The New Sudoku Players' Forum

[bennys]

Code: Select all

 Menneske.no Very Hard #1751664
+-------+-------+-------+
| 2 3 . | 5 4 . | 9 . 8 |
| 9 4 . | . . . | . . 3 |
| 1 8 . | 9 . . | . 4 . |
+-------+-------+-------+
| . 5 4 | . 1 . | 8 9 . |
| . 9 3 | 8 . 4 | 5 . 1 |
| 8 2 1 | 7 9 5 | 3 6 4 |
+-------+-------+-------+
| 3 1 . | 4 . . | . 8 . |
| 4 6 . | . . . | . . 9 |
| 5 7 . | 2 . . | 4 . 6 |
+-------+-------+-------+

+----------------+----------------+----------------+
| 2    3    67   | 5    4    167  | 9    17   8    |
| 9    4    567  | 16   2678 2678 | 126  257  3    |
| 1    8    567  | 9    2367 2367 | 26   4    257  |
+----------------+----------------+----------------+
| 67   5    4    | 36   1    236  | 8    9    27   |
| 67   9    3    | 8    26   4    | 5    27   1    |
| 8    2    1    | 7    9    5    | 3    6    4    |
+----------------+----------------+----------------+
| 3    1    29   | 4    567  679  |*27   8   *25   |
| 4    6    28   |*13  *3578*378  |*127  235  9    |
| 5    7    89   | 2   *38   1389 | 4    13   6    |
+----------------+----------------+----------------+

A = {R8C4,R8C5,R8C6,R9C5}
B = {R7C7,R7C9,R8C7}

1)A and B are almost locked
2)Both of them have 1 as candidate only in R8 (which mean that it cant materialize in both sets)
so R7C5 cant be 5 because then both are locked and that cant be.

[stuartn]

1)A and B are almost locked

How do you mean Bennys?

stuartn

[Jeff]

Excellent observation, bennys.

[bennys]

number of candidates – 1 = number of cells

[tso]

I would call a complex forcing chain. In a simple xy-type forcing chain, each link is a naked single implied by the previous link. In this case, there are links that are naked pairs and naked triples. The chains are still linear (as opposed to a forcing *net*) and are easy enough for humans to follow.

r8c4=1 => r78c7=[naked pair 27] => r7c9=5 => r7c5<>5
r8c4=3 => r8c56+r9c5=[degenerate naked triple 578] => r7c5<>5
Therefore, r7c5<>5.
[edited to correct typo]

[bennys]

If A,B are almost locked and they have a common value (lets say 1) that can appear only on one row(column) for both (the same one for both)
Then for any other value (lets say 5) that is common to both A and B this value cant appear on a cell outside of A and B that can ‘see’ all the candidates for that value in A and B.

[rubylips]

I've just upgraded my solver to handle arguments of this type, which I call 'Conditional Disjoint Subsets', as one of two possible disjoint subsets is created depending upon the value in some cell. This puzzle is particularly interesting as it involves a classic Two-Sector Disjoint Subsets pattern, two conditional patterns and a disjoint subset that's dependent upon a forcing chain. (bennys - you didn't mention that the puzzle is still difficult to solve after your elimination ).

Here's the log for the tricky first six moves:

