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MST

MST

Many of the slides are from Prof. Plaisteds resources at University of North Carolina at Chapel Hill

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Motivation: Minimum

Spanning Trees

To minimize the length of a connecting network, it

never pays to have cycles.

The resulting connection graph is connected,

undirected, and acyclic, i.e., afree tree (sometimes

called simply a tree).

This is the minimum spanning tree orMST

problem.

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Formal Definition of MST

Given a connected, undirected, graph G = (V, E), a

spanning tree is an acyclic subset of edges T Ethat

connects all the vertices together.

Assuming G is weighted, we define the costof a spanning

tree Tto be the sum of edge weights in the spanning tree

w(T) = (u,v)Tw(u,v)

A minimum spanning tree (MST) is a spanning tree of

minimum weight.

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Figure1 : Examples of MST

Not only do the edges sum to the same value, but the

same set of edge weights appear in the two MSTs.

NOTE: An MST may not be unique.

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Steiner Minimum Trees (SMT)

Given a undirected graph G = (V, E)with edge weights and

a subset of vertices V V, called terminals. We wish to

compute a connected acyclic subgraph ofG that includes all

terminals. MST is just a SMT with V=V.

Figure 2: Steiner Minimum Tree

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Generic Approaches

Two greedy algorithms for computing MSTs:

Kruskals Algorithm

Prims Algorithm

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Facts about (Free) Trees

A tree with n vertices has exactly n-1 edges

(|E| = |V| - 1)

There exists a unique path between any two vertices of a

tree

Adding any edge to a tree creates a unique cycle; breaking

any edge on this cycle restores a tree

For details see CLRS Appendix B.5

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Intuition Behind Greedy MST

We maintain in a subset of edgesA, which will initiallybe empty, and we will add edges one at a time, until

equals the MST. We say that a subsetA Eis viable ifA

is a subset of edges in some MST. We say that an edge

(u,v) E-A issafe ifA{(u,v)} is viable. Basically, the choice (u,v) is a safe choice to add so thatA

can still be extended to form an MST. Note that ifA is

viable it cannot contain a cycle. A generic greedyalgorithm operates by repeatedly adding anysafe edge to

the current spanning tree.

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Generic-MST (G, w)

1. A //A trivially satisfies invariant

// lines 2-4 maintain the invariant

2. whileA does not form a spanning tree

3. do find an edge (u,v) that is safe forA

4. AA {(u,v)}

5. returnA //A is now a MST

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Definitions

A cut(S, V-S) is just a partition of the vertices into 2

disjoint subsets. An edge (u, v) crosses the cut if one

endpoint is in Sand the other is in V-S. Given a subset of

edges A, we say that a cut respects A if no edge in Acrosses the cut.

An edge ofEis a light edge crossing a cut, if among all

edges crossing the cut, it has the minimum weight (the

light edge may not be unique if there are duplicate edge

weights).

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When is an Edge Safe?

If we have computed a partial MST, and we wish to

know which edges can be added that do NOT induce

a cycle in the current MST, any edge that crosses a

respecting cut is a possible candidate.

Intuition says that since all edges crossing a

respecting cut do not induce a cycle, then the lightest

edge crossing a cut is a natural choice.

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MST Lemma

Let G = (V, E)be a connected, undirected graph

with real-value weights on the edges. LetA be a

viable subset ofE(i.e. a subset of some MST), let

(S, V-S) be any cut that respectsA, and let (u,v)

be a light edge crossing this cut. Then, the edge

is safe forA.

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Proof of MST Lemma

Must show thatA {(u,v)} is a subset of some

MST

Method:

1. Find arbitrary MST TcontainingA

2. Use a cut-and-paste technique to find another MST T

that containsA {(u,v)}

This cut-and-paste idea is an important proof

technique

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Figure

Figure 3: MST Lemma

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Step 1

Let Tbe any MST forG containingA.

We know such a tree exists becauseA is viable.

If(u, v)is in Tthen we are done.

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Constructing T

If (u, v)is not in T, then add it to T, thus creating a cycle.

Since u and v are on opposite sides of the cut, and since

any cycle must cross the cut an even number of times,

there must be at least one other edge (x, y) in Tthatcrosses the cut.

The edge (x, y)is not inA (because the cut respectsA).By removing (x,y) we restore a spanning tree, T.

Now must show

Tis a minimum spanning tree

A {(u,v)} is a subset ofT

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Conclusion of Proof

Tis an MST: We have w(T) = w(T) - w(x,y) + w(u,v)

Since (u,v) is a light edge crossing the cut, we have

w(u,v) w(x,y). Thus w(T) w(T). So Tis also a

minimum spanning tree.

