The New Sudoku Players' Forum

[ArkieTech]

Code: Select all

 *-----------*
 |..4|..8|...|
 |29.|...|..4|
 |..3|..1|..2|
 |---+---+---|
 |1..|.8.|.3.|
 |..6|.3.|7..|
 |.8.|.7.|..9|
 |---+---+---|
 |7..|9..|1..|
 |4..|...|.97|
 |...|5..|6..|
 *-----------*


Play/Print this puzzle online

[Leren]

Code: Select all

*-----------------------------------------------------------------------*
| 56    f157    4       | 23     29     8       | 39    e1567   16      |
| 2      9     b18-7    | 37     5      6       | 38     17     4       |
| 568   c57     3       | 47     49     1       | 89    d567    2       |
|-----------------------+-----------------------+-----------------------|
| 1      245-7 a2579    | 246    8      2459    | 245    3      56      |
| 59     245    6       | 124    3      2459    | 7      28     158     |
| 3      8      25      | 16     7      245     | 245    16     9       |
|-----------------------+-----------------------+-----------------------|
| 7      2356   258     | 9      246    234     | 1      248    358     |
| 4      12356  125     | 8      126    23      | 25     9      7       |
| 89     123    1289    | 5      124    7       | 6      248    38      |
*-----------------------------------------------------------------------*

(7) r4c3 = r2c3 - (7=5) r3c2 - r3c8 = (5-7) r1c8 = (7) r1c2 => - 7 r2c3, r4c2; stte

Leren

[ArkieTech]

Code: Select all

 *--------------------------------------------------------------------*
 | 56     157    4      | 23     29     8      | 39     1567   16     |
 | 2      9     a78-1   | 37     5      6      | 38     17     4      |
 |b568    57     3      | 47     49     1      | 89     567    2      |
 |----------------------+----------------------+----------------------|
 | 1      2457  *79-25  | 246    8      2459   | 245    3      56     |
 | 59     245    6      | 124    3      2459   | 7      28     158    |
 | 3      8      25     | 16     7      245    | 245    16     9      |
 |----------------------+----------------------+----------------------|
 | 7      2356   258    | 9      246    234    | 1      248    358    |
 | 4      12356  125    | 8      126    23     | 25     9      7      |
 |c89     123   d1289   | 5      124    7      | 6      248    38     |
 *--------------------------------------------------------------------*
7r4c3=(7-8)r2c3=r3c1-(8=9)r9c1-r9c3=9r4c3loop => -25r4c3, -1r2c3; ste


[tlanglet]

Code: Select all

 *--------------------------------------------------------------------*
 | 5*6   c157    4      | 23     29     8      | 39    b1567   16     |
 | 2      9     d178    | 37     5      6      | 38     17     4      |
 |g568    57     3      | 47     49     1      | 89    a567    2      |
 |----------------------+----------------------+----------------------|
 | 1      2457  e2579   | 246    8      2459   | 245    3      56     |
 |f59     245    6      | 124    3      2459   | 7      28     158    |
 | 3      8      25     | 16     7      245    | 245    16     9      |
 |----------------------+----------------------+----------------------|
 | 7      2356   258    | 9      246    234    | 1      248    358    |
 | 4      12356  125    | 8      126    23     | 25     9      7      |
 | 89     123    1289   | 5      124    7      | 6      248    38     |
 *--------------------------------------------------------------------*


Another notational dilemma. Given the initial strong inference, 5r3c8=(5-7)r1c8, my following logic depends of the fact that when r1c8=5 then r1c8<>7 but also that it makes r1c1<>5. I realize it is possible to make a forked graph for the two branches but that is always a pain.

Would an "*" to flag the secondary action be effective? For example,

5r3c8=(5-7)r1c8*=7r1c2-r2c3=(7-9)r4c3=(9-5)r5c1=5r3c1* => r3c2<>5

Ted

[Marty R.]

