Strong -Flow Conjecture for Projective Planar Graphs
Abstract
In 1972, Tutte posed the -Flow Conjecture: that all -edge-connected graphs have a nowhere zero -flow. This was extended by Jaeger et al. (1992) to allow vertices to have a prescribed, possibly non-zero difference (modulo ) between the inflow and outflow. They conjectured that all -edge-connected graphs with a valid prescription function have a nowhere zero -flow meeting that prescription. Kochol (2001) showed that replacing -edge-connected with -edge-connected would suffice to prove the -Flow Conjecture and Lovász et al. (2013) showed that both conjectures hold if the edge connectivity condition is relaxed to -edge-connected. Both problems are still open for -edge-connected graphs.
The -Flow Conjecture was known to hold for planar graphs, as it is the dual of Grötzsch’s Colouring Theorem. Steinberg and Younger (1989) provided the first direct proof using flows for planar graphs, as well as a proof for projective planar graphs. Richter et al. (2016) provided the first direct proof using flows of the Strong -Flow Conjecture for planar graphs. We prove the Strong -Flow Conjecture for projective planar graphs.
1 Introduction
A -flow on a graph is a function that assigns to each edge an ordered pair consisting of a direction, and a value , such that if is the resulting directed graph, then, for each vertex ,
It is easy to see that every graph has a -flow for every value of : set for all . Therefore it is typical to use the following more restrictive concept. A nowhere zero -flow on is a -flow on such that no edge is assigned the value zero. In 1950, Tutte proved that a graph has a nowhere-zero -flow if and only if it has a nowhere-zero -flow, which requires the net flow through each vertex to be exactly zero; see Diestel, [4] for a proof of this equivalence and further background on flows.
Tutte (cf. Bondy and Murty, [1]) conjectured that every -edge-connected graph has a nowhere zero -flow. This is known as the -Flow Conjecture, and while progress has been made for many classes of graphs, it is still an open problem. For planar graphs the -Flow Conjecture is equivalent to Grötzsch’s Theorem, and Steinberg and Younger, [11] provided a direct proof using flows. Steinberg and Younger, [11] also proved that the -Flow Conjecture holds for graphs embedded in the projective plane.
As an extension of -flows, we add a prescription function, where each vertex in the graph is assigned a value in that defines the net flow through the vertex. The prescriptions of the vertices in the graph must sum to zero in . A graph is -connected if for each valid prescription function , has a nowhere-zero -flow achieving . This led Jaeger et al., [6] to pose the following conjecture.
Conjecture 1 (Strong -Flow Conjecture).
Every -edge-connected graph is -connected.
Jaeger, [5] had earlier posed the following weaker conjecture.
Conjecture 2 (Weak 3-Flow Conjecture).
There is a natural number such that every -edge-connected graph has a nowhere zero -flow.
Jaeger, [5] also showed that this conjecture is equivalent to the same statement regarding -connectivity. The Weak 3-Flow Conjecture remained open until Thomassen, [12] proved that sufficed.
Theorem 1.1.
Every -edge-connected graph is -connected.
Lovász et al., [9] extended this to the following result.
Theorem 1.2.
If is a -edge-connected graph, then is -connected.
Kochol, [7] showed that the -Flow Conjecture is equivalent to the statement that every -edge-connected graph has a nowhere zero -flow, so the result of Lovász et al., [9] is one step away from the -Flow Conjecture. As stated, both of these results also make significant steps toward the Strong -Flow Conjecture, as they considered -connectivity. Lai and Li, [8] proved the Strong -Flow Conjecture holds for planar graphs using the duality with graph colouring. Richter et al., [10] provided the first direct proof of this result using flows. Their result is Theorem 1.3.
Theorem 1.3.
Let be a -edge-connected graph embedded in the plane with at most two specified vertices and such that
-
•
if exists, then it has degree , , or , has its incident edges oriented and labelled with elements in , and is in the boundary of the unbounded face,
-
•
if exists, then it has degree and is in the boundary of the unbounded face,
-
•
there are at most two -cuts, which can only be and ,
-
•
if has degree , then does not exist, and
-
•
every vertex not in the boundary of the unbounded face has five edge-disjoint paths to the boundary of the unbounded face.
If has a valid prescription function, then has a valid -flow.
The -Flow Conjecture and the Strong -Flow Conjecture for planar graphs are corollaries of Theorem 1.3.
In this paper we prove the Strong -Flow Conjecture for projective planar graphs, extending Theorem 1.3. This proof requires the preliminary results that appear in de Jong, 2020b [3]. All of these results also appear in the first author’s Ph.D. thesis [2]. The techniques used for the proof in this paper build off those of Thomassen, [12], Lovász et al., [9], Richter et al., [10], and Steinberg and Younger, [11].
In Section 2 we first discuss some of the ideas that will be used throughout this paper.
2 Preliminaries
Basic Definitions
We define an orientation of a graph to be the directed graph obtained by adding a direction to each edge. With reference to determining the -connectivity of with prescription function , an orientation is valid if for each vertex ,
It can be easily seen that this is equivalent to a -flow.
