GATE2010 q28 Revision as of 11:37, 15 July 2014 by Arjun Suresh (talk | contribs) (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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.
2. $6, 6, 6, 6, 3, 3, 2, 2$. Here first vertex has degree $6$, so remove this first vertex, and then subtract $1$ from next $6$ vertices, so we get $5, 5, 5, 2, 2, 1, 2$. Then we get $4, 4, 1, 1, 0, 2$ then $3, 0, 0, -1, 2$. Since degree of a vertex becomes negative, this degree sequence is not possible.

Since (II) comes only in option (A) and (D), and since (A) contains I, which is correct degree sequence, (D) must be right answer. We don't need to check for other sequences.

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