The New Sudoku Players' Forum

[marek stefanik]

Code: Select all

+-------+-------+-------+
| . . . | 1 2 3 | . . . |
| . . 2 | . 6 . | 9 1 . |
| . 7 . | 9 . 4 | . . . |
+-------+-------+-------+
| . . 4 | . . . | . 7 3 |
| . . 9 | . . . | 8 . . |
| 3 6 . | . . . | 5 . . |
+-------+-------+-------+
| . . . | 8 . 5 | . 6 . |
| . 1 8 | . 7 . | 2 . . |
| . . . | 3 1 2 | . . . |
+-------+-------+-------+
...123.....2.6.91..7.9.4.....4....73..9...8..36....5.....8.5.6..18.7.2.....312...

[AnotherLife]

Hi Marek,
As I know your logic, I guessed from the start that this puzzle had to do with some symmetry. Let's place all the 3's into the grid.

Code: Select all

+-------+-------+-------+
| . . . | 1 2 3 | . . . |
| . 3 2 | . 6 . | 9 1 . |
| . 7 . | 9 . 4 | 3 . . |
+-------+-------+-------+
| . . 4 | . . . | . 7 3 |
| . . 9 | . 3 . | 8 . . |
| 3 6 . | . . . | 5 . . |
+-------+-------+-------+
| . . 3 | 8 . 5 | . 6 . |
| . 1 8 | . 7 . | 2 3 . |
| . . . | 3 1 2 | . . . |
+-------+-------+-------+

We can note the following correspondence between the values: 1-1, 2-2, 4-5, 6-7, 8-9. This puzzle has the following isomorphism: the left part of the puzzle (blocks 1, 4, and 7) and the right part (blocks 9, 6, and 3) are reflected relative to two big diagonals, and the central part (blocks 2, 5, and 8) is relected relative to the horizontal line. We can see that no links are destroyed by this permutation. The only cells that stay untouched are r5c4 and r5c6. Now let's assume that the solution is unique, consequently, in the untouched cells we can place only the digits that come into themselves by the isomorphism, so r5c4=2 and r5c6=1, and we get a finish with only singles.

P.S. Thanks for the puzzle. Sorry for somewhat entangled reasoning.

[P.O.]

Code: Select all

after singles and intersections:

45689  4589   56     1      2      3      467    458    45678           
458    3      2      57     6      78     9      1      458             
1568   7      156    9      58     4      3      258    2568           
1258   258    4      256    59     169    16     7      3               
1257   25     9      24567  3      167    8      24     1246           
3      6      17     247    48     178    5      249    1249           
2479   249    3      8      49     5      147    6      1479           
459    1      8      46     7      69     2      3      459             
45679  459    567    3      1      2      47     4589   45789           

6r458c4 => r5c4 <> 7
 r4c4=6 - r4n2{c4 c12} - 157r5c126
 r5c4=6 -
 r8c4=6 - b8n4{r8c4 r7c5} - r6c5{n4 n8} - r3c5{n8 n5} - r2c4{n5 n7}

5r2c149 => r7c12 <> 9
 r2c1=5 - b4n5{r45c1 r45c2} - 49b7p48
 r2c4=5 - r3c5{n5 n8} - r6c5{n8 n4} - r7c5{n4 n9}
 r2c9=5 - b9n5{r89c9 r9c8} - r9n8{c8 c9} - r9n9{c9 c12}
 
6r5c469 => r7c9 <> 9
 r5c4=6 - r8c4{n6 n4} - r7c5{n4 n9}
 r5c6=6 - r5n7{c6 c1} - 24r7c12 - r7c5{n4 n9}
 r5c9=6 - r4c7{n6 n1} - r7n1{c7 c9}
 
ste.


[eleven]

We can note the following correspondence between the values: 1-1, 2-2, 4-5, 6-7, 8-9. This puzzle has the following isomorphism: the left part of the puzzle (blocks 1, 4, and 7) and the right part (blocks 9, 6, and 3) are reflected relative to two big diagonals, and the central part (blocks 2, 5, and 8) is relected relative to the horizontal line.

