What technique can be used to solve this 6x6 sudoku?

Question

Which Sudoku technique can be used to solve this? The screenshot is from the open source Sudoku app by SECUSO (Security Usability Society) for the Android platform.

Answer 1

You can do sudoku to

whatever is in row 4, column 5.

That will eventually place the same digit in a spot where a three cannot go:

After that deduction, the grid will basically fill itself.

The technique that finds relationships like this is called "colouring the pair". When you have a lot of squares around the grid with the same two options only, it's a handy way to give an identity to the two options of the pair. Even if it hadn't conveniently placed a 1-or-3 in a spot that already sees a three, continuing to colour in the "squares that can only be 1 or 3, but cannot be green" in a different colour would have shown that the square in the bottom left corner (r6c1) sees both colours (both numbers of the 1-3 pair) and must therefore be a 2.

Answer 2

Using chess notation (rows 1 to 6 bottom to top and columns a to f, left to right)

.---------------------------------.
| 145  6    14   | 135  2    134  | 6
| 1245 3    124  | 15   6    14   | 5
|----------------+----------------|
|*13   5    6    | 2    4   *13   | 4
| 1234 124  1234 | 6   *13   5    | 3
|----------------+----------------|
| 6    14   134  | 13   5    2    | 2
| 2-13 12   5    | 4   *13   6    | 1
'---------------------------------'
  a    b    c      d    e    f

Double Kite (1 and 3) [* marked cells] => 1 and 3 are false at a1.

due to the linked cells e1,e3,f4,a4 we have a chain of implications: if 1 is false at e1, it is true at e3, false at f4 and true at a4, so either 1 is true at a4 or 1 is true at e1, and (same idea) either 3 is true at a4 a4 or 3 is true at e1.

Another way:

Remote Pair (13) [* marked cells] => 1 and 3 are false at a1. We have either a4=1,e1=3 or a4=3,e1=1 because of the sequence 1-3-1-3 or 3-1-3-1 at the marked cells.

With a1=2 the puzzle is easily solved.

Answer 3

If the upper right hand square is a 3, you can follow around the 1,3 squares until you force the bottom left squares to be a 1,2 pair which forces row 5, column 4 to be a 1. This, of course forces a second 3 to be in the upper right box, which is not allowed. The upper right square becomes a 1,4. This immediately answers the entire 1,3 system, the upper right box and the makes the second row, third column a 2 (by uniqueness).