Graphs without proper subgraphs of minimum degree 3 and short cycles
Abstract
We study graphs on vertices which have edges and no proper induced subgraphs of minimum degree . Erdős, Faudree, Gyárfás, and Schelp conjectured that such graphs always have cycles of lengths for some function tending to infinity. We disprove this conjecture, resolve a related problem about leaftoleaf path lengths in trees, and characterize graphs with vertices and edges, containing no proper subgraph of minimum degree .
1 Introduction
A simple exercise in graph theory is to show that every graph with vertices and at least edges must have an induced subgraph with minimum degree . Moreover, this statement is best possible: there are several constructions with edges which do not have this property. So every graph with vertices and edges must contain an induced subgraph with minimum degree , however this subgraph might be the whole graph. A subgraph of is called proper if . See Figure 1 for two examples of graphs with edges but no proper induced subgraphs of minimum degree . The first of these, has an even stronger property—it has no proper induced or noninduced subgraphs with minimum degree . On the other hand, the second example has a proper noninduced subgraph with minimum degree formed by removing the edge between the two vertices of degree .
In this paper we will study graphs with vertices edges which have no proper induced subgraphs with minimum degree . Following Bollobás and Brightwell [2] we call such graphs degree critical. It is easy to see that graphs with vertices and at least edges contain a proper degree critical subgraph. Erdős (cf [4]) conjectured that they should contain a degree critical subgraph not only on at most , but on at most vertices, for some constant . Degree critical graphs are closely related to several other interesting classes of graphs. For example, they have the property that all their proper subgraphs are 2degenerate (where a graph is defined to be degenerate if it has no subgraph of minimum degree ). Also notice that degree critical graphs certainly have no proper subgraphs with edges. Graphs with edges and no proper subgraphs with edges have a number of interesting properties. They are rigidity circuits: by a theorem of Laman, removing any edge from such a graph produces a graph which is minimally rigid in the plane, i.e., any embedding of it into the plane where the vertices are substituted by joints and the edges by rods produces a rigid structure, but no proper subgraph of has this property. Furthermore, by a special case of a theorem of NashWilliams these graphs are exactly the ones that are the union of two disjoint spanning trees and Lehman’s Theorem characterizes them as the minimal graphs to win the socalled connectivity game on. That is, with two players alternately occupying the edges of , the player playing second is able to occupy a spanning tree.
The study of degree critical graphs was initiated by Erdős, Faudree, Gyárfás, and Schelp [3], where they investigated the possible cycle lengths. They showed that degree critical graphs on vertices always contain a cycle of length , , and , as well as a cycle of length at least , but not necessarily of length more than . Bollobás and Brightwell [2] resolved asymptotically the question of how short the longest cycle length in degree critical graphs can be. They showed that every degree critical graph contains a cycle of length at least and constructed degree critical graphs with no cycles of length more than . Erdős, et al. [3] made the following conjecture about possible cycle lengths in degree graphs.
Conjecture 1.1 (Erdős, Faudree, Gyárfás, and Schelp, [3]).
There is an increasing function such that the following holds such that every degree critical graph on vertices contains all cycles of lengths .
A historical remark must be made here. The exact phrasing of Conjecture 1.1 in [3] is not quite what is stated above. In [3] first a class of graphs, , is defined as “the set of graphs with vertices, edges, and with the property that no proper subgraph has minimum degree .” Then Conjecture 1.1 is stated as “If , then contains all cycles of length at most where tends to infinity.” Notice that the word “induced” is not present in the original formulation. However a careful reading of [3] shows that in that paper “proper subgraph” implicitly must mean “proper induced subgraph”. Indeed many of the constructions given in [3] (such as Examples 1, 2, 3, 5, and 6 on pages 197201) of graphs which have “no proper subgraphs of minimum degree ” actually do have proper noninduced subgraphs with minimum degree . In addition, one can check that all the results and proofs given in [3] concerning graphs with “no proper subgraphs of minimum degree ” hold also for graphs with “no proper induced subgraphs of minimum degree ”. Therefore, it is plausible to assume that the word “induced” should be present in the statement of Conjecture 1.1. This also coincides with the interpretation of the concept in the paper of Bollobás and Brightwell [2].
Consequently throughout most of this paper will study Conjecture 1.1 as it is stated above. However, for the sake of completeness, in Section 4 we will diverge and consider the special case of Conjecture 1.1 when contains neither induced nor noninduced subgraphs with minimum degree .
The main result of this paper is a disproof of Conjecture 1.1. We prove the following.
Theorem 1.2.
There is an infinite sequence of degree critical graphs which do not contain a cycle of length .
In the process of proving this theorem, we will naturally arrive to a question of independent interest, concerning the various leafleaf path lengths (i.e., the lengths of paths going between two leaves) that must occur in a tree. Obviously, if is just a path, then only has a single leafleaf path. However if has no degree vertices, then one would expect to have many different leafleaf path lengths. Of particular relevance to Conjecture 1.1 will be even  trees. A tree is called even if all of its leaves are in the same class of the tree’s unique bipartition and a tree is called a tree if every vertex has degree or . On our way towards the proof of Theorem 1.2 we determine the smallest even number which does not occur as a leafleaf path in every even tree.
Theorem 1.3.

