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Minimum Spanning Tree Algorithms

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Minimum Spanning Tree Algorithms

Prim’s Algorithm

Kruskal’s Algorithm

Runtime Analysis

Correctness Proofs

- A spanning tree for an undirected graph G=(V,E) is a subgraph of G that is a tree and contains all the vertices of G
- Can a graph have more than one spanning tree?
- Can an unconnected graph have a spanning tree?

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(u,v) T w(u,v ) is minimized

Mst T: w( T )=

- The weight of a subgraph is the sum of the weights of it edges.
- A minimum spanning tree for a weighted graph is a spanning tree with minimum weight.
- Can a graph have more then one minimum spanning tree?

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The Problem

- Several pins of an electronic circuit must be connected using the least amount of wire.
Modeling the Problem

- The graph is a complete, undirected graph G = ( V, E ,W ), where V is the set of pins, E is the set of all possible interconnections between the pairs of pins and w(e) is the length of the wire needed to connect the pair of vertices.
- Find a minimum spanning tree.

We will show two ways to build a minimum spanning tree.

- A MST can be grown from the current spanning tree by adding the nearest vertex and the edge connecting the nearest vertex to the MST. (Prim's algorithm)
- A MST can be grown from a forest of spanning trees by adding the smallest edge connecting two spanning trees. (Kruskal's algorithm)

- Tree-vertices: in the tree constructed so far
- Non-tree vertices: rest of vertices

Prim’s Selection rule

- Select the minimum weight edge between a tree-node and a non-tree node and add to the tree

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Select a vertex to be a tree-node

while (there are non-tree vertices) {

if there is no edge connecting a tree node with a non-tree node

return “no spanning tree”

select an edge of minimum weight between a tree node and a non-tree node

add the selected edge and its new vertex to the tree}

return tree

- How is the graph implemented?
- Assume that we just added node u to the tree.
- The distance of the nodes adjacent to u to the tree may now be decreased.
- There must be fast access to all the adjacent vertices.
- So using adjacency lists seems better

- How should the set of non-tree vertices be represented?
- The operations are:
- build set
- delete node closest to tree
- decrease the distance of a non-tree node from the tree
- check whether a node is a non- tree node

- The operations are:

- How should the set of non-tree vertices be represented?
- A priority queue PQ may be used with the priority D[v] equal to the minimum distance of each non-tree vertex v to the tree.
- Each item in PQ contains: D[v], the vertex v, and the shortest distance edge (v, u) where u is a tree node

- This means:
- build a PQ of non-tree nodes with initial values -
- fast build heap O (V )
- building an unsorted list O(V)
- building a sorted list O(V) (special case)

- build a PQ of non-tree nodes with initial values -

- delete node closest to tree (extractMin)
- O(lg V ) if heap and
- O(V) if unsorted list
- O(1) sorted list

- decrease the distance of a non-tree node to the tree
- We need to find the location i of node v in the priority queue and then execute (decreasePriorityValue(i, p)) where p is the new priority
- decreasePriorityValue(i, p)
- O(lg V) for heap,
- O(1) for unsorted list
- O(V ) for sorted list (too slow)

- What is the location i of node v in a priority queue?
- Find in Heaps, and sorted lists O(n)
- Unsorted – if the nodes are numbered 1 to n and we use an array where node v is the v item in the array O(1)
Extended heap

- We will use extended heaps that contain a “handle” to the location of each node in the heap.
- When a node is not in PQ the “handle” will indicate that this is the case
- This means that we can access a node in the extended heap in O(1), and check v PQ in O(1)
- Note that the “handle” must be updated whenever a heap operation is applied

2. Unsorted list

- Array implementation where node v can be accesses as PQ[v] in O(1), and the value of PQ[v] indicates when the node is not in PQ.

Lines 1-5 initialize the priority queue PQ to contain all Vertices. Ds for all vertices except r, are set to infinity.

r is the starting vertex of the TThe T so far is empty

Add closest vertex and edge to current T

Get all adjacent vertices v of u,update D of each non-tree vertex adjacent to u

Store the current minimum weight edge, and updated distance in the priority queue

1. for each uV2.doD [u ] 3. D[ r ] 0

4. PQ make-heap(D,V, {})//No edges5. T 6.7. whilePQ do8.(u,e ) PQ.extractMin() 9. add (u,e) to T10. for each v Adjacent (u )

// execute relaxation11. do if v PQ && w( u, v ) < D [ v ]12. then D [ v ] w (u, v) 13. PQ.decreasePriorityValue( D[v], v, (u,v )) 14. return T // T is a mst.

