1).
Looking at line 3. Empty cells are ABDG.
D&G already have 1. -> One of 3A or 3B is 1.资深,这里又有一个我常用的小技巧。
2).
Because 1A & 2A are locked to 6&8, C has a 3 -> 3A or 3B is 3.
3). from 1) & 2) -> 3A & 3B is locked to 1&3. -> Only "free" cells in this square is 1C & 2C.
4) 2I is 4, plus 3) -> 1C is 4, 2C is 2.