Code: Select all

+----------------+--------------+--------------+
| 56  157   4    | 23  29  8    | 39  1567 16  |
| 2   9     178  | 37  5   6    | 38  17   4   |
| 568 57    3    | 47  49  1    | 89  567  2   |
+----------------+--------------+--------------+
| 1   2457  2579 | 246 8   2459 | 245 3    56  |
| 59  245   6    | 124 3   2459 | 7   28   158 |
| 3   8     25   | 16  7   245  | 245 16   9   |
+----------------+--------------+--------------+
| 7   2356  258  | 9   246 234  | 1   248  358 |
| 4   12356 125  | 8   126 23   | 25  9    7   |
| 89  123   1289 | 5   124 7    | 6   248  38  |
+----------------+--------------+--------------+


Play this puzzle online at the Daily Sudoku site

M-Wing (7=5)r3c2-r3c8=(5-7)r1c8=r1c2=>r2c3<>7

The DP 57 also yields the common outcome, but notating pseudo cells is a challenge.

R13c8[pseudo cell]=16-(16=7)r1c9,r132c8-(7=18)r2c3
R1c2=1-(16=5)r1c91-(5=7)r3c2-(7=18)r2c3=>r2c3<>7

[Leren]

Marty R wrote: M-Wing (7=5)r3c2-r3c8=(5-7)r1c8=r1c2=>r2c3<>7

That chain also qualifies as an M Ring Type B and you can write (7=5)r3c2-r3c8=(5-7)r1c8=r1c2 loop => r1c8 <> 16, r2c3 <> 7, r3c1 <> 5, r4c2 <> 7

I think Danny gave you the link to the M Wing / M Ring exemplars a few days ago.

Leren

[Luke]

Another notational dilemma. Given the initial strong inference, 5r3c8=(5-7)r1c8, my following logic depends of the fact that when r1c8=5 then r1c8<>7 but also that it makes r1c1<>5. I realize it is possible to make a forked graph for the two branches but that is always a pain.

Would an "*" to flag the secondary action be effective? For example,

5r3c8=(5-7)r1c8*=7r1c2-r2c3=(7-9)r4c3=(9-5)r5c1=5r3c1* => r3c2<>5

Ted

I haven't been keeping up, but I remember the old "....chain with memory*...." preface being sufficient and accepted by the majority.

[ronk]

Another notational dilemma. Given the initial strong inference, 5r3c8=(5-7)r1c8, my following logic depends of the fact that when r1c8=5 then r1c8<>7 but also that it makes r1c1<>5. I realize it is possible to make a forked graph for the two branches but that is always a pain.

Would an "*" to flag the secondary action be effective? For example,

5r3c8=(5-7)r1c8*=7r1c2-r2c3=(7-9)r4c3=(9-5)r5c1=5r3c1* => r3c2<>5

Hmm, I'll duck the question by pointing out the grouped strong link (5)r5c1 = (5)r13c1.

5r3c8=(5-7)r1c8=7r1c2-r2c3=(7-9)r4c3=(9-5)r5c1=5r13c1 => r3c2<>5

[tlanglet]

Luke,
Great to see you back posting your words of wisdom flavored with comic relief; this place is not the same without you! My suggested notation was based on the "memory" notion but I obviously forgot the preface; thanks.

Ron,
Also wonderful to note that you are still around to assist us "want-to-be" better sudoku players. In this case, I also found your solution but I was more interested in the notational issue.

Looking forward to more participation from both of you two........

Ted

[JC Van Hay]

Another notational dilemma. Given the initial strong inference, 5r3c8=(5-7)r1c8, my following logic depends of the fact that when r1c8=5 then r1c8<>7 but also that it makes r1c1<>5. I realize it is possible to make a forked graph for the two branches but that is always a pain.

Would an "*" to flag the secondary action be effective? For example,

5r3c8=(5-7)r1c8*=7r1c2-r2c3=(7-9)r4c3=(9-5)r5c1=5r3c1* => r3c2<>5

Ted

A third suggestion : Kraken 5C1 : 5r13c1=(5-9)r5c1=(9-7)r4c3=7r2c3-7r1c2=(7-5)r1c8=5r3c8 That is either r1c1=5 or r3c1=5 or r5c1=5->r3c8=5 :=> -5r3c2 from l-t-r, but from r-t-l ???
A fourth similar one : 5r1c1=*Loop[5r3c1=*(5-9)r5c1=(9-7)r4c3=7r2c3-7r1c2=(7-5)r1c8=5r3c8 @] :=> [5r1c1=5r3c1=5r3c8]-5r3c2 Readable in both directions