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Big Bill · Joined: 30 May 2007 Posts: 5

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

Steve R · Posted: Wed May 30, 2007 3:59 pm Post subject:

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

Marty R. · Posted: Wed May 30, 2007 4:27 pm Post subject: Re: 29 May 2007 VH Help

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

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

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

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

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

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

Earl

Big Bill · Joined: 30 May 2007 Posts: 5

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

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

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.

Angel · Joined: 26 Mar 2006 Posts: 31

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.

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

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).

Marty R. · Posted: Fri Jun 01, 2007 12:02 am Post subject:

RonnyMexico · Joined: 01 Jun 2007 Posts: 2

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.

Marty R. · Posted: Fri Jun 01, 2007 3:51 am Post subject:

RonnyMexico · Joined: 01 Jun 2007 Posts: 2

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.

Big Bill · Joined: 30 May 2007 Posts: 5

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

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

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.

Angel · Joined: 26 Mar 2006 Posts: 31

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