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Given an directed graph (may contain cycles) we have to find total number of simple paths from a fixed source vertex to all other vertices of the graph, i.e.

$$ \text{#(paths from 1 to 2)}+\text{#(paths from 1 to 3)}+\cdots+\text{#(paths from 1 to $n$)}. $$

We have to find this in linear time, as number of vertexes can be as large as 100000. I tried using DFS in this question to generate all the paths and keeping up count of these but this is not fast enough.

Is there any mathematical formula for this?

P.S. I am only interested in total number of paths, not generating all the paths. Each vertex has up to 4 or 5 edges.

For example, if the adjacency matrix of a 4 nodes graph is like this: 

  A B C D
A 0 1 0 0    
B 1 0 1 1    
C 0 1 0 1   
D 0 0 1 0

Then the number of paths are 

from A to B: 1    
from A to C: 2      
from A to D: 2     
therefore 5 paths

So I want the answer to be 5.

The edges may be unidirectional or bidirectional and I want to find only simple paths.

D.W.
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Viplaw Srivastava
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  • Can you credit the original source where you originally encountered this problem? How do you know that the problem can be solved efficiently? – D.W. Mar 07 '19 at 20:44
  • Cross-posted: https://cs.stackexchange.com/q/105176/755, https://stackoverflow.com/q/55000372/781723, https://mathoverflow.net/q/324686/37212. Please do not post the same question on multiple sites. Each community should have an honest shot at answering without anybody's time being wasted. – D.W. Mar 07 '19 at 20:46

1 Answers1

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Suppose you have a polynomial-time algorithm $\mathcal{A}$ to solve this problem, then consider a graph $G$ and arbitrary two vertices $s$ and $t$. We construct a new graph $G'$ by add a new vertex $t'$ into $G$, as well as an edge $(t,t')$. Now we have

\begin{align} \text{#(paths from $s$ to $t$ in $G$)} =&\ \text{#(paths from $s$ to all other vertices in $G'$)}\\ &-\text{#(paths from $s$ to all other vertices in $G$)}. \end{align}

That means, we can use $\mathcal{A}$ to count the number of paths between two vertices in polynomial time. However, the latter problem is #P-hard. So such an algorithm $\mathcal{A}$ does not exist unless $\mathbf{P}=\mathbf{NP}$.

xskxzr
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