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Degree Sequence for Graph - Discrete Mathematics - Solved Homework, Slides of Discrete Mathematics

Shoolini University of Biotechnology and Management SciencesDiscrete Mathematics

During the study of discrete mathematics, I found this course very informative and applicable.The main points in these lecture slides are:Degree Sequence for Graph, Graph Bipartite, Subgraph of Graph, Induced by Vertices, Complimentary Graph, Pair of Nodes, Euler Theorem, Odd Degree, Repeated Edges, Sum of Degrees, Maximal Degree

Typology: Slides

2012/2013

Uploaded on 04/27/2013

atmaja
atmaja 🇮🇳

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CS173: Discrete Mathematical Structures
Spring 2006
Homework #13
Due 07/05/06, 8a
Solutions: 42 points, 3 points for each question.
1. You're given a graph:
a. Give a degree sequence for the graph.
Solution: 3,3,3,3,3,3,3,3.
b. Is the graph bipartite (answer without explanation will not be accepted).
Solution: Yes, blue and red points are the partition.
c. Draw the subgraph of this graph induced by the vertices {E,F,B,C,G}
d. Draw the complimentary graph.
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CS173: Discrete Mathematical Structures Spring 2006 Homework # Due 07/05/06, 8a Solutions: 42 points, 3 points for each question.

  1. You're given a graph: a. Give a degree sequence for the graph. Solution: 3,3,3,3,3,3,3,3. b. Is the graph bipartite (answer without explanation will not be accepted). Solution: Yes, blue and red points are the partition. c. Draw the subgraph of this graph induced by the vertices {E,F,B,C,G} d. Draw the complimentary graph.

e. What is the largest complete subgraph that the graph contains? Solution: Any pair of nodes. The largest cliques size is 2. f. Can you traverse all the vertices starting from A visiting each vertex only once? Solution:Yes, for example: A,B,F,G,C,D,H,E,A g. Can you traverse all the edges visiting each edge exactly once? Solution: No, by Euler theorem at most 2 nodes can have an odd degree.

  1. Draw a graph with the given properties or prove why no such graph exists. a. Simple graph (no loops or repeated edges), six vertices each of degree 2 Solution: loop: 1->2->3->4->5->6->1. b. Simple graph having degree sequence 3,3,3, Solution: this is a complete graph. Just connect all possible vertices with edges. c. Graph with 6 edges and degree sequence 1,2,3,4, Solution: Sum of degrees is always twice the number of edges. This property doesn't hold here, thus the graph is impossible. d. Simple graph with six vertices and degree sequence 2,2,5,5,5, Solution: There are 4 vertices with degree 5, which means they're connected to all the other vertices in graph. Thus the minimal possible degree can be 4, not 2. Impossible e. A simple bipartite graph with 6 nodes and degree sequence 4,3,3,2,2, If an vertex in a bipartite graph has a degree 4, there are at least 4 nodes in the other part of the partition. Thus the bipartite graph must have two groups, one with 2 nodes and one with 4. Note that maximal degree allowed in a part with 4 nodes is 2 (both node on the other side). But one of the vertices with degrees 4,3,3 must be on the side of 4 nodes,