Prim (G )

1. for each uV2. doD [u ] 3. D[ r ] 0

4. PQ make-heap(D,V, {})//No edges5. T

// solution check

7. whilePQ do//Selection and feasibility 8.(u,e ) PQ.extractMin() // T contains the solution so far . 9. add (u,e) to T10. for each v Adjacent (u )11. do if v PQ && w( u, v ) < D [ v ]12. then D [ v ] w (u, v) 13. PQ.decreasePriorityValue

(D[v], v, (u,v) ) 14. return T

1. for each uV2.doD [u ] 3. D[ r ] 0

4. PQ make-PQ(D,V, {})//No edges5. T 6.

7. whilePQ do8.(u,e ) PQ.extractMin() 9. add (u,e) to T10. for each v Adjacent (u )11. do if v PQ && w( u, v ) < D [ v ]12. then D [ v ] w (u, v) 13. PQ.decreasePriorityValue(D[v], v, (u,v)) 15. return T // T is a mst.

Assume a node in PQ can be accessed in O(1)

** Decrease key for v requires O(lgV ) provided the node in heap with v’s data can be accessed in O(1)

Using Extended Heap implementation

Lines 1 -6 run in O (V )

Max Size of PQ is | V |

Count7 =O (V )

Count7(8) = O (V ) O( lg V )

Count7(10) = O(deg(u ) ) =O( E )

Count7(10(11)) = O(1)O( E )

Count7(10(11(12))) = O(1) O( E )

Count7(10(13)) = O( lg V) O( E ) Decrease- Key operation on the extended heap can be implemented in O( lg V)

So total time for Prim's Algorithm is O ( V lg V + E lg V )

What is O(E ) ?

Sparse Graph, E =O(V) , O (E lg V)=O(V lg V )

Dense Graph, E=O(V2), O (E lg V)=O(V2 lg V)

1. for each uV2.doD [u ] 3. D[ r ] 0

4. PQ make-PQ(D,V, {})//No edges5. T 6.7. whilePQ do8.(u,e ) PQ.extractMin() 9. add (u,e) to T10. for each v Adjacent (u )11. do if v PQ && w( u, v ) < D [ v ]12. then D [ v ] w (u, v) 13. PQ.decreasePriorityValue (D[v], v, (u,v)) 15. return T // T is a mst.

Using unsorted PQ

Lines 1 - 6 run in O (V )

Max Size of PQ is | V |

Count7 = O (V )

Count7(8) = O (V ) O(V )

Count7(10) = O(deg(u ) ) =O( E )

Count7(10(11)) = O(1)O( E )

Count7(10(11(12))) = O(1) O( E )

Count7(10(13)) =O( 1) O( E )

So total time for Prim's Algorithm is O (V + V2 +E) = O (V2 )

For Sparse/Dense graph : O( V2 )

Note growth rate unchanged for adjacency matrix graph representation

handle

A

B

C

1

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3

PQ

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1

0, (A, {})

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G

, (B, {})

, (C, {})

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Prim (G, r)1. for each uV2.do D [u ] 3. D[ r ] 0

4. PQ make-heap(D,V, { })5. T

G

A

B

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C 6

B 2

A 2

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A 6

B 5

handle

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Null

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PQ

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(A, {})

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1

, (C, {})

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, (B, {})

7. whilePQ do 8.(u,e) PQ.extractMin()9. add (u,e) to T

2

handle

A

B

C

Null

1

2

PQ

T

(A, {})

A

1

2, (B, {A, B})

2

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G

, (C, {})

2

10. for each v Adjacent (u )11.// relaxation operation

// relaxation

11. do if v PQ && w( u, v ) < D [ v ]12. then D [ v ] w (u, v) 13. PQ.decreasePriorityValue

