dailysudoku.com

[kragzy]

I must be going mad. The first time I solved this puzzle it fell apart in 4.5 minutes and required nothing poor than a couple of hidden pairs. Thinking that I must have jagged the solution in error, I had another attempt and reached this stage:

[prakash]

Kragzy, even with the 47 situation, the puzzle does solve itself quite easily with an XY wing. There must be something here that makes the 47 UR acceptable. One of the reasons why I do not usually use the uniqueness solutions. The next step is the XY wing.

[tlanglet]

A UR is confined to two rows, two columns and two boxes. In this situation the <47> bivalues are in four boxes. No harm, no foul.

I also found a one step xy-wing solution.

Ted

[kragzy]

Ahhhh, thanks folks. I was wrong - I use uniqueness solutions so rarely that I had forgotten about the TWO BOXES ONLY rule. In fact to prove it to myself, I went back and deliberately put the 4s and 7s the other way around and sure enough, the puzzle did not solve.

I was right about one thing though - I am going mad.

[Marty R.]

Alternatively, in kragzy's grid, a W-Wing on the 57s in boxes 25 will finish things off.

[tlanglet]

[daj95376]

[crunched]

I am fully bumfuzzled. I am posting this before reading anybody else's solution. This is as far as I can get. There is no solution I can find...
Damnation for moi!

[sdq_pete]

[Captain Pete]

Or, an XY chain beginning with 47 in C1 ending with 74 in C6.

[daj95376]

[Asellus]

Rather than those roundabout forcing chains (which I don't fancy), I prefer to see that 49 UR as follows.

The induced grouped strong link, (2)r7c23=(5)r1c23, forms a 25 "pseudo-cell" that completes a 258 XY Wing:

UR[(2)r7c23=(5)r1c23] - (5=8)r1c6 - (8=2)r9c6; r9c3<>2

[daj95376]

Yes, my first attempt was excessive. How about ...

[Asellus]

[tlanglet]

[Asellus]

Ted,

Those are good questions... let's see if I can clarify it for you. A bivalue "pseudocell" is just another name for a strong (inference) link between two dissimilar digits. The "psuedocell" name is certainly not necessary and I normally don't use it, which is why I put it in quotes. (I just picked up on it since it was raised recently in other posts.)

Any strong link, to be useful, needs a link from something on one end and a link to something on the other end (and usually both of those links function as weak inference links). In this particular case, the <2>s on one end of the strong link are grouped (in r7 and b7) as are the <5>s on the other end (in r1 and b1).

Using xyz for the XY Wing, the form is (z=x) - (x=y) - (y=z), where "=" denotes the strong links within the 3 bivalue cells and "-" denotes the weak links on the matching digits between the cells. A "pseudocell," however constructed, can replace any (or all) of these bivalues. What this notation does not show are the weak links to the victims between the two z digits (the "pincers"). So, to make it complete, we can write it as a complete loop:
victim<z>s - (z=x) - (x=y) - (y=z) - victim<z>s

For the pseudocell to work ... here, let's say it's in the place of the (z=x) cell ... the weak links on both sides need to be valid. Taking the "x" part, which is <5> in this case, the grouped <5>s can form a weak link to a <5> (or other grouped <5>s, even) provided it shares box 1 or row 1. In this case, the pivot <5> is in row 1; so all is fine.

The "z" part is the <2>s. Note that there are actually two victims: r9c3 and r7c5. The grouped <2>s of the pseudocell can form weak links to a <2> provided that it shares box 7 or row 7. r9c3 shares box 7 and r7c5 shares row 7. Both of these victims also form weak links with the other pincer in r9c6, the "(y=z)" cell, as required (i.e., they are mutual peers of the pincer <2>s, one of which is single and one of which is grouped).

I hope this detailed explanation clarifies things for you. There is no valid elimination of <5> from r7c5 for two reasons: (1) the XY Wing-type structure here has <2>s as pincers, not <5>s; and (2) the <5> in r7c5 cannot "see" the pseudocell grouped <5>s in any case.