The New Sudoku Players' Forum

[danlm]

I did not find any thing... I added only 7 numbers exept the clues...
Please help me!!

Code: Select all

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


Code: Select all

+----------------+------------------+---------------+
| 9    5    4    | 7     2     13   | 8  13    6    |
| 678  3    18   | 1689  168   5    | 4  179   2    |
| 2678 1678 128  | 13689 1368  4    | 5  1379  1379 |
+----------------+------------------+---------------+
| 43   2    7    | 1345  134   9    | 6  1345  8    |
| 438  18   138  | 23456 3467  367  | 9  23457 3457 |
| 5    9    6    | 12348 13478 137  | 27 12347 1347 |
+----------------+------------------+---------------+
| 2367 4    239  | 136   5     1367 | 27 8     79   |
| 1    67   25   | 46    9     8    | 3  24567 457  |
| 3678 678  3589 | 346   3467  2    | 1  45679 4579 |
+----------------+------------------+---------------+

[daj95376]

Withdrawn.

[danlm]

Thank you!

It's hard but I think I understand...

[udosuk]

Okay this is really complicated, but I'll try to demonstrate the trickest move from this position:

Code: Select all

 *--------------------------------------------------------------------*
 | 9      5      4      | 7      2      13     | 8      13     6      |
 | 678    3      18     | 1689   168    5      | 4      179    2      |
 | 2678   1678   128    | 13689  1368   4      | 5      1379   1379   |
 |----------------------+----------------------+----------------------|
 | 34     2      7      | 1345   134    9      | 6      1345   8      |
 | 348    18     138    | 23456  3467   367    | 9      23457  3457   |
 | 5      9      6      | 12348  13478  137    | 27     12347  1347   |
 |----------------------+----------------------+----------------------|
 |-2367   4     *239    |#136    5     #1367   |*27     8     *79     |
 | 1      67    @25     |@46     9      8      | 3      24567 @457    |
 | 3678   678    3589   | 346    3467   2      | 1      45679  4579   |
 *--------------------------------------------------------------------*

(Note: The following blue lines are for experts. If you're a beginner please do scroll down to read the explainations of the logic involved.)

ALS A (*): r7c379={2379} (degree of freedom=1)
ALS B (#): r7c46={1367} (degree of freedom=2)
ALS C (@): r8c349={24567} (degree of freedom=2)
restricted commons between A & B: 3,7
restricted common between A & C: 7
restricted common between B & C: 6
common: 2

Therefore r7c1<>2.

(This is a case of generalised ALS rules, and works because total #restricted commons = total #degrees of freedom - 1.)

A way to see this logic is to visualise what will happen if r7c1=2.
Then r7c379 (A) will become {379} with r7c3=3 and r7c7=7.
Thus r7c46 (B) will become {16} (naked pair).
Also r8c349 (C) will become {456} with r8c4=6.
Then we'll be forced to have two 6s in b8, a contradiction.
Hence r7c1 cannot be 2.

Another way to see it is through branching analysis.
Only these 4 scenarios can happen: either r8c4=6, or r8c9=7, or both, or neither.

If r8c4=6, then r7c34679 will form a naked quint of {12379}, and r7c1<>2.
If r8c9=7, then r7c7=2, and r7c1<>2.
If both of these happen, then r7c1<>2 following either line of reasoning.
If neither of these happens, then r8c4=4, r8c9=5, r8c3=2 and r7c1<>2.

Therefore no matter what r7c1 cannot be 2.

After this it still takes some works (mainly turbot fishes/colouring) to solve the rest of the puzzle.
But at least the Simple Sudoku program is able to figure out those moves.

I know it's not pretty but it's the best I can do for now.
I sincerely hope somebody else can spot a more elegant move to crack this bastard.

[re'born]

udosuk,

I don't have a simpler solution (yet), but I think subset counting is a much faster way to explain your move. You have 8 cells with candidates {1,2,3,4,5,6,7,9} whose max multiplicities are 1,2,1,1,1,1,2,1, respectively, where we note that if r7c7=7, then the max multiplicity of 7 will become 1. Since r7c1 sees all the cells with a 2 in our set and since r7c1=2 => r7c7=7, we see that r7c1=2 would decrease the max multiplicity by 3. Hence we would have only 7 candidates for 8 cells, a contradiction.

Maybe this is easier...then again, maybe not.

[denis_berthier]

I did not find any thing... I added only 7 numbers exept the clues...
Please help me!!

