Given a family of graphs , we say a graph is -saturated if does not contain a subgraph isomorphic to any , but adding any edge to creates a copy of some . The study of -saturated graphs is one of the main topics of interest in extremal combinatorics. Particularly well studied quantities include the extremal number , which denotes the maximum number of edges in an -vertex -saturated graph, and the saturation number , which denotes the minimum number of edges in an vertex -saturated graph. If , then we will often denote simply by .
For a family of graphs , we define the -saturation game as follows. The game is played by two players, Max and Mini. Starting with Max, the two players alternate adding edges to an empty graph on vertices, with the only constraint being that neither player can add an edge that creates a subgraph in . The game ends once the graph becomes -saturated. We define the -game saturation number to be the number of edges in the graph at the end of the game when both players play optimally.
It is worth noting that
The first -saturation game to be considered was the -saturation game, which was introduced by Füredi, Reimer, and Seress  based on another game proposed by Hajnal. Despite being introduced nearly 30 years ago, very little is known about . In their original paper, Füredi, Reimer, and Seress proved that . The only other non-trivial bound for was obtained recently by Biró, Horn, and Wildstrom  who showed that .
The systematic study of saturation games was initiated by Carraher, Kinnersley, Reiniger, and West . In particular, they obtained bounds for the game saturation number of paths and stars, and these results were refined by Lee and Riet . Hefetz, Krivelevich, Naor, and Stojaković  further generalized saturation games to avoiding other graph properties such as colorability, and in particular Keusch  proved asymptotically tight bounds for the game where both players must keep the graph -colorable. Saturation games have also been generalized to other combinatorial structures such as hypergraphs and directed graphs [9, 11].
In view of (1), the problem of determining the order of magnitude of is trivial whenever and have the same order of magnitude, and this will often be the case if contains a tree. Perhaps the simplest non-trivial case of this problem is to try and determine the order of magnitude of when is a family of cycles.
A basic question in this direction that one could ask is: what families of cycles have quadratic game saturation number? It is well known that for all , so by (1) a necessary condition to have is that
consists only of odd cycles. The last author showed that a sufficient condition for a set of odd cycles to have quadratic game-saturation number is to have , in which case we have . Carraher, Kinnersley, Reiniger, and West  showed that where is the set of all odd cycles , though the last author  showed that in general a set of odd cycles containing and need not have game saturation number asymptotic to .
Similarly one could ask for necessary and sufficient conditions for a family of cycles to have linear game saturation number, and for this problem much less is known. Let (respectively, ) denote the set of all cycles (respectively, all odd cycles) of length at least . The following result of Erdős and Gallai shows that a trivial condition to have is for to contain every cycle which is at least as large as some cutoff value .
Theorem 1.1 ([4, Theorem 2.7]).
For all we have
Prior to this work, the only known non-trivial example of a set of cycles with linear game saturation number was the following.
Theorem 1.2 ([12, Theorem 1.4]).
The key idea in the proof of Theorem 1.2 is that Mini is able to play in the -saturation game such that the graph stays -free throughout the game, so the result follows by Theorem 1.1. Our main goal is this paper is to generalize the approach used in Theorem 1.2 to prove that for many more families of cycles .
1.1 Our Results
Our first result shows that if includes roughly half the cycles of length at least as large as some cutoff value , then it has game saturation number which grows linearly with .
Let be a collection of cycles such that , and such that there exists some so that for all , either or . Then
Applying this theorem with in place of immediately gives the following.
Our next result applies to infinite families of cycles which are much sparser than those considered in Theorem 1.3.
For , a family of cycles is said to be -dense if the following properties hold:
The cycles , unless in which case we only require ,
For all , we have ,
For all there exists with and .
Roughly speaking, a family is -dense if it contains no cycle up to length , we have , and the gaps between consecutive cycle lengths in grow no faster than an exponential function in .
If is a -dense family of cycles, then
Let . Then
Note that the gaps between the cycle lengths of this family grow exponentially large. Moreover, because consists only of odd cycles, so in theory its game saturation number could have been much larger than linear.
We will prove Theorem 1.6 by showing that when is a -dense family of cycles, either player can play in the -saturation game so that the graph never contains a cycle of length at most , and given this the result will essentially follow from Theorem 1.1. We will show (see Proposition 3.13) that the bounds in the definition of being -dense are close to best possible for such a strategy to exist.