A vertex in a graph is a directed vertex if all its incident edges are directed. We call this an orientation of . We say that an orientation of extends the orientation of if the direction of the edges at is maintained. In cases involving a directed vertex we take the term valid orientation to include that the orientation extends that of .
We lift a pair of edges and in a graph by deleting and , and adding an edge . We define an edge-cut in to be internal if either or does not intersect the boundary of the specified face(s) of .
Specified Face(s)
First, we consider the idea of a specified face. Theorem 1.3 allows vertices of degree less than on the boundary of the outer face. However, a graph embedded in a higher genus surface does not have a defined outer face, so the result cannot be directly extended.
Let be a graph with a specified face . Let be a graph obtained from by one or more operations. Unless otherwise stated, the specified face is defined as follows:
-
1.
Suppose that is obtained from by deleting or contracting a connected subgraph of that has no edge in common with the boundary of . Then .
-
2.
Suppose that is obtained from by contracting a connected subgraph of that contains the boundary of . Then has no specified face. A face can be chosen arbitrarily; in general we will chose a specified face incident with the vertex of contraction.
-
3.
Suppose that is obtained from by deleting an edge in the boundary of . Let be the other face incident with . Then is the face formed by the union of the boundaries of and (without ).
-
4.
Suppose that is obtained from by deleting a vertex in the boundary of . Let , ,…, be the other faces incident with . Then is the face formed by the union of the boundaries of , ,…,, and (without the edges incident with ).
-
5.
Suppose that is obtained from by contracting a connected subgraph of whose intersection with is a path of length at least one. Then is the face formed by the boundary of without . If the intersection of with consists of more than one path, this contraction can simply be completed in multiple steps.
-
6.
Suppose that is obtained from by lifting a pair of adjacent edges , , where is in the boundary of , is not, and and are consecutive at their common vertex. Let be the other face incident with . Note that is incident with . Let be the other face incident with . Then is the face formed by the union of the boundaries of and (using the lifted edge instead of and ).
When performing these operations we will not explicitly state the new specified face unless necessary.
Edge-Disjoint Paths to the Boundary
As in Richter et al., [10], throughout this paper we are working with graphs for which all vertices not on the boundary of the specified face(s) have at least edge-disjoint paths to the boundary of the specified face(s). We describe here how reductions to the graph affect this condition. The proof of this result is straightforward, and can be found in [2].
Lemma 2.1.
Let be a graph with specified face such that all vertices not on the boundary of have edge-disjoint paths to the boundary of . Let be a graph obtained from by
-
1.
contracting a subgraph of that does not intersect the boundary of to a vertex ,
-
2.
deleting a boundary edge of ,
-
3.
deleting a boundary vertex of ,
-
4.
lifting a pair of adjacent edges , , where is in the boundary of , is not, and and are consecutive at their common vertex, or
-
5.
contracting a subgraph of whose intersection with is a path .
Then all vertices not on the boundary of have edge-disjoint paths to the boundary of .
We therefore only discuss the preservation of this property in cases where Lemma 2.1 does not apply.
Minimal Cuts
Let be a graph with a directed vertex . An edge-cut in with is -robust if and .
Throughout this paper we will perform local reductions on graphs. Many of these reductions will involve considering -robust edge-cuts, either because has a small edge-cut that must be reduced, or because we must verify that the graph resulting from a reduction does not have any small edge-cuts. In all cases, we first consider the smallest possible edge-cuts. Thus, if we consider a -robust -edge-cut in a graph , we may assume that has no -robust at most -edge-cut. Thus it may be assumed that either is connected, or it consists of two isolated vertices whose degrees sum to . The same is true of . Given the sizes of the cuts we consider, generally both and will be connected.
Non-Crossing 3-Edge-Cuts
Let and be distinct edge-cuts in . We say that and cross if , , , and are all non-empty. Throughout this paper we consider graphs that are allowed to have non-crossing -robust -edge-cuts under certain restrictions. The following well-known result allows us to assume that such cuts are non-crossing. A proof can be found in [2].
Lemma 2.2.
Let be an odd positive interger. If is a -edge-connected graph, then any two -edge-cuts in do not cross.
Face Boundaries in the Projective Plane
Let be a graph embedded in the plane, and let be a specified face of . If does not contain a cut vertex, then we may assume that is bounded by a cycle. If is embedded in the projective plane, this is not necessarily true. Suppose that is a vertex that appears more than once in the boundary walk of , and assume that is not a cut vertex. Then there exists a non-contractible curve that passes through only and . Cut along this curve, and draw the graph on the plane. The result is a planar graph with one specified face containing two copies of . See Figure 1 for an illustration. Contract the two copies of to a single vertex. Then is a planar graph with two specified faces, each containing . The following result of de Jong, 2020b [3] guarantees that such a graph has a valid orientation.
Definition 2.3.
An FT graph is a graph embedded in the plane, together with a valid prescription function , such that:
-
1.
is -edge-connected,
-
2.
has two specified faces and , and at most one specified vertex or ,
-
3.
there is at least one vertex in common between and ,
-
4.
if exists, then it has degree , , or , is oriented, and is in the boundary of both and ,
-
5.
if exists, then it has degree and is in the boundary of at least one of and ,
-
6.
has at most one -edge-cut, which can only be or , and
-
7.
every vertex not in the boundary of or has edge-disjoint paths to the union of the boundaries of and .
