I don't know if this puzzle is a repeat, but for sure, I encounteed the very same quagmire in the very same area of the grid before with a NY Times puzzle. How to take advantage of the highly restricted block 2: we already know that the 2 is restricted to the right flank, upper or lower row. But we also know that in c9, we have a 57 pair, so either a 5 or a 7 in row 2. So, joining the right flank of block 2 will either end up being a 25 or 27 pair in the upper and lower rows. It follows that 1 and 8 will be in row 2. Hence 1 in r3c8 and 8 in r1c8. Et voilà ! Still a bit of work, but this does the job. Very clever puzzle. And, as usual, quite doable notation free with good scanning.
Yeah, I do remember , it was in the same area of the grid , it's all coming back to me , u taught us abt this logic , seems that the candidates were also same but it was very long ago.
Tricky and not easy puzzle -349 in col 9, goes to box 6. -Ghost 7s in box7, col 3, so, 7 in r6c1,Thn 26 frm box5 to row 6. -57 pairs in box3,than 469 in box 2, goes to row 2. Now become little soft. Nice tricky puzzle.
