number of totally different path between two nodes in graph theory I have an undirected, unweighted graph representing a network.
I have a starting node and an end one.
My 'network' is reliable if there is no node such that without that node s and t are not reachable i.e. no node is necessary and all nodes have at least one redundant path.
how can I formally verify this?
thank you
 A: If you mean that the network remains connected upon removing any one node, the magic words are "2-vertex-connected graph", or "biconnected graph". An algorithm for determining biconnectivity is described here, though I am sure there are plenty of other places.
A: If you want to calculate number of paths from one node to another ,let say from s to t.then you can follow the following approach.
To reach a node t from s , you need to calculate that how many ways are there to reach t from its adjacent vertices .
means
 PathsTo(s,t)=sum(PathsTo(s,u)) , where u are adjacents vertices of t.
as we can see there is a subproblem optimality and overlapping subproblems, so we can use a DP approach to do this in Linear time.
We can do this by modifying our DFS algorithm.
Psuedocode:
declare an array A of size |V| , memset it to NULL
array represent , number of ways to reach a node v from s.
Paths(s,t):
if(s==t):
 return 1;

else:
 if(A[t]==NULL):
      A[t]=sum(Paths(s,u)) for all adjacent u of t
 else
      return A[t]

Time complexity is O(V+E) same as DFS.