Theorem 2.4.
Every FT graph has a valid orientation.
Chords in the Projective Plane
Let be a graph embedded in the plane and let be a specified face of . Suppose that has a chord with endpoints and . Then there exist subgraphs and of such that and . This is a property that Richter et al., [10] exploited when proving the Strong -Flow Conjecture for planar graphs, and a property that we use throughout this paper.
Now suppose that is a graph embedded in the projective plane with a cycle bounding a closed disk. Let be a chord of . If is contained in an open disk, then is a contractible chord. Otherwise it is a non-contractible chord. This is relevant in the case where the specified face is bounded by a cycle. If has a contractible chord, we may use techniques analogous to those in the plane to reduce the graph. In the case of a non-contractible chord, the graph is not split into subgraphs as it is in the plane, and we require different techniques. We see here that the deletion of such a chord and its endpoints results in a planar graph.
Lemma 2.5.
Let be a graph embedded in the projective plane with a specified face bounded by a cycle. Suppose that is a non-contractible chord of . Let . Then is planarly embedded in the projective plane with one specified face; namely the one containing .
Proof.
Consider the projective plane as represented by a circle of radius , where opposite points are identified. Draw the given embedding of such that the boundary of lies on the unit circle (where the origin is in the specified face), and is contained in the line . The graph does not intersect the line . Cut the projective plane along the line and identify the opposite points on the circle. The result is a planar embedding of where is the outer face. ∎
In practice, we will obtain graphs with up to three vertices of degree three. The following result of de Jong, 2020b [3] shows that such graphs have a valid orientation.
Definition 2.6.
A DTS graph is a graph embedded in the plane, together with a valid -prescription function , such that:
-
1.
is -edge-connected,
-
2.
has a specified face , and at most three specified vertices , , and ,
-
3.
if exists, then it has degree , , or , is oriented, and is on the boundary of ,
-
4.
if or exists, then it has degree and is on the boundary of ,
-
5.
has degree at most where is the number of unoriented degree vertices in ,
-
6.
has at most three -edge-cuts, which can only be , , and , and
-
7.
every vertex not in the boundary of has edge-disjoint paths to the boundary of .
A 3DTS graph is a graph with the above definition, where (6) is replaced by
-
6’.
all vertices other than , , and have degree at least , and if , , and all exist, then every -edge-cut in separates one of , , and from the other two.
Theorem 2.7.
All 3DTS graphs have a valid orientation.
Circulant and Almost Circulant Graphs
In the proof of the Strong -Flow Conjecture for graphs in the projective plane, two specific classes of graphs will arise. Let be an odd integer greater than or equal to , and define to be the circulant graph on vertices with jumps and . Label the vertices . Let be the graph on vertices obtained from by subdividing with a vertex , and adding an edge . We show that both classes of graphs satisfy the Strong -Flow Conjecture.
Lemma 2.8.
Let be an odd integer greater than or equal to . Then and , along with valid prescription functions, have a valid orientation.
Proof.
We obtain a valid orientation by lifting the pair of edges and , and directing and deleting the edges incident with the following vertices in order, to meet each prescription:
This directs the edge , which in turn directs the edge . Since cannot be the only vertex whose prescription is not met, the graph has a valid orientation. ∎
3 Projective Plane
Here we prove that the Strong -Flow Conjecture holds for graphs embedded in the projective plane.
Definition 3.1.
A PT graph is a graph embedded in the projective plane, together with a valid prescription function , such that:
-
1.
is -edge-connected,
-
2.
has a specified face , and at most one specified vertex ,
-
3.
if exists, then it has degree and is in the boundary of ,
-
4.
has at most one -edge-cut, which can only by , and
-
5.
every vertex not in the boundary of has edge-disjoint paths to the boundary of .
We define all -edge-connected graphs on at most two vertices to be PT graphs, regardless of vertex degrees.
A 3PT graph is a graph with the above definition, where (4) is replaced by
-
4’.
all vertices aside from have degree at least , and if exists and is a -edge-cut with , then is contained in an open disk.
Theorem 3.2.
All PT graphs have a valid orientation.
Proof.
Let be a minimal counterexample with respect to the number of edges. If , then consists of only an isolated vertex, and so has a trivial valid orientation. Thus we may assume has at least one edge.
We will establish the following series of properties of .
- PT1:
-
The graph does not contain a loop, unoriented parallel edges, or a cut vertex.
- PT2:
-
The face is bounded by a cycle.
- PT3:
-
There is no contractible chord of incident with a degree or vertex.
We define a Type 1 cut to be an edge-cut that does not intersect the boundary of . We define a Type 2 cut to be an edge-cut that has exactly two edges in the boundary of . We define all remaining edge-cuts (with at least edges in the boundary of ) to be of Type 3.
- PT4:
-
-
1.
The graph does not contain a -robust at most -edge-cut.
-
2.
