I have one question that I would like some assistance on. It dealswith Graphs (Planar Bipartite). I appreciate the help! Question: What is the maximum number of edges that can be in a planargraph that is bipartite? Will rate LIFESAVER! I appreciate the help! Question: What is the maximum number of edges that can be in a planargraph that is bipartite? Will rate LIFESAVER! Solution A planar bipartite graph on v vertices can have at most 2v-4 edges.One can prove this as a simple consequence of the Euler PolyhedralFormula. In graph terms it translates as this: If G is a planargraph with v vertices, e edges, and f faces (the regions formed onthe sphere by the graph drawing) then v-e+f = 2. A proof of thisfact can be found in most texts on Graph theory. Suppose now that a graph G is bipartite and has maximum number ofedges without destroying either planarity or bipartiteness (elsekeep adding more edges). Now the important thing to note is thatevery face of this graph must necessarily have 4 edges. Indeed, ifthere is a cycle of size 2k, then note that we can keep addingextra edges till we only see all faces of side 4. Also, note thatevery edge necessarily is part of two difference faces (\'inner\' and\'outer\', say). This implies the following: Count the number of pairs (F, e) where F is a face and e isan edge in F. Then the number of faces is f, and each of thesefaces has length 4, so the number of such pairs is 4f. But on theother hand, each edge is in 2 different faces, so this number isalso 2e. Hence 4f=2e => e = 2f. Plug this into the Euler formulato get v - e + f = 2 <=> v - e + e/2 = 2 => e/2 = v-2 =>e = 2(v- 2) = 2v - 4..

1. I have one question that I would like some assistance on. It dealswith Graphs (Planar Bipartite).
I appreciate the help!
Question:
What is the maximum number of edges that can be in a planargraph that is bipartite?
Will rate LIFESAVER!
I appreciate the help!
Question:
What is the maximum number of edges that can be in a planargraph that is bipartite?
Will rate LIFESAVER!
Solution
A planar bipartite graph on v vertices can have at most 2v-4 edges.One can prove
this as a simple consequence of the Euler PolyhedralFormula. In graph terms it translates as this:
If G is a planargraph with v vertices, e edges, and f faces (the regions formed onthe sphere by the
graph drawing) then v-e+f = 2. A proof of thisfact can be found in most texts on Graph theory.
Suppose now that a graph G is bipartite and has maximum number ofedges without destroying
either planarity or bipartiteness (elsekeep adding more edges). Now the important thing to note is
thatevery face of this graph must necessarily have 4 edges. Indeed, ifthere is a cycle of size 2k,
then note that we can keep addingextra edges till we only see all faces of side 4. Also, note

2. thatevery edge necessarily is part of two difference faces ('inner' and'outer', say). This implies
the following: Count the number of pairs (F, e) where F is a face and e isan edge in F. Then the
number of faces is f, and each of thesefaces has length 4, so the number of such pairs is 4f. But
on theother hand, each edge is in 2 different faces, so this number isalso 2e. Hence 4f=2e => e =
2f. Plug this into the Euler formulato get v - e + f = 2 <=> v - e + e/2 = 2 => e/2 = v-2 =>e = 2(v-
2) = 2v - 4.