Tight Bounds on the Complexity of Recognizing Odd-Ranked Elements

Let S = 〈s1, s2, s3, ..., sn〉 be a given vector of n real numbers. The rank of z ∈ R with respect to S is defined as the number of elements si ∈ S such that si ≤ z. We consider the following decision problem: determine whether the oddnumbered elements s1, s3, s5, . . . are precisely the elements of S whose rank with respect to S is odd. We prove a bound of Θ(n logn) on the number of operations required to solve this problem in the algebraic computation tree model. Let S = 〈s1, s2, s3, . . . , sn〉 ∈ R n be a given vector. For an arbitrary real z, define the rank of z with respect to S, denoted by rankS(z), as the number of elements of S less than or equal to z. Thus, for instance, the largest element of S has rank n. Let odd(S) denote the set of elements of S whose rank with respect to S is odd. We consider the following problem: determine whether the odd-numbered elements s1, s3, s5, . . . are precisely the elements of S whose rank with respect to S is odd. Without loss of generality, we can assume that n is even because, otherwise, we can append an extra element +∞ without changing the answer. Note that odd(S) has size n/2 if and only if all n values si ∈ S are distinct; hence, the answer is ‘yes’ only if S is a vector of n distinct numbers. We prove matching upper and lower bounds on the number of operations required to solve the problem in the algebraic computation tree (ACT) model (see Ben-Or [1]). The following algorithm solves the problem using O(n logn) comparisons. Sort S ′ = 〈s1, s3, s5, . . . , sn−1〉 in non-decreasing order with an optimal sorting algorithm. Similarly, sort S in non-decreasing order. Then, scan the vector S ′ and the oddnumbered elements of S to decide whether the two are equal. Next, we prove the matching lower bound. For a vector S = 〈s1, s2, s3, . . . , sn〉, let σ(S) denote the permuted vector 〈sσ(1), sσ(2), sσ(3), . . . , sσ(n)〉. We call a permutation σ, where σ(i) is odd if and only if i is odd, a permissible permutation. Lemma 1. There are ((