If contains a -robust at most -edge-cut , then it is either
-
(a)
of Type 1, where is contained in an open disk and the boundary of is in , or
-
(b)
of Type 2, where is contained in an open disk, , and does not have an incident edge in .
-
(a)
-
3.
If has a -robust Type 1 -edge-cut where is contained in an open disk, then the boundary of is in .
-
1.
- PT5:
-
The vertex exists.
Let and be the boundary vertices adjacent to , and let be the remaining vertex adjacent to .
- PT6:
-
Vertices and have degree .
- PT7:
-
Vertex has degree .
- PT8:
-
The edge is a chord.
- PT9:
-
Both and are adjacent to .
The verification of these properties forms the bulk of the proof of Theorem 3.2. Property PT1 is straightforward, and the proof is omitted. It can be read in [2].
PT1.
The graph does not contain a loop, parallel edges, or a cut vertex.
PT2.
The face is bounded by a cycle.
Proof.
Suppose that is not bounded by a cycle. Then there exists a vertex that appears twice on the boundary walk of . By PT1, is not a cut vertex. As shown in Section 2, is a planar graph with two specified faces, each containing , and has at most one vertex of degree . Hence is an FT graph and has a valid orientation by Theorem 2.4, a contradiction. ∎
Claim 3.3.
The graph has no -robust Type 1 cut of size at most where is contained in an open disk and the boundary of is in .
Proof.
Suppose such a exists in . Then by definition, is at least a -edge-cut. Let be the graph obtained from by contracting to a vertex. It is clear that is a PT graph, and thus has a valid orientation by the minimality of . Transfer this orientation to . Let be the graph obtained from by contracting to a vertex. Then is a graph where all vertices have degree at least , with a directed vertex of degree or . Now is planarly embedded by construction, where every vertex adjacent to is on the outer face. Therefore, can be embedded on the plane by inserting into the outer face of . Choose the specified face to be incident with .
If has degree , then is a DTS graph and has a valid orientation by Theorem 2.7. If has degree , delete one boundary edge incident with to form a graph . If has a -robust at most -edge-cut, then has a -robust at most -edge-cut of Type 1, a contradiction. Thus is a DTS graph and has a valid orientation by Theorem 2.7. This yields a valid orientation of , a contradiction. Hence no such cut exists. ∎
Claim 3.4.
The graph has no -robust Type 2 cut of size .
Proof.
Suppose such a cut exists in . Either or is contained in an open disk. Without loss of generality, suppose that is contained in an open disk. Let be the graph obtained from by contracting to a vertex. It is clear that is a PT graph, and thus has a valid orientation by the minimality of . Transfer this orientation to . Let be the graph obtained from by contracting to a vertex . Now is planarly embedded by construction, where every vertex adjacent to is on the outer face. Therefore, can be embedded on the plane by inserting into the outer face of . Choose the specified face to be incident with and all vertices in incident with .
Then is a planar graph with a specified face containing a directed vertex of degree . Hence is a DTS graph and has a valid orientation by Theorem 2.7. This yields a valid orientation of , a contradiction. Hence no such cut exists. ∎
Claim 3.5.
The graph has no -robust Type 2 cut of size where is contained in an open disk and .
Proof.
Suppose such a cut exists in . Let be the graph obtained from by contracting to a vertex. It is clear that is a PT graph, and thus has a valid orientation by the minimality of . Transfer this orientation to . Let be the graph obtained from by contracting to a vertex. Then can be embedded as a planar graph with a specified face containing a directed vertex of degree but no degree vertex. Hence is a DTS graph and has a valid orientation by Theorem 2.7. This yields a valid orientation of , a contradiction. Hence no such cut exists. ∎
In order to complete our analysis of Type 2 cuts, we must consider contractible chords.
PT3.
There is no contractible chord of incident with a degree or vertex.
Proof.
Suppose that such a chord exists where the degree of is or . Let and be subgraphs of such that , , and is contained in an open disk.
Both and are -robust, else has unoriented parallel edges. Both cuts are of Type 2. Therefore, , and so .
It is clear that is a PT graph and has a valid orientation by the minimality of . Transfer this orientation to , and orient . Add a directed edge from to in (in the boundary of ). Then can be embedded as a planar graph with a single specified face, a directed vertex of degree or , and one other possible degree vertex . Hence is a DTS graph and has a valid orientation by Theorem 2.7. This leads to a valid orientation of , a contradiction. ∎
Claim 3.6.
The graph has no -robust Type 2 cut of size where is contained in an open disk; ; and is a single edge that is a boundary edge of .
Proof.
Suppose such a cut exists in . Let be the graph obtained from by contracting to a vertex. Since is a PT graph, it has a valid orientation by the minimality of . Transfer this orientation to . Let be the graph obtained from by contracting to a vertex. Then can be embedded as a planar graph with a specified face containing a directed vertex of degree adjacent to . Since is contained in an open disk, is not incident with a chord by PT3. Let be the graph obtained from by orienting and deleting . Then has a directed vertex of degree and at most one vertex of degree . If contains a -robust at most -edge-cut, then contains an at most -edge-cut of Type 1 or 2, a contradiction. Hence is a DTS graph and has a valid orientation by Theorem 2.7. This yields a valid orientation of , a contradiction. Hence no such cut exists. ∎
Claim 3.7.
