Just as the rule of triples and the rule of interlocking triples generalize the rule of pairs, it's possible to extend the uniqueness/pigeonhole rule to more than one cell.
For pairs, we noted that if N cells in the same group (row, column, or box) contain exactly N values between them, then those values can be eliminated from the rest of the group. This is possible since we know that all of the N values have to live in exactly those N cells.
Recall that the pigeonhole rule is the observation that if there is exactly one place in a group that can hold a particular value, then it must contain that value. We can eliminate all of the other values that we previously thought were possible for that cell.
Suppose that there are exactly two places in a group that can hold two particular values A and B. Then we can eliminate all other values from those cells; if they had any other values, there would be no place for one or both of the values A and B. This works for any number of values: if N values can only live in N cells within a group, all other values can be removed from those N cells.
So the rules of pairs, triples, etc., allow us to remove impossible values from cells outside a particular set of cells. The rule of uniqueness and its more general version that we're discussing here allow us to remove extraneous values from the cells inside the group.
Time from some examples. Consider this puzzle:
Using elimination and uniqeness, we can get this far:
I've circled the places where there are two values that can only be in two cells in a group. The red circles show the only places in row 7 that can hold 1 and 7. The blue circles show the only places in column 3 that can hold 5 and 7. Finally, the green circles show the only places in box 6 that can hold 5 and 6 (note: the same is not true of column 9).
Now we can remove values besides 1 and 7 from the red circles, values besides 5 and 7 from the blue circles, and values beside 5 and 6 from the green circles. The results are shown below:
Another place to apply the rule has come to light. Circled in red are the only places for 4 and 8 in row 7. Removing the additional values, we get to here:
Completing the solution uses the rules of pairs and interlocking triples. The solution (no peeking) is:
I hope you are enjoying these notes. Next time I will start to tackle some more advanced rules that require looking outside of a single group of cells — X-Wings, Swordfish, and Two-or-Three in a Bed (silly names, I know). I'll also discuss some rules that blur the lines between logical solutions and solutions that involve guessing.