The New Sudoku Players' Forum

[Danu Pr]

Please help,, stuck

[eleven]

You missed, that the 8 is locked in (must be in one of) row 2, columns 24 (cannot be in r2c2 and the rest of the row, which is solved).
So you can eliminate it from r1c4.
Then look at the 2 possibilities of r8c3.

[SpAce]

You missed, that the 8 is locked in (must be in one of) row 2, columns 24 (cannot be in r2c2 and the rest of the row, which is solved). So you can eliminate it from r1c4.

I wonder if it was just a careless miss because two other 8s have been eliminated from the same box (how, if not by claiming?). Anyway, none of them are even necessary eliminations, as there's an stte-capable XY-Wing available regardless (or two with the current pencil marks). However, after completing the claiming there's three, and it's usually a good idea to exhaust all basic techniques first anyway.

[Danu Pr]

You missed, that the 8 is locked in (must be in one of) row 2, columns 24 (cannot be in r2c2 and the rest of the row, which is solved).
So you can eliminate it from r1c4.
Then look at the 2 possibilities of r8c3.

yes I miss 8 in r1c4, so I can eliminated it,, but still dont get what your mean for "2 possibilities of r8c3?

[Leren]

..4.9.17.9.6.7.542.572.493..439..65779845621356.3.7894...7.94.5679.4.3.1485.3.7.9

Code: Select all

*-------------------------------------*
| 238 23  4  | 56   9   35  | 1 7  68 |
| 9   13  6  |a18   7   138 | 5 4  2  |
| 18  5   7  | 2   b16  4   | 9 3  68 |
|------------+--------------+---------|
| 12  4   3  | 9    128 18  | 6 5  7  |
| 7   9   8  | 4    5   6   | 2 1  3  |
| 5   6   12 | 3    12  7   | 8 9  4  |
|------------+--------------+---------|
| 123 123 12 | 7   c68  9   | 4 68 5  |
| 6   7   9  | 5-8  4   258 | 3 28 1  |
| 4   8   5  | 16   3   12  | 7 26 9  |
*-------------------------------------*

Not sure what eleven had in mind, but the easiest way to solve this puzzle is via an XY wing in cells abc. It should be easy to see that if a is not 8 then c is 8. If c is not 8 then a is 8.

Either way r8c4 is not 8, so it's 5, and that solves the puzzle. Maybe eleven made a typo and meant r8c4.

Leren

[SpAce]

Not sure what eleven had in mind, but the easiest way to solve this puzzle is via an XY wing in cells abc. It should be easy to see that if a is not 8 then c is 8. If c is not 8 then a is 8.

There are actually two ways to do that, and both work even without completing the claiming. The other XY-Wing with the same elimination uses (16)r9c4 as the b-cell (shown with the original pencil marks, i.e. candidate 8r1c4 in place):

Code: Select all

.--------------.------------------.-----------.
| 238  23   4  |  568   9     35  | 1  7   68 |
| 9    13   6  | a1(8)  7     138 | 5  4   2  |
| 18   5    7  |  2     16    4   | 9  3   68 |
:--------------+------------------+-----------:
| 12   4    3  |  9     128   18  | 6  5   7  |
| 7    9    8  |  4     5     6   | 2  1   3  |
| 5    6    12 |  3     12    7   | 8  9   4  |
:--------------+------------------+-----------:
| 123  123  12 |  7    c6(8)  9   | 4  68  5  |
| 6    7    9  |  5-8   4     258 | 3  28  1  |
| 4    8    5  | b16    3     12  | 7  26  9  |
'--------------'------------------'-----------'

(8=1)r2c4 - (1=6)r9c4 - (6=8)r7c5 => -8 r8c4; stte

Either way r8c4 is not 8, so it's 5, and that solves the puzzle. Maybe eleven made a typo and meant r8c4.