The rest of the paper is organized as follows. In Section 1.2 we establish some basic graph theory notation and facts. In Section 2 we give a short proof of Theorem 1.3. The majority of the paper is dedicated to the proof of Theorem 1.6 in Section 3. Finally, we close with some concluding remarks and open problems in Section 4.
Given a graph , we say that a graph is a subgraph of if and . We say that is an induced subgraph of if . For a set , we define the induced subgraph on to be where . We will often write and to denote the graph obtained by adding or deleting the edge respectively. Given two vertices , a path from to is a geodesic if the path is of length . A path that is a geodesic for some choice of and will just be called a geodesic. The circumference of a graph is the length of the longest cycle in . We say that a path is -avoiding if it does not contain the vertex .
Many of our proofs will rely heavily on the block structure of our underlying graph. We recall that a block of a graph is a maximal subgraph of which is either a or 2-connected, and that the edges of the blocks partition the edges of . By a slight abuse of notation, we will often refer to blocks, vertex sets of blocks, and edge sets of blocks simply as “blocks”.
We recall some basic facts about blocks, and we refer the reader to Diestel  for more information on blocks. Whitney’s Theorem states that a graph on at least vertices is -connected if and only if every pair of vertices is connected by two internally vertex-disjoint paths. Thus given a block on at least three vertices and , there exists a pair of internally disjoint paths from to . A dominating vertex in a block is a vertex adjacent to all other vertices in . We say two blocks are adjacent if their vertex sets have a nonempty intersection. Note that two blocks can overlap in at most one vertex, so two blocks being adjacent is equivalent to them sharing exactly one vertex.
We now move onto some less standard terminology. Given a path with , note that there is a unique list of blocks and integers such that for all and , the edge . We will say is the block geodesic associated with . Note further that if and are two paths, then and are associated with the same block geodesic, which we will call the block geodesic. Furthermore, if the block geodesic consists of blocks, we will say that the block distance from to , denoted , is . In the -saturation game, we let be the graph after the edge has been added in the game. We let be the initial (empty) graph of the game, and we let denote the graph once it has become -saturated.
2 Proof of Theorem 1.3
Proof of Theorem 1.3.
First, we will show that we can play such that at the end of each of our turns the following hold for :
Every block that is not a block contains a triangle, and
There is at most one non-trivial path of blocks, with this path containing at most three edges. When it exists, we denote this path by .
We prove this claim inductively. For the base cases, and have properties (i) and (ii), since there are not enough edges to negate the claim. In the induction we describe moves that maintain properties (i) and (ii) in all possible situations; see also Figure 1. Inductively assume has properties (i) and (ii). Then the only way a block without a triangle could form in would be for this block to consist of the new edge in and some number of blocks from . Since by property (ii) there is only one non-trivial path of blocks, , and as this has length at most , the only blocks that could have formed in in this way are triangle blocks or (a unique) block. If a unique block forms, we can add a diagonal to this block, which then contains a triangle and satisfies property (i), and it is not difficult to see that (ii) is satisfied. Thus we can maintain the desired conditions if contains a non block without a triangle.
With this we can assume that every non block of contains a triangle, so property (i) is satisfied. If contains more than one non-trivial path of blocks, or a path of blocks of length or more, this must have been created by our opponent’s last turn. In either case, we can choose an edge that connects our opponent’s most recent edge to another edge within this block path. This will add a triangle and decrease the length of the longest path of blocks or add to the second non-trivial path of blocks such that it is no longer such a path. In either case, this satisfies property (ii) without violating property (i).
Next, we must handle the case that already satisfies properties (i) and (ii). If there exists a non-trivial path of blocks, then we can add an edge between two vertices on that path such that we create a triangle. The resulting graph satisfies both properties. Thus we may assume has no two adjacent blocks. Any move that does not create a forbidden cycle cannot create a path of blocks on more than three edges, so any move we make satisfies property (ii). Also, such a move can create at most one block which was not present in . If is a block, then property (i) is satisfied. If the vertices of were contained in a single block in , then that block must have contained a triangle, so property (i) is satisfied. Thus must contain the vertices from at least two blocks in . If any of those blocks contained a triangle in , then contains a triangle and property (i) is satisfied. Thus all the blocks from contained in are blocks. However, no two blocks are adjacent in , and it is impossible to form a single block using only non-adjacent blocks and one additional edge. This completes the proof of the claim.
