Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
Drexel University or The Math Forum.
|
|
|
|
Re: a question involving Eular's formula
Posted:
Oct 23, 2012 7:45 PM
|
|
I can't really do subscripts in this e-mail, so for example Ei(i+1)
means E with subscript i,i+1.
Make a new graph by putting the k components in a row -- call them C1,
C2, ... , Ck -- with C1 on the left, C2 second from the left, etc,
and add one new edge E12 from C1 to C2, one E23 from C2 to C3, ... ,
(k-1) in all. Draw each graph without unnecessary edge crossings.
Choose the endpoints of Ei(i+1) so that the endpoint in Ci is on the
right side of Ci and the one in C(i+1) is on the left. The new graph
is connected, has n vertices, e+k-1 edges, and r regions. (If you
draw the edges as I suggested, you don't add any regions. Draw a
sketch to see this.) Then
n - (e+k-1) + r = 2,
giving the result.
WDW
On Tue, Oct 23, 2012 at 5:17 PM, Mike Katherein <ytbsearch@yahoo.com> wrote:
> G is a planar graph that has n vertices, e edges, r regions, and
> k connected components.
>
> Show that the Euler's Formula for G can be written as: n - e + r = k + 1.
> (Remember if G is connected, then k = 1, which means Euler's Formula will be degenerated to n - e + r = 2.)
>
> ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
>
> I am a super-newbie to discrete math, and I don't even know where I should start with this question
>
> I only know that the Eular's theorem for any connected planar graph is n - e + r = 2.
>
> But I don't know how could I transform that formula to n - e + r = k + 1.
>
> any hints or helps will be appreciated, thanks for reading this post
|
|
|
|
