## Anti-Knight X Sudoku

For fans of Killer Sudoku, Samurai Sudoku and other variants

### Anti-Knight X Sudoku

This is a diagonal sudoku with the additional constraint that no cells at a chess-knight move distance may hold the same digit. For instance, if R4C7 = 1, R26C68 & R35C59 <> 1

Code: Select all
.19...........7..2.......1.......18...........34.......4.......5..7...........64.

For lovers of pointing locked candidates, Cross-over, Turbot and other similar seafood
Jean-Christophe

Posts: 149
Joined: 22 January 2006

Nice.........

It probably is not valid but if you could from a starting cell make your knight jump 80 times into completely different cells with a sequence of 9 different boxes each time then you could have a nice DG variant...

tarek
Last edited by tarek on Fri Nov 17, 2006 3:25 am, edited 1 time in total.

tarek

Posts: 3761
Joined: 05 January 2006

Its a nice combination but it seems to me too that there isn't any solution.
The idea of Tarek I can't understand. What is the additional constraint?

Pyrrhon
Pyrrhon

Posts: 240
Joined: 26 April 2006

You would have 9 different groups of 9 different cells......each cell in the group is a knight's jump from another cell in the same group......

tarek

tarek

Posts: 3761
Joined: 05 January 2006

Tarek,

It's maybe possible to make 9 DG each consisting of chain of 9 cells at knight jumps. But some DG will have to share at least 2 cells with a box. eg DG with R1C1 will have to include either R2C3 or R3C2, both in box 1.

Or maybe your idea is to make one big chain of 81 cells at knight jumps with a sequence of warping numbers : 1..9,1..9,... Have no idea if it's feasable, since this is so much constrained. I don't even know if there is such a chain covering all the cells.

Another version : each cell must hold a number consecutive to some other cell at a knight jump. eg if R5C5 = 5, then one of the cells in R37C46 or R46C37 must hold either 4 or 6.

Or else : one and only one consecutive at knight jumps.
Or else : both lower & upper consecutives at knight jumps.
...
Jean-Christophe

Posts: 149
Joined: 22 January 2006

I thought so .....

but the idea of having 9 different boxes sounded possible at first....

so you can start from latin squares & add those 9 groups as opposed to starting from a vanilla sudoku...

tarek

tarek

Posts: 3761
Joined: 05 January 2006

Jean-Christophe wrote:Or maybe your idea is to make one big chain of 81 cells at knight jumps with a sequence of warping numbers : 1..9,1..9,... Have no idea if it's feasable, since this is so much constrained. I don't even know if there is such a chain covering all the cells.

One year (and some) ago, I've pondered about the same ideas... A chain covering all 81 cells (a knight's tour on a 9x9 grid) surely exists, but "a sequence of warping numbers" probably not... I've made a few puzzles about the tours/circuits of wazir (orthongonal king), king and knight with absolute minimum numerical sequence of 81 digits in this thread...
udosuk

Posts: 2698
Joined: 17 July 2005

udosuk,

Thanks for the input. Will give it a try later using less restrictive rules. There is maybe a knight's tour of consecutive numbers where a sequence of 1-2-1 would be allowed provided it does not break the sudoku rules. Now if it takes 100 years to search...

Here is a walkthrough and the solution of my puzzle:
R2C6 = 7 -> R4C5 <> 7
R8C6 = 7 -> R6C5 <> 7
-> R5C5 = 7 (hidden single in N5)
-> R19C19 <> 7
R2C6 = 7 -> R1C8 <> 7
-> R1C7 = 7 (hidden single in R1)
R8C6 = 7 -> R9C2 <> 7
R9C3 = 7 (hidden single in R9)

R1C2 = 1 -> R2C4 <> 1
-> R2C5 = 1 (hidden single in N2)
R9C9 = 1 (hidden single in D\)
1 of R6 locked in R6C14 -> R7C3 <> 1
R6C4 = 1 (hidden single in D/) -> R8C3, R7C6 <> 1
R7C1 = 1, R8C6 = 1, R5C3 = 1

R8C5 = 4 (hidden single in N8)
R6C3 = 4 -> R4C4 <> 4
R1C1 = 4 (hidden single in D\)
4 of R4 locked in R4C69 -> R3C7 <> 4
R4C6 = 4 (hidden single in D/) -> R2C7, R3C4 <> 4
R3C9 = 4, R5C7 = 4, R2C4 = 4

