in Graph Theory edited by
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54 votes
54 votes

If $G$ is the forest with $n$ vertices and $k$ connected components, how many edges does $G$ have?

  1. $\left\lfloor\frac {n}{k}\right\rfloor$
  2. $\left\lceil \frac{n}{k} \right\rceil$
  3. $n-k$
  4. $n-k+1$
in Graph Theory edited by
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4 Comments

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The best answer is a bit misleading. 

First let's see the definitions. 

 

Tree: acyclic, connected and undirected graph.

Forest: collection of 1 or more trees .

Rooted tree: it's a special type of directed acyclic graphs. It has a root node from which there exist unique path to every other vertex. 

The authors says that every tree has a root node which is not a correct statement. 

Concepts like height, root node, depth, children, parents, ancestors etc are for Rooted tree and not tree. A tree is simply a connected , acyclic undirected graph. 

The comment under the best answer by @Vicky rix is a good answer for this question. 

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11 Answers

64 votes
64 votes
Best answer

A forest is a collection of trees. here we are given a forest with $n$ vertices and $k$ components. a component is itself a tree.

Since there are $k$ components means that every component has a root (every tree has one), therefore we have $k$ roots.

Introduction of each new vertex to the forest introduces a single edge to a forest. so for remaining $n-k$ vertices when introduced, to make up to $n$ vertices, contributes to $n-k$ edges.

Hence, ans $=$ option (C) $= (n-k)$

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4 Comments

@Vicky rix

Beautiful explanation.

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Have a doubt..How is it k "connected " components."Connected " implies it is not a forest itself as a forest has disjoint collection of trees.Please anybody explain.
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yes, answer by is the best one !:)

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27 votes
27 votes

A forest is an acyclic graph(with no cycle) , i.e all these components are a tree. With k components there are k roots.

And whenever a new node is added to a tree only a singe edge is introduced.
With k roots , remaining nodes are (n-k) each of which introduces an edge.

Hence, there are $\left(n-k\right)\times 1=\left(n-k\right)$ edges.

edited by
27 votes
27 votes

Another way:

If $1$ edge is removed $2$ components.

If $2$ edges removed $3$ components.

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If $k-1$ edges removed $k$ components.

A tree has $n-1$ edges. So, to make that tree into forest with $k$ components we would remove $k-1$ edges.

Therefore, Total edges in forest (Edges remained) = $n-1-(k-1)=n-k$

7 votes
7 votes
Please verify this method
Answer:

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