Code: Select all

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


I don't have a full solution, but an hxyt-chain of length 7 can eliminate candidate 3 in R5C4.
(solve-sdk-grid (str-cat ?*GridsDir* "danim.sdk"))
***** SudoRules version 12 *****
9..7....6
.3...54.2
.....45..
.27.....8
......9..
5.6......
.4..5..8.
1...983..
.....21..
number 8 : naked-single ==> R1C7 = 8
number 6 : naked-single ==> R4C7 = 6
number 4 : hidden-single-in-row R1 ==> R1C3 = 4
number 2 : hidden-single-in-row R1 ==> R1C5 = 2
number 5 : hidden-single-in-block B1 ==> R1C2 = 5
number 9 : hidden-single-in-block B4 ==> R6C2 = 9
number 9 : hidden-single-in-column C6 ==> R4C6 = 9
number 8 : row R6 interaction with block B5
==> 8 eliminated from the candidates for R5C5
number 8 : row R6 interaction with block B5
==> 8 eliminated from the candidates for R5C4
number 1 : block B4 interaction with row R5
==> 1 eliminated from the candidates for R5C9
number 1 : block B4 interaction with row R5
==> 1 eliminated from the candidates for R5C8
number 1 : block B4 interaction with row R5
==> 1 eliminated from the candidates for R5C6
number 1 : block B4 interaction with row R5
==> 1 eliminated from the candidates for R5C5
number 1 : block B4 interaction with row R5
==> 1 eliminated from the candidates for R5C4
number 7 : c4-chain col-row-bl on cells R6C7-R6C6-R7C6-R9C5
==> 7 eliminated from the candidates for R6C5
number 1 : c4-chain row-bl-col on cells R1C6-R1C8-R3C9-R6C9
==> 1 eliminated from the candidates for R6C6
number 1 : c6-chain on cells R7C4-R7C6-R1C6-R1C8-R3C9-R6C9
==> 1 eliminated from the candidates for R6C4
row R5 : hxyt7-cn-chain on cn-cells C4N2*, C7N2, C1N2, C3N2, C3N5, C3N9 and C3N3* with rows R5, R6, R7, R3, R8, R9 and R7
==> R5 eliminated from the cn-candidates for C4N3
i.e. 3 eliminated from the candidates for R5C4

[ravel]

I know, thats not what udosuk wanted, but maybe danlm wants to see a solution with extended xy-wings. The cells marked with # have a strong (bilocation) link.

Code: Select all

 *--------------------------------------------------------------------*
 | 9      5      4      | 7      2      13     | 8      13     6      |
 | 678    3      18     | 1689   168    5      | 4      179    2      |
 | 2678   1678   128    | 13689  1368   4      | 5      1379   1379   |
 |----------------------+----------------------+----------------------|
 | 34     2      7      | 1345   134    9      | 6      1345   8      |
 | 348    18     138    | 23456  3467   367    | 9      23457  3457   |
 | 5      9      6      | 2348   13478  37     | 27     12347  1347   |
 |----------------------+----------------------+----------------------|
 | 2367   4     #239    | 136    5      1367   |*27     8     *79     |
 | 1      67    #25     | 46     9      8      | 3     #24567  457    |
 | 3678   678   #3589   | 346    3467   2      | 1      45679  4579   |
 *--------------------------------------------------------------------*


Either r7c9=7
or (r7c9=9 => r9c3=9 => r8c3=5 => r8c8=2 =>) r7c7=7
Eliminate 7 from r7c16 and r89c89

Brings you here:

Code: Select all

 *--------------------------------------------------------------------*
 | 9      5      4      | 7      2      13     | 8      13     6      |
 | 678    3      18     | 1689   168    5      | 4      179    2      |
 | 2678   168    128    | 13689  1368   4      | 5      1379   1379   |
 |----------------------+----------------------+----------------------|
 | 34     2      7      | 135    134    9      | 6      1345   8      |
 | 348    18     138    | 2356   346   #367    | 9      23457  3457   |
 | 5      9      6      | 238    1348  #37     |*27     12347  1347   |
 |----------------------+----------------------+----------------------|
 | 236    4      239    | 136    5     #136    |#27     8      79     |
 | 1      7      25     |#46     9      8      | 3     #2456   45     |
 | 368    68     3589   | 346    7      2      | 1      4569   459    |
 *--------------------------------------------------------------------*


Either r8c4=6
or (r8c8=6 => r7c7=2 => r6c7=7 => r6c6=3 => r5c6=7 =>) r7c6 =6
Eliminate 6 from r79c4

