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May 23 vh

 
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nataraj



Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria

PostPosted: Fri May 23, 2008 7:49 am    Post subject: May 23 vh Reply with quote

After basics, an xyz-wing does it. Hint: pivot r9c4: 678, xyz-wing is in col 4 and row 9, it makes r7c4=3 and solves the puzzle

There is also an x-wing (on 8, cols 1 and 6), but it does not solve the puzzle.
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gohast



Joined: 26 Sep 2006
Posts: 18
Location: Dublin, Ireland

PostPosted: Fri May 23, 2008 8:21 am    Post subject: Reply with quote

Very clever idea with the hint in white!
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andras



Joined: 31 Oct 2007
Posts: 56
Location: Mid Wales

PostPosted: Fri May 23, 2008 8:28 am    Post subject: Reply with quote

Indeed, quite a straightforward puzzle. At first I thought that the x-y-z wasn't going to take me very far, but a little more basic clearance solved the thing in no time at all. Smile

John
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nataraj



Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria

PostPosted: Fri May 23, 2008 9:59 am    Post subject: Reply with quote

gohast wrote:
Very clever idea with the hint in white!


Thanks! Very Happy Very Happy
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cgordon



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

PostPosted: Fri May 23, 2008 12:10 pm    Post subject: Reply with quote

Found the same x wing and xyz wing as Nataraj. A good one - the basics didn't seem to go very far. I suspect there are other wings and things - but couldn't see any.
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Marty R.



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

PostPosted: Fri May 23, 2008 3:52 pm    Post subject: Reply with quote

I guess there wasn't the usual variety of techniques here; for me too it was a one-stepper with the XYZ-Wing.
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cgordon



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

PostPosted: Fri May 23, 2008 4:32 pm    Post subject: Reply with quote

It’s too cold for golf today – this is Canada – so I had time to revisit this one.

I looked at the 3 cells in R5 Box5 which contain 5 numbers 25679.

There can’t be a 7 and a 9 because this would leave two naked 5’s in C6.
There can’t be a 6 and a 9 because this would leave a void in R6C4.
There can’t be a 5 and a 6 because we’d see a 7 – 79 – 79 in R5.
There can’t be a 6 and a 7 because we’d see a 5 – 9– 29 – 259 - 59 sequence in R5.
There can’t be a 5 and a 7 because we’d see a 6 – 69 – 269 – 269 - 9 sequence in R5.

So the only possibility left is that the three cells contain a <259>.

Was this a logical procedure or just the application of trial and error?

EDITED NOTE
I erred when I said "There can’t be a 7 and a 9 because this would leave two naked 5’s in C6". But what I could have said is "There can’t be a 7 and a 9 because we’d see a 56 – 6– 26 – 256 - 5 sequence in R5 -
which still leads to the same conclusion.


Last edited by cgordon on Fri May 23, 2008 6:39 pm; edited 3 times in total
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Marty R.



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

PostPosted: Fri May 23, 2008 5:02 pm    Post subject: Reply with quote

Quote:
Was this a logical procedure or just the application of trial and error?


I don't know, but I'm posting a grid for others to ponder.

Code:

+--------------+-----------------+-------------+
| 678 2   1678 | 5    1379  1789 | 79  4  139  |
| 57  4   157  | 2379 12379 6    | 579 8  1359 |
| 9   578 3    | 78   17    4    | 6   2  15   |
+--------------+-----------------+-------------+
| 2   57  4    | 1    8     579  | 3   59 6    |
| 567 3   5679 | 2679 25679 579  | 8   1  4    |
| 1   58  5689 | 69   4     3    | 2   59 7    |
+--------------+-----------------+-------------+
| 357 9   2    | 37   1357  157  | 4   6  8    |
| 358 6   58   | 4    359   589  | 1   7  2    |
| 4   1   78   | 678  67    2    | 59  3  59   |
+--------------+-----------------+-------------+

Play this puzzle online at the Daily Sudoku site
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sdq_pete



Joined: 30 Apr 2007
Posts: 119
Location: Rotterdam, NL

PostPosted: Fri May 23, 2008 7:34 pm    Post subject: Reply with quote

cgordon wrote:
I looked at the 3 cells in R5 Box5 which contain 5 numbers 25679.

The argument would be a bit clearer if you started by saying that the 2 must be included, so that we are talking about 4 candidates (5679) in 2 cells. That gives 6 possibilities and you have excluded 5 of them.

Trial and error? Logically, yes, I suppose. But a neat application of it!

Peter
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Asellus



Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA

PostPosted: Fri May 23, 2008 10:07 pm    Post subject: Reply with quote

Craig,

I can't recreate your arguments. The only combination that is obviously impossible that I can see is {269} because of the r6c4 void. All of the other combinations you've excluded aren't prohibited for the reasons you cite. I had to solve the puzzle almost totally for each of the others before I could find the contradictions:

{256} creates 2 <3>s in c1 and a void at r9c4
{257} creates voids at r1c6 & r3c9 and leaves r1c9 indeterminate
{267} creates voids at r1c6 & r3c4
{279} creates voids at r1c3 & r7c1, leaves r7c1 indeterminate, and creates a UR at r29c79

All of them solve r5 just fine.

In this case, I say definitely trial and error.
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cgordon



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

PostPosted: Sat May 24, 2008 1:39 am    Post subject: Reply with quote

Asellus: Thanks – I hoped you would reply. When I said the three cells in R5 Box5 contained specific numbers – for example 2, 6 and 7 – I also (like an idiot) removed the 6 and 7 from those same three cells. And I did the same with all the other options. So embarrassing!! Many apologies to all!!
Though it did work – with odds of 1 in 3 no less.
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