1 Flows and disjoint noncontractible cycles
For a plane graph , let , , and denote the sets of its vertices, edges, and faces, respectively. For a function , let . Let be a connected plane graph and let be a function such that . Throughout the paper, we view paths in graphs as directed, i.e., having distinguished starting and ending vertices. A linkage in is a system of pairwise edgedisjoint paths in such that

if a path starts in a vertex and ends in a vertex , then and , and

every vertex is the starting or ending vertex of exactly paths from .
Let and be distinct faces of . A cycle in is noncontractible if the open disk in the plane bounded by contains exactly one of the faces and , and contractible otherwise. An circulation is a set of pairwise edgedisjoint noncontractible cycles. The support of a linkage or an circulation is the set of edges of . The goal of this section is to answer the following question: Given the graph , its faces and , and the function as described, what is the maximum size of an circulation whose support is disjoint from the support of some linkage in ? We will not be able to give a full answer, however we will solve the special case arising in the context of Theorem 3, where the function satisfies the condition that and have the same parity for all .
It is more convenient to work in the setting of integral flows. Let us start by giving the necessary notation and describing their relationship to linkages and circulations. Consider an arbitrary orientation of . For a vertex , let and denote the sets of edges entering and leaving , respectively. We drop the subscripts when the directed graph is clear from the context. An (integral) flow in is a function such that for all ; i.e., gives the amount of the flow originating in . The support of the flow is the set of edges of such that is nonzero on the corresponding directed edge of .
Let denote the plane dual of . For a vertex , a face , or an edge , let , , or denote the corresponding face, vertex, or edge of , respectively. For an orientation of and an edge , the edge of is directed so that, looking along the edge in its direction, the edge crosses from left to right. For a function or , let or be the corresponding function such that for every and for every .
Let be a connected plane graph and let be its orientation. For a (possibly closed) walk in and an edge , let denote the number of times traverses in the direction of minus the number of times traverses in the direction opposite to . In particular, if , , and is a path or a cycle, then for every , we have if crosses from right to left (as seen along the direction of ), if crosses from left to right, and if does not cross . For a function , we define . Let denote the function whose value is everywhere.
It is wellknown that a flow can be expressed as a disjoint union of cycles and sourcesink paths. Conversely, we can send a unit of flow along each of the paths and in any direction along each of the cycles to obtain a flow. Hence, the following claim holds.
Observation 8.
Let be a connected plane graph and let be a function such that . Let be a path in from a vertex to a vertex .

If contains a flow , then contains a linkage with . If contains a linkage , then contains a flow with .

For a nonnegative integer , if contains a flow with , then contains an circulation of size with . If contains an circulation of size , then for every integer , contains a flow with and .
Let us now relate the values and for a flow in and distinct paths and in its dual with the same starting and ending vertices. For a closed walk in a plane graph and a face of , let denote the winding number of around . Recall the winding number is defined as follows: Let be any halfline with the starting point inside which intersects only in edges. Then is equal to the number of times intersects from left to right, minus the number of times intersects from right to left. Let us remark that the value of is independent of the exact choice of the halfline . Indeed, the winding number (for ) can also be defined by the following properties.
Observation 9.
Let be an orientation of a connected plane graph and let be a closed walk in .

The outer face of satisfies .

For any edge , .
For a closed walk in a plane graph and a function , let us define
Lemma 10.
Let be a connected plane graph, let be a function such that , and let be a flow in . For any closed walk in , we have , and in particular is the same for all flows. Furthermore, if and are vertices of , and are paths from to in , and is the closed walk obtained as the concatenation of with the reversal of , then .
Proof.
Let be the orientation of with respect to which the flow is defined. Using Observation 9(b), we have
as required. Furthermore, if is the concatenation of and the reversal of , then . ∎
Let be a connected plane graph and let be a function such that . Let and be faces of , and let be a path in from the vertex to the vertex . A flow in is circulationmaximum or circulationminimum if has the maximum or the minimum possible value, respectively, among all flows. In view of Lemma 10, these notions are independent of the choice of the path . We now give a minmax condition for circulationmaximum and minimum flows. A function is even if has the same parity as the degree of for every vertex of .
Lemma 11.
Let be a connected plane graph, let be a function such that , and let and be faces of . Let and be circulationmaximum and minimum flows in , respectively, with maximal supports. Then there exist paths and in from to such that and . Furthermore and are formed by edgesets of forests in , and in particular if is even, then .
Proof.
We prove the claims for ; the claims for follow by the same argument applied to the circulationmaximum flow . Let be any path from to in .
If there existed a cycle with , then note that there exists a flow with support such that , obtained by sending one unit of flow along in an appropriate direction. Then is a flow which contradicts either the assumption that is circulationmaximum (when ) or that is maximal among circulationmaximum flows. Hence, is an edgeset of a forest in . If is even, then each vertex is incident with even number of edges of , and thus the forest cannot have any leaf; this is only possible if .
Hence, we only need to find a path from to such that . Let be the orientation of with respect to which the flow is defined. Let us form an auxiliary directed graph with the vertex set and edge set defined as follows. For each edge ,