The graph has no -robust Type 3 cut of size at most .
Proof.
Suppose that such a cut exists. Without loss of generality, suppose that . Let be the graph obtained from by contracting to a single vertex. Since is a PT graph, it has a valid orientation by the minimality of . Transfer this orientation to . Let be the graph obtained from by contracting to a single directed vertex of degree or . Then appears twice in the boundary walk of . As in the proof of PT2, is a planar graph with two specified faces that both contain the vertex . Since , has no degree vertex. Therefore is an FT graph and has a valid orientation by Claim 2.4. This leads to a valid orientation of , a contradiction. Hence no such cut exists. ∎
We summarise the reducible/irreducible cuts in .
PT4.
-
1.
The graph does not contain a -robust at most -edge-cut.
-
2.
If contains a -robust at most -edge-cut , then it is either
-
(a)
of Type 1, where is contained in an open disk and the boundary of is in , or
-
(b)
of Type 2, where is contained in an open disk, , and does not have an incident edge in .
-
(a)
-
3.
If has a -robust Type 1 -edge-cut where is contained in an open disk, then the boundary of is in .
Proof.
Figure 2 shows the -robust at most -edge-cuts that exist in . The dashed line is used to represent the crosscap. We wish to reduce at low degree vertices in . We establish the existence of and consider its adjacent vertices.
PT5.
The vertex exists.
Proof.
Suppose that does not exist. We prove the following claims:
-
a.
Every vertex on the boundary of has degree .
-
b.
All vertices in are on the boundary of .
-
c.
We have , where is odd.
These properties provide the necessary structure to obtain a contradiction.
Claim PT5a.
Every vertex on the boundary of has degree .
Proof.
Consider the case where the boundary of contains a vertex of degree at least . Orient and delete a boundary edge incident with , and call the resulting graph . Then has at most one degree vertex. Suppose that contains a -robust at most -edge-cut. Then contains a -robust at most -edge-cut, a contradiction. Thus is a PT graph. By the minimality of , has a valid orientation. This yields a valid orientation of , a contradiction. ∎
Claim PT5b.
All vertices in are on the boundary of .
Proof.
Suppose there exists a vertex on the boundary of that has an adjacent vertex not on the boundary of . Then . Let be the edges incident with in order, where and are on the boundary of , and is incident with . Lift the pair of edges , , orient the remaining two edges incident with to satisfy , and delete , calling the resulting graph . Then has at most one degree three vertex (the other endpoint of ). If contains a -robust at most -edge-cut , then contains a -robust at most -edge-cut, or a -robust -edge-cut that uses a boundary edge of and thus is of Type 2 or Type 3, a contradiction (since does not exist). Thus is a PT graph, and so by the minimality of , has a valid orientation. This yields a valid orientation of , a contradiction. ∎
Thus all vertices lie on the boundary of and have degree . Let be the vertices on the boundary of in order.
Claim PT5c.
We have .
Proof.
Consider a vertex . It is clear from the construction that is adjacent to and . Let and be the remaining two vertices adjacent to . Suppose that and are not adjacent. Then is an at most -edge-cut of Type 2. If it is not a -robust -edge-cut, then contains parallel edges, a contradiction. Hence contains a -robust -edge-cut of Type 2, a contradiction. Thus and are adjacent.
Without loss of generality, assume that . Let
If there exists an edge not on the boundary of with both endpoints in , then either contains parallel edges, or contains a contractible chord incident with a degree vertex, a contradiction. The same is true of edges with both endpoints in . Hence every edge not in the boundary of and not incident with has one endpoint in and the other endpoint in . Since every vertex has degree , . If is even, then the non-contractible chords incident with a vertex are parallel edges. Hence is odd and . ∎
By Lemma 2.8, has a valid orientation. ∎
Let , , and be the vertices adjacent to , where and are on the boundary of . Since has no parallel edges, these three vertices are distinct. Note that PT3 implies that is not incident with a contractible chord, so either is not in the boundary of , or is a non-contractible chord.
Claim 3.8.
Let be a 3PT graph where . Then has a valid orientation.
Proof.
Let be a minimal counterexample. If is a PT graph, then has a valid orientation by the minimality of . Thus we may assume that has a -robust -edge-cut where . By the definition of a 3PT graph, is contained in an open disk.
Let be the graph obtained from by contracting to a vertex. Then any -edge-cut in is a -edge-cut in , so is a 3PT graph and has a valid orientation by the minimality of . Transfer this orientation to . Let be the graph obtained from by contracting to a vertex. Then is a 3DTS graph and has a valid orientation by Lemma 2.7. This leads to a valid orientation of , a contradiction. ∎
Claim 3.9.
At least two of , , and have degree .
Proof.
Suppose not. Let be the graph obtained from by orienting and deleting . Then contains at most one vertex of degree . Suppose that contains a -robust at most -edge-cut . Then contains a -robust at most -edge-cut, a contradiction. Hence is a PT graph. Thus by the minimality of , has a valid orientation. This yields a valid orientation of , a contradiction. ∎
Claim 3.10.
