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

 

asked in Graph Theory by Veteran (68.8k points)
edited by | 1.9k views

3 Answers

+29 votes
Best answer

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 , number of edges , 

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

in simple unconnected graph with k component , number of edges ,

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

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

answered by Boss (5.7k points)
selected by
cool...

Very Well Explained .

Please refer to my query.

As this formula is derived for maximum number of edges you can refer this

+24 votes
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
answered by Veteran (15.6k points)
edited by

Please resolve my query.

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.
+10 votes

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

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

To get maximum,  take one vertex each for each component,  except last component.
Now k-1 components has 1 vertex each so no edges.
The last component has  n-(k-1) vertices.
So make last component complete.
It has n-(k-1)  C  2   edges. = (n-k) (n-k+1) / 2  edges.    Very Simple :)

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

answered by Boss (8.9k points)
Nice .... (y)


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