Aptitude
January 31, 2025NTA-UGC-NET 2021 Dec & 2022 June Paper-1
February 2, 2025GATE 2009
| Question 3 |
Which one of the following is TRUE for any simple connected undirected graph with more than 2 vertices?
| No two vertices have the same degree. | |
| At least two vertices have the same degree. | |
| At least three vertices have the same degree. | |
| All vertices have the same degree.
|
Question 3 Explanation:
Method 1:
If all vertices have different degrees, then the degree sequence will be {1,2,3,….n-1}, it will not have ‘n’( A simple graph will not have edge to itself, so it can have edges with all other (n-1) vertices). Degree sequence has only (n-1) numbers, but we have ‘n’ vertices. So, by Pigeonhole principle there are two vertices which has same degree.
Method 2:
A) Consider a triangle, all vertices has same degree, so it is false
C) Consider a square with one diagonal, there are less than three vertices with same degree, so it is false
D) Consider a square with one diagonal, vertices have different degrees. So, it is false.
We can conclude that option B is correct.
If all vertices have different degrees, then the degree sequence will be {1,2,3,….n-1}, it will not have ‘n’( A simple graph will not have edge to itself, so it can have edges with all other (n-1) vertices). Degree sequence has only (n-1) numbers, but we have ‘n’ vertices. So, by Pigeonhole principle there are two vertices which has same degree.
Method 2:
A) Consider a triangle, all vertices has same degree, so it is false
C) Consider a square with one diagonal, there are less than three vertices with same degree, so it is false
D) Consider a square with one diagonal, vertices have different degrees. So, it is false.
We can conclude that option B is correct.
Correct Answer: B
Question 3 Explanation:
Method 1:
If all vertices have different degrees, then the degree sequence will be {1,2,3,….n-1}, it will not have ‘n’( A simple graph will not have edge to itself, so it can have edges with all other (n-1) vertices). Degree sequence has only (n-1) numbers, but we have ‘n’ vertices. So, by Pigeonhole principle there are two vertices which has same degree.
Method 2:
A) Consider a triangle, all vertices has same degree, so it is false
C) Consider a square with one diagonal, there are less than three vertices with same degree, so it is false
D) Consider a square with one diagonal, vertices have different degrees. So, it is false.
We can conclude that option B is correct.
If all vertices have different degrees, then the degree sequence will be {1,2,3,….n-1}, it will not have ‘n’( A simple graph will not have edge to itself, so it can have edges with all other (n-1) vertices). Degree sequence has only (n-1) numbers, but we have ‘n’ vertices. So, by Pigeonhole principle there are two vertices which has same degree.
Method 2:
A) Consider a triangle, all vertices has same degree, so it is false
C) Consider a square with one diagonal, there are less than three vertices with same degree, so it is false
D) Consider a square with one diagonal, vertices have different degrees. So, it is false.
We can conclude that option B is correct.
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