[9]
THEOREM 1
A necessary and sufficient condition for two planar graphs G and 1 G to be duals of 2
each other is as follows. There is a one-to-one correspondence between the edges in G and 1
the edges in G such that a set of edges in 2 G forms a circuit if and only if the corresponding 1
set in G forms a cut-set.2
PROOF
Let us consider a plane representation of a planar graph G. Let us also draw (geometrically) a dual G* of G. Then consider an arbitrary circuit Γ in G. Clearly, Γ will form some closed simple curve in the plane representation of G- dividing the plane into two areas (Jordan curve Theorem). Thus the vertices of G* are partitioned into non-empty, mutually exclusive subsets- one Γ and the other outside.
In other words, the set of edges Γ* in G* corresponding to the set Γ in G is a cut-set in G*. (No proper subset of Γ* will be a cut-set in G*). Likewise it is apparent that corresponding to a cut-set S* in G* there is a unique circuit consisting of the corresponding edge-set S in G such that S is a circuit. This proves the necessity of the theorem.
To prove the sufficiency, let G be a planar graph and let G′ be the graph for which there is a one-to-one correspondence between the cut-sets of G and circuits of G′, and vice-versa. Let G* be a dual graph of G. There is a one-to-one correspondence between the circuits of G′ and cut-sets of G, and also between the cut-sets of G and circuits of G*. Therefore, there is one-to-one correspondence between the circuits of G′ and G*, implying that G′ and G* are 2-isomorphic.
By a theorem, “All duals of a planar graph G are isomorphic; and every graph 2-isomorphic to a dual of G is also a dual of G”, G′ must be a dual of G.
[7]
THEOREM 2
Edges in a plane graph G form a cycle in G if and only if the corresponding dual edges form a bond in G*.
PROOF
Consider D⊆E(G). If D contains no cycle in G, then D encloses no region. It remains possible to reach the unbounded face of G from every face without crossing D. Hence, G*-D*
connected, and D* contains no edge cut.
If D is the edge set of a cycle in G, then the corresponding edge set D* ⊆E(G*) contains all dual edges joining faces inside D to faces outside D. Thus D* contains an edge cut.
If D contains a cycle and more, then D* contains an edge cut and more.
Thus D* is a minimal edge cut if and only if D is a cycle.
Figure:1 [7]
THEOREM 3
The following are equivalent for a plane graph G.
(A) G is bipartite.
(B) Every face of G has even length.
(C) The dual graph G* is Eulerian.
PROOF
A⇒B. A face boundary consists of closed walks. Every odd closed walk contains an odd cycle. Therefore, in a bipartite plane graph the contributions to the length of faces are all even.
B⇒A. Let C be a cycle in G. Since G has no crossings, C is laid out as a simple closed curve; let F be the region enclosed by C. Every region of G is wholly within F or wholly outside F. If we sum the face lengths for the regions inside F, we obtain an even number. Since each face length is even. This sum counts each edge of C once. It also counts each edge inside F twice, since each such edge belongs twice to faces in F. Hence, the parity of the length of C is the same as the parity of the full sum, which is even.
B⇔C. The dual graph G* is connected and its vertex degrees are the face lengths of G.
Figure:2 [12]
THOREM 4
A graph has a dual if and only if it is planar.
PROOF
We need to prove just the “only if” part. That is, we have only to prove that a
non-planar graph does not have a dual. Let G be a non-non-planar graph. Then G contains K or 5 K 3,3
or a graph homeomorphic to either of these. We have already seen that a graph G can have a dual only if every subgraph g of G and every homeomorphic to g has a dual. Thus if we can
show that neither K nor 5 K has a dual, we have proved the theorem. This we shall prove 3,3
by contradiction as follows:
(a) Suppose that K has a dual D. Observe that the cut-sets in 3,3 K correspond to 3,3
circuits in D and vice versa, since K has no cut-set consisting of two edges, D has no 3,3
circuit consisting of two edges. D contains no pair of parallel edges. Since every circuit in
K is of length four or six, D has no cut-set with less than four edges. Therefore, the degree 3,3
of every vertex in D is at least four. As D has no parallel edges and the degree of every vertex is at least four, D must have at least (5×4)/2= 10 edges. This is a contradiction, because K 3,3
has nine edges and so must its dual. Thus K cannot have a dual. Likewise, 3,3
(b) Suppose that the graph K has a dual H. Note that 5 K has (1) 10 edges, (2) no 5
pair of parallel edges, (3) no cut-set with two edges, and (4) cut-sets with only four or six edges. Consequently, graph H must have (1) 10 edges, (2) no vertex with degree less than three, (3) no pair of parallel edges, and (4) circuits of length four and six only. Now graph H contains a hexagon ( a circuit of length six ), and no more than three edges can be added to a hexagon without creating a circuit of length three or a pair of parallel edges. Since both of these are forbidden in H and H has 10 edges, there must be at least seven vertices in at least three. The degree of each of these vertices is atleast three. This leads to H having at least 11 edges. A contradiction.