Probably, but I presumed he also meant the third XY-Wing which is revealed after eliminating 8r1c4:

Code: Select all

.--------------.-----------------.-----------.
| 238  23   4  | a5(6)  9    35  | 1  7   68 |
| 9    13   6  |  18    7    138 | 5  4   2  |
| 18   5    7  |  2     1-6  4   | 9  3   68 |
:--------------+-----------------+-----------:
| 12   4    3  |  9     128  18  | 6  5   7  |
| 7    9    8  |  4     5    6   | 2  1   3  |
| 5    6    12 |  3     12   7   | 8  9   4  |
:--------------+-----------------+-----------:
| 123  123  12 |  7   c(6)8  9   | 4  68  5  |
| 6    7    9  | b58    4    258 | 3  28  1  |
| 4    8    5  |  1-6   3    12  | 7  26  9  |
'--------------'-----------------'-----------'

(6=5)r1c4 - (5=8)r8c4 - (8=6)r7c5 => -6 r3c5,r9c4; stte

One way to see that is considering the two possibilities of (58)r8c4:

(5)r8c4 - (5=6)r1c4
||
(8)r8c4 - (8=6)r7c5

Since at least one of r1c4 or r7c5 must be 6, neither r3c5 nor r9c4 can be 6 (so they're both 1), which solves the puzzle.

[SpAce]

the easiest way to solve this puzzle is via an XY wing

That's somewhat subjective. Personally I think W-Wings are easier to spot, and there's one of those available here as well (actually two, but the other one doesn't solve the puzzle):

Code: Select all

.--------------.-------------------.-----------.
| 238  23   4  |  56    9     35   | 1  7   68 |
| 9    13   6  | a1(8)  7     13-8 | 5  4   2  |
| 18   5    7  |  2     16    4    | 9  3   68 |
:--------------+-------------------+-----------:
| 12   4    3  |  9     128  d1(8) | 6  5   7  |
| 7    9    8  |  4     5     6    | 2  1   3  |
| 5    6    12 |  3     12    7    | 8  9   4  |
:--------------+-------------------+-----------:
| 123  123  12 |  7     68    9    | 4  68  5  |
| 6    7    9  |  58    4     258  | 3  28  1  |
| 4    8    5  | b16    3    c12   | 7  26  9  |
'--------------'-------------------'-----------'

(8=1)r2c4 - r9c4 = r9c6 - (1=8)r4c6 => -8 r2c6; stte

In my subjective opinion, that is probably the easiest way to solve this puzzle (and it also works with the original pencil marks). W-Wings are easier to spot than XY-Wings because they only use two digits and have identical bivalue cells as the end points. All one has to do is to look for such identical bivalue cells and see if they're connected. XY-Wings have one fewer cell but all of its three cells have different contents, so the pattern is not as easy to spot with the naked eye.

(However, in this case there's a cluster of three XY-Wings and just one useful W-Wing, which might make at least one of the XY-Wings jump out more easily. Yet, if that doesn't happen then I think it's a better strategy to look for W-Wings before XY-Wings. That's also what Hodoku does by default, and it also grades XY-Wing slightly harder, which I think is the correct hierarchy for most manual solvers.)

[Danu Pr]

..4.9.17.9.6.7.542.572.493..439..65779845621356.3.7894...7.94.5679.4.3.1485.3.7.9

Code: Select all

*-------------------------------------*
| 238 23  4  | 56   9   35  | 1 7  68 |
| 9   13  6  |a18   7   138 | 5 4  2  |
| 18  5   7  | 2   b16  4   | 9 3  68 |
|------------+--------------+---------|
| 12  4   3  | 9    128 18  | 6 5  7  |
| 7   9   8  | 4    5   6   | 2 1  3  |
| 5   6   12 | 3    12  7   | 8 9  4  |
|------------+--------------+---------|
| 123 123 12 | 7   c68  9   | 4 68 5  |
| 6   7   9  | 5-8  4   258 | 3 28 1  |
| 4   8   5  | 16   3   12  | 7 26 9  |
*-------------------------------------*

Not sure what eleven had in mind, but the easiest way to solve this puzzle is via an XY wing in cells abc. It should be easy to see that if a is not 8 then c is 8. If c is not 8 then a is 8.

Either way r8c4 is not 8, so it's 5, and that solves the puzzle. Maybe eleven made a typo and meant r8c4.

Leren

FINALLY, thanks man you've solved this with your "abc cells", what method it is?

