After using it in the last post on non-positively curved surfaces, I thought it might be nice to give a simple proof of the Okamura-Seymour theorem in the dual setting. This argument arose out of conversations in 2007 with Amit Chakrabarti and former UW undergrad Justin Vincent. I was later informed by Yuri Rabinovich that he and Ilan Newman discovered a similar proof.

Note that establishing a *node-capacitated* version of the Okamura-Seymour theorem was an open question of Chekuri and Kawarabayashi. Resolving it positively is somewhat more difficult.

Theorem 1Given a weighted planar graph and a face of , there is a non-expansive embedding such that is an isometry.

By rational approximation and subdivision of edges, we may assume that is unweighted. The following proof is constructive, and provides an explicit sequence of cuts on whose characteristic functions form the coordinates of the embedding . Each such cut is obtained by applying the following lemma. Note that for a subset of vertices, we use the notation , for the graph obtained from by contracting the edges across

Lemma 2 (Face-preserving cut lemma)Let be a -connected bipartite planar graph and be the boundary of a face of . There exists a cut such that

*Proof:* Fix a plane embedding of that makes the boundary of the outer face. Since is -connected, is a cycle, and since is bipartite and has no parallel edges, .

Consider an arbitrary pair of distinct vertices on . There is a unique path from to that runs along counterclockwise; call this path . Consider a path and a vertex . We say that *lies below* if, in the plane embedding of , lies in the closed subset of the plane bounded by and .

(Note that the direction of is significant in the definition of “lying below,” i.e., belowness with respect to a path in is not the same as belowness with respect to the reverse of the same path in .)

We say that *lies strictly below* if lies below and . We use this notion of “lying below” to define a partial order on the paths in : for we say that is *lower* than if every vertex in lies below .

We now fix the pair and a path so that the following properties hold:

- .
- If are distinct vertices with preceding and , then and .
- If is lower than and , then .

Note that a suitable pair exists because . Finally, we define the cut as follows: does not lie strictly below .

For the rest of this section, we fix the pair , the path and the cut as defined in the above proof.

Lemma 1Let and be distinct vertices on and be such that . Then .

*Proof:* If the lemma holds trivially, so assume . Also assume without loss of generality that precedes in the path . The conditions on imply that all vertices in lie strictly below . Therefore, the path is lower than and distinct from

By property (3), we have , which implies . Since is bipartite, the cycle formed by and must have even length; therefore, .

*Proof:* If there is nothing to prove. If not, we can write

for some where, for all , and . Let and denote the endpoints of with preceding in the path . Further, define and .

By Lemma 1, we have for . Since is a shortest path, we have for . Therefore

The latter quantity is precisely which completes the proof since .

*Proof of Claim 1:* Let be arbitrary and distinct. It is clear that , so it suffices to prove the opposite inequality. We begin by observing that

Let be a path in that achieves the minimum in the above expression. First, suppose . Then we must have . Now, , which implies and we are done.

Next, suppose . Then, there exists at least one vertex in that lies on . Let be the first such vertex and the last (according to the ordering in ) and assume that precedes in the path . Let . Note that may be trivial, because we may have . Now, , whence

This gives

where the first line follows from Eq. (1) and the definition of and the third line is obtained by applying Lemma 2 to the path . If at least one of lies in , then and we are done.

Therefore, suppose . Let . For a path and vertices on , let us use as shorthand for . By property (1), we have and since is bipartite, this means . By property (2), we have and . Using these facts, we now derive

Using this in (**) above and noting that , we get . This completes the proof.

*Proof of Theorem 1:* Assume that is -connected. We may also assume that is bipartite. To see why, note that subdividing every edge of by introducing one new vertex per edge leaves the metric essentially unchanged except for a scaling factor of .

We shall now prove the stronger statement that for every face of there exists a sequence of cuts of such that for all on , we have and that for , . We prove this by induction on .

The result is trivial in the degenerate case when is a single edge. For any larger and any cut , the graph is either a single edge or is -connected. Furthermore, contracting a cut preserves the parities of the lengths of all closed walks; therefore is also bipartite.

Apply the face-preserving cut lemma (Lemma 1) to obtain a cut . By the above observations, we can apply the induction hypothesis to to obtain cuts of corresponding to the image of in . Each cut induces a cut of . Clearly for any . Finally, for any , we have

where the first equality follows from the property of and the second follows from the induction hypothesis. This proves the theorem.