This is no grid isomorphism, which reflects a grid to itself. You cannot combine partly symmetries, you have to make it with one for the whole grid.

[AnotherLife]

Hi Eleven,
It's sad that you don't get my reasoning. It seems that some time ago we understood each other very well. Let me try to explain my solution in another way.

Code: Select all

(*)
+-------+-------+-------+
| . . . | 1 2 3 | . . . |
| . 3 2 | . 6 . | 9 1 . |
| . 7 . | 9 . 4 | 3 . . |
+-------+-------+-------+
| . . 4 | . . . | . 7 3 |
| . . 9 | . 3 . | 8 . . |
| 3 6 . | . . . | 5 . . |
+-------+-------+-------+
| . . 3 | 8 . 5 | . 6 . |
| . 1 8 | . 7 . | 2 3 . |
| . . . | 3 1 2 | . . . |
+-------+-------+-------+

Let A(i,j) be any valid sudoku grid with the above conditions(*). Let us define permutation P and matrix B(i,j)=M(A)(i,j) as follows.
P(1)=1, P(2)=2, P(3)=3, P(4)=5, P(5)=4, P(6)=7, P(7)=6, P(8)=9, P(9)=8
B(i,j)=P(A(10-i,10-j)) for each i=1,..,9 and j=1,2,3,7,8,9
B(i,j)=P(A(10-i,j)) for each i=2,…,8 and j=4,5,6
B(1,4)=1, B(1,5)=2, B(1,6)=3, B(9,4)=3, B(9,5)=1, B(9,6)=2

My statement is that B(i,j) is also a valid sudoku grid with the above conditions(*).
To prove the statement, let us note how mapping M transforms all the houses.

Blocks: b1<->b9, b4<->b6, b7<->b3, b2<->b8, b5<->b5.
Rows: r1<->r9, r2<->r8, r3<->r7, r4<->r6, r5<->r5.
Columns: c1<->c9, c2<->c8, c3<->c7, c4<->c4, c5<->c5, c6<->c6.

As two different digits in a house become two different digits in the transformed house, we get a valid sudoku grid M(A) =B(i,j).
We can also note that B(i,j)=A(i,j) for each predefined value (*).
Now let us assume that the puzzle has only one solution, that is, B=A. Let us take a look at cells r5c4 and r5c6.
A(5,4)=B(5,4)=P(A(5,4)), so A(5,4) can be either 1 or 2, and the same is true of A(5,6). Since A(1,4)=1 and A(9,6)=2, the only possibility is A(5,4)=2 and A(5,6)=1.
From this point, the puzzle is solvable by singles.

[eleven]

Sorry, of course you are right, the mapping you describe is an isomorphism (all cells are mapped to all cells, and with the digit transformation you get the same grid).
I was a bit confused by your reasoning, and i wonder, why i can't find this in my symmetry groups (did i miss another sticks symmetry ?). Have to look into that again later.

[Pat]

Sorry, of course you are right, the mapping you describe is an isomorphism (all cells are mapped to all cells, and with the digit transformation you get the same grid).
I was a bit confused by your reasoning, and i wonder, why i can't find this in my symmetry groups (did i miss another sticks symmetry ?). Have to look into that again later.

i remain confused

is it an isomorphism?

[eleven]

No, it is not. The reason is, that Bogdan's mapping is not possible, i.e. it cannot be reached with equivalence operations.
There might be another trick (like using unavoidable sets), which allows to get to his conclusion, don't know now.

btw, thanks for restoring the hardest thread.

[Pat]

half-turn

swap c46

re-map digits

now, explain r19

[eleven]

Ah yes, so my assumption was right.
Apart from the 123 in r1c456/r9c46, which form an unavoidable (and therefore can be renumbered accordingly, if there are no givens in r5c123), there is always an isomorphism, as Pat described, for puzzles with that properties.
Nice find by Marek. Such puzzles seem to be easy to find.

