IJ wrote:You need row 1, and boxes 8 & 9. It is very neat, but I still doubt anyone could resolve it without the "well if this is 6, then that's a 2, so that's a 7, etc.".
The logic I presented does not start from "well if this is 6, then.." That is T&E. I 'll try to articulate again but may need help if I cant get this across succuinctly.
From the point where IJ and others were stuck, we know Row 1 and Row 9 as well as Column 6 and Column 9 together with boxes 2,3,8 and 9 need the 6 in the right positions.
Now consider the 4 cells r1c6, r1c9, r9c6 and r9c9 forming the infamous X- Wings. Disect the wings to give wing / between r1c9-r9c6 and wing \ between r1c6-r9c9. It is logically clear that the two 6's can only go at the opposite ends of one of the two wings but to make the next logical step it is not necessary to know which wing it is because for either wing, there will be a 6 at the extremes that will place a 6 in the relevant rows, columns and boxes. Note we are not doing any T&E here.
Hence when one homes in on column 9, it is obvious that a 6 will be present
either in r1c9 (and hence r9c6) or in r9c9 (and hence r1c6)only, again we dont care where exactly, thus allowing us to safely and logically conclude even before ascertaining where exactly that 6 is that r7c9 can be 9 only.
I don't agree this is Nishio - this is pure logic.
It is from the 9 in r7c9 that the exact positions of the 6's 2's, 7's and 1's fall in to give that Eureka moment!
I hope that helps.