MST with largest edge

1. Give an example of a weighted undirected graph that is not a tree, but whose minimum spanning tree contains the largest edge in the graph.

         1.1
       0--------0
   6.7/ \        \2.4
     /   \5.2     0
    0     \       /2.3
           0-----0
             3.3

Largest cycle edge excluded

2. Let \(G\) be a weighted, undirected graph. Prove that for any cycle in \(G\), the minimum spanning tree does not include the largest edge in the cycle.

(Common wrong answer on Jamboard.)

Proof considering components (sketch)

• Suppose there exists a minimum spanning tree \(T\) that includes a maximum edge \(e_\text{max} = uv\) of some cycle. Remove \(e_\text{max}\) from \(T\).

• The tree is no longer a minimum spanning tree but instead two components, one containing \(u\) and the other containing \(v\). But since there was a cycle in \(G\) containing \(e_{max}\), there is a path in \(G\) from \(u\) to \(v\). Therefore, some edge \(e\) on this path must connect the two components.

• Exchange \(e\) and \(e_\text{max}\), contradicting minimality.

Disprove with a counterexample

3. Prove or disprove: The minimum spanning tree of a weighted, undirected graph \(G\) includes the smallest edge in every cycle in \(G\).

     (10)     (3)      (11)
e --------- a --- b ----------- d
 \           \   /             /
  \_      (2) \ / (1)        _/
    \          c            /
     \_                   _/
       \                 /
   (16) \_             _/ (15)
          \           /
           \_       _/
             \     /
              \_ _/
                f

Minimum Spanning Tree of a Weighted Graph

• Given a weighted graph \(G\), a minimum spanning tree \(T\) is a subgraph of \(G\) such that:
  • \(T\) contains all the vertices of \(G\), and
  • The sum of weights of the edges of \(T\) is as small as possible.
• An intermediate spanning forest is a subgraph of the MST.

For any intermediate spanning forest \(F\) of a graph \(G\),

• An edge of \(G\) is useless if it is not an edge of \(F\), but both its endpoints are in the same component of \(F\).
• An edge of \(G\) is safe if, in some component of \(F\), it is the minimum-weight edge with exactly one endpoint in the component.

Prim’s Theorem

Theorem. Let \(F\) be an intermediate spanning forest of a weighted graph \(G\), and suppose \(e\) is a safe edge. Then the minimum spanning tree contains \(e\).

• Let \(C\) be the component of \(F\) that \(e\) is “safe” for, so \(e\) has an endpoint in \(C\) and an endpoint in \(G \setminus C\), and \(e\) is the smallest such edge.
• Suppose to the contrary that the minimum spanning tree \(T\) does not contain \(e\). The endpoints \(u\) and \(v\) of \(e\) must be connected by a path in \(T\), and since exactly one of the endpoints of \(e\) is in \(C\), one of the edges \(e'\) of this path must have an endpoint in \(C\) and an endpoint not in \(C\).
• Removing \(e'\) from \(T\) creates a spanning forest with two components, one containing \(u\) and another containing \(v\). Adding \(e\) to this forest connects these components, and so creates a tree. But this new tree has smaller total weight than \(T\), because \(e\) is smaller than \(e'\). That’s a contradiction.

Greedy MST Meta-Algorithm

Let \(G\) be a weighted graph with vertex set \(V\) and edge set \(E\).

MetaMST(V,E)
    F = (V, ∅)
    while F is not a tree
        add a safe edge to F   // NOTE: This choice needs to be specified!!
    return F

(See Jamboard from last time.)

General idea

Use MetaMST, but add all the safe edges inside the while-loop.

MetaMST(V,E)
    F = (V, ∅)
    while F is not a tree
        add a safe edge to F   // NOTE: This choice needs to be specified!!
    return F

In order to analyze the time, we need to specify how the safe edges are located.

Table Groups

Label the vertices with their component number

CountAndLabel(G):                         LabelOneComponent(s,count):
    count <- 0                                Enqueue(s)
    for all vertices v                        while the queue is not empty
        unmark v                                  v <- Dequeue
    for all vertices v                            if v is unmarked
        if v is unmarked                              mark v               // Line #1
            count <- count + 1                        v.comp <- count
            LabelOneComponent(v, count)               for each edge vw 
    return count                                          Enqueue(w)       // Line #2

Suppose \(G\) has \(V\) vertices and \(E\) edges.

1. What is another name for the LabelOneComponent algorithm?

2. How many times does CountAndLabel execute Line #1?

3. How many times does CountAndLabel execute Line #2?

4. What is the running time of CountAndLabel?

Borůvka’s Algorithm

Borůvka(V, E):                              AddAllSafeEdges(E, F, count):
    F = (V, ∅)                                  for i <- 1 to count
    count <- CountAndLabel(F)                       safe[i] <- Null
    while count > 1                             for each edge uv ∈ E
        AddAllSafeEdges(E, F, count)                if comp(u) != comp(v)
        count <- CountAndLabel(F)                       if safe[comp(u)] = Null or w(uv) < w(safe[comp(u)])
    return F                                                safe[comp(u)] <- uv
                                                        if safe[comp(v)] = Null or w(uv) < w(safe[comp(v)])
                                                            safe[comp(v)] <- uv
                                                for i <- 1 to count
                                                    add safe[i] to F   

Borůvka’s Algorithm

Borůvka(V, E):                              AddAllSafeEdges(E, F, count):
    F = (V, ∅)                                  for i <- 1 to count
    count <- CountAndLabel(F)                       safe[i] <- Null
    while count > 1                             for each edge uv ∈ E
        AddAllSafeEdges(E, F, count)                if comp(u) != comp(v)
        count <- CountAndLabel(F)                       if safe[comp(u)] = Null or w(uv) < w(safe[comp(u)])
    return F                                                safe[comp(u)] <- uv
                                                        if safe[comp(v)] = Null or w(uv) < w(safe[comp(v)])
                                                            safe[comp(v)] <- uv
                                                for i <- 1 to count
                                                    add safe[i] to F   

5. What is the running time of the first CountAndLabel call?

6. After the first AddAllSafeEdges call, how many components (at most) are there? (i.e., what is the maximum possible value of count?)

7. What is the worst-case number of iterations of the while-loop?

8. What is the running time of subsequent CountAndLabel calls?

9. Give an upper bound on the running time of AddAllSafeEdges.

Bonus Reading

Read pp. 262-263. Upshot: Borůvka’s Algorithm often beats its worst-case \(O(E \log V)\) running time.