The New Sudoku Players' Forum

[tarek]

Another Easy Toroidal Anti-kNight Sudoku

Text: Show

Code: Select all

3.7.......19......28.........................................18......47.......5.9
 3 . 7 | . . . | . . . 
 . 1 9 | . . . | . . . 
 2 8 . | . . . | . . . 
-------+-------+------
 . . . | . . . | . . . 
 . . . | . . . | . . . 
 . . . | . . . | . . . 
-------+-------+------
 . . . | . . . | . 1 8 
 . . . | . . . | 4 7 . 
 . . . | . . . | 5 . 9

Solution: Show

[Smythe Dakota]

After a few hours, I finally got your last one. Whew! Why do you call these "easy"?

Have you considered the same idea, but on a projective plane or Klein bottle instead of a torus?

Terminology:

A projective plane is the same as a torus, except that opposite edges are glued together in reverse order. For example, instead of r1c3 and r9c3 being vertically adjacent, r1c3 and r9c7 are vertically adjacent. Ditto for rows as well as columns. Think of it as a double Mobius twist (twist it both horizontally and vertically).

A Klein bottle is similar, except that only one pair of edges is Mobius'd. For example, r1 could be glued to r9 in reverse order, while c1 is glued to c9 without the twist.

In the Klein bottle case, you could make it a bit more interesting by not telling us which edge pair (top-bottom or left-right) is Mobius'd and which is straight.

By the way, unlike the torus, it is not possible to embed a true Klein bottle into Euclidean 3-space. The same is true (I think) of a projective plane.

Bill Smythe

[tarek]

After a few hours, I finally got your last one. Whew! Why do you call these "easy"?

Well done. My impression is that these will get easier as you solve more, so that your eyes will automatically look at the grid as a toroid or your brain will memorize the kNight's move for each cell. My next puzzle may be too excessive as it is a Toroidal Anti Root 50 Sudoku [ (7,1) & (5,5)]

Have you considered the same idea, but on a projective plane or Klein bottle instead of a torus?

I just read about them. I liked the Projective plane which should be on the to do list. I'll stick mostly to Anti-King after the next one unless somebody insists on an Eye coordination challenge. The Cylinder, Mobius Strip & Klein bottle are also possible but not as symmetrical

[Smythe Dakota]

.... My next puzzle may be .... a Toroidal Anti Root 50 Sudoku [ (7,1) & (5,5) ] ....

I've actually heard of that Root 50 concept, in another context. A master-level chess player at a tournament showed me a chess problem (white to move and win or something) where the knights were square-root-of-50 knights rather than the standard square-root-of-5 knights.

Bill Smythe

[simon_blow_snow]

Bill, the root-50 leaper is one of the standard features in the JSudoku program, which I believe is what Tarek uses to create these nice puzzles.

Other examples include the giraffe (1,4 leaper), the ostrich (1,5 leaper), the flamingo (1,6 leaper), the zebra (2,3 leaper), and the antelope (3,4 leaper). I think udosuk used to create some very minimal puzzles using the features of the ostrich (down to 1 or 2 given clues for uniqueness).

What I want to ask is: tarek, is there a way to make JSudoku apply the anti-chess properties wrapping around the borders, or we can only make those eliminations manually?

I played your puzzle using the JSudoku program which automatically makes the non-wrapping anti-knight eliminations. Without using any hints from the program I think I solved it within 5 minutes (only singles required). Should take at least twice long if doing it on paper.

[tarek]

I use JSudoku for visual representation of puzzles and to help sometimes in verifying my generated puzzles which I create using my software. JC has definitely implanted most of the Fairy leapers which I've implemented. I have implemented some that are not there (tripper (3,3), (4,4) leaper, root 25, ...) which from a proramming point of view is easy when you've implemented the rest.

JSudoku doesn't support Toroidal, projective plane, Klein bottles nor möbius strips. Your method of solving is the best way to use JSudoku in these circumstances.

I'm not sure if the projective plane will work well with anti chess. I know it will definitely not work with anti-king (r1c1 is a king's move away from r1c1)

[simon_blow_snow]

I'm not sure if the projective plane will work well with anti chess. I know it will definitely not work with anti-king (r1c1 is a king's move away from r1c1)

If you define a "king's move" as "2 single orthogonal steps which does not form a straight line", then r1c1 has 6 distinct targets: r1c2, r2c1, r2c2, r8c9, r9c8, r9c9. r1c1 does not qualify as a target because the only way to construct the 2 orthogonal steps is r1c1-r9c9-r1c1, which in the (imaginary space of) projective plane still form a "straight line".

