假设我们有两种积极的愤怒论点,产生浮动结果。 在确定其中一个论点和增加另一个论点时,由此产生的价值清单是单一模式,然后在下降(可能重复高峰时期的某些结果)。
是否比以一个论点和二元搜索来寻找最高峰结果时使用第二个论点而确定最大程度的更有效方法? 如果是,如何?
我非常怀疑存在有效的算法,因为情况类似:
1 2 3 2 1 2 6 4 3 2 3 4 5 4 3 2 3 4 7 2 1 2 3 2 1
每一行和一栏都是单一模式,但当地最大程度6不是全球最大。 在一个较大的矩阵中,可能会出现多个地方高峰。
There is some information to be gleaned from finding a local maximum, though. If the local maximum is not the global maximum, then the global maximum cannot be elsewhere in the row or column of the local maximum. This effectively splits the matrix into four sub-matrices, on which you could run a recursive algorithm.
Additionally, if you walk along the row and column of a local maximum, and look at the values at each side, you need to find an uphill neighbour in order for the global maximum to potentially be in that quadrant. So you may be able to discard some of the quadrants.
但是,在最坏的情况下,我担心你会重新审视大多数价值观。 我不认为任何东西比对所有各行或所有栏目进行双双双双双双双双双轨搜查更为有效。
决 定
If you had a very efficient algorithm to find the peak, that had to look at only a small number of values, you would not be sure whether the result was the global maximum, or just a local maximum. So you d have to run the algorithm again on the data that you hadn t checked yet. In an NxM matrix, you could have up to N or M (whichever is smaller) local maximums. In order for this to be more efficient than a version of binary search on every row or column (whichever there are fewer of) your peak-finding algorithm would have to be more efficient that O(LogN) or O(LogM).
If you traverse the matrix in any other way than per row or column, you don t know anything about how the values are ordered. If you traverse the matrix by row or column, you know your search space is unimodal, and you can at least leverage that to make your search more efficient.
That means that a form of binary search over the rows or columns (whichever there are fewer of) is the most efficient way to find the global maximum.
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