用伪代码实现介数中心性
时间: 2023-09-12 07:03:40 浏览: 42
伪代码实现介数中心性的基本思路如下:
1. 对于每个节点,计算它的介数中心性
2. 对于每对节点 (s, t),计算它们之间的最短路径数 num_shortest_paths
3. 对于每个节点 v,计算它到所有其他节点的最短路径经过 v 的次数 sigma_v
4. 对于每对节点 (s, t),将 s 的介数中心性增加 (sigma_s * sigma_t / num_shortest_paths)。
下面是伪代码实现:
```
for each node v:
# initialize the betweenness centrality of node v
betweenness[v] = 0
for each node s:
# initialize the number of shortest paths to each node
shortest_paths = {}
for each node v:
shortest_paths[v] = 0
# initialize the distance to each node
distance = {}
for each node v:
distance[v] = -1
# initialize the list of predecessors of each node
predecessors = {}
for each node v:
predecessors[v] = []
# set the distance to the starting node to 0
distance[s] = 0
# create a queue and enqueue the starting node
queue = [s]
while queue is not empty:
# dequeue the next node
v = queue.pop(0)
for each neighbor w of v:
# if w has not been visited before
if distance[w] == -1:
# update its distance and add it to the queue
distance[w] = distance[v] + 1
queue.append(w)
# if the distance to w is the same as the distance to v plus one
if distance[w] == distance[v] + 1:
# increment the number of shortest paths to w
shortest_paths[w] += shortest_paths[v]
# add v to the list of predecessors of w
predecessors[w].append(v)
# initialize the dependencies of each node
dependencies = {}
for each node v:
dependencies[v] = 0
# calculate the dependencies of each node
while predecessors is not empty:
# get the node with the highest distance
max_distance = max(distance.values())
v = [k for k, v in distance.items() if v == max_distance][0]
# remove it from the predecessors list
del predecessors[v]
for each predecessor w of v:
# calculate the dependency of w on v
dependency = (shortest_paths[w] / shortest_paths[v]) * (1 + dependencies[v])
# add it to the dependencies of w
dependencies[w] += dependency
# add it to the betweenness centrality of v
betweenness[v] += dependency
# normalize the betweenness centrality by dividing by the number of pairs of nodes
for each node v:
betweenness[v] /= 2 * (n - 1) * (n - 2)
```
其中,n 表示节点的个数。
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