( D[v], v, (u,v))

handle

A

B

C

Null

1

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PQ

T

(A, {})

1

2, (B, {A, B})

6, (C, {A, C})

A

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G

handle

A

B

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Null

Null

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PQ

T

(A, {})

(B, {A, B})

6, (C, {A, C})

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A

7. whilePQ do 8.(u,e) PQ.extractMin()9. add (u,e) to T

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handle

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Null

Null

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PQ

T

(A, {})

T

(A, {})

(B, {A, B})

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1

A

10. for each v Adjacent (u )11.// relaxation operation

2

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handle

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Null

Null

Null

PQ

T

(A, {})

T

(A, {})

(B, {A, B})

(C, {B, C})

A

7. whilePQ do 8.(u,e) PQ.extractMin()9. add (u,e) to T

2

6

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5

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PQ

T

A

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A

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12

0, (A, {})

, (B, {})

, (C, {})

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G

A

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C 4

B 12

Prim (G, r)1. for each uV2.do D [u ] 3. D[ r ] 0

4. PQ make-PQ(D,V, { })5. T

A 12

C 5

A 4

B 5

PQ

T

(A, {})

A

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A

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12

Null

, (B, {})

, (C, {})

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G

7. whilePQ do 8.(u,e) PQ.extractMin()9. add (u,e) to T

PQ

T

(A, {})

A

A

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12

Null

12, (B, {A, B})

4, (C, {A, C})

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G

10. for each v Adjacent (u )11.// relaxation operation

T

(A, {})

(C, {A, C})

A

12

PQ

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B

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C

Null

Null

12, (B, {A, B})

G

7. whilePQ do 8.(u,e) PQ.extractMin()9. add (u,e) to T

PQ

T

(A, {})

T

(A, {})

(C, {A, C})

A

12

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B

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Null

Null

5, (B, {C, B})

G

10. for each v Adjacent (u )11.// relaxation operation

PQ

T

(A, {})

T

(A, {})

(C, {A, C})

(B, {C, B})

A

12

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5

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B

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C

Null

Null

Null

G

7. whilePQ do 8.(u,e) PQ.extractMin()9. add (u,e) to T

D = [ 0, , …, ]

Prim (G)1. for each uV2.do D [u ] 3. D[ r ] 0

4. PQ make-heap(D,V, { })5. T

PQ = {( 0,(a,)), (,(b,?)), ...(,(h,?))}

T = { }

r

G =

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// relaxation

11. do if v PQ && w( u, v ) < D [ v ]12. then D [ v ] w (u, v) 13. PQ.decreasePriorityValue

( D[v], v, (u,v))

7. whilePQ do 8.(u,e) PQ.extractMin()9. add (u,e) to T10. for each v Adjacent (u )11.// relaxation operation15. return T

D = [ 0,

r

G =

T = {

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PQ = {

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Let G = ( V, E) be a connected, weighted undirected graph. Let T be a promising subset of E. Let Y be the set of vertices connected by the edges in T.

If e is a minimum weight edge that connects a vertex in Y to a vertex in V - Y, then T { e } is promising.

Note: A feasible set is promising if it can be extended to produce not only a solution , but an optimal solution.

In this algorithm: A feasible set of edges is promising if it is a subset of a Minimum Spanning Tree for the connected graph.

T is the promising subset, and e the minimum cost edge of Lemma 1

Let T ' be the MST such that T T '

We will show that if e T' then there must be another MST T" such that T {e} T".

Proof has 4 stages:

1. Adding e to T', closes a cycle in T' {e}.

2. Cycle contains another edge e' T' but e' T

3. T"=T' {e} - {e’ }is a spanning tree

4. T" is a MST

Lemma 1

T

Y

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V -Y

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MST T' but e T'

Lemma 1

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T' {e}

Stage 1

u

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In T' there is a path from uY to v V -Y. Therefore the path must include another edge e' with one vertex in Y and the other in V-Y.

Stage 2

e'

Lemma 1

Y

V -Y

X

T "

e

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X

- If we remove e’ from T’ { e } the cycle disappears.
- T”=T’ {e } - {e’} is connected. T’ is connected. Every pair of vertices connected by a path that does not include e’ is still connected in T”. Every pair of vertices connected by a path that included e’ , is still connected in T” because there is a path in T" =T’ { e } - { e’ } connecting the vertices of e’.

