分层算法的时间复杂性
我正在寻找我将在下面解释的算法的时间复杂性。谢谢您的帮助...。请!
该算法由两个一般步骤组成:
- 获取最近的邻居:
对于每个节点,它仅查看直接邻居,并通过以下公式找到最近的邻居(最短距离)。 在此处输入图像描述
哪个VG是每个节点的总数。摩尔 - 芬罗 Laplacian的反基质
以这种方式继续,直到创建了与最近邻居的节点链为止。最后,当在此链中发现两个彼此接近的节点,并且在群集中彼此之间的距离最少时,链会停止。这两个节点被认为是集群的代表。
- 修剪节点:
在每个群集形成后,根据该群集的两个代表节点计算每个节点的深度,如果每个节点相对于两个代表性节点的深度大于所需的阈值,则该节点被修剪并被视为单成员群集。
I am looking for the time complexity of the algorithm that I will explain below. Thank you for helping me....please!
This algorithm consists of two general steps:
- Get the nearest neighbor :
For each node, it only looks at the direct neighbor and finds the nearest neighbor (shortest distance) through the following formula.
enter image description here
which VG is the total degree of each node .and I is the corresponding element of the Moore-Penrose
inverse matrix of laplacian
It continues in this way until a chain of nodes with the nearest neighbor is created. Finally, the chain stops when two nodes close to each other are found in this chain and have the least amount of distance to each other in the cluster. These two nodes are considered as representatives of the cluster.
- Pruning the node:
After the formation of each cluster, the depth of each node is calculated with respect to the two representative nodes of that cluster, and if the depth of each node with respect to the two representative nodes is greater than the desired threshold, that node is pruned and considered as a single-member cluster.
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