main menu (click on the topics below)
DESCRIPTION
Spanning Trees Minimum Spanning Trees Kruskal’s Algorithm Example Planar Graphs Euler’s Formula. Main Menu (Click on the topics below). Click here to continue. Sanjay Jain, Lecturer, School of Computing. Spanning Trees and Planar Graphs. Sanjay Jain, Lecturer, School of Computing. - PowerPoint PPT PresentationTRANSCRIPT
Spanning Trees Minimum Spanning Trees Kruskal’s Algorithm Example Planar Graphs Euler’s Formula
Main Menu Main Menu (Click on the topics below)
Click here to continue
Sanjay Jain, Lecturer, School of ComputingSanjay Jain, Lecturer, School of Computing
Spanning Trees and Planar GraphsSpanning Trees and Planar Graphs
Sanjay Jain, Lecturer, Sanjay Jain, Lecturer,
School of ComputingSchool of Computing
Spanning TreesSpanning TreesDefinition: A spanning tree for a graph G is a subgraph
of G that
a) contains every vertex of G and
b) is a tree.
.. .
. ..
Spanning TreesSpanning TreesDefinition: A spanning tree for a graph G is a subgraph
of G that
a) contains every vertex of G and
b) is a tree.
.. .
. ..
END OF SEGMENT
PropositionPropositiona) Every connected finite graph G has a spanning tree.b) Any two spanning trees for a graph have the same
number of edges (If G has n vertices, then spanning tree of G has n-1 edges).
Proof: (of a)G is connected. Let G’=G1. If G’ is a tree then we are done.2. Otherwise, delete an edge from a circuit of G’ and go
to 1.
At the end of the above algorithm, G’ will be a spanning tree of G.
END OF SEGMENT
Minimal Spanning Trees.Minimal Spanning Trees.
.
.
..
.Singapore
KL
JakartaLondon
New York
89
76
23
88
34
65
45
Minimal Spanning TreesMinimal Spanning TreesWeighted Graph.Each edge has a weight associated with it.
Minimal spanning tree, is a spanning tree with the minimum weight.
)(
)()(TEe
eweightTweight
Minimal Spanning TreesMinimal Spanning TreesMay not be unique
. .. .
1
1
1
1
Minimal spanning tree can be formed by taking any three edges in the above graph.
END OF SEGMENT
Kruskal’s AlgorithmKruskal’s AlgorithmInput: Graph G, V(G), E(G), weights of edges.Output: Minimal spanning tree of G.Algorithm:1. Initialize T to contain all vertices of G and no edges. Let E=E(G). n= number of vertices in V(G) m=02. While m < n-1 do
2a. Find an edge in E with least weight.
2b. Delete e from E
2c. If adding e to T does not introduce a non-trivial circuit, then add e to the edge set of T
m=m+1
Endif
Endwhile
END OF SEGMENT
ExampleExample
.
..
.Singapore
KL
JakartaLondon
New York
89
76
23
88
34
102
99
.
ExampleExample
.
..
.Singapore
KL
JakartaLondon
New York
89
76
23
88
34
102
99
.
ExampleExample
.
..
.Singapore
KL
JakartaLondon
New York
89
76
23
88
34
102
99
.
ExampleExample
.
..
.Singapore
KL
JakartaLondon
New York
89
76
23
88
34
102
99
.
ExampleExample
.
..
.Singapore
KL
JakartaLondon
New York
89
76
23
88
34
102
99
.
ExampleExample
.
..
.Singapore
KL
JakartaLondon
New York
89
76
23
88
34
102
99
.
ExampleExample
.
..
.Singapore
KL
JakartaLondon
New York
89
76
23
88
34
102
99
.
ExampleExample
.
..
.Singapore
KL
JakartaLondon
New York
89
76
23
88
34
102
99
.
END OF SEGMENT
Planar GraphsPlanar GraphsDefinition: A graph G is planar iff it can be drawn on a
plane in such a way that edges never “cross” (I.e. edges meet only at the endpoints).
.. .
. ..
Plane GraphPlane Graph A drawing of planar graph G on a plane, without any
crossing, is called the plane graph representation of G
.. .. .
. .
.
END OF SEGMENT
Euler’s FormulaEuler’s FormulaTheorem: Suppose G is a connected simple planar graph with n3
vertices and m edges. Then, m 3n-6.
Note that the above theorem is applicable only for connected simple graphs.
END OF SEGMENT
Euler’s Formula ExamplesEuler’s Formula Examples
K5
number of vertices: 5 number of edges : 10
m 3n-6 does not hold.10 3*5-6 =9
So K5 is not planar.
Euler’s Formula ExamplesEuler’s Formula Examples
K4
number of vertices: 4 number of edges : 6 m 3n-6 holds.
6 3*4-6 =6
So K4 may be planar (it is actually planar as we have already seen).
Euler’s Formula ExamplesEuler’s Formula Examples
K3,3
number of vertices: 6 number of edges : 9 m 3n-6 holds.
9 3*6-6 =12
So K3,3 may be planar (however K3,3 is not planar).
END OF SEGMENT
Planar GraphsPlanar Graphs
A graph is planar iff it does not have K5 or K3,3 as a “portion” of it.
There is a linear time algorithm to determine whether a given graph is planar or not. If the graph is planar, then the algorithm also gives a plane graph drawing of it.
END OF SEGMENT
Proof of Euler’s FormulaProof of Euler’s FormulaFaces: The portion enclosed by
edges. The outside is also a face.
Let m be number of edges,n be number of vertices,and f be number of faces.Then:f = m-n+2 ----------- (1) .
..
...
.
..
Proof of Euler’s FormulaProof of Euler’s FormulaFaces: The portion enclosed by
edges. The outside is also a face.
Let m be number of edges,n be number of vertices,and f be number of faces.Then:f = m-n+2 ----------- (1) .
..
...
.
..
Proof of Euler’s FormulaProof of Euler’s FormulaFaces: The portion enclosed by
edges. The outside is also a face.
Let m be number of edges,n be number of vertices,and f be number of faces.Then:f = m-n+2 ----------- (1) .
..
...
.
..
Proof of Euler’s FormulaProof of Euler’s FormulaFaces: The portion enclosed by
edges. The outside is also a face.
Let m be number of edges,n be number of vertices,and f be number of faces.Then:f = m-n+2 ----------- (1) .
..
...
.
..
Proof of Euler’s FormulaProof of Euler’s FormulaFaces: The portion enclosed by
edges. The outside is also a face.
Let m be number of edges,n be number of vertices,and f be number of faces.Then:f = m-n+2 ----------- (1) .
..
...
.
..
Proof of Euler’s FormulaProof of Euler’s FormulaFaces: The portion enclosed by
edges. The outside is also a face.
Let m be number of edges,n be number of vertices,and f be number of faces.Then:f = m-n+2 ----------- (1) .
..
...
.
..
3f/2 m ----------- (2)By substituting (1) in (2) we get3m - 3n + 6 2mor m 3n - 6
END OF SEGMENT