Minimum spanning trees (MST) Def: A spanning tree of a graph G is an - - PowerPoint PPT Presentation

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Minimum spanning trees (MST) Def: A spanning tree of a graph G is an - - PowerPoint PPT Presentation

Feb 04, 2023 204 likes •448 views

Minimum spanning trees (MST) Def: A spanning tree of a graph G is an acyclic subset of edges of G connecting all vertices in G. A sub- forest of G is an acyclic subset of edges of G. Minimum spanning trees (MST) 1 Def: 15 7 8 A spanning

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SLIDE 1

Minimum spanning trees (MST)

Def: A spanning tree of a graph G is an acyclic subset of edges of G connecting all vertices in G. A sub-forest of G is an acyclic subset of edges of G.

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SLIDE 2

Minimum spanning trees (MST)

Def: A spanning tree of a graph G is an acyclic subset of edges of G connecting all vertices in G. A sub-forest of G is an acyclic subset of edges of G. Def: Given is a weighted (undirected) graph G=(V,E,w) where w:E->Reals defines a weight of every edge in E. A minimum spanning tree of G is a spanning tree with the minimum total weight of edges. 15 1 7 6 3 2 2 8 4 5 1 1 7

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SLIDE 3

Kruskal ( G=(V,E,w) )

  • 1. Let T=∅
  • 2. Sort the edges in increasing order of weight
  • 3. For edge e do
  • 4. If T ∪ e does not contain a cycle then

5. Add e to T

  • 6. Return T

Minimum spanning trees (MST) - Kruskal

15 1 7 6 3 2 2 8 4 5 1 1 7

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SLIDE 4

Kruskal ( G=(V,E,w) )

  • 1. Let T=∅
  • 2. Sort the edges in increasing order of weight
  • 3. For edge e do
  • 4. If T ∪ e does not contain a cycle then

5. Add e to T

  • 6. Return T

Minimum spanning trees (MST) - Kruskal

Lemma: Algo is correct.

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SLIDE 5

Kruskal ( G=(V,E,w) )

  • 1. Let T=∅
  • 2. Sort the edges in increasing order of weight
  • 3. For edge e do
  • 4. If T ∪ e does not contain a cycle then

5. Add e to T

  • 6. Return T

Minimum spanning trees (MST) - Kruskal

15 1 7 6 3 2 2 8 4 5 1 1 7 Implementation?

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SLIDE 6

Init (V)

  • 1. for every vertex v do

2. boss[v]=v 3. size[v]=1 4. set[v]={v} Union (u,v)

  • 1. if size[boss[u]]>size[boss[v]] then

2. set[boss[u]]=set[boss[u]] union set[boss[v]] 3. size[boss[u]]+=size[boss[v]] 4. for every z in set[boss[v]] do 5. boss[z]=boss[u]

  • 6. else do steps 2.-5. with u,v switched

Minimum spanning trees (MST) - Kruskal

15 1 7 6 3 2 2 8 4 5 1 1 7 Implementation?

  • Union-Find datastructure
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SLIDE 7

Union (u,v)

  • 1. if size[boss[u]]>size[boss[v]] then

2. set[boss[u]]=set[boss[u]] union set[boss[v]] 3. size[boss[u]]+=size[boss[v]] 4. for every z in set[boss[v]] do 5. boss[z]=boss[u]

  • 6. else do steps 2.-5. with u,v switched

Minimum spanning trees (MST) - Kruskal

15 1 7 6 3 2 2 8 4 5 1 1 7 Analysis of Union-Find Lemma: k Unions take O(k log k) time

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Minimum spanning trees (MST) - Kruskal

15 1 7 6 3 2 2 8 4 5 1 1 7 Analysis of Union-Find Lemma: k Unions take O(k log k) time Corollary: The running time of Kruskal is: O(|E| log |E|) + O(|V| log |V|)

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Prim ( G=(V,E,w) )

  • 1. Let T=∅, H=∅
  • 2. For every vertex v do

3. cost[v]=∞, parent[v]=null

  • 4. Let u be a vertex
  • 5. Update (u)
  • 6. For i=1 to n-1 do
  • 7. u=vertex from H of

smallest cost (remove)

  • Add (u,parent[u]) to T
  • Update(u)
  • Return T

Minimum spanning trees (MST) - Prim

15 1 7 6 3 2 2 8 4 5 1 1 7

Update (u)

  • 1. For every neighbor v of u

2. If cost[v]>w(u,v) then 3. cost[v]=w(u,v), parent[v]=u 4. If v not in H then 5. Add v to H

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Minimum spanning trees (MST) - Prim

Lemma: Prim is correct. Running time:

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SLIDE 11

Single source shortest paths - Dijkstra

15 1 7 6 3 2 2 8 4 5 1 1 7 G=(V,E,w) and a vertex s (w non-negative) shortest paths from s to every other vertex Input: Output: Can use similar idea to Prim?