Vertex does not have degree .
Proof.
Suppose that . Note that is a non-contractible chord of . Let , , , be the edges incident with in order, where and are on the boundary of , and . We may choose the labelling of and so that the -path in the boundary of contains but not .
Suppose that . Then can be redrawn with inside to yield an FT graph with and on both specified faces. Figure 3 shows this drawing. By Theorem 2.4, has a valid orientation, a contradiction.
We may now assume that . Lift the pair , , and orient the remaining edges incident with to satisfy . Orient the remaining edges incident with to satisfy . Delete and , and call the resulting graph . Then is a planar graph. There are three possible degree three vertices in : , , and the vertex incident with (which by assumption are distinct). If contains a -robust at most -edge-cut, then contains a -robust at most -edge-cut, a contradiction to PT4.
If all -robust -edge-cuts in separate two degree vertices, then by Theorem 2.7, has a valid orientation that leads to a valid orientation of , a contradiction. Suppose that contains a -robust -edge-cut that does not separate the degree vertices. We assume that is not an at most -edge-cut. Then is a -edge-cut using edges and . This is a Type 2 cut, for which is in the side not contained in an open disk, a contradiction. Figure 4 shows this cut. Orient a degree three vertex in to obtain a DTS graph. By Theorem 2.7, has a valid orientation. This extends to a valid orientation of , a contradiction. ∎
PT6.
Vertices and have degree .
PT7.
Vertex has degree .
Proof.
Suppose that . Let , , , be the edges incident with in order, where is on the boundary of , and . We prove the following claims
-
a.
Edge is incident with two vertices of degree .
-
b.
Edge is not a chord.
These provide the necessary structure to complete the proof.
Claim PT7a.
Edge is incident with two vertices of degree .
Proof.
Suppose that is not incident with two vertices of degree . Lift the pair , , and orient the remaining edges incident with . Orient the remaining edges incident with , and delete and , calling the resulting graph . Then contains at most one vertex of degree , which is .
We now check that has no -robust - or -edge-cuts. The cases are indicated in italics. We follow a similar process in most future cases.
Suppose that contains a -robust at most -edge-cut . Then contains a -robust at most -edge-cut, a contradiction. Suppose that contains a -robust -edge-cut . Then contains a corresponding -robust cut of size at most . There are two options. First, is of Type 2 and has on the side that is contained in an open disk. We assume that this side is . By Claim 3.8, . Hence . We have , else a smaller cut exists. We conclude that . Let be the maximal connected subgraph of containing but not . Then is an at most -edge-cut of Type 1, a contradiction. This cut can be seen in Figure 5. Second, is of Type 1 in . Then has on the side contained in a disk in , and thus has a valid orientation by Claim 3.8. This leads to a valid orientation of , a contradiction.
Hence is a PT graph, and thus has a valid orientation by the minimality of . This extends to a valid orientation of , a contradiction. ∎
Since is incident with two vertices of degree , is a non-contractible chord. Similarly, must have an analogous incident chord. Neither chord is incident with , since has degree at least . Let and be the vertices adjacent to and respectively via a chord.
Claim PT7b.
Edge is not a chord.
Proof.
Suppose that is a chord. Let be the set of vertices in the interior of the closed disc bounded by , , , and the -subpath of . If is empty, consider the same set with respect to and . Now contains at most edges incident with . Hence it contains at least two edges incident with , since is the only -edge-cut in . Since has no parallel edges, is -robust. This cut can be seen in Figure 6. Let be the minimal -robust edge-cut where and contains at most three edges not incident with .
Contract to a vertex, calling the resulting graph . Then is a PT graph and has a valid orientation by the minimality of . Transfer this orientation to and contract , calling the resulting graph . Delete edges incident with to make the vertex of contraction a degree vertex. Since has no parallel edges, at most one degree vertex results from this process. If has a -robust at most -edge-cut, then was not minimal, a contradiction. ∎
Therefore, and have degree , and has degree .
PT8.
The edge is a chord.
Proof.
Suppose that is not a chord. Let be the edges incident with in order, where , and is on the boundary of . We first prove the following property.
Claim PT8a.
Either is incident with two degree four vertices or is incident with .
Proof.
Suppose that is not incident with two degree four vertices, and is not incident with . Lift the pair , , orient the remaining edges incident with to satisfy , and orient the remaining edges incident with to satisfy . Delete and , calling the resulting graph . Then in , the only possible degree three vertex is . The analysis that has no small cuts is equivalent to that in PT7. Hence is a PT graph, and thus has a valid orientation by the minimality of . This extends to a valid orientation of , a contradiction. ∎
Therefore either is incident with two degree four vertices or is incident with . The same must be true of the corresponding edge incident with . There are three cases.
-
1.
First, suppose that both and are adjacent to . Suppose that is not incident with two degree four vertices. Then orient and delete and to satisfy , and contract the set of vertices to a single degree four vertex, calling the resulting graph . Then has at most one degree three vertex, which is incident with in . If contains a -robust edge-cut of size at most , then contains a -robust edge-cut of size at most that contains a boundary edge incident with , a contradiction. Hence is a PT graph. By the minimality of , has a valid orientation. Transfer this orientation to . Orient the remaining two edges incident with to satisfy , and the remaining two edges incident with to satisfy . Since is satisfied by the orientations of and , the direction of is determined to be the opposite (relative to ) of the direction of . Since cannot be the only vertex whose prescription is not met, this is a valid orientation of , a contradiction.