R6C2 = 3 -> R8C23 <> 3
3 of R8 locked in N9 -> not elsewhere in N9
3 of N7 locked in D/ -> not elsewhere in D/
3 of N7 locked in R7C3, R9C1 -> R9C4 <> 3
3 of C4 locked in R134C4 -> R3C36 <> 3
Turbot Fish on 3 with 3 links R4C4 == D\ == R8C8 .. R8 .. R8C7 == C7 == R2C7
-> 3 either in R4C4 or R2C7 -> R2C3 <> 3
R7C3 = 3 (hidden single in C3)

2 of R1 locked in N2 -> not elsewhere in N2
2 of D/ locked in N7 -> not elsewhere in N7
2 of C3 locked in R34C3 -> R34C1, R4C45, R5C2 <> 2
2 of R4 locked in N4 -> not elsewhere in N4
R9C1 = 2 (hidden single in C1)
2 of R8 locked in R8C78 -> not elsewhere in N9
-> R6C678 <> 2
R5C8 = 2 (hidden single in N6)
R8C7 = 2, R6C5 = 2
R3C3 = 2 (hidden single in D\) -> R1C4 <> 2
R4C2 = 2, R1C6 = 2, R7C4 = 2

R2C7 = 3 (hidden single in C7) -> R13C5 <> 3
R3C1 = 3, R1C4 = 3, R8C8 = 3, R9C5 = 3, R5C6 = 3, R4C9 = 3
R3C2 = 7 (hidden single in C2)
...

Solution:
419382765
685417392
372695814
726954183
951873426
834126957
143269578
568741239
297538641
Jean-Christophe

Posts: 149
Joined: 22 January 2006

Here is another one:

Code: Select all
.3...........3.1..........4.71........2...6........58.9..........5.9...........9.
Jean-Christophe

Posts: 149
Joined: 22 January 2006

Here is a walkthrough and the solution of my puzzle #2:

R2C5 = 3 -> R3C7 <> 3
R3C8 = 3 (hidden single in R3)
R3C9 = 4 -> R4C7, R5C8 <> 4
R4C8 = 4 (hidden single in N6)
R7C1 = 9 -> R5C2, R6C3 <> 9
R6C2 = 9 (hidden single in N4)

R4C3 = 1 -> R5C5, R6C4 <> 1
1 of D/ locked in R8C2, R9C1 -> not elsewhere in N7, R9C4 <> 1
1 of C1 locked in R19C1 -> R9C9 <> 1
1 of C8 locked in R578C8 -> R6C6, R7C9 <> 1
D\, D/ and R8, C1 forms a Generalized X-Wing on 1 -> not elsewhere in R8, C1
1 of C9 locked in N6 -> not elsewhere in N6
-> R5C8 = 7 (naked single)

R6C7 = 5 -> R4C6, R5C5 <> 5
5 of D/ locked in R1C9, R2C8 -> not elsewhere in N3, R1C6 <> 5
5 of C9 locked in R19C9 -> R1C1 <> 5
5 of C2 locked in R235C2 -> R3C1, R4C4 <> 5
D\, C8 and R2, N9 forms a Generalized X-Wing on 5 -> not elsewhere in R2, N9
5 of N1 locked in C2 -> not elsewhere in C2

R5C25 forms a naked Pair on {48} within R5 -> not elsewhere in R5, R28C2, R367C3, R46C4
7 of R6 locked in R6C45 -> R7C3, R8C4 <> 7
R67C3 forms a naked Pair on {36} within C3 -> not elsewhere in C3, R6C14, R7C5, R8C2
R6C1 = 4, R5C2 = 8, R5C5 = 4
R7C2 = 4 (hidden single in R7)

3 of D\ locked in R6C6, R7C7, R9C9 -> R6C9, R8C7, R9C6 <> 3
R6C36 forms a hidden Pair on {36} within R6 -> R6C36 = {36} -> R4C4 <> {36}
6 of N5 locked in C6 -> not elsewhere in C6
R4C479 forms a naked Triplet on {239} within R4 -> not elsewhere in R4

8 of R4 locked in R4C56 -> R2C6, R3C7 <> 8
R34C7, R4C9 forms a generalized naked Triplet on {239} since each cell is a buddy of the other two
-> R2C8 <> {239}

C36 and R6, D/ forms a Generalized X-Wing on 6 -> not elsewhere in R6, D/
-> R2C8 = 5 (naked single)
R3C2 = 5 (hidden single in N1)
R9C9 = 5 (hidden single in C9)
...

Solution:
139452768
864739152
257861934
571986243
382145679
496273581
943527816
615398427
728614395
Jean-Christophe

Posts: 149
Joined: 22 January 2006