By selecting either as the first training example, we could ignore all but the four-word characteristic tuple for both u 1 and u 2 (a saving of seven tuples each); since u 3 and u 4 each have three words in common with the first two candidates, they would no longer depend on their lower four tuples; and although we would also improve the statistics for one of u 5 's tuples (on), nothing could be pruned from u 5 's characteristic tuples.
f backoff (u 1 , C) = f backoff (u 2 , C) = 7 + 7 + 4 + 4 = 22 (see Figure 4).