Hidden Text: Show

Screenshot_2019-02-25-21-12-54-453_com.scn.sudokuchamp.jpg (149.82 KiB) Viewed 947 times

the easiest way to solve this puzzle is via an XY wing

That's somewhat subjective. Personally I think W-Wings are easier to spot, and there's one of those available here as well (actually two, but the other one doesn't solve the puzzle):

Code: Select all

.--------------.-------------------.-----------.
| 238  23   4  |  56    9     35   | 1  7   68 |
| 9    13   6  | a1(8)  7     13-8 | 5  4   2  |
| 18   5    7  |  2     16    4    | 9  3   68 |
:--------------+-------------------+-----------:
| 12   4    3  |  9     128  d1(8) | 6  5   7  |
| 7    9    8  |  4     5     6    | 2  1   3  |
| 5    6    12 |  3     12    7    | 8  9   4  |
:--------------+-------------------+-----------:
| 123  123  12 |  7     68    9    | 4  68  5  |
| 6    7    9  |  58    4     258  | 3  28  1  |
| 4    8    5  | b16    3    c12   | 7  26  9  |
'--------------'-------------------'-----------'

(8=1)r2c4 - r9c4 = r9c6 - (1=8)r4c6 => -8 r2c6; stte

In my subjective opinion, that is probably the easiest way to solve this puzzle (and it also works with the original pencil marks). W-Wings are easier to spot than XY-Wings because they only use two digits and have identical bivalue cells as the end points. All one has to do is to look for such identical bivalue cells and see if they're connected. XY-Wings have one fewer cell but all of its three cells have different contents, so the pattern is not as easy to spot with the naked eye.

(However, in this case there's a cluster of three XY-Wings and just one useful W-Wing, which might make at least one of the XY-Wings jump out more easily. Yet, if that doesn't happen then I think it's a better strategy to look for W-Wings before XY-Wings. That's also what Hodoku does by default, and it also grades XY-Wing slightly harder, which I think is the correct hierarchy for most manual solvers.)

Still i dont understand your described explaination yet. But, thanks anyway. it is my first topic in this forum, so exited with all of your help.
Btw, i want to learn that formula you wrote "(8=1)r2c4 - r9c4 = r9c6 - (1=8)r4c6 => -8 r2c6; stte" etc. Can you please explain it in some words? or some links perhaps, to learn. (w wing, xy wing) . I have google some, but it's hard (for me) to find the pattern in actuall puzzle.

[SpAce]

FINALLY, thanks man you've solved this with your "abc cells", what method it is?

Leren's method was called XY-Wing. Like I mentioned, there were two other XY-Wings available as well. The other method I mentioned was W-Wing. Both of those patterns are described here, for example.

Code: Select all

.--------------.-------------------.-----------.
| 238  23   4  |  56    9     35   | 1  7   68 |
| 9    13   6  | a1(8)  7     13-8 | 5  4   2  |
| 18   5    7  |  2     16    4    | 9  3   68 |
:--------------+-------------------+-----------:
| 12   4    3  |  9     128  d1(8) | 6  5   7  |
| 7    9    8  |  4     5     6    | 2  1   3  |
| 5    6    12 |  3     12    7    | 8  9   4  |
:--------------+-------------------+-----------:
| 123  123  12 |  7     68    9    | 4  68  5  |
| 6    7    9  |  58    4     258  | 3  28  1  |
| 4    8    5  | b16    3    c12   | 7  26  9  |
'--------------'-------------------'-----------'

(8=1)r2c4 - r9c4 = r9c6 - (1=8)r4c6 => -8 r2c6; stte

Still i dont understand your described explaination yet. But, thanks anyway. it is my first topic in this forum, so exited with all of your help.
Btw, i want to learn that formula you wrote "(8=1)r2c4 - r9c4 = r9c6 - (1=8)r4c6 => -8 r2c6; stte" etc. Can you please explain it in some words?

It's called Eureka. It's a simple language to write sudoku chaining logic. The idea is that the chain proves that one or the other end node (or both) must have the indicated value, which can cause eliminations. The W-Wing chain above proves that either r2c4 is 8 or if not, then via the chain r4c6 must be 8 (and vice versa). Either way, r2c6 can't be 8 because it sees both end points (one of which must be 8), so it can be eliminated. The same works for the XY-Wing chains I wrote earlier.

One way to interpret Eureka is as an implication chain. The above chain could thus be read like this:

r2c4<>8 -> r2c4=1 -> r9c4<>1 -> r9c6=1 -> r4c6<>1 -> r4c6=8

(if r2c4 is not 8, then r2c4 is 1, ..., then r4c6 is not 1, then r4c6 is 8)