EDIT: The most general case I need is not a tree but any Directed Acyclic Graph.

I have a directed acyclic graph.

I need to sort it in a list so that in the list every node comes after any node it can walk to in the graph.

So far so good, just use a topological sorting algorithm.

However, I have an additional constraint: Each node in the graph is colored, and I need to find a sorting so that the list is as cohesive as possible. In other words, if you walk the list, you should see a change in colors as few times as possible.

Other thoughts:

  • I need an efficient algorithm. No iterating through all possible sortings.

  • The set of colors is typically much smaller than the set of nodes in the graph.

  • If no efficient exact algorithm is possible, I would be happy with a heuristic one that is likely to produce good results.

  • There is some information that one could potentially use for the heuristic: The colors are also partially ordered, and this ordering of the colors is correlated to the ordering of the nodes. I.e. a node is likely (but not guaranteed) to have a color that is "smaller than" the color of its parent node.

  • You can think of the problem this way: I have a set of jobs, and each job can depend on the output of several other jobs (if x depends on y there is an arrow from x to y). There is an overhead associated with setting up the data needed for each job, but this overhead is significantly reduced if the previous job had the same "color". The whole set of jobs is typically executed many thousand times, so if one can find an not too expensive way to improve the order of the jobs, that will pay off.

So far I have come up with this algorithm which works well most times, but not always:

first pass:

Define the order "<=" so that x <= y if y can walk to x in the graph.

find any topological sort of the nodes and put it in the list L (L is sorted from smallest to largest w.r.t the partial ordering "<=" )

W = []
for x in L:
   find the earliest y in W so that color(x)==color(y) && there is no z in W with z > y and x > z
   if y exists:
       insert x in W after y.
   if y does not exist and there is no z in W with x > z:
       place x at the beginning of W
       place x at the end of W

second pass:

 for each continuous group G of colors in W:
    for x in G in reverse order:
       find the latest y in W so that color(x)==color(y) && there is no z in W with z > x and z < y
       if y exists:
           move x directly before y in W

The second pass may seem a little puzzling, but it is just my experience that it tends to improve the results. I.e. some times it will be able to move one whole group of colors past another group and join a third group of the same color.

I think the reason that the first pass is not sufficient is that there can be complicated interdependencies between the nodes that are sorted in L after the current x you are looking at, and so you can't know the optimal placement of x in the first pass, you just have to go with something.

Any help would be great!

  • $\begingroup$ Reading the begining of the question, I see "tree" and "topological sorting". Can you precise if your graph is directed or not please ? $\endgroup$
    – Optidad
    Jan 30, 2019 at 14:41
  • $\begingroup$ @Vince I think that what is meant is that we have a partial order defined by the transitive closure of the parent relation in a forest. Finding a total order that is consistent with that partial order can be done with a topological sort on the DAG formed by creating edges from parent to child, although there are other methods, of course. $\endgroup$
    – Discrete lizard
    Jan 30, 2019 at 20:23
  • $\begingroup$ Hum okay. This reminds me a very old game with lot of colored tiles. Each turn, the player who has initially one tile chose a new color and transforms all his tiles to this color absorbing the adjacent tiles of this color to grow up his own area. I don't retrieve the name of this game, but if it has an optimal strategy, it fits to this problem. $\endgroup$
    – Optidad
    Jan 30, 2019 at 21:26
  • $\begingroup$ Hi, I realized that the most general case is not a tree, but indeed a DAG. Original post has been edited. $\endgroup$ Jan 31, 2019 at 10:22

1 Answer 1


Suppose you know the list of colors that appear in contiguous parts of an optimal ordering. Then, you can easily find this optimal ordering: simply put as many nodes of the current color in your list as the parent-child order allows and proceed with the next color in your list. This greedy strategy is optimal, because once you have started a color, you will not 'regret' taking a node of the same color later as it adds no cost and only gives more options. If you put all 'free' nodes that already have their parents processed in separate lists, this can be done in $O(n)$, where $n$ is the number of nodes in your forest.

Of course, we do not know this list, so we should figure out how to find it. The number of color lists you would have to check in an uninformed search is $c(c-1)^{k}$, where $c$ is the number of colors you have and $k$ the minimal number of color switches in the final order. If you're lucky, $c$ and $k$ are sufficiently small for a brute-force search and we are done.

However, this will quickly become unfeasible, so I doubt you can afford to spend the $O(c(c-1)^k n)$ time required to do a brute-force search$^1$. Finding an optimal list 'looks hard'. I have no formal proof of e.g. NP-hardness, but it is clear that decisions made early on depend on the structure of the entire forest, so the most natural approach is to try to create subdivisions of by 'cutting' the forest to get subproblems and use e.g. dynamic programming. However, since the nodes can be ordered in many ways, I do not see how to represent these cuts with only polynomially many states. Another hint is that you can see this as a special case of the scheduling problem with partial order requirements.

So, since I think it likely we cannot find an efficient exact solution, you should try an heuristic approach or can perhaps even find an approximation algorithm. I will finish with some observations that may help you in developing such an approach:

  • If we decide on the last color in list first and build the ordering backwards, observe that we can only choose nodes where all descendants have the same color: if not, there is a node of another color that should be later in the list. Since we would select all children of the same color as a parent if we had chosen a front to back order in an optimal ordering, we should select only subtrees of the same color where the root has no parent of the same color. It is possible that for some colors, there are no such subtrees, which means that we have less options for the final color.

  • If there is a color for which you can select all remaining nodes, you should select that color. This may occur quite often if the correlation between the color ordering and node ordering is high.

  • One idea for an heuristic is to select the color that is 'blocking' the most nodes of other colors: you can look at the total number of 'chains' of nodes of the same color that are accessible after removing the color or at the number for a specific color you would like to remove first or the maximum number over all colors.

$^1$: You could be slightly more clever and do recursive backtracking where you can prune some searches, but this has the same worst case running time.

  • $\begingroup$ I am not so sure how similar the approach in the question is to the approach I sketch here. If it turns out to be the same thing, I think this is at least a way to look at the problem that allows more improvements. $\endgroup$
    – Discrete lizard
    Jan 30, 2019 at 20:21
  • 1
    $\begingroup$ Another decent heuristic (for choosing colours in the "forward" direction) is: When you run out of available nodes of the current colour, pick the colour with the most free nodes next. This guarantees the best lower bound on the number of nodes that can be processed before the next colour change is necessary; and, if the colour distribution is reasonably uniform, it yields the highest expected number of nodes too. $\endgroup$ Jan 31, 2019 at 15:06

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