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Clement
Joined: 24 Apr 2006
Posts: 1111
Location: Dar es Salaam Tanzania
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 Posted: Fri Feb 19, 2010 10:31 pm Post subject: Feb 20 VH |
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| W-Wing <57>-7=7-<57> with strong link 7 in row 5 eliminates the 5's in row 7 BOX 9 and r9c2 solving the puzzle. |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sat Feb 20, 2010 2:16 am Post subject: |
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I didn't notice the W-Wing, but the 57s could be used as an M-Wing, although it was a case of simple coloring on 7. Either way, that's all that was needed.
On my 2nd attempt an XY-Wing (275) plus pincer transport did it. |
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arkietech
Joined: 31 Jul 2008
Posts: 1834
Location: Northwest Arkansas USA
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| Posted: Sat Feb 20, 2010 2:18 pm Post subject: |
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| Marty R. wrote: | | a case of simple coloring on 7. |
Or skyscraper 7. |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sat Feb 20, 2010 5:25 pm Post subject: |
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XY-wings: 2-57, -457, 2-57.
Keith |
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cgordon
Joined: 04 May 2007
Posts: 769
Location: ontario, canada
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| Posted: Sat Feb 20, 2010 6:41 pm Post subject: |
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| ... or ER on 7. |
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Louise56
Joined: 21 Sep 2005
Posts: 94
Location: El Cajon, California USA
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| Posted: Sat Feb 20, 2010 7:09 pm Post subject: |
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| Can I remove the 45 in r7c8? Is there a unique rectangle there? |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Sat Feb 20, 2010 7:52 pm Post subject: |
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| Louise56 wrote: | | Can I remove the 45 in r7c8? Is there a unique rectangle there? |
Yes, the Type 1 UR45 in r17c78 deletes both the 4 and 5 in r7c8.
Also note the 6-cell DP on 45 in r167c789. To prevent the DP, r6c89=7, r7c9=7, or r7c9=2.
(7)r6c89 - (7=9)r6c3,
(7)r7c9 - (7=5)r9c9 - (5=9)r9c2 - (9)r6c2 = (9)r6c3,
(2)r7c9 - (2=7)r5c9 - (7=5)r9c9 - (5=9)r9c2 - (9)r6c2 = (9)r6c3.
Thus, all conditions to prevent the deadly pattern force r6c3=9 to complete the puzzle in one step.
Ted |
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Wendy W
Joined: 04 Feb 2008
Posts: 144
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| Posted: Sat Feb 20, 2010 10:37 pm Post subject: |
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| Louise's UR jumped out at me as soon as I finished basics. After that, two xy-wings on 257 solved it. |
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Ema Nymton
Joined: 17 Apr 2009
Posts: 89
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| Posted: Sun Feb 21, 2010 2:19 pm Post subject: |
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2/20/2010
| Code: |
+----------+-------+----------+
| 6 7 8 | 3 2 9 | 5 4 1 |
| 9 4 2 | 5 1 6 | 7 3 8 |
| 1 3 5 | 7 4 8 | 2 6 9 |
+----------+-------+----------+
| 25 1 4 | 8 9 7 | 3 25 6 |
| 27*6 3 | 4 5 1 | 9 8 27A |
| 8 59 79 | 6 3 2 | 1 57 4 |
+----------+-------+----------+
| 57B8 1 | 9 6 3 | 4 27 257C|
| 4 2 6 | 1 7 5 | 8 9 3 |
| 3 59 79 | 2 8 4 | 6 1 57 |
+----------+-------+----------+
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Why does the ABC combination not eliminate the 7 from r5c1?
Ema
~@:o|
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Ema Nymton
Joined: 17 Apr 2009
Posts: 89
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| Posted: Sun Feb 21, 2010 2:34 pm Post subject: |
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Part II
| Code: |
+----------+-------+----------+
| 6 7 8 | 3 2 9 | 5 4 1 |
| 9 4 2 | 5 1 6 | 7 3 8 |
| 1 3 5 | 7 4 8 | 2 6 9 |
+----------+-------+----------+
| 25 1 4 | 8 9 7 | 3 25D6 |
| 27*6 3 | 4 5 1 | 9 8 27A |
| 8 59 79 | 6 3 2 | 1 57 4 |
+----------+-------+----------+
| 57B8 1 | 9 6 3 | 4 27E257C|
| 4 2 6 | 1 7 5 | 8 9 3 |
| 3 59 79 | 2 8 4 | 6 1 57F |
+----------+-------+----------+
|
Why does the ABC combination not eliminate the 7 from r5c1 but the DEF combination does eliminate the '5' from r7c9?
Ema
~@:o|
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sun Feb 21, 2010 4:10 pm Post subject: |
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| Ema Nymton wrote: | .
2/20/2010
| Code: |
+----------+-------+----------+
| 6 7 8 | 3 2 9 | 5 4 1 |
| 9 4 2 | 5 1 6 | 7 3 8 |
| 1 3 5 | 7 4 8 | 2 6 9 |
+----------+-------+----------+
| 25 1 4 | 8 9 7 | 3 25 6 |
| 27*6 3 | 4 5 1 | 9 8 27A |
| 8 59 79 | 6 3 2 | 1 57 4 |
+----------+-------+----------+
| 57B8 1 | 9 6 3 | 4 27 257C|
| 4 2 6 | 1 7 5 | 8 9 3 |
| 3 59 79 | 2 8 4 | 6 1 57 |
+----------+-------+----------+
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Why does the ABC combination not eliminate the 7 from r5c1?
Ema ~@ |
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C can be 2, 5, or 7.
If it is 2, A is 7. If it is 5, B is 7. If it is 7, you cannot make an elimination in R5C1 using only A, B, C.
Keith |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sun Feb 21, 2010 4:15 pm Post subject: |
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| Ema Nymton wrote: | .
Part II
| Code: |
+----------+-------+----------+
| 6 7 8 | 3 2 9 | 5 4 1 |
| 9 4 2 | 5 1 6 | 7 3 8 |
| 1 3 5 | 7 4 8 | 2 6 9 |
+----------+-------+----------+
| 25 1 4 | 8 9 7 | 3 25D6 |
| 27*6 3 | 4 5 1 | 9 8 27A |
| 8 59 79 | 6 3 2 | 1 57 4 |
+----------+-------+----------+
| 57B8 1 | 9 6 3 | 4 27E257C|
| 4 2 6 | 1 7 5 | 8 9 3 |
| 3 59 79 | 2 8 4 | 6 1 57F |
+----------+-------+----------+ |
Why does the ABC combination not eliminate the 7 from r5c1 but the DEF combination does eliminate the '5' from r7c9?
Ema ~@| |
DEF does not eliminate 5 in R7C9.
Ema,
I suggest you get a copy of Sudoku Susser. It explains the "why" of every move.
It is easy to use, and Norm has videos that show some of the finer points.
Keith |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sun Feb 21, 2010 4:55 pm Post subject: |
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| Quote: | | DEF does not eliminate 5 in R7C9. |
May I add the reason is that in this XY-Wing, E is the pivot and D and F the pincers. The pincer number is 5 and no cell with a 5 sees both pincers.
The 5 in r9c2 can be eliminated with a pincer transport, but we're getting ahead of ourselves with that. |
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