Suppose that one has a directed, acyclic graph G, and each vertex $v$ contains a (positive) value $a_v$. Additionally, let $r$ be a constant. For my purposes, $r>1$, but this might not matter. Let $n$ be the number of vertices in G and let $[n]:=\{1,2,\ldots, n\}$.

A topological sort of $G$ is a bijection $G\to[n]$ such that if there is a path from $v$ to $w$, then $\tau(v)<\tau(w)$. Alternately, if we view $G$ as defining a partial ordering, a topological sort is a total ordering extending the partial ordering.

I would like to find $\displaystyle\min_{\tau} \sum_v a_v r^{\tau (v)}$, where we are taking the minimum over all topological sorts of G. It may help to generalize the problem and to finding $\displaystyle\min_{\tau} \sum_v a_v p({\tau (v)})$, where $p:[n]\to \mathbb R$ is a penalty/weight function (perhaps assumed to take positive values and be monotonic).

There are two extreme cases. If $G$ has no edges, we would sort things in ascending or descending order depending on if $r<1$ or $r>1$. If $G$ is already a linear order, there is nothing to be done. Already, the problem seems nontrivial given two disjoint linear orderings, where the problem reduces to the optimal riffle shuffle.

So is there a good algorithm for solving this? I know of some heuristics which help in certain cases, and I can use a bubble-sort type algorithm to get "local" minima, but unless there is a way to recast the problem, I don't see a good way to solve it.

Added later: I want to extend my comment and explain why I view dynamic programming to be insufficient. At best, this will clarify what I'm looking for. At worst, this will reveal a gap in my understanding which someone can clarify.

For there to be a dynamic programming solution, there need to be sub-problems which can be built upon to get a larger solution. For example, when searching for a path through a graph with edge lengths, if a minimal length path passes through a particular vertex, then the path from from the start to that vertex must be of minimal length. If we keep track of all the vertices that can be reached in time less than $t$, then we can ignore all paths through those close vertices which do not begin with a minimal path, and so we need to remember at most one path to any vertex, and at every stage we only need to find the shortest path to an unvisited vertex which is an extension of a known minimal path. This gives $O(n^2)$ storage costs and $O(nm)$ time costs where $n$ is the number of vertices and $m$ is the number of edges.

The obvious sub-problem to use for the problem at hand is that, if we know an initial/terminal segment for an optimal solution, the restriction to the subgraph containing just those elements will yield the same initial/terminal segment. It does not appear that we can say anything stronger. The algorithm this yields is as follows:

  1. Select all vertices with no predecessors, and put each of these singletons into a list of admissible initial segments.
  2. (Definition) For an admissible initial segment of length $n$, we say that an extension of length $n+1$ is admissible if it satisfies both the topological constraints of the directed graph and, if no topologically allowed insertion of the new element has a lower total value. From the collection of all admissible extensions of length $n+1$.
  3. Given the collection of all admissible segments of length $n$, form the collection of all admissible extensions of length $n+1$.
  4. From the collection generated in (3), if any two admissible extensions use exactly the same collection of vertices, remove the segment with a higher associated cost.
  5. Loop through (3) and (4) until you have found the minimal initial segment containing every vertex of the graph.

If the penalty function is monotonic increasing (e.g., if $r>1$), we can improve run time somewhat by adding a heuristic (total cost of adding everything else at the minimum possible distance, ignoring that only one item can be in any particular spot), but even with this improvement, we have the following fundamental problem:

The algorithm doesn't require checking every initial segment, but it does require examining every collection of vertices which could form an allowable initial segment. In the worst case scenario, this is exponential in the number of vertices (though is significantly reduced when there are severe topological constraints). Additionally, in the worst case scenario, the space requirements are on the order of $\binom{n}{n/2}$ where there are $n$ vertices.

The dynamic programming algorithm is still quite an improvement over more naive algorithms, but I would like to find something that runs in polynomial time, or else show that such an algorithm cannot exist.

  • $\begingroup$ Sounds like a dynamic programming algorithm might do the trick. Start by assuming that the last node in the sort is $v$, then build up partial topo-sorts from there, taking at each step the partial sort with the minimal cost function for your best sub-problem solution. $\endgroup$
    – JeremyKun
    Oct 16, 2011 at 19:12
  • 1
    $\begingroup$ @Bean A dynamic programming algorithm would work if I knew that the restriction of an optimal sort to a subgraph was still optimal. This statement is not obvious to me. If it were true, this would allow a divide and conquer + merge approach. In particular, dynamic programming does solve the "two disjoint linear orders" problem. However, without being able to divide and conquer, there appear to be too many possible members for a decent sized initial/final segment of the sort for this to be tractable. $\endgroup$
    – Aaron
    Oct 16, 2011 at 19:44
  • $\begingroup$ And here is a counterexample that the restriction of an optimal solution is optimal. Consider $r=2$ and $(1\to 16)(4)$. The optimal sort is $1,16,4$. However, if we restrict to just $1$ and $4$, the optimal sort is $4,1$. $\endgroup$
    – Aaron
    Oct 16, 2011 at 21:11
  • $\begingroup$ Whether r>1 or r<1 does not matter because the case of r<1 can be handled by considering 1/r and reversing the result. $\endgroup$ Oct 18, 2011 at 12:02
  • $\begingroup$ This is an interesting question. If you start with two linear segments of length $N$, then they can be riffled together in $O(N^2)$ time. The question seems to be whether the resulting order is always stable when a third linear segment is added. I.e., is the optimal order for three linear segments the same as if two linear segments were riffled together and then the third were riffled optimally into the resulting, longer, linear segment? $\endgroup$
    – mjqxxxx
    May 25, 2016 at 17:51

2 Answers 2


Some cases of your problem are solvable by greedy approach. For example $r \ge n$ and $\{\,a_1, a_2, \ldots, a_n\,\} = \{\,1, 2, \ldots, n\,\}$. In this case penalty function tells that the last vertex in our ordering should have the smallest possible value and no matter how all other vertices are placed. Topological order says that it also should have no outgoing arcs. So we should place there a vertex with smallest value among all vertices without outgoing arcs and solve a subproblem. Using heap we can get $O(n \log n + m)$ time for $n$ vertices and $m$ arcs.

In general case the problem seems to be NP-hard. However I failed to prove this fact.

  • $\begingroup$ What makes you think NP-hard? Do you have a polynomial-time algorithm for verifying a solution as optimal? $\endgroup$
    – orlp
    Feb 26, 2018 at 22:30
  • $\begingroup$ @orlp In general NP-hardness of a problem doesn`t mean that the problem belongs to NP. However proper yes/no question for decision vertion of our problem is not whether a provided solution is optimal (because it doesn't look significantly simpler than the original problem), but whether there exists good enough solution, i. e., whether for given $w$ there exists $\tau$ such that $\sum_{v} a_v r^{\tau(v)} \le w$ (and it is easy to verify that answer is yes). $\endgroup$
    – Smylic
    Dec 15, 2019 at 23:42

The generalization of your problem(the one with penalty function) is NP-hard, see the comment to this question.

  • $\begingroup$ Wow, thanks. I realize that the special case that I was initially interested in might still be more tractable, but this is an interesting enough answer and the question has been open for long enough that I will accept it. $\endgroup$
    – Aaron
    Mar 1, 2018 at 6:25

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