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Earl
Joined: 30 May 2007
Posts: 677
Location: Victoria, KS
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 Posted: Fri Apr 16, 2010 1:52 am Post subject: April 16 VH |
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As high as your taxes.
Solution Skyscraper (9)
Early Earl |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Fri Apr 16, 2010 3:43 am Post subject: |
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| There are some other interesting solutions. One is a UR that's a Type 4 but you get get more mileage by using it to create a pincer situation. Another is a flightless M-Wing with transport. There's a standard XY-Wing and a DP that's in three boxes. I didn't notice the solution that you used. |
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arjun
Joined: 28 Sep 2009
Posts: 16
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| Posted: Fri Apr 16, 2010 10:11 am Post subject: |
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This is only the 3rd very hard that I have completed. I don't know the exact notation, but this is what I did.
1. Unique Rectangle with 69
2. Don't remember this step. But I eliminated a "2" in R7C4. It was either a forcing chain or a XYZ wing
3. Another forcing chain - If R9C4 is a 2 or a 4, R7C4 is a 1
4. 249 XYZ wing which means R7C2 is a 6 |
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Willzzz
Joined: 22 Aug 2008
Posts: 5
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| Posted: Fri Apr 16, 2010 10:29 am Post subject: |
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One step
r6c1 can't be a 9 as it would make both r5c6 and r7c6 a 2.
Don't know much about terminology  |
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arjun
Joined: 28 Sep 2009
Posts: 16
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| Posted: Fri Apr 16, 2010 10:33 am Post subject: |
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| Could someone post the grid after basics so I know what other people are talking about? |
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Willzzz
Joined: 22 Aug 2008
Posts: 5
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| Posted: Fri Apr 16, 2010 12:29 pm Post subject: |
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| Arjun, I think your 249 wing solves it on it's own, the earlier steps aren't required. |
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arkietech
Joined: 31 Jul 2008
Posts: 1834
Location: Northwest Arkansas USA
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| Posted: Fri Apr 16, 2010 1:30 pm Post subject: |
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Here is my grid after basics: | Code: | *-----------------------------------------------------------*
| 1 5 6 | 9 3 4 | 7 8 2 |
| 2 3 79 | 78 1 678 | 5 4 69 |
| 8 4 79 | 5 2 67 | 3 19 169 |
|-------------------+-------------------+-------------------|
| 5 2 4 | 6 9 1 | 8 7 3 |
| 679 69 8 | 3 4 27 | 19 129 5 |
| 79 1 3 | 27 8 5 | 6 29 4 |
|-------------------+-------------------+-------------------|
| 469 69 5 | 124 7 29 | 1249 3 8 |
| 3 7 2 | 148 5 89 | 149 6 19 |
| 49 8 1 | 24 6 3 | 249 5 7 |
*-----------------------------------------------------------*
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I don't see a single step solution. Can anyone help?  |
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Steve R
Joined: 24 Oct 2005
Posts: 289
Location: Birmingham, England
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| Posted: Fri Apr 16, 2010 1:55 pm Post subject: |
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| Code: | *----------------------------------------------------------------*
| 1 5 6 | 9 3 4 | 7 8 2 |
| 2 3 79 | 78 1 678 | 5 4 69 |
| 8 4 79 | 5 2 67 | 3 19 169 |
|--------------------+-----------------------+-------------------|
| 5 2 4 | 6 9 1 | 8 7 3 |
| 679 69 8 | 3 4 27 | 19 129 5 |
| 79 1 3 | 27 8 5 | 6 29 4 |
|--------------------+-----------------------+-------------------|
| 469 69 5 | 124 7 pin29 | 1249 3 8 |
| 3 7 2 | 148 5 89 | 149 6 19 |
| pin49 8 1 | piv24 6 3 | 249 5 7 |
*----------------------------------------------------------------* |
The pivot and the pincers of the XY-wing which eliminates 9 from r7c12 are shown.
Steve |
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arkietech
Joined: 31 Jul 2008
Posts: 1834
Location: Northwest Arkansas USA
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| Posted: Fri Apr 16, 2010 3:07 pm Post subject: |
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| Steve R wrote: | The pivot and the pincers of the XY-wing which eliminates 9 from r7c12 are shown.
