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arkietech
Joined: 31 Jul 2008 Posts: 1834 Location: Northwest Arkansas USA
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Posted: Fri Apr 27, 2012 8:27 am Post subject: Vanhegan fiendish 4/27/12 |
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Code: |
*-----------*
|...|..2|7..|
|.3.|.1.|.9.|
|2.8|.3.|4..|
|---+---+---|
|7..|148|...|
|.46|7.5|38.|
|...|963|..4|
|---+---+---|
|..4|.7.|1.3|
|.6.|.8.|.5.|
|..2|3..|...|
*-----------*
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Marty R.
Joined: 12 Feb 2006 Posts: 5770 Location: Rochester, NY, USA
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Posted: Fri Apr 27, 2012 6:04 pm Post subject: |
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M-Wing. R6c3=5-->r1c2=5. R1c2<>1-->r1c3=1; r8c3<>1.
Code: |
+--------------+-------------+--------------+
| 46 159 159 | 468 59 2 | 7 3 68 |
| 46 3 57 | 4568 1 467 | 2568 9 2568 |
| 2 79 8 | 56 3 79 | 4 1 56 |
+--------------+-------------+--------------+
| 7 25 3 | 1 4 8 | 2569 26 2569 |
| 9 4 6 | 7 2 5 | 3 8 1 |
| 158 1258 15 | 9 6 3 | 25 7 4 |
+--------------+-------------+--------------+
| 58 589 4 | 256 7 69 | 1 26 3 |
| 3 6 179 | 24 8 149 | 29 5 279 |
| 15 1579 2 | 3 59 169 | 689 4 6789 |
+--------------+-------------+--------------+
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Play this puzzle online at the Daily Sudoku site |
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arkietech
Joined: 31 Jul 2008 Posts: 1834 Location: Northwest Arkansas USA
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Posted: Fri Apr 27, 2012 6:59 pm Post subject: |
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I finally found one with an e digit Code: |
*-----------------------------------------------------------*
| 46 15-9 159 | 468 59 2 | 7 3 68 |
| 46 3 57 | 4568 1 467 | 2568 9 2568 |
| 2 *79 8 | 56 3 79 | 4 1 56 |
|-------------------+-------------------+-------------------|
| 7 25 3 | 1 4 8 | 2569 26 2569 |
| 9 4 6 | 7 2 5 | 3 8 1 |
| 158 1258 15 | 9 6 3 | 25 7 4 |
|-------------------+-------------------+-------------------|
|*58 x589 4 | 256 7 69 | 1 26 3 |
| 3 6 79-1 | 24 8 149 | 29 5 279 |
|x15 x1579 2 | 3 59 169 | 689 4 6789 |
*-----------------------------------------------------------*
Sue de Cog ab=79 cd=58 e=1 => r1c2<>9, r8c3<>1; stte
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Applying keith's definition
An SdC exists in one box(b7) and one line(c2).
(x) sets up two intersecting sets of exactly five distinct candidates a,b,c,d, and e. One in the box and one in the line.
A cell in the line but not in the box (r2c2) contains only candidates a, b.
A cell in the box but not in the line (r7c1) contains only candidates c, d.
We can eliminate a, b, and e from all other cells in the line, and we can eliminate c, d, and e from all other cells in the box.
I hope I got it right this time.
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DonM
Joined: 15 Sep 2009 Posts: 51
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Posted: Sat Apr 28, 2012 12:34 am Post subject: |
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arkietech wrote: | I finally found one with an e digit Code: |
*-----------------------------------------------------------*
| 46 15-9 159 | 468 59 2 | 7 3 68 |
| 46 3 57 | 4568 1 467 | 2568 9 2568 |
| 2 *79 8 | 56 3 79 | 4 1 56 |
|-------------------+-------------------+-------------------|
| 7 25 3 | 1 4 8 | 2569 26 2569 |
| 9 4 6 | 7 2 5 | 3 8 1 |
| 158 1258 15 | 9 6 3 | 25 7 4 |
|-------------------+-------------------+-------------------|
|*58 x589 4 | 256 7 69 | 1 26 3 |
| 3 6 79-1 | 24 8 149 | 29 5 279 |
|x15 x1579 2 | 3 59 169 | 689 4 6789 |
*-----------------------------------------------------------*
Sue de Cog ab=79 cd=58 e=1 => r1c2<>9, r8c3<>1; stte
|
Applying keith's definition
An SdC exists in one box(b7) and one line(c2).
(x) sets up two intersecting sets of exactly five distinct candidates a,b,c,d, and e. One in the box and one in the line.
A cell in the line but not in the box (r2c2) contains only candidates a, b.
