Largest triangle in convex hull

Question

The question has already been answered, but the main problem I am facing is in understanding one of the answers..

From https://stackoverflow.com/a/1621913/2673063

How is the following algorithm O(n) ?

It states as By first sorting the points / computing the convex hull (in O(n log n) time) if necessary, we can assume we have the convex polygon/hull with the points cyclically sorted in the order they appear in the polygon. Call the points 1, 2, 3, … , n. Let (variable) points A, B, and C, start as 1, 2, and 3 respectively (in the cyclic order). We will move A, B, C until ABC is the maximum-area triangle. (The idea is similar to the rotating calipers method, as used when computing the diameter (farthest pair).)

With A and B fixed, advance C (e.g. initially, with A=1, B=2, C is advanced through C=3, C=4, …) as long as the area of the triangle increases, i.e., as long as Area(A,B,C) ≤ Area(A,B,C+1). This point C will be the one that maximizes Area(ABC) for those fixed A and B. (In other words, the function Area(ABC) is unimodal as a function of C.)

Next, advance B (without changing A and C) if that increases the area. If so, again advance C as above. Then advance B again if possible, etc. This will give the maximum area triangle with A as one of the vertices. (The part up to here should be easy to prove, and simply doing this separately for each A would give O(n2). But read on.) Now advance A again, if it improves the area, etc. Although this has three "nested" loops, note that B and C always advance "forward", and they advance at most 2n times in total (similarly A advances at most n times), so the whole thing runs in O(n) time.

Answer 1

As the author of the answer that is the subject of the question, I feel obliged to give a more detailed explanation of the O(n) runtime.

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Firstly, just as an example, here is a figure from the paper, showing the first few steps of the algorithm, for a particular sample input (a 12-gon). First we start with A, B, C as three consecutive vertices (step 1 in the figure), advance C as long as area increases (steps 2 to 6), then advance B, and so on.

The triangles with asterisks above them are the "anchored local maxima", i.e., the ones that are best for a given A (i.e., advancing either C or B would decrease the area).

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As far as the runtime being O(n): Let the "actual" value of B, in terms of the number of times it's been incremented and ignoring the wrap around, be nB, and similarly for C be nC. (In other words, B = nB % n and C = nC % n.) Now, note that,

1. ("B is ahead of A") whatever the value of A, we have A ≤ nB < A + n

2. nB is always increasing

So, as A varies from 0 to n, we know that nB only varies between 0 and 2n: it can be incremented at most 2n times. Similarly nC. This shows that the running time of the algorithm, which is proportional to the total number of times A, B and C are incremented, is bounded by O(n) + O(2n) + O(2n), which is O(n).