Stage 3

w( e ) w( e' )By the way Prim picks the next edge

Lemma 1

Y

V -Y

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T"

e

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- w( e ) w( e' ) by the way Prim picks the next edge.
- The weight of T", w(T") = w (T' ) + w( e ) - w( e' ) w(T').
- But w (T' ) w(T")because T' is a MST.
- So w (T' ) = w(T")and T" is a MST

Stage 4

Conclusion T { e } is promising

Proof by induction on the set T of promising edges.Base case: Initially, T = is promising.Induction hypothesis: The current set of edges T selected by Prim is promising.Induction step: After Prim adds the edge e,T {e } is promising.Proof: T {e } is promising by Lemma 1.

Conclusion: When G is connected, Prim terminates with |T | = |V | -1, and T is a MST.

Kruskal’s Algorithm

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solution = { } while ( more edges in E) do// Selection select minimum weight edgeremove edge from E// Feasibility if (edge closes a cycle with solution so far)then reject edgeelse add edge to solution// Solution check if |solution| = |V | - 1 return solution

return null // when does this happen?

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1. Sort the edges E in non-decreasing weight2. T 3. For each v V create a set.4. repeat 5.Select next {u,v} E, in order 6.ucomp find (u) 7. vcomp find (v) 8. ifucomp vcomp then8. add edge (u,v) to T9. union ( ucomp,vcomp )10.untilT contains |V | - 1 edges11. returntree T

C= { {a}, {b}, {c}, {d}, {e}, {f}, {g}, {h} }C is a forest of trees.

A

Sorted edges

T

2

A B 2

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B C 5

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B D 7

D

1. Sort the edges E in non-decreasing weight2. T

3. For each v V create a set.

D

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Disjoint data set for G

A

T

Sorted edges

2

A B 2

(A, B)

6

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B C 5

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A C 6

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B D 7

D

5. for each {u,v} in ordered E6. ucomp find (u) 7. vcomp find (v) 8. ifucomp vcomp then9. add edge (v,u) to T10. union( ucomp,vcomp )

Find(A)

Find(B)

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Disjoint data set for G

A

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After merge(A, B)

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T

Sorted edges

2

A B 2

(A, B)

(B, C)

6

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A C 6

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B D 7

D

5. for each {u,v} in ordered E6. ucomp find (u) 7. vcomp find (v) 8. ifucomp vcomp then9. add edge (v,u) to T10. union ( ucomp,vcomp )

Find(B)

Find(C)

A

C

D

B

After merge(A, C)

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A

T

Sorted edges

2

A B 2

(A, B)

(B, C)

6

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5

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A C 6

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B D 7

D

5. for each {u,v} in ordered E6. ucomp find (u) 7. vcomp find (v) 8. ifucomp vcomp then9. add edge (v,u) to T10. union ( ucomp,vcomp )

Find(A)

Find(C)

A

D

B

C

A and C in same set

A

T

Sorted edges

2

A B 2

(A, B)

(B, C)

(B, D)

6

B

B C 5

5

7

A C 6

C

G

B D 7

D

5. for each {u,v} in ordered E6. ucomp find (u) 7. vcomp find (v) 8. ifucomp vcomp then9. add edge (v,u) to T10. union ( ucomp,vcomp )

Find(B)

Find(D)

A

D

B

C

A

After merge

B

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D

Count1 = ( E lg E )

Count2= (1)

Count3= ( V )

Count4 = O( E )

Using Disjoint set-height andpath compression

Count4(6+7+10)=

O((E +V) (V))

Sorting dominates the runtime. We get T( E,V ) = ( E lg E), so for a sparse graph we get ( V lg V)for a dense graph we get( V2lg V2) = ( V2lg V)

Kruskal ( G )1. Sort the edges E in non-decreasing weight2. T 3. For each v V create a set.4. repeat 5.{u,v} E, in order 6.ucomp find (u) 7. vcomp find (v) 8. ifucomp vcomp then9. add edge (v,u) to T10. union ( ucomp,vcomp )11.untilT contains |V | - 1 edges12. returntree T