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Let $G$ be a graph, and let $s$ and $t$ be two vertices of $G$. Can we efficiently sample a shortest $s$-$t$ path uniformly and independently at random from the set of all shortest paths between $s$ and $t$? For simplicity, we can assume $G$ is simple, undirected and unweighted.

Even in many restricted graphs, the number of shortest paths between $s$ and $t$ can be exponential in the size of $G$. Therefore, we would naturally like avoid actually computing all the shortest $s$-$t$ paths. I don't know about the general case, but it seems to me that we can achieve this for some special graph classes.

This feels like something someone must have considered before. Is there any existing research into this, or is this in fact simple to do even for general graphs?

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  • $\begingroup$ Good question Juho. While considering an answer, what do you precisely understand by "sampling a s-t path uniformly at random"? If it suffices for s and t to be picked up randomly the question is trivial so I guess that you mean that all nodes in the shortest path appear with a frequency (ie., probability) which follows a uniform distribution. Or is there any other definition? In particular, for bipartite graphs your question seems to be very easy, isn't it? $\endgroup$ – Carlos Linares López Nov 11 '13 at 20:02
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    $\begingroup$ @CarlosLinaresLópez Consider say the diamond graph, and say $s$ is on the right side of the "vertical edge", and $t$ is on the left side. Now there are 2 shortest paths between $s$ and $t$. The algorithm should return with equal probability either one of these two paths. So $s$ and $t$ are not "picked up randomly", but they are given as input. Does that make it clear? In this sense, I'm not sure if the problem is really easy for bipartite graphs. $\endgroup$ – Juho Nov 11 '13 at 20:20
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    $\begingroup$ @CarlosLinaresLópez In other words, we are given a graph $G$, and two vertices $s,t \in V(G)$. Let $S$ be the set of all shortest paths between $s$ and $t$. Output an element of $S$ uniformly at random. $\endgroup$ – Juho Nov 11 '13 at 20:36
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I am not 100% sure this answer is correct, but here goes:

I think you can reduce this to uniformly random any-paths, from $s-t$, in a DAG with a single source and a single sink.

Given a graph $G$

  1. Make a new empty digraph, $H$.
  2. First: run the BFS part of Dijkstra's shortest-path, starting from $s$, mark all the nodes with their shortest-distance-from-$s$.
  3. Let $d(s,v)$ be the minimum distance from $s-v$; which we know from the BFS step of Dijkstra's shortest-path algorithm.
  4. Then do the next step of Dijkstra's shortest-path algorithm, obtain the shortest-path, store it in $\mathbf p$ (by going backwards from $t$ to $s$).
  5. Now start the following loop; expanation in comments, and below:
    • $q_0=\{t\}$
    • While $q_0 \ne \emptyset$
      • $q_1= \emptyset$
      • For $u \in q_0$
        • So we want to find all possible next nodes for this shortest-subpath from $t-u$
        • For all $\text{edge}(u,v) \in G$ such that $d(s,v) < d(s,u)$
          • $v$ is a neighboring node, with less $d(s,\cdot)$ (it will be $1$ less)
          • Therefore, $t-u-v$ is possible subpath in a shortest path.
          • Put $v \rightarrow H, \text{di-edge}(u,v)\rightarrow H$
          • Now we need to check $v$'s lesser-neighbors next turn.
          • Put $v \rightarrow q_1$
      • Set $q_0$ to $q_1$:
        • $q_0 \leftarrow q_1$

Essentially, I am collecting all possible nodes that can be used in the shortest-path, and placing them in $H$.

More on how this works:

Dijkstra's shortest-path algorithm works by first running a BFS, and marking all the nodes $v\in G$ with their-shortest paths from $s-v$. The next step is to go back from $t-s$, and follow the least-neighboring nodes back.

The thing is, here you can choose any of the least neighboring nodes. What I do here is collect all the least-neighboring nodes each step, which means I account for all the shortest-paths.

Now you quickly think, but hey, why is enumerating them exponential, but my way is not?

The answer is, because I use a set to avoid adding the same nodes twice, I avoid recalculating this for each possible path.

Now we have a DAG that we can traverse in any way from $t-s$, and obtain a shortest-reversed-path from $s-t$. The graph should have $t$ as the only source, and $s$ as the only sink.


If the above is correct, then I think we can take this a step further and solve the problem as follows.

Give each node in the DAG a node-weight; the node-weight will be the number of paths from from that node to $s$. Let us call this $w(v)$.

You can compute these quickly, see Algorithm that finds the number of simple paths from s to t in G.

