COMS 482: unofficial class blog

Minimum Spanning Tree (MST):

Given a connected graph G=(V,E), a spanning tree is a connected subgraph on V. A minimal spanning tree is a spanning tree with the minimal number of edges.

“A spanning tree is a tree.”

To not be a tree, it must (a) be not connected or (b) have a cycle. By definition, (a) is false, because an MST is defined as a connected subgraph. For (b), if a cycle exists we can erase any edge in the cycle and still maintain connectedness.

Each edge has a cost. The minimum spanning tree is a least total cost tree. For example:

MST Algorithm:

• Choose all edges, dropping expensive edges until you have a complete tree
• Choose least cost edges that do not make a cycle (Kruskal’s)
• Build a single tree, always adding least cost edges (Prim’s)

Kruskal’s Algorithm:

Sort edges by cost
T = {}
for e in edges do
    if e does not create a cycle then
        add e to T
    endif
endfor

Prim’s Algorithm:

Choose a start vertex s
T = {s}
While there is a vertex not in T do
    choose least cost edge from T to some vertex not in T
    add this edge to T
endwhile

“Kruskal’s algorithm produces an MST”

Let k₁, … , k_n-1 be edges added by Kruskal’s algorithm in the order added. Assume k₁, … , k_n-1 is not an MST. Then let k_i be the first edge added that makes it impossible to extend k₁, … , k_i to an MST. Consider a true MST that uses k₁, … , k_i-1. Let k₁, … , k_i-1,m_i, …, m_n-1 be the true MST. If we add k_i to k₁, … , k_i-1,m_i, …, m_n-1 (true MST) then we get a cycle. This cycle contains some edge m_j where j >= i. The cycle cannot be all k’s because Kruskal’s algorithm would not have added k.

Consider the graph from adding k_i and removing m_j in k₁, … , k_i-1,m_i, …, m_n-1. We still have a spanning tree, with three cases:

1. |k_i| < |m_j| => Contradiction, because the new tree has less cost than the true MST
2. |k_i| > |m_j| => Contradiction, because m_j would have been chosen by Kruskal's algorithm before k_i.
3. |k_i| = |m_j| => Contradiction, because now both trees are MST and have the same cost, but k_i was chosen to not be in an MST with k₁, … , k_i.

This entry was posted on Monday, January 31st, 2005 at 2:10 pm and is tagged with minimum spanning tree, nbsp nbsp nbsp nbsp nbsp, kruskal s algorithm, connected graph, minimal spanning tree, connected subgraph, spanning trees, single tree, minimal number, k1, vertex, mj, contradiction, lt. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback.

One Response to 'Lecture 4: Minimum Spanning Trees'

1. 
   The COMS 482 Blog » Blog Archive » Final Review: Chapters 1 to 4 said:

   on May 14th, 2005 at 9:42 pm

   […] All of these MST algorithms can be implemented in O(m log n) time. Lecture 4 contains a proof that Kruskal’s algorithm produces an MST, lecture […]

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