graph (ii) shortest path, minimum spanning tree gguy 19-3-2011
TRANSCRIPT
Graph (II)
Shortest path, Minimum spanning tree
GGuy19-3-2011
Review
DFS (Depth-First-Search)Graph traverseTopological sort
BFS (Breadth-First-Search)Finding shortest path
Shortest Path Problem
In a weighted graph GWe want to find a path from s to t,such that the sum of edge weight is minimum
Shortest Path
S
T
3
1
31
32
3
Shortest Path
S
T
3
1
31
32
3
Algorithms
Single source?All pairs?Negative weight edge?Negative cycles?
Dijkstra’s algorithmfor-each v, d[v] ← ∞Q.Insert(s,0)
while not Q.Empty() do(u, w) = Q.ExtractMin()if (visited[u]) continue;d[u] ← w for-each v where (u, v) in E
Q.Insert(v, d[u] + weightuv)
Dijkstra’s algorithm
S
T
3
1
31
32
3
Dijkstra’s algorithm
0
T
3
1
31
32
3
Dijkstra’s algorithm
0
2
T
3
1
31
32
3
Dijkstra’s algorithm
0
2
T
3
1
31
32
3
Dijkstra’s algorithm
0
3
2
T
3
1
31
32
3
Dijkstra’s algorithm
0
3
2
T
3
1
31
32
3
Dijkstra’s algorithm
0
3
3
2
T
3
1
31
32
3
Dijkstra’s algorithm
0
3
3
2
T
3
1
31
32
3
Dijkstra’s algorithm
0
3
3
2
T
3
1
31
32
3
Dijkstra’s algorithm
0
3
3
2
4
3
1
31
32
3
Dijkstra’s algorithm
0
3
3
2
4
3
1
31
32
3
Dijkstra’s algorithm
0
3
3
2
4
3
1
31
32
3
Dijkstra’s algorithm
Implement Q using binary heap (priority queue)At most E edges in the heap
Time complexity: O(E log (E)) = O(E log (V))Space complexity: O(V+E)
Bellman Fordfor-each v, d[v] ← ∞d[s] ← 0
Do V-1 timesfor each edge (u,v)
if (d[u] + weightuv < d[v])d[v] ← d[u] + weightuv
Bellman Ford’s algorithm
S
T
3
1
-21
42
3
Bellman Ford’s algorithm
S
4
3
2
T
3
1
-21
42
3
Bellman Ford’s algorithm
S
3
3
2
2
3
1
-21
42
3
Bellman Ford’s algorithm
S
3
3
2
1
3
1
-21
42
3
Bellman Ford’s algorithm
Assume the shortest path from s->t isP0 P1 P2 … Pm-1
where m < |V| and s is fixed
In the 1st iteration, d[P1] is shortest
In the 2nd iteration, d[P2] is shortest
…In the m-1 iteration, d[Pm-1] is shortest
Bellman Ford’s algorithm
After |V|-1 iterations,If there exists edge (u,v) where
d[u] + weightuv < d[v]
Negative cycle!
Bellman Ford’s algorithm
Time complexity: O(VE)Space complexity: O(V)
Floyd Warshall’s algorithmfor all pairs(i,j), d[i][j] ← ∞
for all edges(u,v), d[u][v] ← weightuv
for k=1 to Vfor i=1 to V
for j=1 to Vd[i][j] =
min(d[i][j], d[i][k]+d[k][j])
Floyd Warshall’s algorithmAssume the shortest path from s->t is
P0 P1 P2 … Pm-1
for any s,t
For example, 5->3->1->2->4->6
i=1: 5->1->6i=2: 5->1->2->6i=3: 5->3->1->2->6i=4: 5->3->1->2->4->6i=5: 5->3->1->2->4->6i=6: 5->3->1->2->4->6
We won’t miss any shortest path!
Floyd Warshall’s algorithm
Time complexity: O(V3)Space complexity: O(V2)
Summary
Negative edges?
Negative cycle?
Time complexity
Dijkstra Single source No No O(E log V)
Bellman Ford Single source Yes Yes O(VE)
Floyd Warshall All pairs Yes Maybe (modified)
O(V3)
Tree
What is a tree?
G is connected and acyclicG is connected and |E| = |V| - 1G is acyclic and |E| = |V| - 1For any pairs in G, there is a unique path
Spanning Tree
3
1
31
32
3
Minimum Spanning Tree
3
1
31
32
3
Kruskal’s algorithm
T ← empty set of edges
Sort the edges in increasing order
For each edge e (in increasing order)if T + e does not contain a cycleadd e to T
Kruskal’s algorithm
How to detect a cycle?
Disjoint Set
Kruskal’s algorithm
If we choose an edge (u,v)Union(u, v)
Kruskal’s algorithm
3
1
21
22
2
Kruskal’s algorithm
3
1
21
22
2
Kruskal’s algorithm
3
1
21
22
2
Kruskal’s algorithm
3
1
21
22
2
Kruskal’s algorithm
3
1
21
22
2
Kruskal’s algorithm
3
1
21
22
2
Kruskal’s algorithm
3
1
21
22
2
Kruskal’s algorithm
3
1
21
22
2
Krustal’s algorithm
Sorting: O(E log V)Select edge: O(E)Check union: O(α(V))
Overall: O(E log V + E α(V))
The Red Rule
The Red Rule states that for any cycle in G, the largest weight edge will NOT be contained in any Minimum Spanning Tree.
Prim’s algorithm
T ← node 1
while size of T < Vchoose a vertex u that is not in Vand the cost adding it to V is minimum
add u to V
Prim’s algorithm
3
1
21
22
2
Prim’s algorithm
3
1
21
22
2
Prim’s algorithm
3
1
21
22
2
Prim’s algorithm
3
1
21
22
2
Prim’s algorithm
3
1
21
22
2
Prim’s algorithm
Use a min heap to maintainDijkstra like implement
Time complexity: O(E log V)