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Thanks Steve -- don't know how I can miss em so easily...  |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Fri Apr 16, 2010 5:19 pm Post subject: |
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| Dan, also note the 69 UR. The DP killers are 7 or 4. The 7 in r5c1 proves 4 in r7c1, killing the 4 in r9c1. |
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arjun
Joined: 28 Sep 2009
Posts: 16
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| Posted: Fri Apr 16, 2010 5:51 pm Post subject: |
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| Marty R. wrote: | | Dan, also note the 69 UR. The DP killers are 7 or 4. The 7 in r5c1 proves 4 in r7c1, killing the 4 in r9c1. |
Doesn't the 69 UR only remove the 9s in r5c1 and r7c1? The puzzle becomes the one below, how does the UR kill the 4 in r9c1?
| Code: | *----------------------------------------------------------------*
| 1 5 6 | 9 3 4 | 7 8 2 |
| 2 3 79 | 78 1 678 | 5 4 69 |
| 8 4 79 | 5 2 67 | 3 19 169 |
|--------------------+-----------------------+-------------------|
| 5 2 4 | 6 9 1 | 8 7 3 |
| 67 69 8 | 3 4 27 | 19 129 5 |
| 79 1 3 | 27 8 5 | 6 29 4 |
|--------------------+-----------------------+-------------------|
| 46 69 5 | 124 7 29 | 1249 3 8 |
| 3 7 2 | 148 5 89 | 149 6 19 |
| 49 8 1 | 24 6 3 | 249 5 7 |
*----------------------------------------------------------------* |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Fri Apr 16, 2010 7:05 pm Post subject: |
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| Quote: | | Doesn't the 69 UR only remove the 9s in r5c1 and r7c1? The puzzle becomes the one below, how does the UR kill the 4 in r9c1? |
Arjun,
The answer is yes when this is played like a conventional Type 4 UR. However, the only ways to kill the Deadly Pattern are a 7 in r5c1 or a 4 in r7c1. If r5c1 =7, by following a chain, it shows that r7c1 must be = 4. So either way, r7c1 must be = 4. |
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arjun
Joined: 28 Sep 2009
Posts: 16
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| Posted: Fri Apr 16, 2010 8:29 pm Post subject: |
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| Marty R. wrote: | | Quote: | | Doesn't the 69 UR only remove the 9s in r5c1 and r7c1? The puzzle becomes the one below, how does the UR kill the 4 in r9c1? |
Arjun,
The answer is yes when this is played like a conventional Type 4 UR. However, the only ways to kill the Deadly Pattern are a 7 in r5c1 or a 4 in r7c1. If r5c1 =7, by following a chain, it shows that r7c1 must be = 4. So either way, r7c1 must be = 4. |
If r5c1=7, r6c1=9, r9c1=4, and r7c1=6. I don't get your point |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Fri Apr 16, 2010 9:32 pm Post subject: |
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| Quote: | | If r5c1=7, r6c1=9, r9c1=4, and r7c1=6. I don't get your point |
This is what I did:
r5c1=7-->r5c6=2-->r7c6=9-->r7c2=6-->r7c1=4.
So the only two ways to kill the DP lead to a 4 in r7c1.
I know your chain works too, and I have asked almost the same type of question that you just did, but it will take someone better at Sudoku theory to explain why mine works. Sorry I couldn't explain in full. |
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Wendy W
Joined: 04 Feb 2008
Posts: 144
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| Posted: Sat Apr 17, 2010 12:22 am Post subject: |
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This one gave me fits -- til I realized that the XY-wing on 249 made TWO eliminations in box 7, not just in column 1. I'm still shaking my head...
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sat Apr 17, 2010 1:22 am Post subject: |
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| Wendy W wrote: | This one gave me fits -- til I realized that the XY-wing on 249 made TWO eliminations in box 7, not just in column 1. I'm still shaking my head...
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Nothing to be embarrassed about, that's how we learn.  |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Sat Apr 17, 2010 2:41 am Post subject: |
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Marty & arjun,
I do not know the "formal/logical/proper" answer to the issue, but I can shed some light on the situation.
Basically, r5c1 can not be equal to 7 as demonstrated by still another chain: (7)r5c1 - (7=2)r5c6 - (2=9)r7c6 - r7c2 = r5c2 - (9=7)r6c1; r5c1<>7
Thus if r5c1=7, then r6c1=7 which is not possible.
The upshot is that the UR69 now becomes a type 1 and r7c1=4 to prevent the deadly pattern.
I was told some time ago without explanation that any path that provided a conclusion, such as the one posted by Marty, is valid. It would be nice to know if that is actually correct and, if so, why.
Ted |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sat Apr 17, 2010 3:49 am Post subject: |
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| Quote: | | I was told some time ago without explanation that any path that provided a conclusion, such as the one posted by Marty, is valid. It would be nice to know if that is actually correct and, if so, why. |
I think it's correct but couldn't articulate why. After all, the basic premise of the simplest forcing chain is that if both parts of an either/or situation yield a common result, then that cell is forced. Obviously, one of those possibilities is invalid, but it works. |
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Mogulmeister
Joined: 03 May 2007
Posts: 1151
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| Posted: Sun Apr 18, 2010 7:32 am Post subject: |
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| Marty R. wrote: | | Quote: | | If r5c1=7, r6c1=9, r9c1=4, and r7c1=6. I don't get your point |
This is what I did:
r5c1=7-->r5c6=2-->r7c6=9-->r7c2=6-->r7c1=4. |
Notice too that Marty's chain is also a simple XY with pincers for 6 on r5c1 and r7c2 |
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