dailysudoku.com

[ZeroAssoluto]

Hi everyone,

[hughwill]

Unusual one using classic methods as one X wing
on 6 opens another X- wing on 6 before revealing
an XY-wing that sets it.

6 X wing- r26
6 X-wing- r57
45-2 XY wing pivot r4c6 r7c8<>2 stte

Hugh

[the_lock_man]

[Earl]

456xyzwing@R5C6: -5R5C4

Early Earl

[Marty R.]

Hugh, that's very unusual for a puzzle to have three standard steps.

Based on previous posts, I'm unable to see the XYZ-Wing or an X-Wing (6) in rows 3 and 7.

[arkietech]

[(5=4)r4c6-(4=2)r4c8-r7c8=2r7c6]-5r7c6; ste

[immpy]

I don't think there is an XYZ-Wing present on 456. The two supposed pincers are made up solely of bi-value 45. One of these bi-value cells should contain the 6 as a candidate, while remaining bi-value, in order for there to be an XYZ-Wing present.

The X-Wing on 6 alluded to in rows 3 and 7 is in the columns 5 and 8.

[mtharp]

Folks, I am new to this, but it seems like there is an UR on 45 in r45c16 with diagonal candidates (3,6). I would like to solve this without using chains, which seems to me like trial and error.

I know it can be solved with X-wings.

Am hoping for a solution based on the UR.

immpy, your last instruction on URs was particularly helpful.

[Marty R.]

[mtharp]

Marty, thanks.

You wrote:

Well, the 6 in r5c6 proves a 3 in r6c7=> -3r4c7,r6c1

If you would, please help me understand your logic outside of using a trial and error chain.

Again thanks for the response. Sorry to be so dense.

Mike

[immpy]

Hello Mike, and thanks for your nice comment. I am just beginning to grasp some of these variations on the UR myself, so I will do my best to explain with my (limited) knowledge.

Sadly, I don't think this one will fit into any of the 6 UR patterns, based on what I understand at this point.

It is ALMOST a Type 6 UR (45) with the extra (non UR) candidates being diagonal from each other.....BUT....EITHER the 4 OR the 5 would have to be the only possibilities for BOTH of their rows AND columns in order for the Type 6 UR to be present. The 4 and 5 simply occur too often in both of the rows and columns of this rectangle.

But IF it would have been present, a Type 6 UR not only eliminates one of the candidates (in this case either 4 or 5) from both of the cells which contain the extra non-UR candidates, it results in TWO placements, as well. In this case both of the bi-value cells of 45 (which make up the opposite diagonal) would become either the 4 or the 5, based on what we eliminated from the cells with the extra candidates.

That's a lot to think about....but I hope the bulb is becoming illuminated.
Thanks again Mike.

cheers.....immp

[mtharp]

immp, thanks for the type 6 UR lesson. It is starting to make sense. The operative word being "starting".

Is there another way to tackle this UR without using chains?

Mike

[Marty R.]

[immpy]

Sorry Mike, there is no UR present (on 45) to be tackled. Those X-wings you mentioned knowing about (I believe there are two of them on the 6) do just fine on their own. Sometimes a technique we favor is simply not present, no matter how close it may appear to being one.
Interestingly, there is a Type 5 UR (on34) r46c17 with extra candidates occurring in the two cells in r4. As per the Type 5 UR parameters the 3s occur in ONLY the four cells that make up the UR, so the 4s could be eliminated from the two cells with extra candidates in r4 which comprise the UR. But that only results in two eliminations, which lead to nothing else, and one would still need the X-wings on 6 to solve the puzzle. An example of a fruitless UR, one could say. I would ignore this one, anyway, as it seems vague, at best.

cheers...immp

[Marty R.]

[mtharp]

immp & Marty, I really appreciate the time you folks spend bringing us slow folks up to speed.

Thank you both.

Mike

[Marty R.]