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29 May 2007 VH Help

 
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Big Bill



Joined: 30 May 2007
Posts: 5

PostPosted: Wed May 30, 2007 3:38 pm    Post subject: 29 May 2007 VH Help Reply with quote

I am stuck. Did anyone get a solution for the 29 May VH.

This is where I am and the hint suggests a 2 in C1, C5.

[code]
+------------+--------------+------------+
| 4 6 78 | 27 28 3 | 9 15 15 |
| 3 9 2 | 1 5 78 | 78 6 4 |
| 5 78 1 | 6 4 9 | 378 2 37 |
+------------+--------------+------------+
| 9 4 378 | 5 38 678 | 2 13 16 |
| 27 1 6 | 2379 239 4 | 5 8 37 |
| 8 237 5 | 237 1 67 | 367 4 9 |
+------------+--------------+------------+
| 6 38 389 | 4 7 5 | 1 39 2 |
| 1 5 39 | 39 6 2 | 4 7 8 |
| 27 27 4 | 8 39 1 | 36 359 56 |
+------------+--------------+------------+


Thanks Bill
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Steve R



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

PostPosted: Wed May 30, 2007 3:59 pm    Post subject: Reply with quote

There is an xy-wing pivoted on r4c3 with candidates (37). The pincers are r4c5, candidates(38), and r1c3, candidates(78).

Whether the pivot contains 3 or 7, one of the pincers must contain 8. 8 can therefore be eliminated from r1c5.

Steve
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Marty R.



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

PostPosted: Wed May 30, 2007 4:27 pm    Post subject: Re: 29 May 2007 VH Help Reply with quote

Big Bill wrote:
I am stuck. Did anyone get a solution for the 29 May VH.

This is where I am and the hint suggests a 2 in C1, C5.

Code:

+------------+--------------+------------+
| 4  6   78  | 27   28  3   | 9   15  15 |
| 3  9   2   | 1    5   78  | 78  6   4  |
| 5  78  1   | 6    4   9   | 378 2   37 |
+------------+--------------+------------+
| 9  4   378 | 5    38  678 | 2   13  16 |
| 27 1   6   | 2379 239 4   | 5   8   37 |
| 8  237 5   | 237  1   67  | 367 4   9  |
+------------+--------------+------------+
| 6  38  389 | 4    7   5   | 1   39  2  |
| 1  5   39  | 39   6   2   | 4   7   8  |
| 27 27  4   | 8    39  1   | 36  359 56 |
+------------+--------------+------------+



Thanks Bill


I don't know if I reached this same position, but I never spotted an XY-Wing. Alternatively, strong links (skyscraper) on 7 in rows 1 and 4 eliminate the 7 from r2c6.
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Earl



Joined: 30 May 2007
Posts: 677
Location: Victoria, KS

PostPosted: Wed May 30, 2007 4:48 pm    Post subject: May 29 puzzle Reply with quote

I am stuck before Bod and the x-y wing. which I can see if I can get tht far.

I dont know how to get rid of the 2 in R1C8, and/or the 25 in R2C8.
Why must R2C5 be a 5?

Thanks
Earl
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Earl



Joined: 30 May 2007
Posts: 677
Location: Victoria, KS

PostPosted: Wed May 30, 2007 4:55 pm    Post subject: May 29 puzzle Reply with quote

Bilol,
The 8 in R6C1 elminates your 8 in R4C3, which sets up an x-y wing R1C3, R3C3, R3C5, eliminating the 8 in RIC5. But I dont know how you got R2C5 to be a 5.

Thanks Earl
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Earl



Joined: 30 May 2007
Posts: 677
Location: Victoria, KS

PostPosted: Wed May 30, 2007 5:29 pm    Post subject: May 29 puzzle Reply with quote

Igmore my previous posts about May 29 puzzle.
I found my studpid error. Senior moment.

Earl
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Big Bill



Joined: 30 May 2007
Posts: 5

PostPosted: Thu May 31, 2007 5:43 pm    Post subject: Reply with quote

Steve and Marty,
Those are two techniques that I have not seen before. I am familiar with the XY and XYZ wings but I have not seen a pivoted XY wing. I can follow your explaination but do you have a website that explains it.

I have never seen a strong link (skyscraper) and could not even follow the explaination. Marty, do you know of a website or could you give me an more detailed explaination.

Thanks,

Bill
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cgordon



Joined: 04 May 2007
Posts: 769
Location: ontario, canada

PostPosted: Thu May 31, 2007 6:55 pm    Post subject: Reply with quote

I found this one very challenging. Contrary to above solutions I figure there is no 5 in R1C8 - since it is eliminated by a skyscraper on 5s in C4 and C9. In fact my col 8 ended up like this:

+-----+-----+--------+
| - - - | - - - | - 12 - |
| - - - | - - - | - - - |
| - - - | - - - | - 23 - |
+-----+-----+--------+
| - - - | - - - | - 13 - |
| - - - | - - - | - - - |
| - - - | - - - | - - - |
+-----+-----+--------+
| - - - | - - - | - 39 - |
| - - - | - - - | - - - |
| - - - | - - - | - - - |
+-----+-----+--------+

.... so I figured my xy was the 12: my xz the 23 and yz the 13. Thus the 3 is eliminated from the 39. It worked - so I'd be upset if I got the right answer for the wrong reasons.
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Angel



Joined: 26 Mar 2006
Posts: 31

PostPosted: Thu May 31, 2007 7:34 pm    Post subject: Reply with quote

Given that you had the situation you posted, I would say you got the right answer - but surely this is simply a naked triple. I don't think you need to try and find an xy-wing here.

