The New Sudoku Players' Forum

[MikeF]

Can anyone please suggest a way to solve this?

*-----------*
|5..|7..|.8.|
|...|..5|..3|
|..2|1.6|.5.|
|---+---+---|
|93.|...|.1.|
|82.|95.|..7|
|.4.|...|.9.|
|---+---+---|
|...|2.9|5..|
|75.|...|...|
|21.|5.4|..6|
*-----------*


{5} {69} {1346} {7} {2349} {23} {12469} {8} {1249}
{146} {6789} {1468} {48} {2489} {5} {124679} {2467} {3}
{34} {789} {2} {1} {3489} {6} {479} {5} {49}
{9} {3} {57} {468} {2468} {278} {268} {1} {258}
{8} {2} {16} {9} {5} {13} {346} {346} {7}
{16} {4} {57} {368} {12368} {12378} {2368} {9} {258}
{346} {68} {3468} {2} {13678} {9} {5} {347} {148}
{7} {5} {34689} {368} {1368} {138} {123489} {234} {12489}
{2} {1} {389} {5} {378} {4} {3789} {37} {6}

[bennys]

Code: Select all

+----------------------+----------------------+----------------------+
| 5      69   <1346>   | 7      2349   23     | 12469  8      1249   |
|<146>   6789 <1468>   |<48>    2489   5      | 124679 2467   3      |
|<34>    789    2      | 1      3489   6      | 479    5      49     |
+----------------------+----------------------+----------------------+
| 9      3      57     | 468    2468   278    | 268    1      258    |
| 8      2     *16     | 9      5      13     | 346    346    7      |
| 16     4      57     | 368    12368  12378  | 2368   9      258    |
+----------------------+----------------------+----------------------+
| 346    68     3468   | 2      13678  9      | 5      347    148    |
| 7      5      34689  | 368    1368   138    | 123489 234    12489  |
| 2      1      389    | 5      378    4      | 3789   37     6      |
+----------------------+----------------------+----------------------+


 Because of the 16 in R5C3 we have   R5C3 or R5C3 must be 3,4 or 8
 but  if R2C1 = 4 then    R3C1 = 3 and  R2C4 = 8
 So R2C1 <> 4


After that you can use colours and then its easy.

[angusj]

After that you can use colours and then its easy.

Hmm ... my take is -
After that there are 4 xy-wings, a naked quad and another forcing chain. So all in all, a very hard puzzle .

[bennys]

For some reason I felt that it will be easy but you are right there are some steps that need work.

[QBasicMac]

For some reason I felt that it will be easy

Errk! I use T&E on such puzzles and even T&E was tough.

Mac

[MikeF]

Code: Select all

+----------------------+----------------------+----------------------+
| 5      69   <1346>   | 7      2349   23     | 12469  8      1249   |
|<146>   6789 <1468>   |<48>    2489   5      | 124679 2467   3      |
|<34>    789    2      | 1      3489   6      | 479    5      49     |
+----------------------+----------------------+----------------------+
| 9      3      57     | 468    2468   278    | 268    1      258    |
| 8      2     *16     | 9      5      13     | 346    346    7      |
| 16     4      57     | 368    12368  12378  | 2368   9      258    |
+----------------------+----------------------+----------------------+
| 346    68     3468   | 2      13678  9      | 5      347    148    |
| 7      5      34689  | 368    1368   138    | 123489 234    12489  |
| 2      1      389    | 5      378    4      | 3789   37     6      |
+----------------------+----------------------+----------------------+


 Because of the 16 in R5C3 we have   R5C3 or R5C3 must be 3,4 or 8
 but  if R2C1 = 4 then    R3C1 = 3 and  R2C4 = 8
 So R2C1 <> 4


After that you can use colours and then its easy.

Many thanks for the replies.

bennys, that's a neat strategy, which I haven't seen before. Does it have a name? Do you find that it is often helpful? And I'm wondering what exactly is the pattern to look for in applying this strategy?

Thanks again.

Mike

[angusj]

bennys, that's a neat strategy

It might be neat, but it doesn't make sense (to me at least).

A more logical explanation as to why r2c1 can't be 4 is:
if r2c1 = 4 then r2c4=8, r3c5<> 8, r3c2=8, r7c2=6, r7c1<>6, r6c1=6, and finally r2c1=1 which is a contradiction. Therefore r2c1 <> 4.

