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23.1 Growing a minimum spanning tree

23.1-1

Let $(u, v)$ be a minimum-weight edge in a connected graph $G$. Show that $(u, v)$ belongs to some minimum spanning tree of $G$.

Theorem 23.1 shows this.

Let $A$ be the empty set and $S$ be any set containing $u$ but not $v$.

23.1-2

Professor Sabatier conjectures the following converse of Theorem 23.1. Let $G = (V, E)$ be a connected, undirected graph with a real-valued weight function $w$ defined on $E$. Let $A$ be a subset of $E$ that is included in some minimum spanning tree for $G$, let $(S, V - S)$ be any cut of $G$ that respects $A$, and let $(u, v)$ be a safe edge for $A$ crossing $(S, V - S)$. Then, $(u, v)$ is a light edge for the cut. Show that the professor's conjecture is incorrect by giving a counterexample.

Let $G$ be the graph with $4$ vertices: $u, v, w, z$. Let the edges of the graph be $(u, v), (u, w), (w, z)$ with weights $3$, $1$, and $2$ respectively.

Suppose $A$ is the set $\{(u, w)\}$. Let $S = A$. Then $S$ clearly respects $A$. Since $G$ is a tree, its minimum spanning tree is itself, so $A$ is trivially a subset of a minimum spanning tree.

Moreover, every edge is safe. In particular, $(u, v)$ is safe but not a light edge for the cut. Therefore Professor Sabatier's conjecture is false.

23.1-3

Show that if an edge $(u, v)$ is contained in some minimum spanning tree, then it is a light edge crossing some cut of the graph.

Let $T_0$ and $T_1$ be the two trees that are obtained by removing edge $(u, v)$ from a $\text{MST}$. Suppose that $V_0$ and $V_1$ are the vertices of $T_0$ and $T_1$ respectively.

Consider the cut which separates $V_0$ from $V_1$. Suppose to a contradiction that there is some edge that has weight less than that of $(u, v)$ in this cut. Then, we could construct a minimum spanning tree of the whole graph by adding that edge to $T_1 \cup T_0$. This would result in a minimum spanning tree that has weight less than the original minimum spanning tree that contained $(u, v)$.

23.1-4

Give a simple example of a connected graph such that the set of edges $\{(u, v):$ there exists a cut $(S, V - S)$ such that $(u, v)$ is a light edge crossing $(S, V - S)\}$ does not form a minimum spanning tree.

A triangle whose edge weights are all equal is a graph in which every edge is a light edge crossing some cut. But the triangle is cyclic, so it is not a minimum spanning tree.

23.1-5

Let $e$ be a maximum-weight edge on some cycle of connected graph $G = (V, E)$. Prove that there is a minimum spanning tree of $G' = (V, E - \{e\})$ that is also a minimum spanning tree of $G$. That is, there is a minimum spanning tree of $G$ that does not include $e$.

Let $A$ be any cut that causes some vertices in the cycle on once side of the cut, and some vertices in the cycle on the other. For any of these cuts, we know that the edge $e$ is not a light edge for this cut. Since all the other cuts won't have the edge $e$ crossing it, we won't have that the edge is light for any of those cuts either. This means that we have that e is not safe.

23.1-6

Show that a graph has a unique minimum spanning tree if, for every cut of the graph, there is a unique light edge crossing the cut. Show that the converse is not true by giving a counterexample.

Suppose that for every cut of $G$, there is a unique light edge crossing the cut. Let us consider two distinct minimum spanning trees, $T$ and $T'$, of $G$. Because $T$ and $T'$are distinct, $T$ contains some edge $(u, v)$ that is not in $T'$. If we remove $(u, v)$ from $T$, then $T$ becomes disconnected, resulting in a cut $(S, V - S)$. The edge $(u, v)$ is a light edge crossing the cut $(S, V - S)$ (by Exercise 23.1-3) and, by our assumption, it's the only light edge crossing this cut. Because $(u, v)$ is the only light edge crossing $(S, V - S)$ and $(u, v)$ is not in $T'$, each edge in $T'$ that crosses $(S, V - S)$ must have weight strictly greater than w$(u, v)$. As in the proof of Theorem 23.1, we can identify the unique edge $(x, y)$ in $T'$ that crosses $(S, V - S)$ and lies on the cycle that results if we add $(u, v)$ to $T'$. By our assumption, we know that $w(u, v) < w(x, y)$. Then, we can then remove $(x, y)$ from $T'$ and replace it by $(u, v)$, giving a spanning tree with weight strictly less than $w(T')$. Thus, $T'$ was not a minimum spanning tree, contradicting the assumption that the graph had two unique minimum spanning trees.

Here's a counterexample for the converse:

Here, the graph is its own minimum spanning tree, and so the minimum spanning tree is unique. Consider the cut $(\{x\}, \{y, z\})$. Both of the edges $(x, y)$ and $(x, z)$ are light edges crossing the cut, and they are both light edges.

23.1-7

Argue that if all edge weights of a graph are positive, then any subset of edges that connects all vertices and has minimum total weight must be a tree. Give an example to show that the same conclusion does not follow if we allow some weights to be nonpositive.