if , then we include in the reversal of ,

if , then we include in the edge , and

if , then we include both and its reversal in .
The graph is drawn in the plane in the same way as , with the two opposite edges in the last case drawn close next to each other. Let us remark that is defined so that sending a unit of flow along any directed cycle in and adding the flow to results in a valid flow in .
Let be the set of faces for which there exists a path in from to such that all edges of intersect from left to right. If , then by the definition of we conclude that , and thus we can set .
Hence, suppose that . Let be the graph obtained from by deleting all vertices and edges that are only incident with faces in , and let be the face of containing . Since , the boundary of the face contains an noncontractible cycle . All edges of separate a face in from a face not in , and thus all of them are in directed from right to left (looking from ). Let be defined as follows: For each edge corresponding to an edge , we let if and are directed in the same way and otherwise. On all other edges , we set . Then is a flow with , and for every by the definition of the graph . Hence, is a flow with , contradicting the assumption that is circulationmaximum. ∎
The proof of Lemma 11 can be easily turned into an efficient algorithm to find the flows. For a function , let denote .
Lemma 12.
Let be a connected plane graph, let be an even function such that , and let and be distinct faces of . There exists an algorithm with time complexity which finds circulationmaximum and minimum flows in with support , or decides that contains no flow.
Proof.
It suffices to find an circulationmaximum flow, an circulationminimum flow is obtained using the same algorithm to find a circulationmaximum flow. We can assume that is the outer face of . It is possible to find a flow in (or to decide that no flow exists) in time using FordFulkerson algorithm. Since is even, the complement of is Eulerian; we can partition it into cycles in linear time and add an arbitrary flow on each of the cycles to . Hence, from now on is a flow with support .
In time we construct the auxiliary graph as described in the proof of Lemma 11 (note that is an orientation of , since ). We initialize and as the subgraph of drawn in the boundary of the face . Then, we repeat the following steps:

While there exists an edge of directed from left to right as seen from ,

if is a bridge in (i.e., it is incident with from both sides), then delete from , otherwise

let be the facial walk of the face of not contained in , delete from , and add to .


If became the outer face of , then stop; otherwise, remove from all the edges not incident with its outer face, and all isolated vertices.

Note that became a cycle in the previous step (since all its edges are incident with both and the outer face). Add two units of flow on the edges of the cycle to , and reverse the edges in and .
Note that the part (i) can be implemented efficiently by maintaining a queue of edges directed from left to right, updated whenever is added to in step (i) or when the direction of edges are altered in step (iii). Since each edge is added to at most twice (for each face of it is incident with) and deleted at most once, the total time complexity of (i) over the whole run of the algorithm is . As for the step (iii), note that the direction of each edge is changed at most once, and thus the total time complexity of (iii) over the whole run of the algorithm is also . ∎
Let be a connected plane graph and let be a function such that . For a cycle in , we define to be the set of faces of contained in the open disk in the plane bounded by , we let be the set of the corresponding vertices of , and we let . Note that if is even, then for every cycle in . Let and . For an edge of , we define , , and (i.e., is viewed as a cycle of length two). For a cycle or edge , we let . The maxflow mincut theorem implies the following.
Observation 13.
Let be a connected plane graph and let be a function such that . Then contains a flow (or equivalently, a linkage) if and only if for every cycle in .
If for every cycle in , then we say that the function is feasible. For a set of cycles and edges of , we let . For an edge contained in cycles and edges from , we define . If and are faces of , we say that is connecting if contains a path from to . We say that is laminar if for any cycles , the open disks in the plane bounded by and are either disjoint or one is a subset of the other.
If is feasible, then let denote the maximum integer such that there exists a linkage and an circulation of size in with disjoint supports. We are now ready to prove the main result of this section, the minmax theorem for . The argument used to prove the part (a) is based on the idea of Seymour [16] for commodity flows.
Theorem 14.
Let be a connected plane graph, let and be faces of , and let be a feasible even function such that . Let and be circulationmaximum and minimum flows in , respectively. Then the following claims hold.

