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[cgordon]

That has to be the easiest "very hard" puzzle ever. The XYZ wing step is reached in no time at all - and at this stage, there are few unsolved cells left.

[keith]

There is an unneeded Unique Swordfish, an X-wing, and a BUG. If you do not use the BUG, there is a choice of a couple of XY-wings or an XYZ-wing that will solve it.

Keith

[sdq_pete]

Certainly easy up to the sticking point, which, as you say, was at an advanced stage. That point for me was the following and I found this a bit tricky:

[keith]

In the position posted above, there is an X-wing on <5> in C16.

Note the "Unique Swordfish" on <45> in R289. Then neither of R9C25 can be <45>. They must be <4> because of the strong link on <4> in R9 - then neither of them is <5>. These eliminations are made also by the X-wing.

Keith

[cgordon]

For all my bragging about how easy this was - I seem to have screwed up here. My correct solution has a 7 in R9C1. But surely the 567 in R5CI form an XYZ with the 67 in R6C1 and 57 in R9C6. I'm new to these XYZ wings but shouldn't the above XYZ eliminate the 7 in R9C1. I am so confused.

[sdq_pete]

I suppose the simple way forward is the X-wing mentioned by Keith which somehow fell on my blind spot. Your potential XYZ-wing doesn't work, I fear. Like a number of potential XYZ-wings, the triplet is there but all three cells - the pivot and both pincers - must be able to see a victim and this we don't have. In the case of your proposed XYZ-wing, I actually don't see this at all since the pivot (R1C5) can't even see R9C6. I thought at first you meant the 57 in R5C6. But here too there is in fact no victim that all three cells can see.

So I think it's either one of Keith's routes or my elimination of the 6 in R6C2.

Peter

[George Woods]

Don't know if it counts as an XY wing , but I used the 67 at r7c4 if 7 then r7c3 is 2, and if 6 then r4c5 is 2 so we have an XY wing (in my book) eliminating the 2 at r4c3 and giving a solution.

Incidentally Draw/play gives solution of 5 in Box2 and shortly after a 5 in box5 - seems to me more difficult than my way!....

[Marty R.]

[cgordon]

But surely that's guessing? I still don't see why the 7 isn't eliminated from R9C1 - since this cell "sees" the 7s in R5C1, R6C1 and R9C6. That's my understanding of how XYZ wings work.

[keith]

cg,

No. Here is how XYX-wings work:

There is a root cell, XYZ. It sees two other cells, XZ and YZ. The root cell has three possibilities:

It is Z.
It is X, in which case XZ is Z.
It is Y, in which case YZ is Z.

So, any other cell which sees all three cells in the XYZ-wing cannot be Z.

In this example, R9C1 is the root cell <567>, R9C6 is <57>, R6C1 is <67>. An XYZ-wing, yes. But, it makes no eliminations.

Keith

[Marty R.]

I solved it as follows, maybe differently only because of differences in the sequences we use to look for things.

[cgordon]

I can see that either R4C5 or R8C5 has to be a 5 - which eliminates the 5 from R1C5. But I can't see how the rest proceeds without a guess or assumption - whether on 5s or 6s. I thought these things could be solved without guessing.

[Marty R.]

[Mogulmeister]

cg - if you want to see a graphic example of a skyscraper look at the one that is illustrated in Mesmin's puzzle in this thread

http://www.dailysudoku.com/sudoku/forums/viewtopic.php?t=1860&sid=9cb62644c9357068f48772c4a472f413

The skyscraper is in pink, the eliminated 5's in orange.

Hope that helps.