partition a sequence of 2n real numbers so that

Question

I\'m currently reading The Algorithm Design Manual and I\'m stuck on this exercise.

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Take a sequence of 2n real numbers as input. Design an O(n log n) algorithm tha

Answer 1

Given the input array

x₀ x₁ x₂ ... x_2n

use merge sort to sort it in O( n log n) time to produce a sorted list

x_a0 x_a1 ... x_a2n

where the indices a_i indicate the permutation that you have performed on the initial list to obtain the second list.

I claim then that the pairing that produces the minimum maximum sum over all pairings is the following pairing:

(x_ai, x_a2n-i) for i = 0, 1, ..., n. That is, you group the highest valued number with the lowest valued number available.

Proof

Proceed by induction, for the case 2n=2 this is obvious.

Without loss of generality consider that the input is a list of sorted numbers (since if it is not then sort them)

x₀ x₁ x₂ ... x_2n.

Consider the pairing of x_2n with any number, then clearly the minmum sum of this pairing is achieved with (x_2n, x₀).

Now consider the pairing of x_2n-1, either (x_2n,x₀),(x_2n-1,x₁) is the pairings that produce the minumim max sum, or (x_2n,x₁),(x_2n-1,x₀) is, or both are. In the latter case our choice doesn't matter. In the seconds to last case, this is impossible (think about it.) In the general case if we proceed inductively by this process, when we are looking for a pair for x_2n-k, x_k is the lowest unused value that we can pair up with, however; suppose instead we pair up x_k with some other x_2n-j for j < k, to try to get a lower minum sum. This is impossible as, x_2n-j + x_k >= x_2n-k + x_k, so at most we can only achieve the same minimum sum.

This means that choosing (x_2n-k,x_k) gives us the minimum pairings.