First, we show that the subset of edges of minimum total weight that connects all the vertices is a tree. To see this, suppose not, that it had a cycle. This would mean that removing any of the edges in this cycle would mean that the remaining edges would still connect all the vertices, but would have a total weight that's less by the weight of the edge that was removed. This would contradict the minimality of the total weight of the subset of vertices. Since the subset of edges forms a tree, and has minimal total weight, it must also be a minimum spanning tree.

To see that this conclusion is not true if we allow negative edge weights, we provide a construction. Consider the graph $K_3$ with all edge weights equal to $-1$. The only minimum weight set of edges that connects the graph has total weight $-3$, and consists of all the edges. This is clearly not a $\text{MST}$ because it is not a tree, which can be easily seen because it has one more edge than a tree on three vertices should have. Any $\text{MST}$ of this weighted graph must have weight that is at least $-2$.

23.1-8

Let $T$ be a minimum spanning tree of a graph $G$, and let $L$ be the sorted list of the edge weights of $T$. Show that for any other minimum spanning tree $T'$ of $G$, the list $L$ is also the sorted list of edge weights of $T'$.

Suppose that $L'$ is another sorted list of edge weights of a minimum spanning tree. If $L' \ne L$, there must be a first edge $(u, v)$ in $T$ or $T'$ which is of smaller weight than the corresponding edge $(x, y)$ in the other set. Without loss of generality, assume $(u, v)$ is in $T$.

Let $C$ be the graph obtained by adding $(u, v)$ to $L'$. Then we must have introduced a cycle. If there exists an edge on that cycle which is of larger weight than $(u, v)$, we can remove it to obtain a tree $C'$ of weight strictly smaller than the weight of $T'$, contradicting the fact that $T'$ is a minimum spanning tree.

Thus, every edge on the cycle must be of lesser or equal weight than $(u, v)$. Suppose that every edge is of strictly smaller weight. Remove $(u, v)$ from $T$ to disconnect it into two components. There must exist some edge besides $(u, v)$ on the cycle which would connect these, and since it has smaller weight we can use that edge instead to create a spanning tree with less weight than $T$, a contradiction. Thus, some edge on the cycle has the same weight as $(u, v)$. Replace that edge by $(u, v)$. The corresponding lists $L$ and $L'$ remain unchanged since we have swapped out an edge of equal weight, but the number of edges which $T$ and $T'$ have in common has increased by $1$.

If we continue in this way, eventually they must have every edge in common, contradicting the fact that their edge weights differ somewhere. Therefore all minimum spanning trees have the same sorted list of edge weights.

23.1-9

Let $T$ be a minimum spanning tree of a graph $G = (V, E)$, and let $V'$ be a subset of $V$. Let $T'$ be the subgraph of $T$ induced by $V'$, and let $G'$ be the subgraph of $G$ induced by $V'$. Show that if $T'$ is connected, then $T'$ is a minimum spanning tree of $G'$.

Suppose that there was some cheaper spanning tree than $T'$. That is, we have that there is some $T''$ so that $w(T'') < w(T')$. Then, let $S$ be the edges in $T$ but not in $T'$. We can then construct a minimum spanning tree of $G$ by considering $S \cup T''$. This is a spanning tree since $S \cup T'$ is, and $T''$ makes all the vertices in $V'$ connected just like $T'$ does.

However, we have that

$$w(S \cup T'') = w(S) + w(T'') < w(S) + w(T') = w(S \cup T') = w(T).$$

This means that we just found a spanning tree that has a lower total weight than a minimum spanning tree. This is a contradiction, and so our assumption that there was a spanning tree of $V'$ cheaper than $T'$ must be false.

23.1-10

Given a graph $G$ and a minimum spanning tree $T$, suppose that we decrease the weight of one of the edges in $T$. Show that $T$ is still a minimum spanning tree for $G$. More formally, let $T$ be a minimum spanning tree for $G$ with edge weights given by weight function $w$. Choose one edge $(x, y) \in T$ and a positive number $k$, and define the weight function $w'$ by

$$ w'(u, v) = \begin{cases} w(u, v) & \text{ if }(u, v) \ne (x, y), \\ w(x, y) - k & \text{ if }(u, v) = (x, y). \end{cases} $$

Show that $T$ is a minimum spanning tree for $G$ with edge weights given by $w'$.

Let $x(T) = \sum_{(x, y) \in T} w(x, y)$. We have $w'(T) = w(T) - k$. Consider any other spanning tree $T'$, so that $w(T) \le w(T')$.

If $(x, y) \ne T'$, then $w'(T') = w(T') \ge w(T) > w'(T)$.

If $(x, y) \in T'$, then $w'(T') = w(T') - k \ge w(T) - k = w'(T)$.

Either way, $w'(T) \le w'(T')$, and so $T$ is a minimum spanning tree for weight function $w'$.

23.1-11 $\star$

Given a graph $G$ and a minimum spanning tree $T$, suppose that we decrease the weight of one of the edges not in $T$. Give an algorithm for finding the minimum spanning tree in the modified graph.

If we were to add in this newly decreased edge to the given tree, we would be creating a cycle. Then, if we were to remove any one of the edges along this cycle, we would still have a spanning tree. This means that we look at all the weights along this cycle formed by adding in the decreased edge, and remove the edge in the cycle of maximum weight. This does exactly what we want since we could only possibly want to add in the single decreased edge, and then, from there we change the graph back to a tree in the way that makes its total weight minimized.