Code: Select all

1. Consider the chain r1c8-1-r1c6~1~r9c6-1-r9c8.
The cell r9c8 must contain the value 1 if the cell r1c8 doesn't.
Therefore, these two cells are the only candidates for the value 1 in Column 8.
- The moves r2c8:=1 and r8c8:=1 have been eliminated.
Consider the chain r1c6-1-r1c8-1-r9c8-1-r9c6.
The cell r9c6 must contain the value 1 if the cell r1c6 doesn't.
Therefore, these two cells are the only candidates for the value 1 in Column 6.
- The moves r2c6:=1 and r8c6:=1 have been eliminated.
The cells r8c7 and r8c8 contain 1 value from {1,3} and 1 value from {2,5,7}.
The values 1 and 3 occupy 2 of the cells r8c7, r8c8 and r8c4 in some order.
- The moves r8c5:=3 and r8c6:=3 have been eliminated.
Consider the cell r8c4.
When it contains the value 1, the value 7 in Row 8 must appear in the cells r8c3, r8c6 or r8c7.
When it contains the value 3, the value 7 in Box 8 must appear in the cells r8c6 or r9c5.
Whichever value it contains, the cell r8c5 cannot contain the value 7.
When it contains the value 1, the value 8 in Row 8 must appear in the cells r8c3, r8c6 or r8c7.
When it contains the value 3, the value 8 in Box 8 must appear in the cells r8c6 or r9c5.
Whichever value it contains, the cell r8c5 cannot contain the value 8.
- The moves r8c5:=7 and r8c5:=8 have been eliminated.
The value 5 is the only candidate for the cell r8c5.
2. The cell r2c8 is the only candidate for the value 5 in Column 8.
3. The cell r3c3 is the only candidate for the value 5 in Row 3.
4. The cell r7c9 is the only candidate for the value 5 in Row 7.
5. Consider the chain r3c5~2~r3c9-2-r4c9-2-r4c6-2-r5c5.
When the cell r3c5 contains the value 2, so does the cell r5c5 - a contradiction.
Therefore, the cell r3c5 cannot contain the value 2.
- The move r3c5:=2 has been eliminated.
Consider the chain r8c8-2-r5c8-2-r4c9-2-r4c6-3-r4c4-3-r8c4-3-r8c8.
The cell r8c8 must contain the value 3 if it doesn't contain the value 2.
Therefore, these two values are the only candidates for the cell r8c8.
- The move r8c8:=7 has been eliminated.
The value 7 in Column 7 must lie in Box 9.
- The moves r2c7:=7 and r3c7:=7 have been eliminated.
Consider the chain r4c4-3-r8c4-3-r8c8-2-r5c8-2-r5c5.
The cells r4c4 and r5c5 contain one value from the set {3,2} and the value 6.
The value 6 occupies 1 of the cells r4c4 and r5c5.
- The move r4c6:=6 has been eliminated.
Consider the chain r1c6-1-r9c6-1-r8c4-1-r8c7-7-r8c6.
When the cell r1c6 contains the value 7, so does the cell r8c6 - a contradiction.
Therefore, the cell r1c6 cannot contain the value 7.
- The move r1c6:=7 has been eliminated.
The values 2, 3, 7 and 8 occupy the cells r2c5, r2c6, r3c5 and r3c6 in some order.
- The moves r2c5:=6, r2c6:=6, r3c5:=6 and r3c6:=6 have been eliminated.
The cell r3c7 is the only candidate for the value 6 in Row 3.
6. The values 1, 6 and 9 occupy the cells r1c6, r7c6 and r9c6 in some order.
- The moves r7c6:=7, r9c6:=3 and r9c6:=8 have been eliminated.
Consider the chain r3c6-2-r3c9-2-r4c9-2-r4c6.
The cell r4c6 must contain the value 2 if the cell r3c6 doesn't.
Therefore, these two cells are the only candidates for the value 2 in Column 6.
- The move r2c6:=2 has been eliminated.
The values 1, 2, 3, 6 and 9 occupy the cells r1c6, r3c6, r4c6, r7c6 and r9c6 in some order.
- The move r3c6:=7 has been eliminated.
Consider the cell r3c5.
When it contains the value 7, the value 2 in Column 5 must appear in the cells r5c5 or r7c5.
When it contains the value 3, the value 2 in Box 2 must appear in the cells r1c6, r2c4 or r3c6.
Whichever value it contains, the cell r2c5 cannot contain the value 2.
- The move r2c5:=2 has been eliminated.
The cell r2c7 is the only candidate for the value 2 in Row 2.


[bennys]

Here is some XY-Wing version with almost locked set

Code: Select all

+----------------------+----------------------+----------------------+
| .      x      .      | .      .      .      |  52     .      .     |
| .      .      .      | .      .      .      |  .      .      .     |
| .      .      .      | .      .      .      |  .      .      .     |
+----------------------+----------------------+----------------------+
| .      .      .      | .      .      .      |  .      .      .     |
| .      .      .      | .      .      .      |  .      .      .     |
| .      .      .      | .      .      .      |  .      .      .     |
+----------------------+----------------------+----------------------+
| .      51     .      | .      .      .      |  1234   134    .     |
| .      .      .      | .      .      .      |  234    .      .     |
| .      .      .      | .      .      .      |  .      .      .     |
+----------------------+----------------------+----------------------+

here x cant be 5

[ronk]

r8c4=1 => r78c7=[naked pair 27] => r7c9=5 => r5c5<>5
r8c4=3 => r8c56+r9c5=[degenerate naked triple 578] => r5c5<>5
Therefore, r5c5<>5.

I assume r5c5 is a typo, and should be r7c5.

[Jeff]

I would call a complex forcing chain. In a simple xy-type forcing chain, each link is a naked single implied by the previous link. In this case, there are links that are naked pairs and naked triples. The chains are still linear (as opposed to a forcing *net*) and are easy enough for humans to follow.

In a chain which involves naked pair of naked triple, the implication simply bifurcate or trifurcate into branches. So, it is perfectly OK to called it a chain, ie. a chain with poly implications.