Mirrored from **Sudopedia**, the Free Sudoku Reference Guide

In Sudoku Variants with additional diagonals like Sudoku-X and Argyle Sudoku, there are some techniques that are very similar to Crossover and Double Crossover.

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The pink cells are the only cells with the candidate 1 in the 2 diagonals of this Sudoku-X. Because the candidate 1 is not present in the common cell **r5c5**, the pink cells must contain exactly 2 occurrences of the digit 1. Due to the fact that the green cell at **r2c7** sees (i.e., is weakly linked to) the cells **r2c2**, **r2c8** and **r7c7**, a 1 at this position would constrain the 2 occurrences of the digit 1 in the diagonals to **r4c46**, which is impossible, since they are peers of each other.

Similarly, a 1 in the green cell at **r4c7** would constrain the 2 occurrences of the digit 1 in the diagonals to **r2c28**, which is again impossible.

Therefore, with respect to the triple crossover we can safely eliminate the candidate 1 from the green cells.

*Note:* This scenario is referred to as a *triple* crossover because each of the green cells sees 3 pink cells.

There is also another variant of this technique:

The logic here is analogous to that described in the first example. The pink cells represent the possible candidate positions for digit 1 in row 6 and row 8. Since these two groups are disjoint (i.e., do not overlap), they must contain 2 occurrences of the digit 1. If **r3c7** or **r5c5** would contain a 1, this would force the two 1s in these two rows into **r68c5** and **r68c7**, respectively.

Therefore, with respect to the triple crossover we can safely eliminate the candidate 1 from the green cells.

A similar logic still works here, where the green cell sees 4 pink cells (hence the term "*quadruple* crossover"):

Or in its diagonal variant:

The 2 houses marked in yellow (row 2 and D/), taken together, must contain 2 occurrences of the digit 1, since a 1 cannot exist within the overlapping region (**r2c8**). If the green cell at **r9c9** were to contain a 1, it would constrain both occurrences to **r23c7**, which is impossible, since they both belong to the same column. We can therefore safely eliminate the 1 from **r9c9**.

If we have more diagonals we get more complicated crossovers. Here a 10-tuple crossover in a sudoku with the bold X pattern (6 diagonals).

And a 13tuple crossover in a Argyle Sudoku:

In this more complicated example, the three boxes with the yellow background are disjoint, and must therefore contain 3 occurrences of the digit 1, which are locked in the cells marked in light and dark pink. The green cell sees all light pink cells. Therefore, if the green cell were to contain a 1, it would constrain the three 1s in boxes 1, 3 and 5 to the cells marked in dark pink. However, this is impossible, because the dark pink cells can only accommodate 2 of these 3 occurrences in total, due to the fact that **r2c9** and **r6c5** see each other.

From the above examples, we can formulate the general principle as follows:

"*If all cells with a certain candidate in N pairwise disjoint, locked, groups (N > 1) that are not weakly linked to a certain green cell are located in (N - 1) pairwise disjoint groups and the green cell isn't in these N groups, then the green cell cannot have this candidate.*"

*Note:* To be disjoint is not really necessary. The technique is still valid if all common cells belonging to a pair of groups either do not have the candidate (as is the case in some of the examples above), or are weakly linked to the green cell. In the latter case, if one of the common cells were to contain the digit in question, the fact that it sees the green cell would directly prevent a 1 from existing in the green cell, even though the logic described above would break down.