibilities, so maybe the rooks in the corners should just be predefined and rule 3 tossed.
motris

Posts: 71
Joined: 13 March 2006

Can queens, and rooks, move 8 squares (e.g. from r1c1 to r1c9)? In chess, they can move at most 7 squares.

Bill Smythe
Smythe Dakota

Posts: 546
Joined: 11 February 2006

motris wrote:Well, at the very least there are two answers with the reflection along the main diagonal also working. ....

There is at least one other way to generate additional solutions, too. Reflect along the other long diagonal (r1c9-r9c1), and replace all silver queens with gold queens and vice versa, and all blue queens with green queens and vice versa.

I suspect the problem is defective, or was not translated properly from the site from which it came.

Incidentally, for those of us with black-and-white printers, the stated problem is equivalent to a vanilla sudoku with the following additional rules:

A. Diagonal constraints apply, on all diagonals, to all digits except 1 and 9 (the "rooks").

B. The four corner squares must all contain the digits 1, 2, 8, 9.

C. Within each box, either a 1 attacks (rookwise) either an 8 or a 9 or both, or a 9 attacks either a 1 or a 2 or both, or both.

Somebody on this forum told me that diagonal constraints cannot apply to all nine digits (in an otherwise vanilla sudoku), otherwise there will be no solution. (He may, however, have been talking about wrap-around diagonals, e.g. r1c5-r5c9 and r6c1-r9c4 glued together.) That may be what motivated this problem, whatever it is supposed to be.

Bill Smythe
Smythe Dakota

Posts: 546
Joined: 11 February 2006

Thomas, thanks for your observation. Of course there are at least 2 solutions with the reflection about the main diagonal. The example is wrong and we will see if this is the only mistake they've made.

I guess rule 3 is there combined with rule 4 to make sure one can only achieve one solution isomorphically. An interesting problem is how many pieces do we need in the initial setup to ensure a unique solution. For example, how many solutions does the following 5-piece setup give?

Smythe Dakota wrote:Can queens, and rooks, move 8 squares (e.g. from r1c1 to r1c9)? In chess, they can move at most 7 squares.

In fairy chess variants on 10x10 boards the queens/rooks can move 10 squares, so I guess the common sense is that they are not distance-limited on larger-sized boards.

Smythe Dakota wrote:There is at least one other way to generate additional solutions, too. Reflect along the other long diagonal (r1c9-r9c1), and replace all silver queens with gold queens and vice versa, and all blue queens with green queens and vice versa.

I suppose the point of the initial setup is that you cannot remove the pieces already "on board" and replace with other pieces? But then I again I might be wrong. One of the great enjoyment in chess is during a certain stage, one can swap the positions of the white and black queens, and see how the game will turn. (Is it the standard practice in USA too?)

Smythe Dakota wrote:I suspect the problem is defective, or was not translated properly from the site from which it came.

I'm inclined to believe you on this. My inferior level of English is probably the culprit here.

Smythe Dakota wrote:Somebody on this forum told me that diagonal constraints cannot apply to all nine digits (in an otherwise vanilla sudoku), otherwise there will be no solution. (He may, however, have been talking about wrap-around diagonals, e.g. r1c5-r5c9 and r6c1-r9c4 glued together.) That may be what motivated this problem, whatever it is supposed to be.

I suspect that "somebody" was either myself or Condor from New Zealand, who just came out to post again after a long period of inactivity. I don't think the wrap-around diagonals are relevant to the "diagonal constraints" though.

Again, thanks for your insights guys!
udosuk

Posts: 2698
Joined: 17 July 2005

Well, after further tampering with the puzzle, it seems the number of pieces in the initial setup to guarantee a unique solution varies quite a bit with the rules.

For example, if we drop rules 3 & 4, it takes at least 8 pieces for a unique solution:
1. Each block must contain a set of 9 distinct pieces.
2. Each piece must not attack another piece identical to itself.

Where if we enlarge rule 4 with 4 additional conditions, we can get a unique solution for as few as ONE set piece:
1. Each block must contain a set of 9 distinct pieces.
2. Each piece must not attack another piece identical to itself.
3. Only black or white pieces can stand on the 4 corner squares.
4. Within each block:
1. at least one rook must attack one or both pieces of the opposite colour,
2. the blue and green queens must attack each other,
3. the blue and gold queens must attack each other,
4. the green and silver queens must attack each other,
5. the gold and white queens must attack each other.

I don't suppose anyone can solve it manually though...
udosuk

Posts: 2698
Joined: 17 July 2005

Okay, my inspiration is to create a Sudoku Variant puzzle with only ONE given using some fairly general rules. I'm not surprised few are interested, as the puzzle is not likely to be human solvable. (I know making human solvable puzzles is definitely not my strength.) Anyhow, here is my attempt to make the rules even more general (rule 3 [corners] is dropped):

Final version ()
1. Each piece must not attack another piece identical to itself.
2. Within each block:
1. there must be 9 distinct pieces,
2. the green queen must attack the gold queen,
3. the gold queen must attack the black queen,
4. the black queen must attack the white queen,
5. the white queen must attack the silver queen,
6. the silver queen must attack the blue queen,
7. one or both rook(s) must attack the opposite queen(s).
Then ANY ONE of the 16 pieces in this setup gives a unique solution:

The Answer is different to the original one above.

udosuk

Posts: 2698
Joined: 17 July 2005

udosuk wrote:
Smythe Dakota wrote:.... There is at least one other way to generate additional solutions, too. Reflect along the other long diagonal (r1c9-r9c1), and replace all silver queens with gold queens and vice versa, and all blue queens with green queens and vice versa.

I suppose the point of the initial setup is that you cannot remove the pieces already "on board" and replace with other pieces? ....

I wasn't. From the original example, if you reflect along r1c9-r9c1 AND interchange silver with gold AND interchange blue with green, you'll end up with all the givens on the same squares, but with a different final solution. The interchanges compensate for the reflection.

Bill Smythe
Smythe Dakota

Posts: 546
Joined: 11 February 2006

Smythe Dakota wrote:From the original example, if you reflect along r1c9-r9c1 AND interchange silver with gold AND interchange blue with green, you'll end up with all the givens on the same squares, but with a different final solution. The interchanges compensate for the reflection.

Have you checked that the alternative solution wasn't identical to the one proposed by motris?
udosuk

Posts: 2698
Joined: 17 July 2005

udosuk wrote:.... Have you checked that the alternative solution wasn't identical to the one proposed by motris?

No, but that would mean the original solution had an interesting, but weird, sort of pseudo-symmetry, wherein switching blues with greens, etc, AND reflecting diagonally, resulted in all entries (not just givens) remaining unchanged. Whew!

Bill Smythe
Smythe Dakota

Posts: 546
Joined: 11 February 2006

udosuk's intuition with this other reflection is correct. Each of the pieces (except for the red which matches itself) is paired in the initial solution when mirroring through the center of the board. The blues and greens are all the same when reflected through the center. The silvers and yellows are the same. The white/black queens and white/black rooks are the same as well. As a result, the other diagonal reflection proposed plus piece swap is equivalent to the first diagonal reflection.

I guess to say this in a more mathematically based way, "inverting the pieces" on this board is the same as a reflection through the center (r5c5). Such a transformation could also be accomplished by reflecting through two perpindicular axes, such as the two diagonals. So, inverting + using one main diagonal reflection, equals the solution posed by doing just the other main diagonal reflection. So, to this point, I still just have the 2 valid grids for the first puzzle.
motris

Posts: 71
Joined: 13 March 2006