Path cover with paths of bounded length, in the plane
I have a weighted, undirected, Euclidean complete graph $G$, a special vertex $r$, and an upper bound $b$. I want to find a minimum-cost path cover that covers all vertices of $G$, subject to the restriction that every path must have length at most $b$ and every path must start at the vertex $r$. In other words, I want to minimize the sum of the lengths of the paths, subject to the requirement that the paths cover every vertex at least once and each path starts at $r$ and has bounded length.
Are there any algorithms for this problem that will be effective on medium-size graphs (with, say, at most 200 vertices)? This problem smells likely to be NP-hard; if so, are there exact algorithms that are more efficient than naive exhaustive search or heuristics that are likely to work well in practice?
Recall that the length of a path is the sum of the lengths of the edges it contains. Also, a path cover is a collection of paths such that every vertex belongs to at least one path. A Euclidean complete graph is a graph where each vertex represents a point in the plane, there is an edge between every pair of vertices, and the length of that edge whose length is given by the Euclidean distance between those two points. (It follows that, without loss of generality, we can assume no two paths have any vertex in common other than $r$.)
This question is motivated by Location Selection Algorithm in Solar Engineering.
algorithms graphs heuristics weighted-graphs set-cover
D.W.♦D.W.
$\begingroup$ This is certainly NP-hard: given any Euclidean Longest Path problem, add a new vertex $r$ sufficiently far away from every vertex, and make $b$ big enough to allow all vertices to be contained in a single path. $\endgroup$ – j_random_hacker Dec 14 '16 at 15:14
$\begingroup$ Idea: If you have a pair of vertices $u, v$ such that $u$ is the nearest neighbour of $v$, then there is an optimal solution in which $u$ is not the endpoint of a path. (Suppose it was: Then delete any single edge adjacent to $v$ (there could be either 1 or 2 of these) and connect $v$ to $u$ instead for a solution with cost at least as good.) $\endgroup$ – j_random_hacker Dec 14 '16 at 16:13
This smells like a problem that might be NP-hard, so you'll probably have to use one of the standard methods for dealing with intractability: e.g., devise a heuristic, or formulate it so you can apply an existing solver.
I suggest that you try formulating it as a mixed integer linear programming (ILP) problem, and if that's too slow, try simulated annealing. I'll sketch both approaches.
Integer programming
We'll think of our task as selecting a subset of the edges, so that they can be pieced together into paths as required. Introduce a zero-or-one variable for each edge to indicate whether or not it is selected. In other words, for each pair of vertices $i,j$, introduce a zero-or-one variable $x_{i,j}$, with the intended meaning that $x_{i,j}=1$ implies the edge $(i,j$) is included.
We'll add a bunch of inequalities that capture the requirement that the selected edges form a bunch of paths of the desired form:
Each vertex (other than $r$) has exactly one edge out of it: $\sum_j x_{i,j} = 1$ for each $i \ne r$.
Each vertex (other than $r$) has at most one edge into it: $\sum_i x_{i,j} \le 1$ for each $j \ne r$.
$r$ has no edges coming out of it: $x_{r,j} = 0$ for all $j$.
For each vertex $i$ introduce a variable $y_i$ (not restricted to integer values), with the intended meaning that $y_i$ captures the distance from $r$ to $i$ along the harness that connects the two. We'll require that no harness is longer than the limit: $y_i \le b$ for all $i$. Also $y_r = 0$.
Finally, we'll ensure the consistency of the $x$'s and $y$'s by adding
$$y_i \ge y_j + \ell(i,j) - (b + \ell(i,j)) (1-x_{i,j})$$
for all $i,j$, where $\ell(i,j)$ is the length of the edge $(i,j)$. (Notice how this works: if edge $(i,j)$ is not selected, i.e., $x_{i,j}=0$, this inequality implies only $y_i \ge 0$; but if edge $(i,j)$ is selected, this inequality ensures that $i$'s distance from $r$ is $j$'s distance from $r$ plus the length of the edge between $i$ and $j$.)
Now ask an ILP solver to minimize the objective function $\sum_{i,j} \ell(i,j) x_{i,j} $, subject to these inequalities. The solution will correspond the optimal solution to the original problem.
You'll have to test to see whether existing off-the-shelf ILP solvers (e.g., CPLEX) are able to solve this ILP instance efficiently. You end up with an ILP instance with about $|V|^2$ variables and inequalities. If it takes too long, some solvers will have an option to specify a "timeout", so they give you the best solution they've found so far within the specified time, even if it isn't necessarily the globally optimal solution.
Some minor optimizations are possible. For instance, we can omit any edge $(i,j)$ whose length exceeds $b$ (eliminating the variable $x_{i,j}$ from the linear program, as it is forced to be zero). Also, for any pair of edges $(i,j),(k,l)$ that "cross" (i.e., their line segments intersect), we know they can't both be simultaneously selected, i.e., we can add the inequality $x_{i,j}+x_{k,l}\le 1$. You probably don't want to add all such inequalities, but you could try adding a small number (e.g., for all pairs of short edges). You can also add the inequality $y_i \ge \ell(r,i)$ for all $i$, in case that helps the solver.
This approach is also fairly flexible: for instance, we can limit the number of paths that are allowed to be used (by adding an inequality to upper-bound $\sum_i x_{i,r}$) or make other minor adjustments. You can limit the number of edges allowed in each path by adding variables $z_i$ that count the number of hops between $i$ and $r$ along the harness connecting the two, and adding the inequality $z_i \ge z_j + 1 - K (1-x_{i,j})$ where $K$ is a suitably large constant (larger than the maximum number of edges per path) and then adding inequalities to upper-bound all of the $z_i$'s.
Simulated annealing
Another option is to use some form of local search, such as simulated annealing. The idea is that we start from a candidate solution and at each step try making a small, randomly selected, local change and check whether it is an improvement or not.
Part of the art of simulated annealing is to choose a set of allowable changes; at each step, you'll randomly select from this set to obtain your candidate change. In this case, one reasonable class of "changes" is: Pick a pair of vertices (probably ones that aren't too far apart), and swap which paths they are in. For example, suppose I've selected vertex $b$ and vertex $e$, where the path containing $b$ looks like $\cdots \to a \to b \to c \to \cdots$ and the path containing $e$ looks like $\cdots \to d \to e \to f \to \cdots$. Then after the swap operation you'll end up with the two paths $\cdots \to a \to e \to c \to \cdots$ and $\cdots \to d \to b \to f \to \cdots$. You can then evaluate whether this increased or decreased the objective function. You might want to experiment with the distribution on pairs of vertices (do you try to make nearby vertices more likely to be selected than far-apart ones?).
There may be other operations that would be useful to add and make simulated annealing work better. Selecting an appropriate set of operations and choosing the parameters for simulated annealing is largely an art, and may require some experimentation. It's hard to predict how well this will work without trying it.
$\begingroup$ "This smells like a problem that might be NP-hard" -- did you see my comment giving a simple reduction from Euclidean Longest Path? Is there a problem with it? $\endgroup$ – j_random_hacker Dec 16 '16 at 1:57
$\begingroup$ @j_random_hacker, I missed it earlier, but now that I read it, it looks like solid to me! Nice reduction. $\endgroup$ – D.W.♦ Dec 16 '16 at 5:17
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