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Minimum Spanning Tree (MST)

- Input
- A (simple) graph G = (V,E)
- Edge cost cij for each edge e E
- Optimization Problem
- Find a minimum-cost spanning tree
- Spanning tree: a set of |V|-1 edges T such that each vertex is incident to at least one edge in T, and T contains no cycles.

MST Example: Some Feasible Spanning Trees

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cost = 17

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cost = 16

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cost = 14

cost = 16

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MST Optimality Conditions: Path Optimalityc25 > c12

c25 > c15

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c34 > c23

c34 > c12

c34 > c15

c34 > c45

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c24 > c12

c24 > c15

c24 > c45

Path Optimality Condition

- A spanning tree T is a minimum spanning tree if and only for every out-of-tree edge (i,j), cij cuv for every in-tree edge (u,v) on the path from i to j in T .
- This is clearly a necessary condition for a MST. If an out-of-tree edge (i, j) has a lower cost than any in-tree edge (u,v) on the path from i to j in T, then we can improve T by replacing edge (u,v) with edge (i, j).

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Path Optimality: Necessity6

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Replacing in-tree edge (2,4) with out-of-tree edge (4,5) would decrease the cost of this tree. Therefore, it cannot be optimal.

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c25 >c12

c24 > c12

c45 > c12

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MST Optimality Conditions: Cut Optimality6

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Removing an in-tree edge (u,v) creates a cut in the tree.

For any out-of-tree edge (i,j) crossing the cut cij cuv.

Cut Optimality

- Given a spanning tree T, let C[u,v] be the set of edges in the cut formed by removing in-tree edge (u, v) from T.
- A spanning tree T is a MST if and only if every in-tree edge is a minimum cost edge in C[u,v].

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Cut Optimality: Necessity6

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The in-tree edge (2, 4) is not a minimum cost edge in the cut formed by removing it from T.

If we remove in-tree edge (2, 4), we can restore connectivity by adding out-of-tree edge (4, 5).

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Cut Optimality: Necessity6

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If we remove in-tree edge (2, 4), we can restore connectivity by adding out-of-tree edge (4, 5).

Since the resulting tree has a lower total cost, the first tree could not have been optimal.

Sufficiency of Cut and Path Optimality

- It is easy to see that any MST must satisfy the cut and path optimality conditions.
- If a tree T happens to satisfy these conditions, does that imply that T is a MST?

Sufficiency of Cut Optimality

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G

T*

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- Let T* be a MST of G.
- Suppose that TC satisfies the cut optimality condition.
- Show that TC is also a MST.

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TC

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Sufficiency of Cut Optimality

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T*

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Since T* is a MST, it must satisfy the path optimality condition.

So, c25 c24 (and c25 c45).

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Since TC satisfies the cut optimality condition, c25 c24 (and c25 c35 ).

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TC

This implies c25= c24.

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Replacing (2,4) with (2,5) in T* creates a new MST, T**.

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Sufficiency of Cut Optimality

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T**

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Since T** is a MST, it must satisfy the path optimality condition.

So, c23 c35 (and c23 c25).

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Since TC satisfies the cut optimality condition, c23 c35 (and c25 c13 ).

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TC

This implies c23= c35.

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Replacing (3,5) with (2,3) in T** creates a new MST, T***.

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Sufficiency of Cut Optimality

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T***

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Since TC is identical to T***, it is optimal.

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This argument can be formalized to apply to the general case and show that any tree that satisfies the cut optimality condition must be optimal.

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Thus, cut optimality is a sufficient condition for a tree to be a MST.

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Kruskal’s Algorithm (Path Optimality)

- F := E
- T := {}
- Repeat Until |T| = |V| - 1
- Begin
- Select (i,j) F such that cij = min(cuv: (u,v) F)
- F := F \ {(i,j)}
- If T {(i,j)} does not contain a cycle then
- T := T {(i,j)}
- End

Can be implemented in O(|E|+|V| log |V}) time plus the time for sorting the edges.

Testing for Cycles

- Let GT be the subgraph of G induced by the set of edges in T.
- As edges are added to T, the algorithm creates a forest (i.e., a collection of trees).
- Each tree in the forest forms a connected component of GT.
- By keeping track of which component each node is in, we can quickly, and easily determine whether or not adding a new edge to T will create a cycle.

Testing for Cycles

Initialize component[i] = 0 for all i V.

When edge (i,j) is inspected, there are 5 cases to

- component[i] = component[j] = 0

Add (i,j) to T; (i,j) becomes a new component of GT.

- component[i] = 0, component[j] > 0.

Add (i,j) to T; vertex i will go into the same component as j.

- component[i] > 0, component[j] = 0.

Add (i,j) to T; vertex j will go into the same component as i.

- component[i] > 0, component[j] > 0, component[i] component[j]

Add (i,j) to T; merge components.

- component[i] = component[j] > 0

Adding (i,j) would create a cycle.

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Kruskal’s Algorithm: Example 22

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(2, 3) creates a cycle because vertices 2 and 3 are in the same connected component.

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Kruskal’s Algorithm: Example 22

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(2, 4) does not create a cycle because vertices 2 and 4 are in different connected components.

Prim’s Algorithm (Cut Optimality)

- Choose any node v in V.
- S1 := {v}
- T := {}
- k := 1
- Repeat Until |T| = |V| - 1
- Begin
- Select a minimum cost edge (i,j) in the cut [Sk,V\{Sk}]
- T := T {(i,j)}
- If i Sk then Sk+1 := Sk {j}
- Else If j Sk then Sk+1 := Sk {i}
- k := k+1
- End

Can be implemented in O(|E|+|V| log |V}) time.

Directed Spanning Tree (DST)

- Input
- A network (directed graph) G = (N,A)
- Arc cost cij for each edge arc (i,j) A
- A designated root node s N
- Optimization Problem
- Find a minimum-cost directed-out tree T = (N, A*) rooted at node s.
- Directed-Out Tree: T = (N, A*) contains a unique, directed path from s to every other node in N (every other node in the tree has in-degree = 1).

Is DST a (Pure) Network Flow Problem?

- Suggestion:
- Let b(s) = |N| - 1
- Let b(i) = -1 for i in N\{s}
- Let ij = 0
- Let uij = |N|
- Each node gets 1 unit of flow
- Total cost of flow is minimized
- Arcs carrying flow form a directed spanning tree

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