dailysudoku.com

nataraj · Posted: Fri May 23, 2008 7:49 am Post subject: May 23 vh

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.

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

Very clever idea with the hint in white!

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

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

nataraj · Posted: Fri May 23, 2008 9:59 am Post subject:

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

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.

Marty R. · Posted: Fri May 23, 2008 3:52 pm Post subject:

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

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

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.

Marty R. · Posted: Fri May 23, 2008 5:02 pm Post subject:

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

Asellus · Posted: Fri May 23, 2008 10:07 pm Post subject:

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.

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

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.