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### Ehsan_sShuvo's blog

By Ehsan_sShuvo, history, 3 years ago, ,

Could anyone please explain me clearly why we have to do

low[vertex]=min(low[vertex],dis[i]);

Yeah!I know that is for lowest discovery time determination.But my question is that all the time we put minimum time in the current node comparing with its adjacent nodes,but not increase the time.My question is two or more nodes cant visit at a time.But lowest time show that,it can do that.But how? Could anyone please clear me out?Actually,i read more than 5/6 resources,but i couldn't get that.There all says about subtree,ancestor,blah blah,but my memory couldn't make a sense for that,what they actually want to say.

Following is the psuedocode of finding out articulation point.

time = 0
function DFS(adj[][], disc[], low[], visited[], parent[], AP[], vertex, V)
visited[vertex] = true
disc[vertex] = low[vertex] = time+1
child = 0
for i = 0 to V
if visited[i] == false
child = child + 1
parent[i] = vertex
DFS(adj, disc, low, visited, parent, AP, i, n, time+1)
low[vertex] = minimum(low[vertex], low[i])
if parent[vertex] == nil and child > 1
AP[vertex] = true
if parent[vertex] != nil and low[i] >= disc[vertex]
AP[vertex] = true
else if parent[vertex] != i
low[vertex] = minimum(low[vertex], disc[i])

• -11

 » 3 years ago, # |   0 My question is two or more nodes cant visit at a time.But lowest time show that,it can do that.But how? lowest time of a node denotes the discovery time of oldest ancestor which we can reach from this node. Two or more nodes can have the same oldest ancestor and hence same lowest time. You can read my blog if it is still not clear to you.
•  » » 3 years ago, # ^ |   0 Basically,i read your tutorial last night which i have found in hackerearth. but,thanks.Basically,i got that after watching a long tutorial what you said.BTW,thanks. Any link could you provide for Bridge?
•  » » 5 months ago, # ^ |   0 Great explanation
 » 3 years ago, # |   0 take a look on https://visualgo.net/en/dfsbfs i think u will be able to understand