GATE CSE 1991 | Question: 01,xv

Question

The maximum number of possible edges in an undirected graph with $n$ vertices and $k$ components is ______.

Comments

I first saw Tendua’s explanation(Scroll down) then only the proofs made sense. thanks.

Answer 1

$N$ vertices and  $K$ components.
(Component means they are non connected subgraphs)

We want maximum edges in total graph. How to get maximum edges ?

To get maximum,  take one vertex each for each component,  except last component.
Now $k-1$ components have $1$ vertex each and so no edges.
The last component has  $n-(k-1)$ vertices.
So make the last component complete.
i.e., It has ${}^{n-(k-1)}  C_ 2 =\frac{ (n-k) (n-k+1) }{ 2}$  edges.  

Must do a similar model qsn on forest:   https://gateoverflow.in/580/gate1992_03-iii

Comments

Nice .... (y)

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nice way. it is easy to visualize by taking example like if we take 6 vertex and 2 componenet. there r multiple way to draw the graph like (1,5) (2,4) (3,3)... so on in braket 1st no. shows vertex in first component like this if we draw this we see that maximum condition occurs only when all componets have 1 vertex and 1 component have all remaining vertices.

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Component of a GRAPH – Subgraph.

Component of a FOREST – Tree.

These two definitions are the keys to solving this question and the question mentioned in the link.

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Good

Answer 2

Hopefully it should be clear that in any such graph all components will be complete, i.e., have all possible edges. Thus the only remaining question is how large each component should be?

If there are two components with $a$ and $b$ vertices, $a>1, b> 1$, then together they can have at most

$\binom{a}{2} + \binom{b}{2} = \frac{1}{2} \left(a^2 - a + b^2 - b \right)$ edges.

However, if we place all but one of the vertices in a single component, we could have

$\binom{a+b-1}{2} + \binom{1}{2} = \frac{1}{2} \left(a+b-1\right)\left(a + b - 2\right)$
$=\frac{1}{2} \left(a^2 + 2ab -3a + b^2 - 3b + 2 \right)$ edges.

Subtracting the first quantity from the second gives

$\frac{1}{2}\left(\left(2ab - 3a - 3b +2\right) - \left(-a - b \right)\right) \\=ab-a-b+a \\= (a-1)(b-1) \text{ which is } > 0$

Hence it is better not to have two components with multiple vertices.

This leaves us with the answer that all components should have one vertex except one, which will have $n-k+1$  vertices, for a total of $\binom{n-k+1}{2}$ edges.

In simple connected graph, with number of edges as $e,$ we have

$(n-1) \leq e \leq n. \frac{(n-1)}{2}$

In simple disconnected graph with $k$ components and number of edges as $e,$ we have

$(n-k) \leq e \leq (n-k).\frac{(n-k+1)}{2 }$

Note: Put $k=1$ then it will be connected graph .

Reference @ http://www.quora.com/What-is-the-maximum-number-of-edges-in-graph-with-n-vertices-and-k-components
Another read @ http://stackoverflow.com/questions/24003861/maximum-number-of-edges-in-undirected-graph-with-n-vertices-with-k-connected-com

Comments

nice explanation @mithlesh

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Your comment saved me. Thank you saumya mishra :)

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Excellent answer.

This is a better answer with a little bit flavor of the proof.

Answer 3

easy one.
if u want maximum number of nodes and k vertices then u should have k-1 components. with only one vertices and only one should contain remaining vertex ,

like

if i take 6 vertex and have to make 4 component then i will make the  4 component in this way ,

1 vertex 1 vertex 1 vertex 3 vertex.

and now i will make the last one complete graph . then it will have maximum number of edges.

so if u have n vertices and k component then just give (k-1) component one vertex and the remaining will be (n-(k-1)) now make that bigger one a complete graph . i.e ( n-k+1) ( n-k-1+1)/2 complete graph formula . n(n-1)/2 = (n-k)(n-k+1)/2

Comments

Please resolve my query.

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Here, they have asked maximum number of edges. The number of edge in any graph with n vertices and k components lies between   n-k and (n-k+1)(n-k)/2. We have to keep the number of vertices in each component such that the total number of edges is maximum.

Answer 4

when graph do not contain self loops and is undirected then the maximum no. of edges are-

(n-k+1)(n-k)/2

It is because maximum number of edges with n vertices is n(n-1)/2.

Now for example, if we are making an undirected graph with n=2 (4 vertices) and there are 2 connected components i.e, k=2, then first connected component contains either 3 vertices or 2 vertices, for simplicity we take 3 vertices (Because connected component containing 2 vertices each will not results in maximum number of edges). These 3 vertices must be connected so maximum number of edges between these 3 vertices are 3 i.e, (1->2->3->1) and the second connected component contains only 1 vertex which has no edge. So the maximum number of edges in this case are 3. This implies that replacing n with n-k+1 in the formula for maximum number of edges i.e, n(n-1)/2 will results in (n-k+1)(n-k)/2 which is maximum number of edges that a graph of n vertices with k connected component can have.