'Length' of a path is the number of edges in the path.
Given a source and a destination vertex, I want to find the number of paths form the source vertex to the destination vertex of given length k.
We can visit each vertex as many times as we want, so if a path from a
to b
goes like this: a -> c -> b -> c -> b
it is considered valid. This means there can be cycles and we can go through the destination more than once.
Two vertices can be connected by more than one edge. So if vertex a
an vertex b
are connected by two edges, then the paths , a -> b
via edge 1 and a -> b
via edge 2 are considered different.
Number of vertices N is <= 70, and K, the length of the path, is <= 10^9.
As the answer can be very large, it is to be reported modulo some number.
Here is what I have thought so far:
We can use breadth-first-search without marking any vertices as visited, at each iteration, we keep track of the number of edges 'n_e' we required for that path and product 'p' of the number of duplicate edges each edge in our path has.
The search search should terminate if the n_e
is greater than k, if we ever reach the destination with n_e
equal to k, we terminate the search and add p
to out count of number of paths.
I think it we could use a depth-first-search instead of breadth first search, as we do not need the shortest path and the size of Q used in breadth first search might not be enough.
The second algorithm i have am thinking about, is something similar to Floyd Warshall's Algorithm using this approach . Only we dont need a shortest path, so i am not sure this is correct.
The problem I have with my first algorithm is that 'K' can be upto 1000000000 and that means my search will run until it has 10^9 edges and n_e the edge count will be incremented by just 1 at each level, which will be very slow, and I am not sure it will ever terminate for large inputs.
So I need a different approach to solve this problem; any help would be greatly appreciated.
Given G=(V,E) obtain G′ from G by replacing each node by a clique of size N=nc where n=|V| and c≫1. For each simple path of length ℓ in G there are roughly (N!)ℓ paths in G′. So if G has an s−t Hamiltonian path, there will be at least (N!)
Because there are exactly eight paths of length four from a to d. By inspecÅon of the graph, we see that a,b,a,b,d; a,b,a,c,d; a,b,d,b,d; a,b,d,c,d; a,c,a,b,d; a,c,a,c,d; a,c,d,b,d; and a, c, d, c, d are the eight paths of length four from a to d.
In general, the number of directed k-vertex paths (k≥2) in Kn is n×(n−1)×⋯×(n−k+1), this is the number of sequences of length k without repeated entries. Thus the number of undirected k-vertex paths is 12n×(n−1)×⋯×(n−k+1).
As i realised from the comments, Dijkstra's algorithm doesn't work for unweighted graphs.
So, here's a nifty graph theory trick that I remember for this one.
Make an adjacency matrix A
. where A[i][j]
is 1 if there is an edge between i
and j
, and 0 otherwise.
Then, the number of paths of length k
between i
and j
is just the [i][j]
entry of A^k.
So, to solve the problem, build A
and construct A^k using matrix multiplication (the usual trick for doing exponentiation applies here). Then just look up the necessary entry.
EDIT: Well, you need to do the modular arithmetic inside the matrix multiplication to avoid overflow issues, but that's a much smaller detail.
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