There is an integer such that every even  tree with contains leafleaf paths of lengths .

There is an infinite family of even  trees , such that contains no leafleaf path of length .
Part of Theorem 1.3 will be used to construct our counterexample to Conjecture 1.1, while part shows that our method, as is, can not deliver a stronger counterexample. Hence it would be interesting to determine the shortest cycle length which is not present in every sufficiently large degree critical graph. Theorem 1.2 shows that this number is at most , while Erdős et al. [3] showed that it is at least . They also mention that their methods could be extended to work for . In Section 5 we verify their statement, by giving a short proof that every degree critical graph must contain .
Finally, we revisit Conjecture 1.1 with the word “induced” removed from the definition of degree critical. We characterize all vertex graph with edges and no proper (not necessarily induced) subgraph with minimum degree and show that the conjecture is true for them in a much stronger form.
Theorem 1.4.
Let be a graph with vertices, edges and no proper subgraph with minimum degree . Then is pancyclic, that is, it contains cycles of length for every and .
Theorem 1.4 will follow from a structure theorem which we shall prove about graphs with vertices, edges and no proper (not necessarily induced) subgraphs with minimum degree . It will turn out that there are only two particular families of graphs satisfying these conditions. One of them is the family of wheels and the other is a family of graphs obtained from a wheel by replacing one of its edges with a certain other graph.
The structure of this paper is as follows. In Section 2 we construct our counterexamples to Conjecture 1.1 via proving part of Theorem 1.3 and Theorem 1.2. In Section 3 we study necessary leafleaf path lengths in even  trees and prove part Theorem 1.3. In Section 4 we prove the weakening of Conjecture 1.1 when the word “induced” is removed from the definition. In Section 5 we show that degree critical graphs on at least vertices always contain a sixcycle. In Section 6 we make some concluding remarks and pose several interesting open problems raised naturally by our results. Our notation follows mostly that of [1].
2 Counterexample to Conjecture 1.1
The goal of this section is to prove Theorem 1.2. First we need some preliminary results about 13 trees.
Given a tree , define to be the graph formed from by adding two new vertices and , the edge as well as every edge between and the leaves of . See Figure 2 for an example of a graph .
Notice that if is a  tree then is degree critical. In the case when is an even  tree, the cycles of have nice properties.
Lemma 2.1.
Let be an even tree. Then the following hold:

The graph contains a cycle of length contains a leafleaf path of length .

