dailysudoku.com

[KSipher]

I am also stuck on this puzzle. I would appreciate a few pointers from this point...

Thanks for your help
Kathy

[TKiel]

Jeff,

To solve the puzzle, look in Boxes 3 & 6 and re-read sdq_pete's post above, specifically the last line. No W-wing needed, but I've tried to explain it below.

[cgordon]

[sdq_pete]

[TKiel]

[eddieg]

And might I add please to my request. Note that I am a few days late in getting to this puzzle (darn work and family getting in way of hobbies).


"Cells r3c8 & r9c7 (marked W) each have the same two candidates . Cells marked A & a are strong links on <3>. One of A or a must be <3>. If A is <3>, then r9c7 is <1>. If a is <3>, then r3c8 is <1>. In either case, r3c7 can't be <1>. That's a W-wing. "

Okay, I follow the explanation somewhat, but when I put this into practice, my mind does not say that cells marked A and a are strong links on <3>, thus ...., etc. I quess I am looking for a deeper explanation of W-Wings are this strong link relationship. Can you expand upon this are point me somewhere else.

And, if I make the cognizant deduction that either A or a has to be a <3>, which allows removing the <3> from R4C9, which opens up the XY Wing with a pivot at R1C9, this eliminates the <1> from R4C7 and R6C7 which quickly solved the puzzle for me. I get everything, just not the strong link on 3 statement.

TIA.

[Marty R.]

[nataraj]

[eddieg]

ding, ding, ding - That's the sound of the bell in my head going off in case you're not familiar with the reference.

Now I see that a '3' must be in the top row of Box 4 that eliminates the '3' from R4C9, and the rest falls into place.

As the character in Napolean Dynamite often says in his ackward style, "Idiot", but this time it is me referring to myself. Or I could have referred to an old Saturday Night Live character, who after getting his facts wrong and making a big stink over nothing and having his errors pointed out to him and his stating, simply states "Never Mind".

[Jeff]

Many thanks for all the great posts. Somehow I missed the XY wing centered on R1C9.

[Asellus]

I want to stress one point regarding the W-Wing discussion above:

Using nataraj's model with the two remote {a,b} bivalue cells that can each see the ends of a strong link on "b", we know only that one of the bivalue cells is not "b" and thus must be "a". We don't know the status of the other bivalue cell: it can still be either "a" or "b".

In other words, at least one of the bivalue cells is "a". It's possible that they both are. The elimination of "a" from the buddy cells of the two bivalue cells is, of course, just as justified.

The reason to point this out is so that logical errors aren't made by those who try to carry the concept further.

[KSipher]

You all are the greatest. I finally solved this puzzle, altho I have to go back and reread all the answers so I can understand W wings...

Thanks!
Kathy

[mahesh]

Hi,

My first post here. an interesting forum. I got stuck in exactly the spot that Kathy was at.

"6" can be both in R6C2 or R5C3. I don't see how you guys (including the solver-hint) came up with R5C3).

Can someone explain in simple terms.

Thanks,

Mahesh.

[TKiel]

Mahesh,

Are you familiar with an XY-wing? Three cells with a candidate pattern of XY-YZ-ZX. The YZ cell is often called the pivot. That cell must share a group (row/box/column) with each of the other cells, which are often called pincers. Any cell/cells that share a group with both of the pincers, can have the X candidate excluded. Why? If the pincers aren't the X candidate, one must be Z and one must be Y, which leaves no candidates for the pivot cell.

There's one with the pivot in r9c3 (3,9), pincers in r5c3 (6,9) & r7c2 (3,6). The candidate pattern would be (6,3)(3,9)(9,6). Any cell that 'sees' (shares a group) with both r5c3 & r7c2 can have the 6 excluded. In this case there are 6 cells that 'see' both of those cells, two of which contain (6) and one of which is r6c2. I'll leave it to you to find the other.

[mahesh]

Tracy,

Awesome. Thank you so much for the kind explanation. I do see this principle clearly now.

Howeever, there's still a problem in the puzzle below. I suppose, the pattern is more complicated than the XY wing.

R6C2: (6,3)
R9C3: (3,9)
R5C3: (9,6,3)

Now, why do you not consider the 3 in R5C3.

Thanks,

Mahesh.

[TKiel]

Mahesh,

Must be a mis-communication somewhere.

The image of Kathy's grid you included in your post does not show a <3> in r5c3, it shows only <69> and r6c2 contains <69>, not <63>.
Are we still talking about the same puzzle?

[mahesh]

Tracy,

Here is my problem. I decided to post here using Kathy's grid because that is point where I got stuck. At that stage, the solver filled in a 6 in R5C3. I could'nt really figure out why it was doing so.

Your explanation is nice. But is really dependent on eliminating 2&3 from the candidates for R5C3. And the solver/Kathy somehow eliminate those.
Don't know why.

I wish there were a simpler method to solving these without the extraordinary book-keeping.

[nataraj]

[TKiel]

nataraj

I think mahesh was remarking about how hard keeping track of the candidate profiles can be (when using the DS program the candidates must be manually excluded from each affected cell when a cell is solved), not how hard it is to solve the puzzle.

mahesh,

That clears some things up. Here is the candidate grid after placing all the singles from the start of the puzzle.

[mahesh]

And the diabolical ones, it is hoped, shall burn in hell..

Although not the inferal type, I do find "hard" questions and those at lower levels relatively easy. I have never used paper to solve any of those.

However, the VH, I cannot seem to make progress. It seems the only way out is to write out everything and go through the techniques outlined by Kathy and others in the forum.

Anyway, back to more earthly matters....

cheers.

-m.

[quote="nataraj"]