Code: Select all

 *--------------------------------------------------------------------*
 | 9      5      4      | 7      2      13     | 8      13     6      |
 | 678    3      18     | 1689   168    5      | 4      179    2      |
 |#2678   168    128    | 13689  1368   4      | 5      1379   1379   |
 |----------------------+----------------------+----------------------|
 | 34     2      7      | 135    134    9      | 6      1345   8      |
 | 348    18     138    | 2356   346    367    | 9      23457  3457   |
 | 5      9      6      | 238    1348   37     | 27     12347  1347   |
 |----------------------+----------------------+----------------------|
 |#236    4      239    | 13     5     #136    | 27     8      79     |
 | 1      7     *25     |*46     9      8      | 3      2456  *45     |
 | 368    68     3589   | 34     7      2      | 1      4569   459    |
 *--------------------------------------------------------------------*


Either r3c1=2
or (r7c1=2 => r7c6=6 => r8c4=4 => r8c9=5 =>) r8c3=2
Eliminate 2 from r3c3

[udosuk]

Thanks for the moves, re'born, Denis & ravel.

I can surely see the logic of re'born's subset counting approach, but I would often prefer more formulated proofs then word-proofs like that. I guess it's a matter of personal taste.

I know, thats not what udosuk wanted, but maybe danlm wants to see a solution with extended xy-wings. The cells marked with # have a strong (bilocation) link...

Actually I think they're good logical moves, but I don't understand what they have to do with "xy-wings"? In (basic) xy-wings you have a pivot/pilot and 2 wings. Can you specify those parts in your "extended" versions?

Code: Select all

 *--------------------------------------------------------------------*
 | 9      5      4      | 7      2      13     | 8      13     6      |
 | 678    3      18     | 1689   168    5      | 4      179    2      |
 |#2678   168    128    | 13689  1368   4      | 5      1379   1379   |
 |----------------------+----------------------+----------------------|
 | 34     2      7      | 135    134    9      | 6      1345   8      |
 | 348    18     138    | 2356   346    367    | 9      23457  3457   |
 | 5      9      6      | 238    1348   37     | 27     12347  1347   |
 |----------------------+----------------------+----------------------|
 |#236    4      239    | 13     5     #136    | 27     8      79     |
 | 1      7     *25     |*46     9      8      | 3      2456  *45     |
 | 368    68     3589   | 34     7      2      | 1      4569   459    |
 *--------------------------------------------------------------------*


Either r3c1=2
or (r7c1=2 => r7c6=6 => r8c4=4 => r8c9=5 =>) r8c3=2
Eliminate 2 from r3c3

Here in the 2nd branch you just proved r7c1=2 => r8c3=2, hence two 2s in b7. So you could just have eliminated 2 from r7c1 as well.

And playing along your path above, from this original position:

Code: Select all

 *--------------------------------------------------------------------*
 | 9      5      4      | 7      2      13     | 8      13     6      |
 | 678    3      18     | 1689   168    5      | 4      179    2      |
 | 2678   1678   128    | 13689  1368   4      | 5      1379   1379   |
 |----------------------+----------------------+----------------------|
 | 34     2      7      | 1345   134    9      | 6      1345   8      |
 | 348    18     138    | 23456  3467   367    | 9      23457  3457   |
 | 5      9      6      | 12348  13478  137    | 27     12347  1347   |
 |----------------------+----------------------+----------------------|
 |-2367   4      239    |*136    5     *1367   |*27     8     *79     |
 | 1      67    *25     |*46     9      8      | 3      24567 *457    |
 | 3678   678    3589   | 346    3467   2      | 1      45679  4579   |
 *--------------------------------------------------------------------*

If r7c1=2 => [r7c46]=6 & [r7c79]=7 => r8c49=[45] => r8c3=2
=> contradiction (two 2s in b7).

Isn't this way shorter than anything we've posted above?

[ravel]

Phew, i was really tired yesterday. I wanted to say "extended xy-chain", which is my name for xy-chains, where also bilocation links are used. And fixed on closing the last one i totally missed the contradiction r7c1=2 => r8c3=2.
Your contradiction chain is hard to spot using a grouped link combined with multiple inference. The notation is a bit confusing and you dont need r7c9:
r7c1=2 => (r7c4|6=6 => r8c4=4) => r7c7=7 => r8c9=5 => r8c3=2

[udosuk]

Thanks for the clarification. So let's drop r7c9 and rephrase my contradiction chain:

r7c1=2 => r7c7=7 & r8c3=5 => r8c9=4 => r8c4=6 => contradiction (no 6 on r7)