For every path from to in , .

for every connecting set of cycles and edges of .

There exists a laminar connecting set of cycles and edges of such that .
Furthermore, there exists an algorithm with time complexity which given , , , and returns

a linkage and an circulation of size in with disjoint supports, and

a laminar connecting set of cycles and edges of such that .
Proof.
Suppose that is a linkage and is an circulation in such that and is maximum, i.e., equal to . For any cycle in , the set forms an edgecut in separating from . Clearly, contains at least paths with one end in and the other hand in , and thus all but at most edges of are intersected by paths in .
Consider now an connecting set of cycles and edges of . Let be the multigraph obtained from by giving each edge the multiplicity (i.e., when constructing , the cycles of are made edgedisjoint and edges of are turned into cycles by increasing the multiplicity of edges). According to the previous paragraph, all but at most edges of are intersected by paths in . All vertices of have even degree, and thus each component of is edgeconnected. Since is connecting, it follows that contains two edgedisjoint paths from to , and thus every noncontractible cycle in intersects at least two edges of . Since the supports of and are edgedisjoint, we conclude that . Therefore, (b) holds.
Let be any path from to in . By Observation 8, there exists a flow and a flow in with , , and . In particular, the supports of and are disjoint, and thus and are flows in . Since and are circulationmaximum and minimum, respectively, we have and . Consequently,
On the other hand, by Lemma 11, since is even we can assume that . Let and . Since , all values of and
are odd, and thus all values of
and are integers. Consequently, is a flow and is a flow in . Furthermore, observe that . By Observation 8, we conclude that contains a linkage and an circulation with disjoint supports such that . Therefore, . Combining the inequalities, we conclude that (a) holds. Note that circulationmaximum and minimum flows with maximal supports can be found in time using the algorithm of Lemma 12, and they can be converted into a linkage and an circulation of size in with disjoint supports in time as described in this paragraph.Finally, let us prove the part (c). By Lemma 11, there exist paths and in from to such that and . Let be the closed walk obtained as the concatenation of with the reversal of . By Lemma 10, we have , and thus
(1) 
For an integer , let be the set consisting of edges of such that is incident with faces and satisfying and (one can view as assigning heights to faces of , and then corresponds to the contour lines at height very slightly less than ). Let be the set of edges of through that both and pass in the same direction. We claim that can be chosen to consist of the edges of and of the cycles into which the sets naturally decompose. Let us describe the construction precisely.
By Observation 9(b), for an edge of incident with faces and and letting ,

if , then belongs to exactly one of the sets , namely to ,

if , then belongs exactly to two of the sets, namely to and , and

if or , then does not belong to any of the sets.
It follows that
(2) 
Let denote the subgraph of with the edge set and the vertex set consisting of the vertices incident with the edges of . Observe that if is a face of and faces and of satisfy , then either both and , or both and . Let and denote the sets of faces of for that the former or the latter, respectively, holds. For , let denote the subgraph of drawn in the boundary of . Note that faces in only share edges with faces in and viceversa. Consequently, , and the graphs for are pairwise edgedisjoint and edgeconnected. For , let denote the set of connected blocks of ; since is edgeconnected and all its edges are incident with , is a set of cycles. For , let if is contained in the open disk of the plane bounded by , and let otherwise. Since , for each face we have
(3) 
Let and . Clearly, is connecting, since . Observe that if and , then there exists a face such that , and thus the set is laminar. It remains to argue that .
Let be the minimum of . Note that for any ,
(4) 
By (4) and (3), and using the fact that , we have
and thus by (1),
(5) 
By (2), we have
and the definition of gives
By (5), we conclude that
By (b), we conclude that , and thus (c) holds. Furthermore, observe that this construction of the set from the flows and can be performed in time . ∎
For the coloring applications, we need the following consequence.
Corollary 15.
There exists an algorithm with time complexity which given a connected plane graph , faces and of , a path from to in , an even function such that , and an integer returns one of the following:

A flow in such that and , or

a cycle in such that , or

a laminar connecting set of cycles and edges of with , and flows and in such that and .
Proof.
If is not feasible, then a straightforward modification of FordFulkerson algorithm returns a cycle in such that . We can return such a cycle and stop.
Hence, suppose that is feasible. Let be a
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