BTW, I haven't actually done the original puzzle yet - so apologies if jumping in here at your end move is inappropiate.
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cgordon



Joined: 04 May 2007
Posts: 769
Location: ontario, canada

PostPosted: Thu May 31, 2007 9:47 pm    Post subject: Reply with quote

I'm not sure what a naked triple is - at least, not as far as Sudoku is concerned. However, I believe there were other 2s, 3s and 9s, in the column - I only showed the pairs - so there was no straightfoward elimination (I hope).
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Marty R.



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

PostPosted: Fri Jun 01, 2007 12:02 am    Post subject: Reply with quote

Quote:
I have never seen a strong link (skyscraper) and could not even follow the explaination. Marty, do you know of a website or could you give me an more detailed explaination.


Bill, I'll try.

Code:
+------------+--------------+------------+
| 4  6   78  | 27   28  3   | 9   15  15 |
| 3  9   2   | 1    5   78  | 78  6   4  |
| 5  78  1   | 6    4   9   | 378 2   37 |
+------------+--------------+------------+
| 9  4   378 | 5    38  678 | 2   13  16 |
| 27 1   6   | 2379 239 4   | 5   8   37 |
| 8  237 5   | 237  1   67  | 367 4   9  |
+------------+--------------+------------+
| 6  38  389 | 4    7   5   | 1   39  2  |
| 1  5   39  | 39   6   2   | 4   7   8  |
| 27 27  4   | 8    39  1   | 36  359 56 |
+------------+--------------+------------+


Note the 7s in rows 1 and 4. Each row has only two possibilities and one possibility in each row shares a column, column 3. Certainly both rows can't have their 7 in column 3, therefore, one or both of the other possibilities (r1c4 and r4c6) must be true. Thus, any cells that see both r1c4 and r4c6 may not contain a 7. Strong links covers more territory; this is the "skyscraper" form of strong links, the most common form that most of us see.

I hope this is clear enough.
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RonnyMexico



Joined: 01 Jun 2007
Posts: 2

PostPosted: Fri Jun 01, 2007 1:00 am    Post subject: Reply with quote

I completed this one via the XY Wing, but I was not familiar with this skyscraper method until now. I think I get it, but I have two questions for Marty R., just to make sure I understand correctly.

First, you say "certainly both rows can't have their 7 in column 3, therefore, one or both of the other possibilities (r1c4 and r4c6) must be true." Isn't it required that only one of those possibilities were true? If both were true, column 3 would be devoid of 7s.

Second, in your earlier post you said the skyscraper eliminated the 7 from r2c6. Doesn't it also eliminate the 7s in r5c4 and r6c4, as those two also see both r1c4 and r4c6? Just making sure, thanks.
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Marty R.



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

PostPosted: Fri Jun 01, 2007 3:51 am    Post subject: Reply with quote

RonnyMexico wrote:
I completed this one via the XY Wing, but I was not familiar with this skyscraper method until now. I think I get it, but I have two questions for Marty R., just to make sure I understand correctly.

First, you say "certainly both rows can't have their 7 in column 3, therefore, one or both of the other possibilities (r1c4 and r4c6) must be true." Isn't it required that only one of those possibilities were true? If both were true, column 3 would be devoid of 7s.

Second, in your earlier post you said the skyscraper eliminated the 7 from r2c6. Doesn't it also eliminate the 7s in r5c4 and r6c4, as those two also see both r1c4 and r4c6? Just making sure, thanks.


Ronny,

You are correct on both questions. In the first case, I just wasn't paying attention to the number of 7s in the column.

In the second case, I only mentioned the elimination from r2c6 because the solving of that cell is what opens the puzzle up, but it probably would have been more correct to mention all the cells involved.

And welcome to the forum.
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RonnyMexico



Joined: 01 Jun 2007
Posts: 2

PostPosted: Fri Jun 01, 2007 4:16 am    Post subject: Reply with quote

Thanks for the response and welcome. I was pretty sure that was the case, I just wanted to be certain that I understood this method. Thanks again.
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Big Bill



Joined: 30 May 2007
Posts: 5

PostPosted: Fri Jun 01, 2007 7:09 am    Post subject: Reply with quote

Marty,
Thanks for the excellent explanation. I see exactly what you are talking about and the logic behind it.
And thanks to Ronny for the additional clarification.

Now another new technique to master.

Thanks to all.

Bill
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cgordon



Joined: 04 May 2007
Posts: 769
Location: ontario, canada

PostPosted: Fri Jun 01, 2007 5:27 pm    Post subject: Reply with quote

Your right Angel. Ignore my earlier dumb post. If there's a 12, a 13 and a 23 in a column - there obviously ain't no 1, 2 or 3 elsewhere in that column.
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Angel



Joined: 26 Mar 2006
Posts: 31

PostPosted: Fri Jun 01, 2007 6:04 pm    Post subject: Reply with quote

I'm glad we agree! I was a bit worried that I'd misunderstood what you were saying and had misled you.
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