This strategy works with single numbers. We've already used closed loops of conjugate pairs to find things like X-Wings and Sword-Fish. X-Wings contains 4 cells in a perfect rectangle. Sword-Fish requires 6 or 9 cells in a grid. This strategy works with odd numbers of pairs in a loop starting with 5. There are several varieties depending on how 'perfect' the loop is.
Let us use the words perfect pair instead of conjugate pair to mean any number that exists only twice in one unit (row, column or box). This means we can use imperfect to mean a number that occurs three or more times in a unit. (Obviously if it only occurred once it would solve that cell).
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 rowR5C5 - R7C5 - along the columnR7C5 - R7C9 - along the rowR7C9 - R4C9 - along the columnTo 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.
To repeat, In an actual Sudoku there can never be a closed loop of five perfectly paired cells. And that is exactly where the solving technique lies. Any such structure must have one or more cells that disrupt the perfect pairings. We can refer to the cells which prevent one of the pairings from being perfect as guardians. Here's the trick: logically, one or more of the guardians must contain the candidate number. If none of the guardian cells were real, then the pairings would all be perfect and, as was already noted, that is flat-out impossible in a valid Sudoku. Accordingly, we can make the following assertions:
If there is only one guardian cell, the candidate number can be installed in that cell.
If there is more than one guardian, any cell that is seen by all the guardian cells cannot contain the candidate number; hence
If all the guardian cells are in a single column, row or block of the Broken Wing, the candidate can be erased from both the Broken Wing cells in that column, row or block. Type 1 - Single Guardians The variants of this strategy depend on how many Imperfect connections there are in the loop. To achieve one guardian there must be four perfect pairs and one Imperfect connection. Figure 1 illustrates this. That one guardian is the cell that disrupts the 5-loop from being perfect.
Type 2 - Double Guardians
Let us use the words perfect pair instead of conjugate pair to mean any number that exists only twice in one unit (row, column or box). This means we can use imperfect to mean a number that occurs three or more times in a unit. (Obviously if it only occurred once it would solve that cell).
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 rowR5C5 - R7C5 - along the columnR7C5 - R7C9 - along the rowR7C9 - R4C9 - along the columnTo 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.
To repeat, In an actual Sudoku there can never be a closed loop of five perfectly paired cells. And that is exactly where the solving technique lies. Any such structure must have one or more cells that disrupt the perfect pairings. We can refer to the cells which prevent one of the pairings from being perfect as guardians. Here's the trick: logically, one or more of the guardians must contain the candidate number. If none of the guardian cells were real, then the pairings would all be perfect and, as was already noted, that is flat-out impossible in a valid Sudoku. Accordingly, we can make the following assertions:
If there is only one guardian cell, the candidate number can be installed in that cell.
If there is more than one guardian, any cell that is seen by all the guardian cells cannot contain the candidate number; hence
If all the guardian cells are in a single column, row or block of the Broken Wing, the candidate can be erased from both the Broken Wing cells in that column, row or block. Type 1 - Single Guardians The variants of this strategy depend on how many Imperfect connections there are in the loop. To achieve one guardian there must be four perfect pairs and one Imperfect connection. Figure 1 illustrates this. That one guardian is the cell that disrupts the 5-loop from being perfect.
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 rowR8C3 - R7C2 - within the boxThis 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.
Type 3 - Disruptive Guardians
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 columnR7C4 - R7C7 - along the rowThis 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.
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