O(n) 最坏情况下的 2D 寻峰算法?

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问题描述

我正在学习麻省理工学院的本课程.在第一堂课中，教授提出了以下问题:-

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

二维数组中的峰值是一个值，使得它的所有 4 个邻居都小于或等于它，即.对于

A peak in a 2D array is a value such that all it's 4 neighbours are less than or equal to it, ie. for

a[i][j] 为局部最大值，

a[i+1][j] <= a[i][j]
&& a[i-1][j] <= a[i][j]
&& a[i][j+1] <= a[i][j]
&& a[i+1][j-1] <= a[i][j]

现在给定一个 NxN 二维数组，在数组中找到一个峰值.

Now given an NxN 2D array, find a peak in the array.

通过迭代所有元素并返回一个峰值，这个问题可以在 O(N^2) 时间内轻松解决.

This question can be easily solved in O(N^2) time by iterating over all the elements and returning a peak.

然而，它可以通过使用分而治之的解决方案优化在 O(NlogN) 时间内解决这里.

However it can be optimized to be solved in O(NlogN) time by using a divide and conquer solution as explained here.

但是他们说存在一个O(N)时间算法来解决这个问题.请建议我们如何在 O(N) 时间内解决这个问题.

But they have said that there exists an O(N) time algorithm that solves this problem. Please suggest how can we solve this problem in O(N) time.

PS(对于那些懂python的人)课程人员已经解释了一个方法here(问题 1-5.寻峰证明)并且还在他们的问题集中提供了一些 python 代码.但是所解释的方法完全不明显并且很难破译.python 代码同样令人困惑.所以我复制了下面代码的主要部分给那些了解python并且可以从代码中知道使用的是什么算法的人.

PS(For those who know python) The course staff has explained an approach here (Problem 1-5. Peak-Finding Proof) and also provided some python code in their problem sets. But the approach explained is totally non-obvious and very hard to decipher. The python code is equally confusing. So I have copied the main part of the code below for those who know python and can tell what algorithm is being used from the code.

def algorithm4(problem, bestSeen = None, rowSplit = True, trace = None):
    # if it's empty, we're done
    if problem.numRow <= 0 or problem.numCol <= 0:
        return None

    subproblems = []
    divider = []

    if rowSplit:
        # the recursive subproblem will involve half the number of rows
        mid = problem.numRow // 2

        # information about the two subproblems
        (subStartR1, subNumR1) = (0, mid)
        (subStartR2, subNumR2) = (mid + 1, problem.numRow - (mid + 1))
        (subStartC, subNumC) = (0, problem.numCol)

        subproblems.append((subStartR1, subStartC, subNumR1, subNumC))
        subproblems.append((subStartR2, subStartC, subNumR2, subNumC))

        # get a list of all locations in the dividing column
        divider = crossProduct([mid], range(problem.numCol))
    else:
        # the recursive subproblem will involve half the number of columns
        mid = problem.numCol // 2

        # information about the two subproblems
        (subStartR, subNumR) = (0, problem.numRow)
        (subStartC1, subNumC1) = (0, mid)
        (subStartC2, subNumC2) = (mid + 1, problem.numCol - (mid + 1))

        subproblems.append((subStartR, subStartC1, subNumR, subNumC1))
        subproblems.append((subStartR, subStartC2, subNumR, subNumC2))

        # get a list of all locations in the dividing column
        divider = crossProduct(range(problem.numRow), [mid])

    # find the maximum in the dividing row or column
    bestLoc = problem.getMaximum(divider, trace)
    neighbor = problem.getBetterNeighbor(bestLoc, trace)

    # 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)

    # 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

    # figure out which subproblem contains the largest number we've seen so
    # far, and recurse, alternating between splitting on rows and splitting
    # on columns
    sub = problem.getSubproblemContaining(subproblems, bestSeen)
    newBest = sub.getLocationInSelf(problem, bestSeen)
    if not trace is None: trace.setProblemDimensions(sub)
    result = algorithm4(sub, newBest, not rowSplit, trace)
    return problem.getLocationInSelf(sub, result)

#Helper Method
def crossProduct(list1, list2):
    """
    Returns all pairs with one item from the first list and one item from
    the second list.  (Cartesian product of the two lists.)

    The code is equivalent to the following list comprehension:
        return [(a, b) for a in list1 for b in list2]
    but for easier reading and analysis, we have included more explicit code.
    """

    answer = []
    for a in list1:
        for b in list2:
            answer.append ((a, b))
    return answer

最坏情况下的

O(n) 最坏情况下的 2D 寻峰算法?

问题描述

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