by ravel » Fri Jul 07, 2006 11:54 am
In block 9 the 2 must be in c7, therefore r3c7<>2.
In columns 3 and 6 the 3 can only be in rows 4 and 6 (x-wing). Therefore r6c7<>3.
This leaves a naked pair 58 in r36c7.
Then r9c9 is the only place for an 8 in block 9.
Without x-wing:
In block 7 3 must be in column 1, therefore r5c1<>3.
Then in row 5 3 must be in block 6, therefore not in r6c7 (same block).