We now assume that is incident with two degree four vertices. The same must be true of the corresponding edge incident with . Let these vertices be and respectively. Suppose that and are not adjacent. Let and be the components of , where the labelling is chosen so that . Consider the cut . If contains a single vertex, then has parallel edges, a contradiction. Hence is -robust. Note that is a Type 2 cut and has size at most . This cut is shown in Figure 8.
Figure 8: PT8: Small Type 2 cut (1). Contract to a vertex, calling the resulting graph . It is clear that is a PT graph. Thus has a valid orientation by the minimality of . Transfer this orientation to and contract calling the resulting graph . Note that can be embedded in the plane with a directed vertex of degree at most . Delete the edges in the cut that are incident with , calling the resulting graph . Then is planar, has at most one vertex of degree , and has a directed vertex of degree at most . If has a -robust at most -edge-cut, then has a -robust at most -edge-cut, which is of Type 1, or Type 2 with in the side not in an open disk, a contradiction. Hence is a DTS graph and has a valid orientation by Theorem 2.7. This leads to a valid orientation of , a contradiction.
Hence we may assume that and are adjacent. Note that has two components: with neighbours in among , , and , and , with neighbours in among , , , and . If is a 2-robust cut, then it has size at most and is of Type 1, a contradiction. If contains a single vertex, then has parallel edges incident with , a contradiction. Hence and are adjacent to . This graph is shown in Figure 9.
Figure 9: PT8: and adjacent to Let . If is a 2-robust cut, then it has size and is of Type 3, a contradiction. If , then this vertex is repeated in the boundary walk of , so the boundary of is not a cycle, a contradiction. Hence . Lift the pair of edges , and orient , , , , in order (each having at least two unoriented edges). This determines the direction of , and since cannot be the only vertex to not meet its prescription, the result is a valid orientation of , a contradiction.
-
2.
Now suppose that and the corresponding edge incident with each have two endpoints of degree , and respectively. Then and are on the boundary of . Then contains a -robust edge-cut of size at most (containing , one or two edges incident with , and one or two edges incident with ). If the labelling is chosen so that , then the graph obtained by contracting to a single vertex is planar, so by Claim 3.5, has a valid orientation.
-
3.
Finally, suppose without loss of generality that has two endpoints of degree , and is adjacent to . Let be the other endpoint of . Let and be the edges incident with that are not or , where is on the boundary of . Assume that does not have two incident vertices of degree . Orient and delete and to satisfy , and contract the set of vertices to a single vertex of degree , calling the resulting graph . If contains a -robust at most -edge-cut, then contains a -robust at most -edge-cut containing a boundary edge incident with , a contradiction. Then is a PT graph, so by the minimality of , has a valid orientation. Transfer this orientation to . Orient the remaining two edges incident with to satisfy . Since is satisfied by the orientations of and , the direction of is determined to be the opposite (relative to ) of the direction of . Since cannot be the only vertex whose prescription is not met, this is a valid orientation of , a contradiction.
We now assume that has two incident vertices of degree . Then is a chord. Let be the other endpoint of . Consider the cut where , , and is connected and maximised. If contains only one vertex, then has parallel edges, a contradiction. Hence is -robust. Note that is a Type 2 cut and has size at most . This cut is shown in Figure 10.
Figure 10: PT8: Small Type 2 cut (2). Contract to a vertex, calling the resulting graph . It is clear that is a PT graph. Thus has a valid orientation by the minimality of . Transfer this orientation to and contract calling the resulting graph . Note that is planar and has a directed vertex of degree at most . Delete two consecutive edges incident with where one is a boundary edge, calling the resulting graph . Then is planar and has at most one vertex of degree . If has a -robust at most -edge-cut, then has a -robust at most -edge-cut, either of Type 1, or Type 2 where is on the side not contained in an open disk, a contradiction. Hence is a DTS graph and has a valid orientation by Theorem 2.7. This leads to a valid orientation of , a contradiction. ∎
We are left with the case where is a chord. Let the edges incident with be in order, where , and is a boundary edge of . Suppose that is not incident with two degree four vertices, and is not incident with . Lift the pair , , orient the remaining edges incident with to satisfy , and orient the remaining edges incident with to satisfy . Delete and , calling the resulting graph . Then in , the only degree three vertex is . The argument that contains no small cuts is equivalent to previous arguments. Hence is a PT graph. By the minimality of , has a valid orientation. This extends to a valid orientation of , a contradiction.
Therefore is incident with either two degree four vertices or with . The same must be true of the corresponding edge incident with .
PT9.
Both and are adjacent to .
Proof.
