The New Sudoku Players' Forum

[ArkieTech]

Code: Select all

 *-----------*
 |...|824|...|
 |..5|...|9..|
 |.3.|...|.4.|
 |---+---+---|
 |6..|..7|..2|
 |9..|...|..4|
 |8..|6..|..3|
 |---+---+---|
 |.2.|...|.7.|
 |..6|...|8..|
 |...|593|...|
 *-----------*


Play/Print this puzzle online

[Leren]

Code: Select all

*--------------------------------------------------------------------------------*
| 17      1679    179      | 8       2       4        | 1356    1356    1567     |
|b1247    14678   5        | 137     1367    16       | 9       126     1678     |
|c127     3       178      | 179     1567    1569     |d126     4       1678     |
|--------------------------+--------------------------+--------------------------|
| 6       145     134      | 1349    13458   7        | 15      1589    2        |
| 9       157     1237     | 123     1358    1258     | 1567    1568    4        |
| 8       1457    1247     | 6       145     1259     | 157     159     3        |
|--------------------------+--------------------------+--------------------------|
| 35      2       149      | 14      68      68       | 1345    7       159      |
| 35      149     6        | 1247    147     12       | 8       135     159      |
|a147     1478    1478     | 5       9       3        |e126-4   126     16       |
*--------------------------------------------------------------------------------*

Purple Cow: (4)r9c1 = (4-2) r2c1 = r3c1 - r3c7 = r9c7 => r9c7 <4>; stte

Leren

[tlanglet]

First I found
ANS(9=14)r7c34-4r7c7=(4*-2)r9c7=r9c8-r2c8=(2-4)r2c1=4r9c1- 4r9c7* Conflict => r7c34 <>14

Then I realized
2r9c7=2r9c8-r2c8=(2-4)r2c1=4r9c1 => r9c7<>4

Ted

[eleven]

Not a nice puzzle to play on paper - and not a nice one to look for 1-steppers. I wonder, if anything not very difficult is possible, which does not place the 4 in r7c7.
Yet another way to do that is with the almost pair 17 in r13c1. The pair leads to 4 in r9c1, a 2 in r3c1 to a 2 in r9c7.

[daj95376]

Not a nice puzzle to play on paper - and not a nice one to look for 1-steppers. I wonder, if anything not very difficult is possible, which does not place the 4 in r7c7.
Yet another way to do that is with the almost pair 17 in r13c1. The pair leads to 4 in r9c1, a 2 in r3c1 to a 2 in r9c7.

This is the simplest 1-stepper solution that my solver found without using <4> in r7c7 or resulting in r9c7<>4.

Code: Select all

 after basics
 +-----------------------------------------------------------------------+
 |  17     1679   179    |  8      2      4      |  1356   1356   1567   |
 | c1247   14678  5      |  137    1367   16     |  9     b126    1678   |
 |  127    3      178    |  179    1567   1569   |  126    4      1678   |
 |-----------------------+-----------------------+-----------------------|
 |  6      145    134    |  1349   13458  7      |  15     1589   2      |
 |  9      157    1237   |  123    1358   1258   |  1567   1568   4      |
 |  8      1457   1247   |  6      145    1259   |  157    159    3      |
 |-----------------------+-----------------------+-----------------------|
 |  35     2      149    |  14     68     68     |  1345   7      159    |
 |  35     149    6      |  1247   147    12     |  8      135    159    |
 | d147    1478   1478   |  5      9      3      |  1246e a126e  a16e    |
 +-----------------------------------------------------------------------+
 # 137 eliminations remain

 (16=2)r9c89 - r2c8 = (2-4)r2c1 = r9c1 - (4=126)r9c789  =>  r7c79,r8c89,r9c123<>1


[eleven]

Nice, thank you.