GATE CSE 2003 | Question: 40

Question

A graph $G=(V,E)$ satisfies $\mid E \mid \leq 3 \mid V \mid  - 6$. The min-degree of $G$ is defined as $\min_{v\in V}\left\{ \text{degree }(v)\right \}$. Therefore, min-degree of $G$ cannot be 

• A. $3$
• B. $4$
• C. $5$
• D. $6$

Comments

What about k33 it is also fails the face contain cycle of three ? Answer should be A.

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By handshaking lemma :

let K be the min degree of vertices then  $k*\left | V \right | < = 2 * |E|$

putting $ |E| <= 3 |V| -6$ we get

$|V| <= \frac{-12}{(k-6)}$

Hence putting option D ie 6 is not possible :)

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According to the handshaking lemma,we have–

Sum of degree of all vertices =2* |E|

⇒Sum of min-degree of all vertices ≤ 2 *|E| (If we replace degree by min-degree) …i)

⇒Sum of avg-degree of all vertices =2 *|E| (If we replace degree by avg-degree)  ….ii)

⇒Sum of max-degree of all vertices  $\geq$2 *|E| (If we replace degree by max-degree)  …iii)

We need eqn i) in this question.

Sum of min-degree of all vertices ≤ 2 *|E| (If we replace degree by min-degree) 

(Sum of min-degree of all vertices)/2 ≤ |E| 

⇒(Min-degree * V)/2 ≤ |E|                  ⇒(Min-degree * V)/2 ≤ 3|V|-6[given condition]

⇒ (6*V)/2≤ 3|V|-6   ⇒ 3*V/2≤ 3|V|-6    ⇒ 0≤ -6 which is false 

Hence correct answer is option D.

Answer 1

Say every vertex has a minimum degree, therefore, least number of edges that will be in the graph is given by the handshaking lemma as $\frac{min \times |v|}{2}$

But the maximum number of edges for such a graph is defined in the question as $3\times |v| - 6$

Putting the minimum number of edges obtained by handshaking lemma in the given inequality, we get:

$\frac{min \times |v|}{2} \leq 3\times |v| - 6$

$\quad \frac{6 \times |v|}{2} \leq 3\times |v| - 6\text{; putting min degree as 6}$

$\;3 \times |v| \leq 3\times |v| - 6$

$\qquad\;\;0 \leq - 6 $

which is definitely inconsistent.

Hence, answer = option (D)

Comments

@Satbir sir

i correct myself I misinterpret the above statement 

the correct statement is following:

"Every planar graph contains at least one vertex with degree at most 5."

   

 

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Dear Sir

Thanks for this ans. I just wanted to know why are we assuming minimum degree of all the vertices to be same, can’t every vertex have diff min degrees.

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Can someone please to me,  how other options are not satisfying the equation?

Answer 2

Let the min-degree of $G$ be $x$. Then $G$ has at least $\left[|v|\times \frac{x}{2}\right]$ edges.

$\left[|v|\times \frac{x}{2}\right]\leq \left[\left(3\times |v|\right) -6\right]$

for $x=6$, we get $0 \leq {-6}$, Therefore, min degree of $G$ cannot be $6$.

Correct answer is (D).

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$\text{An alternative approach,}$

Let the min_degree of a graph be $\text{'x'}$ , then

 $x\leq \left(\frac{2e}{n}\right)$,
given  , $e\leq \left(3n - 6\right)$          $\text{\{it will be planner graph\}}$
put the value of $e$ ,then min_degree will be ,

 $x\leq \frac{(2(3n-6))}{n}$
 $x\leq \frac{\left(6n - 12\right)}{n}$
 $x\leq \left( \frac{6n}{n} - \frac{12}{n} \right)$
 $x\leq \left(6 - \frac{12}{n}\right)$ ,
 when number of vertices is more , then value of

$\left(\frac{12}{n}\right)$ will be less , $\left(\frac{12}{n} = 0.000001 \text{assume}\right)$ ,

then min_degree will be ,

$x\leq \left(6 - 0.000001\right)$
$x\leq 5.999999$  , max value
$x\leq \text{floor value}\left(5.9999999\ldots \right)$
$x = 5$ , maximum value of min_degree of defined graph (i.e. planner graph)

Comments

ans is D or C??

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If we put x=4 then

|v|*4/2<=3|v|-6

-1<= -6 becomes false

Plz explain.

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@ diksha,

if you put x=4 then equation becomes -|V| <= -6 not -1 <= -6.

Answer 3

Given that, |E| ≤3 |V| − 6

We know that the inequality, min-degree ≤ 2|E| / |V| .

Put the value of |E| into inequality.

min-degree ≤ 6 - 12 / |V|

let min-degree is m.

m ≤ 6 - 12 / |V|

So our eqn will become , |V| ≥ 12 / (6-m) .

Apply hit and try to check all the options.

Option (a) 3   ,put m=3 then  |V| ≥ 4 .Yes, it is possible.

Option (b) 4 ,put m=4 then  |V| ≥ 6 .Yes ,it is possible.

Option (c) 5  ,put m=5 then  |V| ≥ 12 .Yes, it is possible.

Option (d) 6   ,put m=6 then  |V| ≥ 12/0 . It is NOT possible due to Divide by Zero or indeterminant state.

Therefore, min-degree of G cannot be 6.

The correct answer is (D) 6

Comments

No need to use hit and trial.

m ≤ 6 - 12 / |V|   , Now since |V| is a positive number, 12/|V| will be positive as well hence,  6- 12/|V| will be less than 6 [since we are deducting some positive number from 6 so, result will be definitely less than 6]

So, m ≤ (something less than 6)  implies----> m<6  

Answer 4

Summation of degree = 2* no. of edges

                                   = 2( 3V -6)

Let min degree be d, then

d*V<= 2*(3V-6)

d*V<=6V-12

Taking V on left side-

(d-6)*V<= -12

Multiplying both side by (-1)

(6-d)*V>= 12

Minimum value of d should be 6 to make left side positive as no. of Vertices can't be negative.

Thus minimum degree, d=6.

Comments

If d is less than 6 then also left side is a positive no ??