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+27 votes

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
in Graph Theory by Boss (17.5k points)
edited by | 5.1k views
Havel-Hakimi Algorithm.

6 Answers

+27 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:

  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 Loyal (5.7k points)
edited ago by
You have not applied the algorithm correctly while tracing it for option 2...(5,5,5,2,2,2,1)-> (4,4,1,1,1,1)
yes now edited.

For this part-

"Option II) 6,6,6,6,3,3,2,2 → 5,5,5,2,2,1,2 ( arrange in ascending order) → 5,5,5,2,2,2,1 → 4,4,1,1,1,1 → 3,0,0,0,1 this type of graph can not possible  so its not a graphical."

why we have  5,5,5,2,2,1,2  --> last 2 should be 1


@ Regina Phalange

in option 2) 6,6,6,6,3,3,2,2 here max degree is 6 so we apply theorem upto sixth position.

in option 1) 7,6,5,4,4,3,2,1 here max degree is 7 so we apply theorem upto seventh position..

hope u got it now .. :) 


This might help ....

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.

+25 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. :)

by Boss (19.9k points)
edited by
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?
+5 votes

D is correct ans

by Active (4k points)
0 votes
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
by Boss (11.6k points)
0 votes
ans d , 2nd and 4th not graphical using havell hakimi and maximum degree should not be equal to total no of vertices
by (29 points)
–3 votes
A is the correct option step1:arrange the degree in descending order Step2:consider the 1st element and decrease each element by 1 Step3:continue this for all elements Step4:the option having even number of 1s is the correct option
by Boss (14.4k points)
Answer is D) try to do it with Havell Hakimmi theorem when you  wil apply theorem on choice II and IV


in II you will left with vertex one of degree 3 ie. 3,0,0 which is not possible

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