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[ArkieTech]

Code: Select all

 *-----------*
 |79.|.6.|.51|
 |8..|...|..6|
 |..3|...|2..|
 |---+---+---|
 |...|4.2|...|
 |3..|...|..8|
 |...|1.7|...|
 |---+---+---|
 |..8|...|4..|
 |6..|...|..2|
 |93.|.1.|.85|
 *-----------*


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[Leren]

Code: Select all

*-----------------------------------------------------------------------*
| 7      9     c24D     | 23     6      34      | 8      5      1       |
| 8      1245   1245    | 2579   24579  1459    | 379    3479   6       |
|b145C   6      3       | 5789   4579   14589   | 2      479    479     |
|-----------------------+-----------------------+-----------------------|
|a15     1578   69      | 4      38     2       | 135679 13679  379     |
| 3      1247   1247    | 569    59     569     | 17     1247   8       |
| 245B   2458   69      | 1      38     7       | 3569   23469  349     |
|-----------------------+-----------------------+-----------------------|
|a125A   1257   8       | 235679 2579   3569    | 4      13679  379     |
| 6      1457   1457    | 35789  4579   34589   | 1379   1379   2       |
| 9      3      47-2    | 267    1      46      | 67     8      5       |
*-----------------------------------------------------------------------*

wxyz-wing (aabc): (2=15) r47c1 - (15=4) r3c1 - (4=2) r1c3 => -2 r9c3; stte
or
m-wing (ABCD): -2 r7c1 = (2-4) r6c1 = r3c1 - (4=2) r1c3 => -2 r9c3; stte

Leren

[daj95376]

(2=15) r47c1 - (15=4) r3c1 - (4=2) r1c3 => -2 r9c3; stte

Hmmm! I understand what Leren is doing, but I still have issues with Eureka notation for ALS structures. In this case, reading the chain from r-to-l seems awkward for the strong inference in r3c1. This is a case where I'd use a 3-cell ALS.

(2=4)r743c1 - (4=2)r1c3 => -2 r9c3; stte

On second thought, why can't it be a 4-cell LS that spans more than one house/unit?

(2=2)r743c1,r1c3 => -2 r9c3; stte

I think that I've just discovered a new way (for me) to write an XY-Chain.

[7b53]

wxyz-wing (aabc): (2=15) r47c1 - (15=4) r3c1 - (4=2) r1c3 => -2 r9c3; stte

I found it easier to follow, starting from a bivalue cell
(2=4)r1c3 - (4=15)r34r1 - (15=2)r7c1 => r9c3<>2
-or-
just pick one cell from column 1 to form a locked set {1,5} with r4c1.
digit 2 will be at (r1c3 or r7c1)