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Guardians From sudokuwiki.org, the puzzle solver's site |
| 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 |