Знаходження принаймні двох шляхів однакової довжини у спрямованому графіку

Припустимо, у нас є спрямований графік та два вузли$G=(V,E)$$A$ and $B$. I would like to know if there are already algorithms for calculating the following decision problem:

Are there at least two paths between $A$ and $B$ of the same length?

How about the complexity? Can I solve it in polynomial time?

I would like to add a new constrain on the graph, maybe the problem is more solvable. On adjacency matrix, every column is not empty. So, every node has at least one arrow on input and there is also at least one node connected to itself. So if the node is the $i$-th node, then $(i,i)$ is an edge in the graph.

Consider a graph $G$, we want to know if there are two different paths from $A$ to $B$ of the same length. What to do? Simple: code two paths in one. Define graph $G'$ with vertices $V \times V \times \{0,1\}$. You make a step in $G'$ by making two independent steps in $G$. The additional bit tells you whether the two paths have already split from each other.

Formally, there is an edge $(i,j,e) \to (i',j',e')$ in $G'$ iff $i \to i'$, $j \to j'$ in $G$ and $e'=e ∨ (i,i')≠(j,j')$.

The algorithm checks if there is a path $(A,A,0)$ to $(B,B,1)$ in $G'$, which is $O(V^4)$, or something like $O((V+E)^2)$.

If you agree this algorithm is correct then, as a consequence, the path in $G'$ is of length at most $2n^2$, therefore potential "path collisions" must occur latest at that length. You can get a $O(V^{\omega} \log V)$ algorithm from this observation, where $\omega$ is matrix multiplication complexity (ask if you need a spoiler...).

I strongly feel there is a $O(V+E)$ algorithm, which uses more of the structure of the problem.

Probably i've an answer for this problem, but i'm not sill sure that it works.

It is not important to "find" the two path, the only important thing is to "know" if they exist or not. I don't think that this is an NP complete problem.

So, take the adjacency matrix $\textbf A$. We can easily suppose that it is filled with 0,1 value. (0 = no edge; 1 = there is an edge) Let's use the following algebra with 3 values (0,1,2), where everything works as usual except: $2+\text{<something>} = 2$; $2*\text{<whatever greater than 0>} = 2$

So, if there are two paths of same length from $i,j$ i'm expecting that there is a value $p$ such that $(\textbf A^p)_{i,j} = 2$.

Let $n$ is the number of vertex in the graph (or, let's say, that $\textbf A$ has dimension $n\times n$). I don't know the value of $p$, but if i iterate $\textbf A$ by multiplying with itself for at most $n^2$ i should find the answer. (so, $p<n^2$) (the sense is that i check $\text A$, then i check $\text A^2$, then i check $\text A^3$ and so on...)

here is my argumentation:

• if the two paths are simple paths, well, it works; if there are, at most i have to iterate $n$ times.
• If there is at least one neasted cycle or there is a path with two cycles, well, i have to find the least common multiple (LCM). $n^2$ is a bigger value for sure and in less than $n^2$ times if there is i should have to find them.
• If the two paths are two distinct paths, both with one cycle, then it's more or less similar to finding a solution for this two equation: $\alpha + \beta m = \gamma + \delta k$, where $m$ and $k$ are the length of these two distinct cycles. Matrix multiplication $\text A^q$, as defined above, says "is there a path from $i$ to $j$ whose length is $q$?" So, if $\textbf A^q$ is grater than $1$ it means that there are more paths leading from $i$ to $j$. By iterating the matrix $n^2$ times we pass through all the possible combination of $\delta$ and $\beta$. Indeed, $LCM(a,b)$ is defined as $(a*b)/GCD(a,b)$ and no cycle can be greater than $n$.

I stop to iterate once i found $(\textbf A^p)_{i,j} = 2$.

am i wrong?