Assume we play according to this strategy throughout the game, which implies that every block of is a block or contains a triangle. We claim that this implies that is -free. If some block of contained a cycle of length and a triangle , then we will show that this block also contains a cycle of length for some which contains the three vertices in the triangle as consecutive vertices. This implies that also contains a cycle of length , which contradicts being -free.
Suppose the triangle intersects in two or three vertices, not necessarily sequentially along . This yields two or three paths along between each of the vertices of that it intersects. One of these paths must have length at least , call it with endpoints say , without loss of generality. Then the cycle formed by and the edges , give the desired cycle.
Next, suppose that and have exactly one vertex in common, say without loss of generality. Since is 2-connected, there exists some shortest path in from a vertex in to a vertex of . Since is a shortest path, it contains at most one vertex from and , say these vertices are , which must be the endpoints of . Then and split into two paths, one of which will have length at least . This long path, together with and the edges and , gives the desired cycle.
Finally, suppose that and are disjoint. Then since is -connected, we can find two paths, disjoint from each other and internally-disjoint with both and , connecting distinct vertices of the triangle to distinct vertices of the cycle. This again partitions into two paths, one of which will have length at least , and using this together with the two paths from and the third vertex of gives the desired cycle.
Thus is -free. Using Theorem 1.1 gives
3 -Dense Families of Cycles
In this section, we consider the cycle saturation game for families of cycles with sufficiently small of gaps between consecutive forbidden cycle lengths.
Let us briefly outline the strategy we use in the -saturation game when is a -dense family of cycles. Ideally we would like to play such that at the end of each of our turns, is a connected graph where every block is dominated, i.e. contains a dominating vertex , and moreover is the unique vertex of which is closest to some special vertex . Assuming this holds, if our opponent connects two vertices and at blocks distance from each other, then the path through their dominating vertices creates a cycle of length . We further require that each dominating vertex be the endpoint of a path of length in another block, which will allow us to extend this path through dominating vertices in such a way that we can actually get any cycle length between and , with at least one of these lengths forbidden by assumption of being -dense. Thus distant blocks can not be connected to one another. This will imply that has small circumference, and hence relatively few edges.
It remains to describe how we can maintain this ideal structure (or at least something close to it). If our opponent ever connects an isolated vertex to one of these blocks , then we can try to make adjacent to the dominating vertex of . If this is impossible then it will turn out that ’s neighbor dominates the block and is the end of a long path in , so this new block maintains the desired properties. The issue will be when our opponent connects two isolated vertices. To maintain connectedness we are forced to make one of these vertices adjacent to . By doing this repeatedly the opponent can create some non-desirable structures, but we can at least maintain that any that do appear are incident to .
3.1 -Fantastic Graphs
We now move onto our formal definitions. Throughout the remainder of this section, we will assume we are playing the -saturation game for a -dense family of cycles . To define the structure that we wish to maintain in this game, we work with a graph with a specified vertex . We say a block in is rooted if there exists a vertex that is a dominating vertex in , and for which for all . When is rooted, we say is the root of and that roots the block . Note that if and is rooted, then is the root. Furthermore, note that if every block in has a root, then every vertex is in exactly one block for which it is not the root, namely the first block in the block geodesic. If is a vertex that roots every block containing it except for one block , then we will call this block the stem of and denote it by .
We say a vertex in a rooted block is finished if is the endpoint of a path of length at least in the induced subgraph . We say the block is finished if all vertices in are finished. If a vertex or block is not finished, we call it unfinished. We say a block is nearly -dominated if contains , and all but one vertex in is adjacent to . In this case we define (even though is technically not a root of the block).
Two blocks and are together called an -umbrella if the following hold:
The vertex dominates ,
The block is a block,
The blocks and intersect at a vertex ,
For any other block , if intersects , then they intersect at ,
The vertex in has degree 1, and we will refer to this vertex as the handle of the -umbrella.
We say that an -umbrella is finished if the unique neighbor of the handle is finished in , and unfinished otherwise.
See Figure 2 for an illustration of an -umbrella. Note that if and constitute an -umbrella, then and .
We are now ready to state the definition of a -fantastic graph, which is the main structural tool we will need in this section.
Given a graph with specified vertex , let be the subgraph induced by the set of vertices which are contained in a block which is nearly -dominated or part of an unfinished -umbrella, and if no such blocks exist set . Let be the subgraph induced by the set consisting of and all of the vertices contained in the blocks of not contained in . We say is -fantastic whenever has the following properties:
Every block in is rooted,
If a vertex is the root of some block in , then is finished in its stem .
