The New Sudoku Players' Forum

[pjb]

My line of thinking was simply that if 2 is true in r7c9, then the DP results, so the 2 must be false. This is due to the SL of 2 in c6 and r7, and the SL of 8 in c9 and r8. The same is true for the 2s at r8c9 and r8c6. What I missed was that if the 8 at r7c6 is true, the DP also results, so (8)r7c6 is also false.

Phil

[Marty R.]

A good study in Notation.

Code: Select all

 *--------------------------------------------------------------------*
 | 3467   1346   1467   | 8      46     5      | 2      14     9      |
 | 46     1469   1469   | 12     246    7      | 8      5      3      |
 | 5      2      8      | 3      9      14     | 147    1467   167    |
 |----------------------+----------------------+----------------------|
 | 678    168    1267   | 5      28     3      | 179    12679  4      |
 | 346    5      1246   | 12     7      9      | 13     8      126    |
 | 9      138    127    | 6      248    14     | 1357   127    1257   |
 |----------------------+----------------------+----------------------|
 | 48     489    459    | 7      1     *28     | 6      3     *258    |
 | 1      7      56     | 9      3     *268    | 45     24    *258    |
 | 2      68     3      | 4      5      68     | 179    179    17     |
 *--------------------------------------------------------------------*
[(2=5)r8c78-5r78c9
           -(5=28)r8c36]->[dp:28r78c69]->2r8c8; ste


I'm sure it is indeed a good study, but a little complicated for me.

Let me try a rewrite, changing the last term.

[(2=5)r8c78-5r78c9
-(5=28)r8c36]->[dp:28r78 (2r78c9=8r78c9)]contradiction->2r8c8; ste

Does that make sense? The contradiction, of course, is that if you combine the 1st and last terms, there are no 2s left in box 9.

Why is the last term preceded by ->?

[eleven]

Marty, a unique puzzle cannot have a U4 pattern

Code: Select all

. . 1 | 2 . .
. . 2 | 1 . .
. . . | . . .

in the solution, if not one of the numbers are given.
So you can eliminate each candidate, which leads to this pattern. In this case the three 2's and the 8 would force it, so they can be eliminated.

[eleven]

Another way to look at it, is that 2r8c8 is the only "external" that is required.

Ah yes, of course. Since there is no external candidate in r7c6, also removing the externals of r8&b9 (or r8&c9) leaves (hidden) pairs in the UR cells (and a deadly pattern).

[David P Bird]

Here's another view:
(45=6)r8c37 - (6)r9c2 = (6-8)r9c6 = (8)XWing:r78c69 -UR- (2)Xwing:r78c68 = (2)r8c8 => r8c8 <> 4

.

[ArkieTech]

A good study in Notation.

Code: Select all

 *--------------------------------------------------------------------*
 | 3467   1346   1467   | 8      46     5      | 2      14     9      |
 | 46     1469   1469   | 12     246    7      | 8      5      3      |
 | 5      2      8      | 3      9      14     | 147    1467   167    |
 |----------------------+----------------------+----------------------|
 | 678    168    1267   | 5      28     3      | 179    12679  4      |
 | 346    5      1246   | 12     7      9      | 13     8      126    |
 | 9      138    127    | 6      248    14     | 1357   127    1257   |
 |----------------------+----------------------+----------------------|
 | 48     489    459    | 7      1     *28     | 6      3     *258    |
 | 1      7      56     | 9      3     *268    | 45     24    *258    |
 | 2      68     3      | 4      5      68     | 179    179    17     |
 *--------------------------------------------------------------------*
[(2=5)r8c78-5r78c9
           -(5=28)r8c36]->[dp:28r78c69]->2r8c8; ste


I'm sure it is indeed a good study, but a little complicated for me.

Let me try a rewrite, changing the last term.

[(2=5)r8c78-5r78c9
-(5=28)r8c36]->[dp:28r78 (2r78c9=8r78c9)]contradiction->2r8c8; ste

Does that make sense? The contradiction, of course, is that if you combine the 1st and last terms, there are no 2s left in box 9.

Why is the last term preceded by ->?

Marty,

What I am trying to notate is if r8c8 is not 2 then you have the following deadly pattern:

Code: Select all

 *--------------------------------------------------------------------*
 | 3467   1346   1467   | 8      46     5      | 2      14     9      |
 | 46     1469   1469   | 12     246    7      | 8      5      3      |
 | 5      2      8      | 3      9      14     | 147    1467   167    |
 |----------------------+----------------------+----------------------|
 | 678    168    1267   | 5      28     3      | 179    12679  4      |
 | 346    5      1246   | 12     7      9      | 13     8      126    |
 | 9      138    127    | 6      248    14     | 1357   127    1257   |
 |----------------------+----------------------+----------------------|
 | 48     489    459    | 7      1     (28)    | 6      3     (28)    |
 | 1      7      6      | 9      3     (28)    | 5      4     (28)    |
 | 2      68     3      | 4      5      68     | 179    179    17     |
 *--------------------------------------------------------------------*


so r8c8 must be a 2