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be the degree sequence of any graph?
 
be the degree sequence of any graph?
 
 
I. 7, 6, 5, 4, 4, 3, 2, 1
+
I. $7, 6, 5, 4, 4, 3, 2, 1$
 
 
II. 6, 6, 6, 6, 3, 3, 2, 2
+
II. $6, 6, 6, 6, 3, 3, 2, 2$
 
 
III. 7, 6, 6, 4, 4, 3, 2, 2
+
III. $7, 6, 6, 4, 4, 3, 2, 2$
 
 
IV. 8, 7, 7, 6, 4, 2, 1, 1
+
IV. $8, 7, 7, 6, 4, 2, 1, 1$
 
 
  
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<ol>
 
<ol>
 
<li>First arrange degree sequence in decreasing order.</li>
 
<li>First arrange degree sequence in decreasing order.</li>
<li>Remove 1st vertex, and let its degree be k, then subtract 1 from next k vertices.</li>
+
<li>Remove $1^{st}$ vertex, and let its degree be $k$, then subtract $1$ from next $k$ vertices.</li>
<li>If all vertices have degree 0, then answer is yes i.e. given degree seqeunce can be a degree sequence for a graph.  
+
<li>If all vertices have degree $0$, then answer is yes i.e. given degree sequence can be a degree sequence for a graph.  
If any vertex has degree &lt; 0, then answer is no, otherwise repeat  
+
If any vertex has degree &lt; $0$, then answer is no, otherwise repeat  
 
step 2.</li>
 
step 2.</li>
 
</ol>
 
</ol>
 
So we check each degree sequence given in question :  
 
So we check each degree sequence given in question :  
 
<ol>
 
<ol>
<li>7, 6, 5, 4, 4, 3, 2, 1. Here first vertex has degree 7, so remove this first vertex, and then subtract 1 from next 7
+
<li>$7, 6, 5, 4, 4, 3, 2, 1$. Here first vertex has degree $7$, so remove this first vertex, and then subtract $1$ from next $7$
vertices, so we get 5,4,3,3,2,1,0. Then we get 3,2,2,1,0,0, then 1,1,0,0,0, and then 0,0,0,0. So answer is yes.</li>
+
vertices, so we get $5,4,3,3,2,1,0$. Then we get $3,2,2,1,0,0$ then $1,1,0,0,0$ and then $0,0,0,0$. So answer is yes.</li>
 
</ol>
 
</ol>
 
Since (I) comes only in one option i.e. option <b>(A)</b>, it is correct answer. We don't need to check for other sequences.
 
Since (I) comes only in one option i.e. option <b>(A)</b>, it is correct answer. We don't need to check for other sequences.

Revision as of 13:43, 29 June 2014

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?

I. $7, 6, 5, 4, 4, 3, 2, 1$

II. $6, 6, 6, 6, 3, 3, 2, 2$

III. $7, 6, 6, 4, 4, 3, 2, 2$

IV. $8, 7, 7, 6, 4, 2, 1, 1$


(A) I and II

(B) III and IV

(C) IV only

(D) II and IV

Solution by Happy Mittal

This can be solved using havel hakimi theorem, which says :

  1. First arrange degree sequence in decreasing order.
  2. Remove $1^{st}$ vertex, and let its degree be $k$, then subtract $1$ from next $k$ vertices.
  3. If all vertices have degree $0$, then answer is yes i.e. given degree sequence can be a degree sequence for a graph. If any vertex has degree < $0$, then answer is no, otherwise repeat step 2.

So we check each degree sequence given in question :

  1. $7, 6, 5, 4, 4, 3, 2, 1$. Here first vertex has degree $7$, so remove this first vertex, and then subtract $1$ from next $7$ vertices, so we get $5,4,3,3,2,1,0$. Then we get $3,2,2,1,0,0$ then $1,1,0,0,0$ and then $0,0,0,0$. So answer is yes.

Since (I) comes only in one option i.e. option (A), it is correct answer. We don't need to check for other sequences.




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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?

I. 7, 6, 5, 4, 4, 3, 2, 1

II. 6, 6, 6, 6, 3, 3, 2, 2

III. 7, 6, 6, 4, 4, 3, 2, 2

IV. 8, 7, 7, 6, 4, 2, 1, 1


(A) I and II

(B) III and IV

(C) IV only

(D) II and IV

Solution by Happy Mittal[edit]

This can be solved using havel hakimi theorem, which says :

  1. First arrange degree sequence in decreasing order.
  2. Remove 1st vertex, and let its degree be k, then subtract 1 from next k vertices.
  3. If all vertices have degree 0, then answer is yes i.e. given degree seqeunce can be a degree sequence for a graph. If any vertex has degree < 0, then answer is no, otherwise repeat step 2.

So we check each degree sequence given in question :

  1. 7, 6, 5, 4, 4, 3, 2, 1. Here first vertex has degree 7, so remove this first vertex, and then subtract 1 from next 7 vertices, so we get 5,4,3,3,2,1,0. Then we get 3,2,2,1,0,0, then 1,1,0,0,0, and then 0,0,0,0. So answer is yes.

Since (I) comes only in one option i.e. option (A), it is correct answer. We don't need to check for other sequences.




blog comments powered by Disqus