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[immpy]

Hello all, enjoy the puzzle...

[Mogulmeister]

After basics:

[TomC]

I think I was on the same lines as you Mogulmeister, but saw it differently

If r8c4=8 then r7c2=7 leaves no space for a <7> in box 8

[TomC]

There is a <68> in r1 and c1, so r1c1<>6 as then no <8> in box 8

[Mogulmeister]

There is a pleasing symmetry about that pattern with a lovely simplicity to it. As you say Tom, a 6 in r1c1 triggers the 8 in both <68> outliers and box 8 is toast for 8s.

As always, you can plug 6 into r1c1 and follow a loop round and it removes itself from r1c1.

+6r1c1-(6=8)r9c1-(8)r9c46=(r8c4)-(8=6)r1c4 so r1c1 < > 6

[TomC]

I agree, this is a very good looking contradiction

It stems from the pattern of 8's in box 8

First, just give the two <68> pairs in r1 and c1 names, say a and b (not really needed but makes my point cleaner)

Look at box 8 and the position of the 8's in that box. They make an "L" shape which shows that wherever you put an 8 it will remove an 8 in a or b. But crucially, it also shows that a and b cannot both be 8's as this gives the contradiction.

With this information, the <568> in r1c1 pops up to force the contradiction and solve the puzzle.

[Mogulmeister]

Nice.