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.
On Tue, Oct 23, 2012 at 5:17 PM, Mike Katherein <firstname.lastname@example.org> 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