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|In Figure 1 we have highlighted the number 3. Amongst all the candidate threes is a loop of five 3's. They form four perfect pairs:
R5C7 - R5C5 - along the row
R5C5 - R7C5 - along the column
R7C5 - R7C9 - along the row
R7C9 - R4C9 - along the column
To close the loop we have an imperfect triplet in the sixth box.
The question is: can a closed loop of five candidate cells be constructed with each cell perfectly-paired in two ways with the next linking cells in the loop? The answer is no. Such a formation is impossible in a Sudoku puzzle. In such a loop, if you "placed" a candidate in any one of the cells and followed the consequences around the loop, you'd generate an automatic contradiction - forcing the number to disappear entirely from a row, cell or block, or to appear twice in a single line or block, depending on how you proceed.
(Note: You have to turn off Simple Colouring, Remote Pairs, XY-Chain, BUG and Forcing Chains).
Guardian 1: Load Example or : From the Start
|Type 2 - Double Guardians
In Figure 2 we have highlighted the number 7. Amongst all the candidate 7's is a loop of five 7's. There are two imperfect connections in the loop:
R8C3 - R8C9 - along the row
R8C3 - R7C2 - within the box
This gives us two guardian 7's in R7C1 and R8C7 marked in red squares. Whatever cells these two can both 'see' we can eliminate the 7 from them. Since in this example they form the opposite corners of a rectangle we can safely remove the 7 from R7C7 marked in a red circle. The other corner, R8C1,
contains a solved square.
Solving R7C7 allows us to complete the puzzle using other strategies.
Guardian 2: Load Example or : From the Start
|Type 3 - Disruptive Guardians
In Figure 3 we have highlighted the number 1. Amongst all the candidate 1's is a loop of five 1's. There are two imperfect connections in the loop:
R2C4 - R7C4 - along the column
R7C4 - R7C7 - along the row
This gives us two guardian 1's in R3C4 and R7C6 marked in red squares.
Whatever cells these two can both 'see' we can eliminate the 1 from them. Like in the example above, they form the opposite corners of a rectangle but the difference is that we're eliminating a 1 that's actually part of the loop. This is perfectly legitimate and follows from Rule 3 described above. The elimination occurs because R7C4 can be seen by both guardians.
Guardian 3: Load Example or : From the Start