By Harry Y. F. Lam
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Technical Report CMS-TR97-08, Oxford Brookes University, UK. 5 Sorting pairwise sums Introduction Let A be some linearly ordered set and (⊕) :: A → A → A some monotonic binary operation on A, so x ≤ x ∧ y ≤ y ⇒ x ⊕ y ≤ x ⊕ y . Consider the problem of computing sortsums :: [A] → [A] → [A] sortsums xs ys = sort [x ⊕ y | x ← xs, y ← ys] Counting just comparisons, and supposing xs and ys have the same length n, how long does sortsums xs ys take? Certainly O(n 2 log n) comparisons are suﬃcient. There are n 2 sums and sorting a list of length n 2 can be done with O(n 2 log n) comparisons.
Using the fact that log E (n) = Ω(n 2 log n), we conclude that at least this number of comparisons is required. Now for the meat of the exercise. Lambert’s algorithm depends on two simple facts. Deﬁne the subtraction operation ( ) :: A → A → A by x y = x ⊕ negate y. 2), which we leave as an exercise, requires all the properties of an Abelian group. 2) says that the latter problem is, in turn, reducible to the problem of sorting subtractions over a single list. 2) are used. ]] where Label a is a synonym for (a, (Int, Int)).
Thinking that saddleback search was the gold standard for this problem, I steered them away from pursuing the thought. Only afterwards did I wonder whether they might have had a point. Apart from describing a new algorithm for an old problem, I think that two other methodological aspects are worthy of note. First, formal program calculation is heavily inﬂuenced by the available computational methods of the target language. While nobody would say it was elegant, Mary’s ﬁnal program is simple enough, given recursion and list concatenation as basic constructs, but would be more diﬃcult to express with just arrays and loops.