Sudoku Logic II

This morning's Sudoku puzzle illustrates an interesting sequence of logic that arises in perhaps 1 puzzle in 10.  The partially completed puzzle is shown.  The logic concerns the contents of the squares A1, A3, E1, and E3.

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)?