Each vertex roots at most one unfinished block in ,
The vertex is adjacent to at most one vertex of degree in .
We note that the graph consists of along with a set of isolated vertices. Note that in a slight abuse of notation, we will sometimes use and to refer to and , respectively.
3.2 Preliminary work towards -Dense results
We will ultimately show that for any -dense family of cycles , either player can play the -saturation game such that the graph is -fantastic at the end of their turns. To this end, throughout this subsection we let refer to a -fantastic graph, and we consider , , , and as defined above. We also let refer to any -free graph with specified vertex , which will usually be thought of as for some edge .
Let be a -free graph with specified vertex . If is a block of that is either rooted or nearly -dominated, then the longest -avoiding path in is of length at most .
Assume to the contrary that contains an -avoiding path . If the edges and are in , then is a cycle of length , which contradicts the fact that is -free. This implies that must be nearly -dominated, , and exactly one of the edges or are not in . We assume without loss of generality . Since is non-trivial and -connected, must have at least two neighbors in , and specifically at least one neighbor with . If for any , then is a cycle of length , a contradiction. If for some , then is a cycle of length in , a contradiction. We conclude that for all rooted and nearly -dominated blocks , we have that does not contain an -avoiding path of length . ∎
The circumference of any -fantastic graph is at most .
Let be a -fantastic graph, and note that every block of a -fantastic graph is either rooted or nearly -dominated. If had a cycle of length , then this cycle would have to be in some block because cycles are -connected and blocks are maximal -connected subgraphs. Thus, would necessarily contain an -avoiding path of length , a contradiction with Lemma 3.3. So, the circumference of is at most . ∎
We continue our discussion on the structural properties of -fantastic graphs with the following two lemmas regarding which path lengths are attainable between specific vertices.
Let be -fantastic. If a vertex is finished in a rooted block and , then for each with there is a path of length in .
If the result is immediate, so assume . Since is finished there exists an -avoiding path in . If then the path works. If and for any , then is a path of length . If with , then we use . In each case we find a path of length , thus completing the proof. ∎
Recall that we denote the block distance from vertex to vertex as , a vertex can only have one stem, and that roots all other blocks containing it.
Let be -fantastic, and let with be such that . Then there is a path of length in for every with
Moreover, if we can do this for all with
Let be the block geodesic, and observe that these blocks are all rooted since . For , let be the vertex in both and , and note that e.g. as otherwise would be a shorter path of blocks from to . We will first prove, regardless of if or , that for all there is a path of length
and moreover the path ends in a geodesic.
Indeed, starting from we transverse a geodesic. If , then by definition , i.e. is the unique block containing which is not the root of. The uniqueness of this block implies , and by Property 3.2 this means that is finished in . Thus we can apply Lemma 3.5 to find a path of length from to in . If , then and a symmetric argument gives the same conclusion. Continuing in this manner, Lemma 3.5 gives for each a path of length from to in , and once we reach we can traverse a geodesic to complete the path. This proves the claim.
By choosing the appropriate values for the ’s, we can find a path of length for every with
To finish the proof for the case, it suffices to show and . The first part is immediate, and the second part follows by considering the path after deleting duplicated vertices if, say, for some .
In the case where , we can again apply the claim to find paths of length for all with and which end in the edge . Note that does not root (since ), and thus it must root so it will be be finished in . By Lemma 3.5 we can replace the edge with a path of length for any , thus allowing us to build a path of every length with . Since and , the result follows.
We show that we can always find long paths between pairs of vertices in adjacent blocks, except for one exceptional case.
Let be -fantastic. Let with , and let be the block geodesic. Then there is a path of length in , unless roots and and both and are unfinished.
Let . Note that being -fantastic implies that roots at least one of these blocks, and that due to and having block distance . First consider the case that roots both and . We claim that either or are finished. If , then this is true by assumption, and if , then one of the blocks or must be finished by Property 3.2, so one of or must be finished, and we assume without loss of generality that is finished in . Then by Lemma 3.5, there is a path of length which can be extended using the edge to a path of length that ends at , so we are done in this case.
The next two lemmas will help us in situations in which our opponent plays an edge that is incident with a nearly -dominated block.
Let be a -free graph with specified vertex . If is a nearly -dominated block and is the vertex not adjacent to , then is -free.