A cell in the box but not in the line (r7c1) contains only candidates c, d.
We can eliminate a, b, and e from all other cells in the line, and we can eliminate c, d, and e from all other cells in the box.
I hope I got it right this time.
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Good find on a SDC with 5 candidates in 2 cells. The candidates used and the eliminations shown are accurate. However, the definition given above applies to only one form of SDC wherein there are 5 candidates (abcde) in 3 cells on the line in a box, a cell with candidates ab on the line outside the box and a cell cd in the box outside the line. In that case: 'We can eliminate a, b, and e from all other cells in the line, and we can eliminate c, d, and e from all other cells in the box.'
In the present case, there are 5 candidates (15789) in 2 cells on the line in the box, a cell on the line outside the box with (79) and 2 cells in the box that collectively and accurately contain (158) to satisfy this form of SDC, but in this case, we can eliminate a and b (not e) from all other cells in the line, and we can eliminate c, d, and e from all other cells in the box.
Of course, there can be other variations of SDC wherein there are 4 or 6 candidates in 2 cells or 7 candidates in 3 cells on the line in the box and so on.
Last edited by DonM on Sat Apr 28, 2012 1:15 am; edited 1 time in total |
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DonM
Joined: 15 Sep 2009 Posts: 51
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Posted: Sat Apr 28, 2012 12:38 am Post subject: |
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DonM wrote: | arkietech wrote: | I finally found one with an e digit Code: |
*-----------------------------------------------------------*
| 46 15-9 159 | 468 59 2 | 7 3 68 |
| 46 3 57 | 4568 1 467 | 2568 9 2568 |
| 2 *79 8 | 56 3 79 | 4 1 56 |
|-------------------+-------------------+-------------------|
| 7 25 3 | 1 4 8 | 2569 26 2569 |
| 9 4 6 | 7 2 5 | 3 8 1 |
| 158 1258 15 | 9 6 3 | 25 7 4 |
|-------------------+-------------------+-------------------|
|*58 x589 4 | 256 7 69 | 1 26 3 |
| 3 6 79-1 | 24 8 149 | 29 5 279 |
|x15 x1579 2 | 3 59 169 | 689 4 6789 |
*-----------------------------------------------------------*
Sue de Cog ab=79 cd=58 e=1 => r1c2<>9, r8c3<>1; stte
|
Applying keith's definition
An SdC exists in one box(b7) and one line(c2).
(x) sets up two intersecting sets of exactly five distinct candidates a,b,c,d, and e. One in the box and one in the line.
A cell in the line but not in the box (r2c2) contains only candidates a, b.
A cell in the box but not in the line (r7c1) contains only candidates c, d.
We can eliminate a, b, and e from all other cells in the line, and we can eliminate c, d, and e from all other cells in the box.
I hope I got it right this time.
|
Good find on a SDC with 5 candidates in 2 cells. The candidates used and the eliminations shown are accurate. However, the definition given above applies to only one form of SDC wherein there are 5 candidates (abcde) in 3 cells on the line in a box, a cell with candidates ab on the line outside the box and a cell cd in the box outside the line. In that case: 'We can eliminate a, b, and e from all other cells in the line, and we can eliminate c, d, and e from all other cells in the box.'
In the present case, there are 5 candidates (15789) in 2 cells on the line in the box, a cell on the line outside the box with (79) and 2 cells in the box that collectively and accurately contain (158) to satisfy this form of SDC, but in this case, we can eliminate a and b (not e) from all other cells in the line, and we can eliminate c, d, and e from all other cells in the box.
Of course, there can be other variations of SDC wherein there are 4 or 6 candidates in 2 cells or 7 candidates in 3 cells on the line in the box and so on. |
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arkietech
Joined: 31 Jul 2008 Posts: 1834 Location: Northwest Arkansas USA
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Posted: Sat Apr 28, 2012 1:01 am Post subject: |
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DonM wrote: | In the present case, there are 5 candidates (15789) in 2 cells on the line in the box, a cell on the line outside the box with (79) and 2 cells in the box that collectively and accurately contain (158) to satisfy this form of SDC, but in this case, we can eliminate a and b (not e) from all other cells in the line, and we can eliminate c, d, and e from all other cells in the box. |
Thanks, you are right. This(getting a good definition) is quite complex. Back to the drawing board.
The e's being removed have to "see" all the e's in the set in the block...?? going to take some thought. |
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SudoQ
Joined: 02 Aug 2011 Posts: 127
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Posted: Sat Apr 28, 2012 7:51 am Post subject: |
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You can also show that r8c3<>1 like this:
r3c2=x -> r8c3=x (x=7/9)
/SudoQ |
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