recategorized by
2,949 views

3 Answers

4 votes
4 votes

Answer will be (A)

for 3 nodes in a simple connected undirected graph degree can be 2,1,1

for 4 nodes in a simple connected undirected graph degree can be 3,2,2,1

By these degree we can draw a simple connected undirected graph (havel hakimi theorem)

1 votes
1 votes
min degree of node is 1

max degree= n-1

so we have n vertices

2 vertices will have some degree acc to pigeon hole principle

so max no of nodes with distinct degree= n-1
0 votes
0 votes
In a simple graph degree of each node can not be distinct. There have to be at least 2 nodes with the same degree. for e.g if i have  a simple graph of 4 nodes, and i want each node to have a distinct degree then the degree sequence has to be : 3 2 1 0

but degree sequence 3 2 1 0 is impossible because if a node is having 0 degrees then the graph is not connected . also from havel hakimi we can prove that this sequence is not possible : -

3 2 1 0 -> 0 1 0 -1 => now -1 came so as per havel hakimi this degree sequence can not be possible for a simple graph.

any distinct degree sequence will result in the same.. hence for n nodes at most n-1 nodes can have distinct degrees with min 2 nodes having same degrees .

Related questions

1 votes
1 votes
2 answers
3
Akanksha Kesarwani asked Jan 22, 2016
2,849 views
A simple undirected graph ‘X’ has 10 vertices. If ‘X’ has 5 equally sized connected components, the maximum number of edges in graph ‘X’ is _________.
2 votes
2 votes
0 answers
4
Parshu gate asked Nov 11, 2017
1,170 views
Let G be a planar Graph Such that every phase is bordered by exactly 3 edges which of the following can never be value for X(G) a)2 b)3 C)4 d)none of these