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SLIDE 12

Dijkstra ( G=(V,E,w), s )

  • 1. Let H=∅
  • 2. For every vertex v do

3. dist[v]=∞

  • 4. dist[s]=0
  • 5. Update (s)
  • 6. For i=1 to n-1 do
  • 7. u=extract vertex from H
  • f smallest cost
  • 8. Update(u)
  • Return dist[]

15 1 7 6 3 2 2 8 4 5 1 1 7

Update (u)

  • 1. For every neighbor v of u

2. If dist[v]>dist[u]+w(u,v) then 3. dist[v]=dist[u]+w(u,v) 4. If v not in H then 5. Add v to H

Single source shortest paths - Dijkstra

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SLIDE 13

15 1 7 6 3 2 2 8 4 5 1 1 7

Single source shortest paths - Dijkstra

Lemma: Dijkstra is correct. Running time:

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SLIDE 14

15 1 7 6 3 2 2 8 4 5 1 1 7

All pairs shortest paths – Floyd-Warshall

G=(V,E,w), w non-negative shortest paths between all pairs of vertices Input: Output:

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SLIDE 15

15 1 7 6 3 2 2 8 4 5 1 1 7

All pairs shortest paths – Floyd-Warshall

Idea 1:

  • Use Dijkstra from every vertex

G=(V,E,w), w non-negative shortest paths between all pairs of vertices Input: Output:

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SLIDE 16

15 1 7 6 3 2 2 8 4 5 1 1 7

All pairs shortest paths – Floyd-Warshall

Idea 1:

  • Use Dijkstra from every vertex

Idea 2:

  • How about dynamic programming?

G=(V,E,w), w non-negative shortest paths between all pairs of vertices Input: Output:

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SLIDE 17

15 1 7 6 3 2 2 8 4 5 1 1 7

All pairs shortest paths – Floyd-Warshall

the length of the shortest path from i to j using

  • nly vertices ≤ k

Heart of the algorithm: S[i,j,k] =

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SLIDE 18

15 1 7 6 3 2 2 8 4 5 1 1 7

All pairs shortest paths – Floyd-Warshall

Heart of the algorithm: S[i,j,k] = S[i,j,k] = How to compute S[i,j,k] ? the length of the shortest path from i to j using

  • nly vertices ≤ k
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SLIDE 19

Floyd-Warshall ( G=(V,E,w) )

  • 1. For i=1 to |V| do

2. For j=1 to |V| do 3. S[i,j,0] = w(i,j)

  • 4. For k=1 to |V| do

5. For i=1 to |V| do 6. For j=1 to |V| do 7. S[i,j,k] = min { 8. S[i,j,k-1], 9. S[i,k,k-1]+S[k,j,k-1] } 10.Return ?

15 1 7 6 3 2 2 8 4 5 1 1 7

All pairs shortest paths – Floyd-Warshall

S[i,j,k] = w(i,j) if k = 0 min { S[i,j,k-1], S[i,k,k-1] + S[k,j,k-1] } if k > 0

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SLIDE 20

Input: directed G=(V,E,w) and a vertex s Output:

  • FALSE if exists reachable

negative-weight cycle,

  • distance to every vertex, otherwise.
  • 1

7 6 3 2 2

  • 7

4 5 1 1

  • 3

Single source shortest paths – Bellman-Ford

4

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SLIDE 21

Input: directed G=(V,E,w) and a vertex s Output:

  • FALSE if exists reachable

negative-weight cycle,

  • distance to every vertex, otherwise.
  • 1

7 6 3 2 2

  • 7

9 5 1 1

  • 3

Single source shortest paths – Bellman-Ford

4

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Bellman-Ford ( G=(V,E,w), s )

  • 1. For every vertex v

2. d[v] = ∞

  • 3. d[s]=0
  • 4. For i=1 to |V|-1 do

5. For every edge (u,v) in E do 6. If d[v]>d[u]+w(u,v) then 7. d[v]=d[u]+w(u,v)

  • 8. For every edge (u,v) in E do

9. If d[v]>d[u]+w(u,v) then 10. Return NEGATIVE CYCLE 11.Return d[]

  • 1

7 6 3 2 2

  • 7

8 5 1 1

  • 3

Single source shortest paths – Bellman-Ford

4

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SLIDE 23

Bellman-Ford ( G=(V,E,w), s )

  • 1. For every vertex v

2. d[v] = ∞

  • 3. d[s]=0
  • 4. For i=1 to |V|-1 do

5. For every edge (u,v) in E do 6. If d[v]>d[u]+w(u,v) then 7. d[v]=d[u]+w(u,v)

  • 8. For every edge (u,v) in E do

9. If d[v]>d[u]+w(u,v) then 10. Return NEGATIVE CYCLE 11.Return d[]

  • 1

7 6 3 2 2

  • 7

8 5 1 1

  • 3

Single source shortest paths – Bellman-Ford

Lemma: Bellman-Ford is correct. Running time: 4