2D peak finding algorithm in O(n) worst case time?

Question

I was doing this course on algorithms from MIT. In the very first lecture the professor presents the following problem:-

A peak in a 2D array is a value such that al

Answer 1

To see thata(n):

Calculation step is in the picture

To see algorithm implementation:

1) start with either 1a) or 1b)

1a) set left half, divider, right half.

1b) set top half, divider, bottom half.

2) Find global maximum on the divider. [theta n]

3) Find the values of its neighbour. And record the largest node ever visited as the bestSeen node. [theta 1]

# update the best we've seen so far based on this new maximum
if bestSeen is None or problem.get(neighbor) > problem.get(bestSeen):
    bestSeen = neighbor
    if not trace is None: trace.setBestSeen(bestSeen)

4) check if the global maximum is larger than the bestSeen and its neighbour. [theta 1]

//Step 4 is the main key of why this algorithm works

# return when we know we've found a peak
if neighbor == bestLoc and problem.get(bestLoc) >= problem.get(bestSeen):
    if not trace is None: trace.foundPeak(bestLoc)
    return bestLoc

5) If 4) is True, return the global maximum as 2-D peak.

Else if this time did 1a), choose the half of BestSeen, go back to step 1b)

Else, choose the half of BestSeen, go back to step 1a)

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To see visually why this algorithm works, it is like grabbing the greatest value side, keep reducing the boundaries and eventually get the BestSeen value.

# Visualised simulation

round1

round2

round3

round4

round5

round6

finally

For this 10*10 matrix, we used only 6 steps to search for the 2-D peak, its quite convincing that it is indeed theta n

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By Falcon