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

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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?

- $7, 6, 5, 4, 4, 3, 2, 1$
- $6, 6, 6, 6, 3, 3, 2, 2$
- $7, 6, 6, 4, 4, 3, 2, 2$
- $8, 7, 7, 6, 4, 2, 1, 1$

- I and II
- III and IV
- IV only
- II and IV

39 votes

Best answer

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:

- $7,6,5,4,4,3,2,1 \rightarrow 5,4,3,3,2,1,0 \rightarrow 3,2,2,1,0,0 \rightarrow 1,1,0,0,0 \rightarrow 0,0,0,0$ so it is graphical.
- $6,6,6,6,3,3,2,2 \rightarrow 5,5,5,2,2,1,2$ ( 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.
- $7,6,6,4,4,3,2,2 \rightarrow 5,5,3,3,2,1,1 \rightarrow 4,2,2,1,1,0 \rightarrow 1,1,0,0,0 \rightarrow 0,0,0,0$ so it is graphical.
- $8,7,7,6,4,2,1,1$, 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**)

@Arjun Sir.

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

0

edited
May 12
by Akshatgupta0698

Hey guys, I think for 2^{nd} 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

0

24 votes

The answer is clearly D.

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. :)