[MikeF]

bennys, that's a neat strategy

It might be neat, but it doesn't make sense (to me at least).

A more logical explanation as to why r2c1 can't be 4 is:
if r2c1 = 4 then r2c4=8, r3c5<> 8, r3c2=8, r7c2=6, r7c1<>6, r6c1=6, and finally r2c1=1 which is a contradiction. Therefore r2c1 <> 4.

OK, I can follow that angusj.

I'm wondering, though, why is that "more logical"?

[angusj]

I'm wondering, though, why is that "more logical"?

Code: Select all

Because of the 16 in R5C3 we have   R5C3 or R5C3 must be 3,4 or 8

How can R5C3 be 1 or 6 and R5C3 be 3,4 or 8 ???

[bennys]

Actually i had typo but it looks to me that Mike understood that.
Here is how it should be.

Code: Select all

+----------------------+----------------------+----------------------+
| 5      69   <1346>   | 7      2349   23     | 12469  8      1249   |
|<146>   6789 <1468>   |<48>    2489   5      | 124679 2467   3      |
|<34>    789    2      | 1      3489   6      | 479    5      49     |
+----------------------+----------------------+----------------------+
| 9      3      57     | 468    2468   278    | 268    1      258    |
| 8      2     *16     | 9      5      13     | 346    346    7      |
| 16     4      57     | 368    12368  12378  | 2368   9      258    |
+----------------------+----------------------+----------------------+
| 346    68     3468   | 2      13678  9      | 5      347    148    |
| 7      5      34689  | 368    1368   138    | 123489 234    12489  |
| 2      1      389    | 5      378    4      | 3789   37     6      |
+----------------------+----------------------+----------------------+


 Because of the 16 in R5C3 we have   R1C3 or R2C3 must be 3,4 or 8
 but  if R2C1 = 4 then    R3C1 = 3 and  R2C4 = 8
 So R2C1 <> 4


Regarding what is more logical?
I don't know but i prefer my explanation because i feel it gives better understanding to what is going on but I am not objective.

[angusj]

Because of the 16 in R5C3 we have R1C3 or R2C3 must be 3,4 or 8

That would only be true if R5C3 was both 1 and 6 which of course it can't.

[bennys]

It looks to me that you have problem with understanding what OR is.
We are talking on one of them.
Here the Example again in simplified way

Code: Select all

+----------------------+
| <1685>  <167>  <16>  |
|                      |
|                      |
+----------------------+


Because of the 16 in R1C3
R1C1 OR R1C2 must be 7 or 8 or 5.

[angusj]

OK. I'm really not trying to be difficult, I just still don't get it.

Edit: OK, I finally do get it. You logic is impeccable. Sorry for being dense .

[bennys]

Being clear is not one of my strong suits.

[Jeff]

I am amazed with the fact that benny's technique described here, DanO's technique described under the thread of "More forcing chains"and Sue De Coq's technique described under the thread of "Two-Sector Disjoint Subsets" did not receive enough applause and attentions that they truly deserve.

I like these techniques very much because:

• The deductions can be done simply by inspection without filtering, graphing, or tabling.
• The same deductions, if expressed in terms of a forcing chain or forcing net, would involve 3 implications or more; which is not easy to identify manually by any other currently available non T&E human executable techniques.
• They are non T&E human executable technique.

With DanO's and bennys' technique, deduction is through locating a target cell and then identifying a surrounding forcing net to make the target cell a non-cell.

With Sue De Coq's technique, deduction is through locating 2 or 3 target cells and then identifying 2 bivalue cells, in the same units of the target cells, which are mutual exclusive subsets of the target cells.

In my opinion, these techniques are in the same category as xyz-wing, which should be applied before any attempt is made to search for double implication chains. In fact, xyz-wing, wxyz-wing and their derivatives are just subsets of benny's technique. Well done again, bennys.

These techniques should be given proper names. I am surely not the one to decide, but would like to open for suggestions and hereby submitting my recommendations as follows.

DanO's and Benny's technique: "Non-Cell Contradiction" as the forcing net would force the target cell into a non-cell.

Sue De Coq's technique: "Advanced Naked pairs" as the target cells can be seen as a bivalue cell that forms a naked pair with the other bivalue cells.