By looking at the 3's and 7's in columns D and F, it is easy to see that the squares E1 and E3 must contain 3 and 7. Row 3 is also missing 3 and 7, so that A3 must also contain 3 or 7.
Now think of E1, E3, and A3 as forming three corners of the rectangle A1-E1-E3-A3. Whenever you have four squares in this arrangement, and three allow only the same two numbers (3 and 7 in this case), then the requirement that the puzzle must have a unique solution means that the 4th corner (A1) cannot be either of the two numbers. For, if it did, then both (A1,E1,E3,A3)=(3,7,3,7) and (A1,E1,E3,A3)=(7,3,7,3) would solve the puzzle. Caution: this rule only works if the 4 corners are contained within only 2 3X3 blocks of the puzzle. If all 4 corners are in different 3X3 blocks, the logic is not correct.
So, A1 cannot be either 3 or 7. Now look more closely at the A1-C3 block. Neither B1 nor C2 can be 3 either. Therefore the only place to place a 3 in the A1-C3 block is at A3. It follows that E3=7, E1=3 as well.
Later in the same puzzle, a similar situation came up again (very unusual):
H1, H9, and I9 only allow 7 and 8. What number goes in the fourth corner of the rectangle (I1)?
