I have an undirected, unweighted graph, and I'm trying to come up with an algorithm that, given 2 unique nodes on the graph, will find all paths connecting the two nodes, not including cycles. Here's an illustration of what I'd like to do: Graph example
Does this algorithm have a name? Can it be done in polynomial time?
Thanks,
Jesse
Suresh suggested DFS, MRA pointed out that it's not clear that works. Here's my attempt at a solution following that thread of comments. If the graph has $m$ edges, $n$ nodes, and $p$ paths from the source $s$ to the target $t$, then the algorithm below prints all paths in time $O((np+1)(m+n))$. (In particular, it takes $O(m+n)$ time to notice that there is no path.)
The idea is very simple: Do an exhaustive search, but bail early if you've gotten yourself into a corner.
Without bailing early, MRA's counter-example shows that exhaustive search spends $\Omega(n!)$ time even if $p=1$: The node $t$ has only one adjacent edge and its neighbor is node $s$, which is part of a complete (sub)graph $K_{n-1}$.
Push s on the path stack and call search(s):
path // is a stack (initially empty)
seen // is a set
def stuck(x)
if x == t
return False
for each neighbor y of x
if y not in seen
insert y in seen
if !stuck(y)
return False
return True
def search(x)
if x == t
print path
return
seen = set(path)
if stuck(x)
return
for each neighbor y of x
if y not in path:
push y on the path
search(y)
pop y from the path
Here search does the exhaustive search and stuck could be implemented in DFS style (as here) or in BFS style.
return after print path. and 2. there is nothing to prevent circular endless loop between two nodes on one edge. Can be fixed by checking if the next node is already on the stack. This will require more changes to the way things are pushed and popped into the stack.
$\endgroup$
if y not in path: instead of if y not in seen: in the search function.
$\endgroup$
There is an easy way to partition the set of $s$-$t$ paths in a graph $G$. Fix an edge $tt'$ in $G$. Let $P_1$ be the set of paths from $s$ to $t$ which use the edge $tt'$, and let $P_2$ be the set of paths from $s$ to $t$ in $G-tt'$. Then $P_1 \cap P_2 = \emptyset$ and the set of $s$-$t$ paths $P = P_1 \cup P_2$. Moreover, there is a one to one correspondence between the set of paths $P_1$ and the set of $s$-$t'$ paths in the graph $G-t$.
Thus, we get an easy recursive algorithm to find the set of paths $s$-$t$ paths in a graph $G$. Pick an edge $tt'$ incident the vertex $t$ and recursively calculate the sets $P_1$ and $P_2$. With a small amount of pre-processing, we can ensure that the runtime is $O(m(p+1))$ where $m$ is the number of edges and $p$ is the number of $s$-$t$ paths.
To make the recurrence relation on the runtime work, consider the following. We can test in time $O(m)$ if a given graph $G$ and pair of vertices $s$ and $t$ if $G$ has 0, exactly one, or at least two distinct $s$-$t$ paths. To see this, simply find the block decomposition of the graph and, and check if there is any non-trivial block between $s$ and $t$ in the tree.
We can push this slightly farther. Given an instance of the problem $G$ and vertices $s$ and $t$, we can reduce the problem in time $O(m)$ to a graph $\bar{G}$ and vertices $x$ and $y$ such that for all edges $xx'$ incident to $x$, we have that
To see this, again using the block decomposition, we contract any bridge in the graph and delete edges not contained in any $s$-$t$ path. As above, this can be done in $O(m)$ time.
We give an $O(m(p-1))$ time algorithm to find the set of $s$-$t$ paths in a given graph $G$ with at least two $s$-$t$ paths.
Step 1 can be performed in $c'm$ operations for some constant $c'$. The initial check in steps $2$ and $3$ can be performed in $c'(m-1)$ steps. If we must recursively run the algorithm, this will require another $c(m-1)(p_i - 1)$ operations for $i = 1, 2$ in Steps 2 and 3, respectively. As $p_i \ge 1$, we can bound the work in each of Steps 2 and 3 by $c'm + cm(p_i - 1)$. Thus, the total number of operations is at most $3c'm + c(m)(p_1 +p_2 - 2) \le cm(p-1)$ if we choose $c \ge 3c'$.
Let $G=(V,E)$ be a graph.
$FindPaths(p,f)$ prints all paths which end in $f$ and can be obtained by adding nodes to path $p$. $p$ is for path, $f$ is for final (node).
Def $FindPaths(p,f)$:
Let $x$ be the last node of $p$.
For each edge $xy$ for some $y$ in $E$
$\ \ \ \ $If $y$ is not in $p$
$\ \ \ \ $$\ \ \ \ $If $y=f$
$\ \ \ \ $$\ \ \ \ $$\ \ \ \ $Print $p-y$
$\ \ \ \ $$\ \ \ \ $Else
$\ \ \ \ $$\ \ \ \ $$\ \ \ \ $$FindPaths(p-y,f)$
If $s$ is the start node and $t$ is the ending node, run $FindPaths(s,t)$.
You can represent path and edges as strings. To check if a node is in a path $p$ you just have to check whether the string contains the character that represents the node. To get the final node of a path use the function to get the last character of a string.
EDIT: My answer is not math research level, but introduction to programming.
For shortest paths look at the Wikipedia page of the Floyd–Warshall algorithm
and also:
"An algorithm for computing all paths in a graph" by Lars-Erik Thorelli in BIT Numerical Mathematics, July 1966, Volume 6, Issue 4, pp 347-349
If I'm not mistaken, I think an adaptation of a dynamic programming all-pairs-shortest-path algorithm (like the Floyd-Warshall algorithm, considering edge weights of 1) might find all paths. Consider the following scheme to find the total number of all paths leading from u to v (or, in fact, from any start node to any destination):
A matrix $M_1$ is initialized as the adjacency matrix of the graph. That is, $M_1[u,v]$ containes the number of simple paths of length at most 1 from u to v. After that, for all $i$ from 2 to the number of nodes the matrix $M_i$ is updated as follows: $M_i[u,v]$ equals the sum of the entries $M_{i-1}[u,w]$ for all nodes w adjacent to v. Hence, $M_i[u,v]$ containes the number of simple paths of length at most i from u to v. This scheme runs in time polynomial in the input size, and it can also be modified easily to also store the actual paths in time (and space) polynomial in the output size.
