Closest pair of points (CLRS pg 1043): Running time of splitting a sorted array into two sorted arrays

Question

In finding the closest pair of points in O(nlgn) time, the pseudocode for splitting a sorted list into two sorted lists (CLRS 3rd ed pg 1043) is said to run in O(n) time.

However, this assumes that line 4 runs in constant time, which I find hard to believe (I'd assume it runs in O(lgn) time if it were stored as a binary tree, giving a total running time of O(nlgn).

Y is a sorted array, YL and YR are the two new sub-arrays. PL is a subset of Y in random order, and YL is the same subset, but in sorted order.

Where am I going wrong with my reasoning?

Answer 1

For simplicity sake we're assuming the list is of integers and not strings or integers which can complicate things greatly here.

There are two calculations to consider here:

1. for loop: This runs for length of Y times, which I'm assuming is N here
2. the tricky part - comparison of Y[i] with PL(Note: the comparison of two numbers is constant if we consider them to be of word size). Now, accessing Y[i] is constant since we're dealing with Random Access Machines. However, to compare it with an array PL of length, say, k will take k time. If this k is very small and independent of the size of input array Y, this ideally would be constant.

To write it with greater precision would mean you consider the time taken for k comparisons (length of PL) and hence, the total time of this pseudo code would be O(Nk). But, if the assumptions that k is random and independent of N hold true, it really is O(N)

Answer 2

I don't know how it is supposed to work in the book, but thinking about the way the algorithm looks like, you can come up with the following idea:

• Y[i], X[i], YL[i], XL[i], YR[i] and XR[i] are integers, corresponding to the index of the ith-point (so you just have to store some global array which, given the index, returns the x or y coordinate).
• PL[i] is a boolean, true if the i-th point is in the left part, false otherwise.

At each recursion step, you can compute PL[i] using y coordinates (O(n) time). Then you separate the set of points in two sets "left" and "right" using the algorithm from the book, replacing the line if Y[i] in PL by if PL[Y[i]] (such access is O(1), so in overall we get O(n)).

This has O(n) time complexity and uses O(n) memory.

Thus the closest pair problem is solved that way in T(n) = O(n log n).