If adding the edge creates a cycle in , this cycle would be contained in the vertices of , which implies that there must be a path of length at least in , and hence an -avoiding path of length in , but this contradicts Lemma 3.3. ∎
We remind the reader that only unfinished -umbrellas are in , which in particular means the unique neighbor of the handle in an -umbrella of must be unfinished. We also recall for , we require that both and are in for to be -dense (but for we do not require ).
Let be -fantastic for . Let be distinct vertices such that one of the following holds:
and are in distinct nearly -dominated blocks,
and are both handles in distinct -umbrellas, or
One of is in a nearly -dominated block while the other is the handle in an -umbrella.
Let be the block containing the edge in , and let be the vertices such that . If is -free, then both and are also -free.
Towards a contradiction, we may assume contains a cycle from , so there is an path in of length with such that , say the path . We can assume without loss of generality that the vertices and are both in the same nearly -dominated block or -umbrella in (and consequently the same is true for and ).
We claim that this path contains the vertices and in that order. If not, then is either completely contained inside a nearly -dominated block, which contradicts Lemma 3.8, or is completely contained inside an -umbrella in . By the definition of , the neighbor of in this -umbrella must be unfinished, but is an -avoiding path of length , giving us a contradiction and proving the claim.
Let denote the sets of vertices in the nearly -dominated block or -umbrella in containing and , and let denote the set of vertices in the nearly -dominated block or -umbrella in containing and . Then and for some , and note that while .
We now claim that we can find a cycle of length in of the form for some vertex . Indeed, if is an -umbrella, we can choose to be the unique neighbor of in , then by the definition of , we have that , so is such a cycle. Thus we may assume that is a nearly -dominated block. Since is -connected and non-trivial, has a neighbor that is not (nor since ). We consider two cases based on .
Case 1: The vertex for some . Note that since . This implies that , and so is an edge. Then is a in , a contradiction to .
Case 2: The vertex for any . Since , . Then is a in with and at distance along the cycle.
Thus, we have exhibited a cycle as claimed. This implies that . Since , we have . We conclude and that and are defined. We now show that contains either a or a , which will give us a contradiction. Indeed, since , we have that is adjacent to at least one vertex with , so is a in , where , proving the claim and completing the proof. ∎
Our next lemma characterizes what moves within the cycle saturation game will leave the graph -fantastic. For the rest of the section, we refer to a legal move as an allowable move in the -saturation game for the implied family which is -dense.
If is -fantastic, and are vertices such that is a legal move, then is also -fantastic. Further, does not contain any nearly -dominated blocks which were not in .
Note that adding an edge within a block does not interfere with any of the properties of being -fantastic nor create a nearly -dominated block. Thus we can assume . If , by Lemma 3.6 there is a path of length for every with , which implies that would complete a cycle in , a contradiction. Thus, we may assume .
Since is a legal move there is no path of length . By Lemma 3.7, both and are adjacent to . Then still dominates the block resulting from adding and (in particular meaning it is not nearly -dominated), and all the properties of being -fantastic are retained as desired. ∎
The following lemma will allow us to focus our attention only on those cases in which our opponent makes a move that results in a graph that is not -fantastic.
If is -fantastic but not -saturated, then there exists a legal move such that is -fantastic. Moreover, if has at most one nearly -dominated block, then can be chosen so that has no nearly -dominated blocks.
First assume contains a nearly -dominated block with the vertex in this block not adjacent to . By Lemma 3.8 we can add , which creates a rooted block with root . This makes -fantastic with no nearly -dominated blocks. Thus we may assume that contains no nearly -dominated blocks.
Assume has an isolated vertex . If is adjacent to a vertex of degree , say , then is a legal move making an -umbrella. Otherwise is a legal move. Thus we may assume contains no isolated vertices.
Suppose contains an -umbrella in , say with handle and its unique neighbor, and recall that cannot be finished. Thus can be added without creating a forbidden cycle or nearly -dominated block. We can then assume that contains no -umbrellas in , and consequently that is trivial.
Since is trivial and there are no isolated vertices, we must have . By hypothesis there exists a legal move involving two vertices of , and by Lemma 3.10 and any such move leaves the graph -fantastic, so we are done. ∎
3.3 Main Results for -Dense Families
We are now ready to prove our main structural result for this section.
Let be a -dense set of cycles for some . Then either player can play the -saturation game such that at the end of each of their turns, the graph is -fantastic. Moreover, if , then that player can guarantee that the graph contains no nearly -dominated blocks at the end of each of their turns.