[tarek]

because the only way to construct the 2 orthogonal steps is r1c1-r9c9-r1c1, which in the (imaginary space of) projective plane still form a "straight line".

I'm not sure that the rules about straight lines apply here. Because this Projective plane from what I read can't be projected into the real 3 space. I still think that you need to take a 90 degree turn to get from r9c9 to the next r1c1.
Also, where is the next stop of a straight line that starts at r1c1 and passes orthogonally to the neighbouring r9c9? If it's a straight line there should be one answer!

My Euler Spectacles are making me dizzy now

[Smythe Dakota]

.... r1c1 does not qualify as a target [ of r1c1 ] because the only way to construct the 2 orthogonal steps is r1c1-r9c9-r1c1, which in the (imaginary space of) projective plane still form a "straight line".

I don't agree. There are two ways to get from r1c1 to r9c9 orthogonally -- one cell up, or one cell left. If you go one way and come back the other, surely it is reasonable to consider these two steps as orthogonal to each other.

Bill Smythe

[tarek]

These Anti-chess pieces have caused the solver to report incompatibilities on the 9x9 Projective plane board.

Code: Select all

Leaper (x,y) Example
--------------------
Anti-King (1,0) and (1,1) r1c1
Anti-Fers (1,1) r1c1
Anti-Tripper (3,3) r2c2
Anti-Camel (1,3) r1c2
Anti-Ostrich (1,5) r1c3
Anti-Root50 (1,7) and (5,5) r1c4

[Smythe Dakota]

Not surprising. There seems to be some kind of weird "fold" at the corners of the projective plane. Each 3x3 "carpet", instead of consisting of nine cells, in the corners seems to consist of only 7. And the center square is adjacent to itself, and two of the other "nine" coincide, etc etc etc.

Maybe it would be useful, instead of starting with a large 9x9 square, to start with a large circle, divided radially (9 pie shapes) and concentrically (9 circles), and to glue each edge piece to its opposite number. Except that there is no opposite number, since 9 is odd.

Maybe divide the big circle into 18 pie shapes and 9 concentric circles, and consider each little "trapezoid" to be identified with its opposite number. Then it's 9x9 again. That would be more in line with the "points at infinity" and "line at infinity" concept of projective geometry, wherein you start with the Euclidean plane and throw in a point at infinity for each direction, i.e. a point at infinity is defined as an equivalence class of parallel lines. In this way, topologically, the lines no longer look like lines, but rather like circles.

I dunno.

Bill Smythe

[tarek]

I'm inclined to keep it simple for the time being (if I can use the word simple to describe these) ....
Anti-Knight should be possible with the 9x9 Projective plane & i'm generating some.

I'm tempeted now to support the Klein bottle ...
I will be declaring the sides which form a cylinder & the ones wich form a mobius strip ...
If I'm going to hide which sides are which, I will then test that a 90 degree turn of the sides will yield an invalid puzzle or elso we'll end up with 2 solutions (Well, that would be nice twist if such a puzzle exists).

tarek

[Smythe Dakota]

It would be hilarious if you could come up with a Klein bottle that has exactly one vertical and one horizontal solution, with the two solutions different.

Or, one that has multiple solutions, both vertical and horizontal, but only one which works both ways.

Bill Smythe

[tarek]

It would be hilarious if you could come up with a Klein bottle that has exactly one vertical and one horizontal solution, with the two solutions different.

Or, one that has multiple solutions, both vertical and horizontal, but only one which works both ways.

Hilarious as it may sound ... The solver can be programmed to handle it.

Your 1st Scenario (2 puzzles that are 90 degree symmetry isomorphs each with a single solution that is not 90 degree isomorph of the other); is possible with 2 passes of the solver over the puzzle and the 90 degrees version.

Your Second Scenario is actually 1 puzzle with double the constraints. Each 1/2 portion may or may not be enough to solve the puzzle. This needs more work but is achievable.

The same thing can be said about combining the projective plane with the toroid.

I need to post the Anti-Root50 Toroidal 1st then an Anti-kNight projective plane before dedicating some time for this.

tarek

[Smythe Dakota]

Well, then, how about a puzzle which has multiple solutions in each of the following nine modes:

vanilla anti-Knight
cylinder anti-Knight, vertical
cylinder anti-Knight, horizontal
Mobius strip anti-Knight, vertical
Mobius strip anti-Knight, horizontal
torus anti-Knight
projective plane anti-Knight
Klein bottle anti-Knight, Mobius vertical
Klein bottle anti-Knight, Mobius horizontal

-- and, in fact, multiple solutions in any combination of up to eight of the above -- but only one solution that works for all nine.



Bill Smythe