The graph contains a cycle of length contains two vertexdisjoint leafleaf paths and such that or contains a leafleaf path of length .
Proof.
For (i), let be a cycle in . Notice that since is an even tree, is bipartite. So must contain the edge and hence must be a leafleaf path of length as required. For the converse, notice that any path of length between leaves and can be turned into a cycle of length by adding the vertices and as well as the edges of .
For (ii), let now be a cycle in . If then is a leafleaf path in of length . Now suppose that both . Notice that since is even, all leafleaf paths in have even length. Therefore, all cycles containing the edge in must have odd length, and hence does not contain . Thus consists of two vertexdisjoint leafleaf paths such that their lengths sum to , as required. For the converse, first notice that any leafleaf path of length can be turned into a cycle of length in by adding the vertex and the edges between the endpoints of and . Also, any two vertexdisjoint leafleaf paths of length with endpoints and of length with endpoints and can be turned into a cycle of length in by adding the vertices and , and the edges , and of . ∎
We say that a rooted binary tree is perfect if all nonleaf vertices have two children and all rootleaf paths have the same length (or, alternatively if where is the depth of ). Given a sequence of positive integers , we define a tree as follows. First consider a path on vertices with vertex sequence . For each satisfying , add a perfect rooted binary tree of depth with root vertex . For and add two perfect rooted binary trees each: trees and of depths with root vertices and , respectively and trees and of depths with root vertices and , respectively. Finally, for each , we add the edges , as well as the edges , , , and . See Figure 3 for an example of a graph .
Notice that for any sequence of positive integers, the tree is a  tree. We will mainly be concerned with oddeven sequences, that is, sequences for which for all (that is, is even is even). It turns out that for oddeven sequences the leafleaf path length of the tree are easy to characterize.
Lemma 2.2.
Let be an oddeven sequence. Then we have the following:

The tree contains no leafleaf path of odd length. In particular, is an even tree.

For every integer , , the tree contains a leafleaf path of length .

For , the tree contains a leafleaf path of length if and only if either or there are two distinct integers and such that .

For every , the tree contains a leafleaf path of length if and only if there are two distinct integers and such that .
Proof.
Leafleaf paths of can be classified based on their intersection with the path . Note that this intersection is always a (potentially empty) path.
If the intersection is empty then the path is a leafleaf path of a perfect binary tree of depth for some , and hence its length is for some , .
If the intersection is a single vertex, then this vertex must be either or . Then the path is a leafleaf path going through the root in one of the perfect binary trees on and of depths and , respectively, and hence its length is or , respectively.
If the intersection is a segment for some , then the path has length . This implies the “only if ” part of (iii) and (iv). Note also that all these paths have even length ( is even because is an oddeven sequence), and so (i) holds.
For (ii) and the “if” part of (iii) and (iv) one must only note that a perfect tree of depth contains a leafleaf path of every even length and hence all leafleaf pathlengths given by the classification can actually be realized. ∎
We now produce a sequence of integers such that for every , the tree will not have leafleaf paths of length .
2.1 avoiding sequences
We will be concerned with twosided sequences of positive integers. Again, we say that such a sequence is an oddeven sequence if for all .
Definition 2.3.
Let be a positive even integer. A twosided sequence of positive integers is called avoiding if for all and if for every , , we have .
