A Graph G - V- E satisfies -E- 3-V-6- The min-degree of G is defined as minv-V degree v- Therefore- min-degree of G cannot be

Given nbsp; nbsp; nbsp;|E|≤ 3|V| 6 nbsp; nbsp; nbsp;...(i) Let nbsp;δ be the minimum degree of the graph. nbsp; Now nbsp; nbsp;δ canno ...

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Standard XII

Chemistry

Difference between Internal Energy and Enthalpy

A Graph G = V...

Question

A Graph G = (V, E) satisfies $| E | \leq 3 | V | - 6.$ The min-degree of G is defined as $min v \in V$ {degree (v)}. Therefore, min-degree of G cannot be

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Solution

The correct option is D 6
Given $| E | \leq 3 | V | - 6$ ...(i)

Let $δ$ be the minimum degree of the graph.

Now $δ$ cannot exceed the average degree of the graph.

So, $δ \leq \frac{2 | E |}{| V |}$ ...(ii)

Substitute eq. (i) in eq. (ii) and get

$δ \leq \frac{2}{| v |} (3 | V | - 6)$

$\Rightarrow δ \leq 6 - \frac{12}{| V |}$

Clearly the minimum degree has to be less than 6 always and hence cannot be equal to 6.

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A Graph G = (V, E) satisfies |E|≤3|V|−6. The min-degree of G is defined as minv∈V {degree (v)}. Therefore, min-degree of G cannot be

A Graph G = (V, E) satisfies $| E | \leq 3 | V | - 6.$ The min-degree of G is defined as $min v \in V$ {degree (v)}. Therefore, min-degree of G cannot be