A {(u,v)} T: Remember that (x, y) is not inA. Thus

A T - {(x, y)}, and thus

A {(u,v)} T - {(x, y)} {(u,v)} = T

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MST Lemma: Reprise

Let G = (V, E)be a connected, undirected graph with

real-value weights on the edges. LetA be a viable subset

ofE(i.e. a subset of some MST), let (S, V-S) be any cut

that respectsA, and let (u,v) be a light edge crossing thiscut. Then, the edge is safe forA.

Point of Lemma: Greedy strategy works!

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Basics of Kruskals Algorithm

Attempts to add edges toA in increasing order of weight(lightest edge first)

If the next edge does not induce a cycle among the current

set of edges, then it is added to A. If it does, then this edge is passed over, and we consider

the next edge in order.

As this algorithm runs, the edges ofA will induce a forest

on the vertices and the trees of this forest are mergedtogether until we have a single tree containing all vertices.

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Detecting a Cycle

We can perform a DFS on subgraph induced by the edgesofA, but this takes too much time.

Use disjoint set UNION-FIND data structure. This

data structure supports 3 operations:Create-Set(u): create a set containing u.Find-Set(u): Find the set that contains u.Union(u, v): Merge the sets containing u and v.

Each can be performed in O(lg n) time.

The vertices of the graph will be elements to be stored inthe sets; the sets will be vertices in each tree ofA (storedas a simple list of edges).

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MST-Kruskal(G, w)

1. A // initially A is empty

2. for each vertex v V[G] // line 2-3 takes O(|V|) time

3. do Create-Set(v) // create set for each vertex

4. sort the edges ofEby nondecreasing weight w5. for each edge (u,v) E, in order bynondecreasing weight

6. do if Find-Set(u) Find-Set(v) // u&v on different trees

7. then AA {(u,v)}

8. Union(u,v)

9. returnA

Total running time is O(|E| lg |E|)=O(|E| lg |V|).

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Example: Kruskals Algorithm

Figure 4: Kruskals Algorithm

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Analysis of Kruskal

Lines 1-3 (initialization): O(V)

Line 4 (sorting): O(E lg E)

Lines 6-8 (set-operation): O(E log E)

Total: O(E log E)

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Correctness of Kruskal

Idea: Show that every edge added is a safe edge forA

Assume (u, v) is next edge to be added toA. Will not create a cycle

LetAdenote the tree of the forestA that containsvertex u. Consider the cut (A, V-A). This cut respectsA (why?)

and (u, v) is the light edge across the cut (why?)

Thus, by the MST Lemma, (u,v) is safe.

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Intuition behind Prims Algorithm

Consider the set of vertices Scurrently part of the tree, andits complement (V-S). We have a cut of the graph and the

current set of tree edgesA is respected by this cut.

Which edge should we add next? Light edge!

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Basics of Prim s Algorithm

It works by adding leaves on at a time to the current

tree.

Start with the root vertex r(it can be any vertex).At any

time, the subset of edgesA forms a single tree. S= vertices

ofA.

At each step, a light edge connecting a vertex in Sto a

vertex in V- Sis added to the tree.

The tree grows until it spans all the vertices in V.

Implementation Issues:

How to update the cut efficiently?

How to determine the light edge quickly?

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Implementation: Priority Queue

Priority queue implemented using min-heap can support

the following operations in O(lg n) time: Insert (Q, u, key): Insert u with the key value key in Q

u = Extract_Min(Q): Extract the item with minimum key value

in Q

Decrease_Key(Q, u, new_key): Decrease the value ofus key

value to new_key

All the vertices that are notin the S(the vertices of the

edges inA) reside in a priority queue Q based on a keyfield. When the algorithm terminates, Q is empty. A =

{(v, [v]): v V - {r}}

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Example: Prims Algorithm

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MST-Prim(G, w, r)

1. QV[G]

2. for each vertex u Q // initialization: O(V) time

3. do key[u]

4. key[r]0 // start at the root

5. [r]NIL // set parent ofrto be NIL

6. while Q // until all vertices in MST

7. do uExtract-Min(Q) // vertex with lightest edge

8. for each v adj[u]

9. do ifv Q and w(u,v) < key[v]

10. then [v]u

11. key[v]w(u,v) // new lighter edge out ofv

12. decrease_Key(Q, v, key[v])

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Analysis of Prim

Extracting the vertex from the queue: O(lg |V|)

For each incident edge, decreasing the key of the

neighboring vertex: O(lg |V|)

The other steps are constant time.

The overall running time is, where

T(n) = uV(lg |V| + deg(u)lg |V|)

= uV(1+ deg(u))lg |V|

= lg |V| (|V| + 2|E|) = O(|E| lg |V|)

Essentially same as Kruskals

If use Fibonacci Heap, O(|E| + |V| lg |V|).

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Correctness of Prim

Again, show that every edge added is a safe edge forA

Assume (u, v) is next edge to be added toA.

Consider the cut (A, V-A).

This cut respectsA (why?) and (u, v) is the light edge across the cut (why?)

Thus, by the MST Lemma, (u,v) is safe.