Consider the case where is incident with two degree four vertices. Then is a chord. Let be the other endpoint of . Then is adjacent to on the boundary of , else has a -robust at most -edge-cut of Type 2, where is not in the side contained in a disk, a contradiction. This cut is shown in Figure 11. If the corresponding edge incident with is a chord, the same is true. Then has a -robust at most -edge-cut, a contradiction. This cut can also be seen in Figure 11. Thus we may assume that and are adjacent.
Let , , , be the edges incident with in order, where , , and is a boundary edge of . Now is not incident with two vertices of degree , else has either a -robust at most -edge-cut, or parallel edges. Also, cannot be adjacent to via parallel edges. Orient and delete and to satisfy , and contract the set of vertices to a single vertex of degree , calling the resulting graph . Now has only one possible degree three vertex, in it is incident with . If contains a -robust edge-cut of size at most , then contains a -robust edge-cut of size at most containing a boundary edge incident with , a contradiction. Thus is a PT graph, and has a valid orientation by the minimality of . Transfer this orientation to . Orient the remaining edges incident with . Since and satisfy , is known to have the opposite direction (relative to ) from . Since cannot be the only vertex whose prescription is not met, this is a valid orientation of . ∎
The remaining case is that both and are adjacent to . Let be the path on the boundary of from to that includes , and let be the path on the boundary of from to that includes . Let and be the subsets of vertices in and respectively that have an adjacent vertex of degree at least that is not .
If , then let be such that and is minimised. If , then let be such that and is minimised. Without loss of generality, suppose that . Let be the edges incident with in order, where if , and (with the convention that and if necessary), and otherwise, and . At least one of and is incident with a degree vertex, that is not , by definition.
If, for example, , then is adjacent to and , is adjacent to , is adjacent to and is adjacent to . Likewise is adjacent to and , to , to , and to . More generally, the set defined next, consists of those vertices whose adjacencies are determined in this fashion.
If , let . Otherwise, let . Orient and delete and to satisfy , and contract the set of vertices to a single vertex of degree . Call the resulting graph . By definition, has at most one degree three vertex. If has a -robust at most -edge-cut, then has a corresponding -robust at most -edge-cut that does not separate from , a contradiction. Hence is a PT graph. By the minimality of , has a valid orientation. Transfer this orientation to .
Suppose that . Orient and delete the two remaining unoriented edges incident with the following vertices in order:
There is only one unoriented edge at (), which by construction must have the opposite direction (relative to ) from . Since cannot be the only vertex whose prescription is not met, this is a valid orientation for .
Suppose that . Orient and delete the two remaining unoriented edges incident with the following vertices in order:
There is only one unoriented edge at (), which by construction must have the opposite direction (relative to ) from . Since cannot be the only vertex whose prescription is not met, this is a valid orientation for .
4 Discussion
The following result (the Strong -Flow Conjecture for projective planar graphs) is a direct corollary of Theorem 3.2.
Theorem 4.1.
Let be a -edge-connected graph embedded in the projective plane. Then is -connected.
The case where the prescription function is such that for all , is the 3-Flow Conjecture for projective planar graphs, shown by Steinberg and Younger, [11].
Unlike in the plane, in the projective plane (Theorem 3.2) we do not allow a directed vertex. In the plane this is necessary, in order to reduce small edge-cuts. In the projective plane we utilise the property that one of the two graphs resulting from contracting the sides of an edge-cut is a planar graph, and thus are able to apply planar results.
In fact, adding a directed vertex of degree to the result is not possible. Consider a graph with specified face such that every vertex is on the boundary of . Let be a degree vertex on the boundary of , and a degree vertex on the boundary of , adjacent to via a non-contractible chord. Let . Let the two paths between and on the boundary of be and where and all vertices in aside from and have degree . We assume that for all , is adjacent to , , , and , where we set and . Finally, and are adjacent. We let be . See Figure 12.
Suppose that and all edges incident with are directed out from . Let . Let . Let . Finally, let all other vertices have prescription . We have
Therefore is a valid prescription function for .
Lemma 4.2.
The graph does not have a valid orientation that meets .
Proof.
The edges incident with meet the prescription of , , , and , hence each of these vertices has all remaining edges directed in or all remaining edges directed out. Since and are adjacent, one has all edges directed in and the other has all edges directed out. Consider . The edges and are directed one in and one out of . Thus the remaining edges incident with : and , are directed into to satisfy .
We show that for all , and are directed out of , and that for all , and are directed out of . We consider the sequence . The property is known to hold for . Let be the first vertex in this sequence for which this property does not hold.
Suppose that for some . Then by definition, the property holds for . Hence and are directed out of . This satisfies the prescription of and , which are adjacent, and so the remaining three edges at one of these vertices are directed in, and the remaining three edges at the other are directed out. Both are adjacent to , so the edges and are directed one in and one out of . Hence the remaining two edges incident with are directed out of , as required.
Now suppose that for some . Then by definition, the property holds for . Hence and are directed out of . This satisfies the prescription of and , which are adjacent, and so the remaining three edges at one of these vertices are directed in, and the remaining three edges at the other are directed out. Both are adjacent to , so the edges and are directed one in and one out of . Hence the remaining two edges incident with are directed out of , as required.
Therefore, is directed out of , and is directed into . Hence no direction of meets . Thus has no valid orientation that meets . ∎
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