Once we have the node-weight, we can uniformly pick a path by:

  • Layout the DAG as a level-structure (for visualization)
  • At each level, choose an arbitrary ordering between the nodes ie. a notion of "left to right".
  • Traversing down the DAG: at each step $i$, $i \in \left[1,\left|\mathbf p\right|\right]$ (where $|\cdot|$ means size-of, in this case, the length of the shortest-path):
    • Let $u_i$ be the current node (starting at $t$)
    • Add up all the weights of the children of $u_i$, and using an RNG, choose one child node, $v_i$, uniformly between the weighted children.
    • Set $u_{i+1} = v_i$, and go to the next step
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  • $\begingroup$ The level-structure, and notion of left-to-right were part of my initial attempt to simply generate $r\in \left[0,w(t)\right)$, and choose a path that way, but I didn't figure that out, so you can safely ignore them. $\endgroup$ – Realz Slaw Nov 11 '13 at 22:26
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    $\begingroup$ This answer looks great! I love the ideas! I tried to write it out in a slightly different way (in my answer), as a test of my understanding. In any case, I just wanted to share my appreciation for this lovely answer! $\endgroup$ – D.W. Nov 12 '13 at 0:18
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Here is a solution based upon the ideas in Realz Slaw's answer. It is basically a re-exposition of his ideas that might be clearer or easier to follow. The plan is that we will proceed in two steps:

  1. First, we will build a graph $S$ with the following property: any path from $s$ to $t$ in $S$ is a shortest path from $s$ to $t$ in $G$, and every shortest path from $s$ to $t$ in $G$ is also present in $S$. Thus, $S$ contains exactly the shortest paths in $G$: all the shortest paths, and nothing more. As it happens, $S$ will be a DAG.

  2. Next, we will sample uniformly at random from all paths from $s$ to $t$ in $S$.

This approaches generalizes to an arbitrary directed graph $G$, as long as all edges have positive weight, so I'll explain my algorithm in those terms. Let $w(u,v)$ denote the weight on the edge $u \to v$. (This generalizes the problem statement you gave. If you have an unweighted graph, just assume every edge has weight 1. If you have an undirected graph, treat each undirected edge $(u,v)$ as the two directed edges $u\to v$ and $v\to u$.)


Step 1: extract $S$. Run a single-source shortest-paths algorithm (e.g., Dijkstra's algorithm) on $G$, starting from source $s$. For each vertex $v$ in $G$, let $d(s,v)$ denote the distance from $s$ to $v$.

Now define the graph $S$ as follows. It consists of every edge $u \to v$ such that (1) $u \to v$ is an edge in $G$, and (2) $d(s,v) = d(s,u) + w(u,v)$.

The graph $S$ has some convenient properties:

  • Every shortest path from $s$ to $t$ in $G$ exists as a path in $S$: a shortest path $s=v_0,v_1,v_2,\dots,v_k=t$ in $G$ has the property that $d(s,v_{i+1})=d(s,v_i)+w(v_i,v_{i+1})$, so the edge $v_i \to v_{i+1}$ is present in $S$.

  • Every path in $S$ from $s$ to $t$ is a shortest path in $G$. In particular, consider any path in $S$ from $s$ to $t$, say $s=v_0,v_1,v_2,\dots,v_k=t$. Its length is given by the sum of the weights of its edges, namely $\sum_{i=1}^k w(v_{i-1},v_i)$, but by the definition of $S$, this sum is $\sum_{i=1}^k (d(s,v_i)-d(s,v_{i-1})$, which telescopes to $d(s,t)-d(s,s)=d(s,t)$. Therefore, this path is a shortest path from $s$ to $t$ in $G$.

  • Finally, the absence of zero-weight edges in $G$ implies that $S$ is a dag.

Step 2: sample a random path. Now we can throw away the weights on the edges in $S$, and sample a random path from $s$ to $t$ in $S$.

To help with this, we will do a precomputation to compute $n(v)$ for each vertex $v$ in $S$, where $n(v)$ counts the number of distinct paths from $v$ to $t$. This precomputation can be done in linear time by scanning the vertices of $S$ in topologically sorted order, using the following recurrence relation:

$$n(v) = \sum_{w \in \text{succ}(v)} n(w)$$

where $\text{succ}(v)$ denotes the successors of $v$, i.e., $\text{succ}(v) = \{w : v \to w \text{ is an edge in $S$}\}$, and where we have the base case $n(t)=1$.

Next, we use the $n(\cdot)$ annotation to sample a random path. We first visit node $s$. Then, we randomly choose one of the successors of $s$, with successor $w$ weighted by $n(w)$. In other words:

choosesuccessor(v):
    n = 0
    for each w in succ(w):
        n = n + n(w)
    r = a random integer between 0 and n-1
    n = 0
    for each w in succ(w):
        n = n + n(w)
        if r < n:
            return w

To choose a random path, we repeatedly iterate this process: i.e., $v_0=s$, and $v_{i+1} =$ choosesuccessor$(v_i)$. The resulting path is the desired path, and it will be sampled uniformly at random from all shortest paths from $s$ to $t$.

Hopefully this helps you understand Realz Slaw's solution more easily. All credit to Realz Slaw for the beautiful and clean solution to this problem!


The one case this doesn't handle is the case where some edges have weight 0 or negative weight. However, the problem is potentially not well-defined in that case, as you can have infinitely many shortest paths.

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  • $\begingroup$ Glad you took the time to fully get my answer; I wasn't sure it is correct. Now I am vindicated :D. $\endgroup$ – Realz Slaw Nov 11 '13 at 23:05

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