Note that and are both trivially -fantastic and do not contain nearly -dominated blocks. Now let us assume that is -fantastic for some , and if that further contains no nearly -dominated blocks. We will show that we can play such that is -fantastic, unless is already -saturated. In the analysis that follows, when we will not verify that our own move does not create a new nearly -dominated block, but it is easy to verify that the only time our strategy has our move creating such a block is when in Case 5c.
We consider cases based on the edge added at time which we denote by . We also let denote the set of isolated vertices of .
Case 1: . We play the edge , which creates an -umbrella in and maintains being -fantastic.
Case 2: . First note that we do not need to consider the case when since as well.
Case 2a: . If is contained inside a nearly -dominated block, then this block becomes rooted with root , so is -fantastic and we are in Case 0. The only other possibility is that is the handle of an -umbrella since all other vertices in are adjacent to . As such, adding the edge causes this -umbrella to become a single block which is rooted with root , so the graph is still -fantastic and we are in Case 0.
Case 2b: . If was not adjacent to a degree vertex in , then we are still -fantastic and in Case 0. Otherwise, by Property 3.2 there is exactly one other degree vertex adjacent to , and adding the edge creates a block rooted at , leaving the graph -fantastic.
Case 3: and . This adds an unfinished block rooted at . We only consider the cases where is the root of another unfinished block (which violates Property 3.2), and the case where is an unfinished vertex (which violates Property 3.2). In any other case, remains -fantastic since the Proprieties 3.2 and 3.2 cannot be affected by a new unfinished block, and thus we are in Case 0.
Case 3a: is the root of an unfinished block . We add an edge from to some unfinished vertex . This does not create any cycle of length at least since was unfinished, so this is a legal move. Now is only adjacent to at most one unfinished block again, and Property 3.2 holds. Since is adjacent to the root of this block, Property 3.2 holds as well.
Case 3b: is an unfinished vertex in some block . We add the edge . This does not create a cycle of length at least since was unfinished, so this is a legal move. The resulting block is rooted with root , so properties 3.2 and 3.2 hold in . Thus, is -fantastic.
Case 4: and .
Case 4a: is in a nearly -dominated block . We add the edge containing missing from , which is legal by Lemma 3.8. This creates an -umbrella, and thus is -fantastic.
Case 4b: is in an -umbrella consisting of blocks and with . Let be the handle and its unique neighbor (possibly with or ). Then by the definition of , is unfinished. Then we can add the edge as this does not create a cycle in due to the fact that is unfinished. Then becomes a block rooted at , and so if is the block containing , and constitute an -umbrella, and thus is -fantastic.
Case 5: .
Case 5a: and are in the same nearly -dominated block or the same -umbrella in . If and are both in a nearly -dominated block, then this block remains nearly -dominated. If and are both in the same -umbrella, this either remains an -umbrella or if either or was the handle in , this becomes a nearly -dominated block. In either case is -fantastic, so we are in Case 0.
Case 5b: is in an -umbrella and is not the handle. By Case 5a, we may assume that is not in the same -umbrella as . If is in a nearly -dominated block or is the handle of an -umbrella, then the block containing in is nearly -dominated, so by Lemma 3.8, we can add the edge to turn this block into a rooted block with root , which results in an -umbrella so is -fantastic.
It remains to consider when is also in an -umbrella but not the handle. Note that the addition of the edge forms a block rooted at , adjacent to two rooted blocks in . Let and be the handles of the original -umbrellas containing and respectively, and let and be the neighbors of and . If either or are unfinished in , then we can add the edge or creating an -umbrella and leaving -fantastic. If both and are finished in , then all these blocks are in , so is -fantastic and we are in Case 0.
Case 5c: The conditions of Case 5a and Case 5b are not met. Then one of the following holds:
and are in distinct nearly -dominated blocks,
and are both handles of distinct -umbrellas, or
One of is in a nearly -dominated block while the other is a handle of an -umbrella.
Note that if , then we do not have nearly -dominated blocks, and and cannot both be handles of -umbrellas as would create a , which is forbidden. Hence, we may assume . Let be the block containing in , and note that is only adjacent to other blocks at . By Lemma 3.9, in all cases we can add an edge containing and some other vertex of , which will turn into a nearly -dominated block, leaving -fantastic. Since this case only happens with , we do not create -dominated blocks when , as desired.
Case 6: and .
Case 6a: The block distance . By Lemma 3.6 we have a path of length for every , and thus contains a for every satisfying
If , then since , we have that contains cycles of every length with . In particular it contains , a contradiction, so we may assume