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April 16 VH

 
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Earl



Joined: 30 May 2007
Posts: 625
Location: Victoria, KS

PostPosted: Fri Apr 16, 2010 1:52 am    Post subject: April 16 VH Reply with quote

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

PostPosted: Fri Apr 16, 2010 3:43 am    Post subject: Reply with quote

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

PostPosted: Fri Apr 16, 2010 10:11 am    Post subject: Reply with quote

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

PostPosted: Fri Apr 16, 2010 10:29 am    Post subject: Reply with quote

One step
r6c1 can't be a 9 as it would make both r5c6 and r7c6 a 2.

Don't know much about terminology Smile
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arjun



Joined: 28 Sep 2009
Posts: 16

PostPosted: Fri Apr 16, 2010 10:33 am    Post subject: Reply with quote

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

PostPosted: Fri Apr 16, 2010 12:29 pm    Post subject: Reply with quote

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

PostPosted: Fri Apr 16, 2010 1:30 pm    Post subject: Reply with quote

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     |
 *-----------------------------------------------------------*
I don't see a single step solution. Can anyone help? Confused
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Steve R



Joined: 24 Oct 2005
Posts: 289
Location: Birmingham, England

PostPosted: Fri Apr 16, 2010 1:55 pm    Post subject: Reply with quote

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

PostPosted: Fri Apr 16, 2010 3:07 pm    Post subject: Reply with quote

Steve R wrote:
The pivot and the pincers of the XY-wing which eliminates 9 from r7c12 are shown.
Thanks Steve -- don't know how I can miss em so easily... Sad
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Marty R.



Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA

PostPosted: Fri Apr 16, 2010 5:19 pm    Post subject: Reply with quote

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

PostPosted: Fri Apr 16, 2010 5:51 pm    Post subject: Reply with quote

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

PostPosted: Fri Apr 16, 2010 7:05 pm    Post subject: Reply with quote

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

PostPosted: Fri Apr 16, 2010 8:29 pm    Post subject: Reply with quote

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
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Location: Rochester, NY, USA

PostPosted: Fri Apr 16, 2010 9:32 pm    Post subject: Reply with quote

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

PostPosted: Sat Apr 17, 2010 12:22 am    Post subject: Reply with quote

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...
Embarassed
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Marty R.



Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA

PostPosted: Sat Apr 17, 2010 1:22 am    Post subject: Reply with quote

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...
Embarassed

Nothing to be embarrassed about, that's how we learn. Very Happy
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tlanglet



Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills

PostPosted: Sat Apr 17, 2010 2:41 am    Post subject: Reply with quote

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

PostPosted: Sat Apr 17, 2010 3:49 am    Post subject: Reply with quote

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: 703

PostPosted: Sun Apr 18, 2010 7:32 am    Post subject: Reply with quote

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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