In order to check if an oddeven sequence with for all is avoiding, consider the graph of the sequence. Call a point in conflict with another point , , if . Notice that the points in conflict with a fixed point lie on the two diagonal lines and . Since being in conflict is a symmetric relation we can say that we blame a conflict on the point with lower first coordinate (the first coordinates of points in conflict cannot be equal). Then the points , whose conflicts with are blamed on lie on the single line . Indeed, the first coordinates of a point on the other diagonal line is at most , hence these conflicts are not blamed on . We define the fault line of the point to be the line . From the above discussion we obtain the following proposition.
Proposition 2.4.
A sequence is avoiding if, and only if, there do not exist two distinct indices and such that lies on the fault line of .
It is useful to note that all the points on the line have the same fault line .
Theorem 2.5.
There is a avoiding oddeven sequence.
Proof.
Let be the periodic sequence of period consisting of repetitions of
We claim is a avoiding oddeven sequence. It is clearly an oddeven sequence, and for all . We prove that it is avoiding by showing that in the graph of this sequence, no point lies on the fault line of another point. Then Proposition 2.4 implies the theorem.
Figure 4 is a snapshot of two periods of the graph. The points on the graph are black circles, and the fault lines are drawn in red. Note that points on a line parallel to the line “” have the same fault line, and that this fault line crosses when the second coordinate is .
From the picture we see that no point of the sequence lies on a fault line of another point, implying that is indeed avoiding. ∎
Proof of part (ii) of Theorem 1.3.
Let be the first terms (starting at ) of the avoiding sequence produced by Theorem 2.5. The tree is a 13tree for any sequence by construction. Since is an oddeven sequence, is also an even tree by part (i) of Lemma 2.2. The tree contains no leafleaf paths of length , since and part (iv) of Lemma 2.2 tells us that a leafleaf path of length exists only if there are distinct and such that , which is not case since is avoiding. ∎
Proof of Theorem 1.2.
We let be the graph constructed from the tree given by part (ii) of Theorem 1.3. Since is a 13tree, the graph is degree critical, as required. Since is an even 13tree, we can use part (i) of Lemma 2.1 and the fact that does not contain a leafleaf path of length to conclude that contains no cycle of length . ∎
3 Possible leafleaf path lengths in even  trees
In this section we prove part (i) of Theorem 1.3. We first need a lemma about possible lengths of leafleaf paths in binary trees which have no short rootleaf paths.
Lemma 3.1.
Let be an even rooted binary tree and let be the length of its shortest rootleaf path. Then contains leafleaf paths of lengths .
Proof.
The proof is by induction on . The statement is certainly true for . Let now and let and be the children of of the root .
Suppose first that in one of the subtrees and , rooted at and , respectively, the shortest rootleaf path is of length as well. In this case we can apply induction to this subtree and find in it a leafleaf paths of all length . The leafleaf path of the subtree are of course leafleaf paths of , so we are done in this case.
Otherwise, the length of the shortest rootleaf path of both subtrees and are (the subtrees cannot contain a shorter rootleaf path, because itself does not contain a rootleaf path shorter than ). Then by induction there are leafleaf path of all length in both of these subtrees and hence also in . To construct a leafleaf path of length in let be a path between and a leaf of of length , and be a path between and a leaf of of length . Then the path formed by joining and to using the edges and is a leafleaf path in of length . ∎
The following proposition shows that finding which leafleaf paths lengths always occur in sufficiently large trees is equivalent to finding the for which avoiding sequences exist.
Proposition 3.2.
Let be a positive integer. The following are equivalent.