Code: Select all

 +-------+-------+-------+
 | . . 5 | 1 2 . | 6 . . |
 | . . . | . . 6 | . . 1 |
 | . . . | . 7 . | . . 4 |
 +-------+-------+-------+
 | . 1 2 | . 4 . | 5 . 8 |
 | . . . | . . . | . . . |
 | 9 . 4 | . 5 . | 2 1 . |
 +-------+-------+-------+
 | 5 . . | . 6 . | . . . |
 | 1 . . | . . 7 | . . . |
 | . . 7 | . 1 2 | 4 . . |
 +-------+-------+-------+

[AnotherLife]

Eleven, your puzzle is even harder than Marek's. It is rated SER 9.3 and it needs dynamic forcing chains if you are to you solve it by standard methods. But if you set r5c5=3, r5c4=2, and r5c6=1, its rating falls to SER 6.6.

[eleven]

Yes, because of the symmetry r46c6 cannot b3 3, which makes it easy.

Code: Select all

+-------------------+-------------------+-------------------+
| 4     3789  5     | 1     2     389   | 6     3789  379   |
| 2     3789  389   | 4     389   6     | 3789  5     1     |
| 368   3689  1     | 389   7     5     | 389   2     4     |
+-------------------+-------------------+-------------------+
| 367   1     2     | 67    4    *39    | 5     379   8     |
| 3678  5     368   | 2     389   1     | 379   4     3679  |
| 9     368   4     | 67    5    *38    | 2     1     367   |
+-------------------+-------------------+-------------------+
| 5     2     389   | 389   6     4     | 1     3789  379   |
| 1     4     3689  | 5     389   7     | 389   3689  2     |
| 368   3689  7     | 389   1     2     | 4     3689  5     |
+-------------------+-------------------+-------------------+

[Edit:] But to prove the symmetry, it needs to be shown, that r1c6 and r9c4 have to be the same number - which is not trivial

As a quick substitute a ER 9,2:

Code: Select all

 +-------+-------+-------+
 | . 4 . | 1 2 3 | 6 . . |
 | 8 . . | . . 4 | . 2 . |
 | . . . | . 6 . | . . . |
 +-------+-------+-------+
 | 3 . . | . . 9 | 1 . . |
 | . 5 . | . . . | . 4 . |
 | . . 1 | . . 8 | . . 3 |
 +-------+-------+-------+
 | . . . | . 7 . | . . . |
 | . 2 . | . . 5 | . . 9 |
 | . . 7 | 3 1 2 | . 5 . |
 +-------+-------+-------+

[denis_berthier]

Code: Select all

 +-------+-------+-------+
 | . . 5 | 1 2 . | 6 . . |
 | . . . | . . 6 | . . 1 |
 | . . . | . 7 . | . . 4 |
 +-------+-------+-------+
 | . 1 2 | . 4 . | 5 . 8 |
 | . . . | . . . | . . . |
 | 9 . 4 | . 5 . | 2 1 . |
 +-------+-------+-------+
 | 5 . . | . 6 . | . . . |
 | 1 . . | . . 7 | . . . |
 | . . 7 | . 1 2 | 4 . . |
 +-------+-------+-------+

Eleven, your puzzle is even harder than Marek's. It is rated SER 9.3 and it needs dynamic forcing chains if you are to you solve it by standard methods.

Eleven's puzzle is in W11, which is admittedly very hard (and harder than marek's one, in W7), but it needs neither dynamic nor forcing chains, let alone a combination of them.

[AnotherLife]

In terms of YZF_Sudoku and HoDoKu, whips are special cases of dynamic forcing chains (forcing nets), so from a manual solver's point of view Eleven's puzzle needs dynamic forcing chains.

[denis_berthier]

In terms of YZF_Sudoku and HoDoKu, whips are special cases of dynamic forcing chains (forcing nets), so from a manual solver's point of view Eleven's puzzle needs dynamic forcing chains.

No comment needed.