###### GATE 2007

October 24, 2023###### UGC NET CS 2018-DEC Paper-2

October 25, 2023# GATE 2007

Question 23 |

Which of the following graphs has an Eulerian circuit?

Any k-regular graph where k is an even number. | |

A complete graph on 90 vertices. | |

The complement of a cycle on 25 vertices.
| |

None of the above. |

Question 23 Explanation:

Two necessary condition for the existence of Eulerian circuits is

→ all vertices in the graph have an “even degree”.

→ And the graph must be corrected.

Now in option (C) it is saying that the complement of a cycle on 25 vertices without complement the degree of each vertex is 2.

Now since there are 25 vertices, so maximum degree of each vertex will be 24 and so in complement of cycle each vertex degree will be 24 – 2 = 22.

There is a theorem which says “G be a graph with n vertices and if every vertex has a degree of atleast n-1/2 then G is connected.”

So we can say that complement of cycle with 25 vertices fulfills both the conditions, and hence is Eulerian circuit.

→ all vertices in the graph have an “even degree”.

→ And the graph must be corrected.

Now in option (C) it is saying that the complement of a cycle on 25 vertices without complement the degree of each vertex is 2.

Now since there are 25 vertices, so maximum degree of each vertex will be 24 and so in complement of cycle each vertex degree will be 24 – 2 = 22.

There is a theorem which says “G be a graph with n vertices and if every vertex has a degree of atleast n-1/2 then G is connected.”

So we can say that complement of cycle with 25 vertices fulfills both the conditions, and hence is Eulerian circuit.

Correct Answer: C

Question 23 Explanation:

Two necessary condition for the existence of Eulerian circuits is

→ all vertices in the graph have an “even degree”.

→ And the graph must be corrected.

Now in option (C) it is saying that the complement of a cycle on 25 vertices without complement the degree of each vertex is 2.

Now since there are 25 vertices, so maximum degree of each vertex will be 24 and so in complement of cycle each vertex degree will be 24 – 2 = 22.

There is a theorem which says “G be a graph with n vertices and if every vertex has a degree of atleast n-1/2 then G is connected.”

So we can say that complement of cycle with 25 vertices fulfills both the conditions, and hence is Eulerian circuit.

→ all vertices in the graph have an “even degree”.

→ And the graph must be corrected.

Now in option (C) it is saying that the complement of a cycle on 25 vertices without complement the degree of each vertex is 2.

Now since there are 25 vertices, so maximum degree of each vertex will be 24 and so in complement of cycle each vertex degree will be 24 – 2 = 22.

There is a theorem which says “G be a graph with n vertices and if every vertex has a degree of atleast n-1/2 then G is connected.”

So we can say that complement of cycle with 25 vertices fulfills both the conditions, and hence is Eulerian circuit.

Subscribe

Login

0 Comments