15,187 views

The degree sequence of a simple graph is the sequence of the degrees of the nodes in the graph in decreasing order. Which of the following sequences can not be the degree sequence of any graph?

1. $7, 6, 5, 4, 4, 3, 2, 1$
2. $6, 6, 6, 6, 3, 3, 2, 2$
3. $7, 6, 6, 4, 4, 3, 2, 2$
4. $8, 7, 7, 6, 4, 2, 1, 1$
1. I and II
2. III and IV
3. IV only
4. II and IV

Which of the following sequences can not be the degree sequence of any graph?

that’s what the question is also asking for right?

Yes

A degree sequence $d_1,d_2,d_3\ldots d_n$ of non negative integer is graphical if it is a degree sequence of a graph. We now introduce a powerful tool to determine whether a particular sequence is graphical due to Havel and Hakimi

Havel–Hakimi Theorem :

→ According to this theorem, Let $D$ be the sequence $d_1,d_2,d_3 \dots dn$ with $d_1 \geq d_2 \geq d_3 \geq \ldots d_n$ for $n \geq 2$ and $d_i \geq 0$.
→ If each $d_i = 0$ then $D$ is graphical
→ Then $D_0$ be the sequence obtained by:
→ Discarding $d_1$, and
→ Subtracting $1$ from each of the next $d_1$ entries of $D$.
→ That is Degree sequence $D_0$ would be : $d_2-1, d_3-1, \ldots , d_{d_1+1} -1, \ldots , d_n$
→ Then, $D$ is graphical if and only if $D_0$ is graphical.

Now, we apply this theorem to given sequences:

1. $7,6,5,4,4,3,2,1 \rightarrow 5,4,3,3,2,1,0 \rightarrow 3,2,2,1,0,0 \rightarrow 1,1,0,0,0 \rightarrow 0,0,0,0$ so it is graphical.
2. $6,6,6,6,3,3,2,2 \rightarrow 5,5,5,2,2,1,2$ ( arrange in descending order) $\rightarrow 5,5,5,2,2,2,1 \rightarrow 4,4,1,1,1,1 \rightarrow 3,0,0,0,1$ we cannot continue to get all $0's$, so it is not graphical.
3. $7,6,6,4,4,3,2,2 \rightarrow 5,5,3,3,2,1,1 \rightarrow 4,2,2,1,1,0 \rightarrow 1,1,0,0,0 \rightarrow 0,0,0,0$ so it is graphical.
4. $8,7,7,6,4,2,1,1$, here degree of a vertex is $8$ and total number of vertices are $8$,  which is impossible, hence it is not graphical.

Hence, only option (I) and (III) are graphic sequence and answer is option-(D)

by

That's how you write a complete answer.

Written in the best possible way.

@Arjun Sir.

In point $\mathbf{II}$ there is a correction. It should be arrange in $\mathbf{descending}$ order.

edited

Hey guys, I think for 2nd option algo is applied like this, please correct if I am wrong,

6,6,6,6,3,3,2,2 → 5,5,5,2,2,1,1 → 4,4,1,1,0,0 → 3,0,0,0,0. [Wrong way]

I got my mistake here. so just updating. I didn't followed algo correctly.

di'=di for i = 1 to n-dn-1

di'=di'-1 for i= n-dn to n-1, where n-> no. of vertices and,

dn -> max degree among degree sequence

You can eliminate the last sequence i.e 4th one as... the total number of vertices is 8 and the maximum degree given is 8 too. which isn't possible at all. The maximum degree you can have out of 8 vertices is 7.

Now coming to the method for solving such questions is through Havel-Hakimi Algorithm.

you can implement it by following one simple video. Here it is. :)

YES (D) option is the correct.!!
Is this in the syllabus?
a big yes
We eliminated option D only because the graph is a simple graph. Were it a non-simple graph, then D won't be wrong, right?

D is correct ans

I) 5433210 = > 322100=> 11000=> 0000 so option i  correct eliminates A

ii)66663322 = > 555221 =>44110=>3000 so not graphic sequence eliminates B and C so D is answer