I believe this is true but have not been able to get a formal proof for either. But is it true that any minimum spanning tree is reachable by applying Kruskal's algorithm? Similarly, is this true for Prim's algorithm?

EDIT: To be more precise, I want to know if given an MST for a connected, undirected, weighted graph, is it guaranteed that there is a sequence of steps using Kruskal or Prim which generates this MST. E.g. There are different choices for Kruskal when there are multiple edges with the same weight. Similarly for Prim.

As indicated by Raphael's comment and j_random_hacker's comment, the answer is positive. In fact, any MST is reachable by any MST algorithm with some minor exceptions.

For a graph $G$, two weight functions on all edges (to real numbers) are defined as (weakly) comparison-compatible (to each other) if we can extend the strict weak ordering on the edges induced by either weight function to the same strict total order. That is, we can devise consistent tie-breaking rules with each weight function so that the result of comparison of any two edges by one weight function and its tie-breaking rules is the same as the result by the other weight function and its tie-breaking rules.

Lemma 1: Let $w_1$ and $w_2$ be two weight functions. The following five statements are equivalent to each other.

  1. $w_1$ and $w_2$ are comparison-compatible.
  2. In any set of edges, there is a common lightest edge by $w_1$ and by $w_2$.
  3. In any set of edges, there is a common heaviest edge by $w_1$ and by $w_2$.
  4. There exists a weight function $w_3$ that assign distinct weights to distinct edges such that $w_3$ is comparison-compatible to $w_1$ and to $w_2$.
  5. for any edge $e_1$ and $e_2$ such that $e_1$ is lighter than $e_2$ by one weight function, $e_1$ is as light as or lighter than $e_2$ by the other weight function.

Proof of lemma 1 is left as an easy exercise.

Lemma 2: Let two weight functions $w_1$ and $w_2$ be such that if $e_1\lt_{w_1}e_2$, then $e_1\lt_{w_2}e_2$. Then they are comparison-comptible.

(Hint to) Proof: One approach is to verify the condition 5 of lemma 1. Another approach is to verify the condition 2 of lemma 1 by showing that in any set of edges, a lightest edge by $w_2$ is also a lightest edge by $w_1$,

A comparison-based algorithm on a graph $G$ is defined as comparison-compatible if for any two (weakly) comparison-compatible weight functions on all edges, we can run the algorithm with one weight function in a way that can be repeated without any change with the other weight function. In particular, those two runs of the algorithm will have the same output.

Please note that most if not all MST algorithms can be stated in two flavours. The first flavour assumes that distinct edges of $G$ have distinct weights, which is used when the main concern is to find one MST (which is in fact also the unique MST). The second flavour allows distinct edges of $G$ to have same weights. Of course in this answer, where the main concern is to find all MSTs, we will only care MST algorithms in the second flavour.

A comparison-compatible MST algorithm can find all MSTs.

To prove the above proposition, we just need to adapt slightly the section, "Kruskal can find every MST" in the computation of the number of MSTs. For any MST $m$ of $G$ with weight function $w_1$, choose a positive weight that is lighter than the absolute difference between any pair of unequal edge weights. If we subtract that weight from the weight of each edge in $m$ without changing weights of other edges, we obtain a new weight function, say, $w_2$. If edge $e_1$ is lighter than $e_2$ by $w_1$, $e_1$ must be lighter than $e_2$ by $w_2$ as well. By lemma 2, $w_1$ and $w_2$ are comparison-compatible. Note that $m$ is the unique MST with $w_2$. (One way to show this uniqueness is to prove that whenever the weight of one MST edge is reduced, any MST with the new weight function must contain that edge.) So any run of the algorithm on $w_2$ will find $m$. Because the algorithm is comparison-compatible, we can run the algorithm the same way with $w_1$ or with $w_2$. Since that run will find the unique MST, $m$ with $w_2$, it will find $m$ as well with $w_1$.

Every MST algorithm is comparison-compatible

The above proposition sounds overbroad. Well, by every MST algorithm, I mean any general comparison-based MST algorithm that I have seen, excluding those apparently pathological ones such as wrong or having unnecessary steps. Since a comparison-compatible MST algorithm can find all MSTs, every MST algorithm can find all MSTs. In particular, each of the four classic MST algorithms, namely Borůvka's, Prim's, Kruskal's and reverse-delete algorithms can find all MSTs.

Here are three more comparison-compatible MST algorithms.

  • the delete-heavy-edge algorithm. Start with all edges. Repeatedly find a cycle and remove one of its heaviest edge until no cycle remains.
  • the add-non-heavy-edge algorithm. Start with the empty set. Iterate through all edges. Each edge is added and, if a cycle is formed, remove one of it heaviest edges.
  • the replace-by-lighter-edge algorithm. Start with any spanning tree $T$. Find the cycle in $T$ plus an edge $e$ not in $T$. If an edge $t$ in that cycle is heavier than $e$, remove $t$ from $T$ and add $e$ to $T$. Repeat until we cannot reduce the weight of $T$ any more.

The following MST algorithm is not comparison-compatible.

  • the degree-biased Kruskal's algorithm, which is Kruskal's algorithm with the following modification. Suppose when we are going to remove an edge with the minimum weight from $S$ as in the wikipedia's description of Kruskal's algorithm, we have multiple edges with the minimum weight to choose. The edge we choose to remove will be an edge whose sum of degrees of its two vertices is the largest among all choices. This algorithm cannot be comparison-compatible since it does not find the MST $\{ab,bc,cd\}$ of the graph with vertices $a,b,c$, and edge $ab$ of weight $1$ and edges $bc, cd, db$ of weight $2$. This algorithm is considered pathological because of that unnecessary modification.

Please note that all eight MST algorithms mentioned above are comparison-based.

How to show an algorithm is comparison-compatible?

I will use Kruskal's algorithm as an example. Here is the description of Kruskal's algorithm (in the second flavour) on a weighted undirected connected graph $G$.

  • create a graph $F$ that has the same vertices as $G$ but without edges. So $F$ is a forest of separated trees of a single vertex.
  • create a set $S$ containing all the edges in the graph.
  • while $S$ is nonempty and $F$ is not yet spanning
    • select an edge with minimum weight from $S$.
    • remove that edge from $S$.
    • If that edge connects two different trees then add it to the forest $F$, combining two trees into a single tree
  • output $F$.

Let $w_1$ and $w_2$ be two comparison-compatible weight functions on $G$. Kruskal's algorithm involves the weight function only in the step that "select an edge with minimum weight from $S$". The condition 2 of lemma 1 tells us that we can select a common lightest edge in this step. We can then check easily by induction that we can run all steps in the same way with $w_1$ and with $w_2$. So Kruskal's algorithm is comparison-compatible.

An algorithm is comparison-compatible if, in loose terms, it can be described in three kinds of steps.

  1. do something that does not involve weight.
  2. select an edge with the minimum weight in a given set of edges
  3. select an edge with the maximum weight in a given set of edges

This sufficient condition covers Borůvka's, Prim's, Kruskal's, reverse-delete, delete-heavy-edge and add-non-heavy-edge algorithm. Note that in order to fit this sufficient condition, we may have to change certain descriptions of an algorithm without affecting the set of reachable MSTs. Because of the exception of the degree-biased Kruskal's algorithm being comparison-compatible, let me emphasize this sufficient condition is described in loose terms

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.