There is an integer such that every even 13tree of order at least contains a leafleaf path of length .

There exists no avoiding oddeven sequence .
Proof.
Let us assume first that (i) holds with integer . Let be an arbitrary oddeven sequence such that . Notice that since is even if and only if is even, there are infinitely many indices for which . Therefore we can choose two indices and such that and , . Then by parts (iii) and (iv) of Lemma 2.2, the tree has a leafleaf path of length if and only if there are two distinct indices and such that holds. On the other hand notice that by part (i) of Lemma 2.2, is an even 13 tree and hence, since its order is at least , does have a leafleaf path of length . That is, there do exist indices such that holds, implying that is not avoiding.
Now assume that (ii) holds. Let us define , where . Let be an arbitrary even  tree of order at least . We will show that contains a leafleaf path of length . Since is a tree of maximum degree at most on vertices it must contain a path with vertices. Let be the subtree of consisting of the connected component of containing and let be the length of the shortest path from to a leaf of . Note that is an oddeven sequence, because is an even tree.
Suppose first that we have . Choose an index such that holds and let . Then is a binary tree rooted at the neighbour of , with no rootleaf paths shorter than , so Lemma 3.1 gives us a leafleaf path of length .
Suppose now that we have . Since , the Pigeonhole Principle implies that there must be indices such that all hold. Consider now the infinite periodic sequence
denoted by . This is an oddeven sequence as the sequence was oddeven. By our assumption is not avoiding. But , so there must be indices such that . Since the sequence is positive we must have and by periodicity we can assume that . The way we chose and ensures that for every between and , so we also have . We can now find a leafleaf path in of length by concatenating a shortest path from to a leaf of , the path between and and a shortest path from to a leaf of . ∎
We now proceed to prove part (i) of Theorem 1.3. We do this by showing that part (ii) of Proposition 3.2 holds for .
Theorem 3.3.
There is no avoiding oddeven sequence.
Proof.
Consider an oddeven sequence with for all . Assume that it is avoiding. As in the proof of Theorem 2.5, we will consider the graph of and consider fault lines. In this case, the fault line of a point is the line . Since is avoiding, Proposition 2.4 implies that no point of the graph lies on the fault line of another point of the graph. Notice however, that a point of the form , which by definition lies on its own fault line, is not itself a barrier to a sequence being avoiding.
We start with some lemmas about configurations of fault lines that lead to contradictions. We will actually deal with a slight generalization of fault lines, which we call excluded lines. An excluded line is defined to be any line of the form with even, that does not contain a point in the graph of , except possibly the point with second coordinate . Since is an oddeven sequence, for any point in the graph of the sequence, the integer is even. Hence every fault line of the sequence is also an excluded line.
In the following discussion lines of slope whose intercepts differ by exactly are called consecutive. We start with a trivial observation.
Lemma 3.4.
There cannot be four consecutive excluded lines for .
Proof.
If there were four excluded lines , , , and , where is even, then all of the points with even coordinate at most on the line are on one of these lines. Hence would be on an excluded line, a contradiction.
∎
This easily leads to the next lemma.
Lemma 3.5.
There cannot be three consecutive excluded lines for .
Proof.
If there were three consecutive excluded lines , , and , where is even, then must be equal to as all the other even values at most would put on one of the three lines. Similarly, we must have , and hence we have fault lines and . This forces , giving also the fault line . Now can only be or to avoid the original three fault lines, but it clearly cannot be , since that would put on its fault line. But if , then its fault line is , and there would be consecutive fault lines , contradicting Lemma 3.4.
∎
A few more lemmas of this sort will be useful for the proof.
Lemma 3.6.
There cannot be three excluded lines of the form , , and (with even).
Proof.
If this were the case, then this would force , which results in the fault line . If , then there would be three consecutive excluded lines , which would contradict Lemma 3.5. This forces , which results in the fault line . Similarly, in order to avoid a third consecutive fault line , we must have and , resulting in the fault lines and , respectively. This leaves us with no valid choices for , since a value of would create a third consecutive fault line , a value of would create a third consecutive fault line , and a value of or would put on the fault line of a previous point. Therefore, this configuration cannot occur.
∎
Lemma 3.7.
There cannot be three excluded lines of the form , , and (with even).
Proof.
If this were the case, then this would force , which results in the fault line . If , there would be three consecutive excluded lines , contradicting Lemma 3.5. So we must have , resulting in the fault line (the other values of would put on an excluded line). If , then we would have the configuration of fault lines forbidden by Lemma 3.6, so we must have , resulting in the fault line . But then we have the three consecutive fault lines , also a contradiction.
∎
All previous lemmas pave way for our final technical lemma:
Lemma 3.8.
There cannot be consecutive excluded lines for .
Proof.
Suppose there were two consecutive excluded lines and for some even . Consider the possible values for . It cannot be , since this is on the excluded line . It cannot be , as this this would create a third consecutive excluded line . It also cannot be , because this would create the fault line , contradicting Lemma 3.6. Thus, we must have , which means we have the fault line . We also must have , since values or would put a point of the graph on one of the excluded lines, and value would yield the fault line , contradicting Lemma 3.6. Thus, we also have the fault line .
Now consider . It cannot be or , since these would put a point of the graph on an excluded line. It cannot be , otherwise it would create the fault line , and we would have three consecutive fault lines contradicting Lemma 3.5. It cannot be , for if it were, there would be the three fault lines in contradiction with Lemma 3.7. Hence we have and the fault line . Similarly, we must have , since or put it on a fault line, would create three consecutive fault lines , and would create the configuration of fault lines forbidden by Lemma 3.7. Therefore, there is also the fault line . But now every possible value for leads to a contradiction. If it is , , or , then it is on a fault line. If it is or , it creates a fault line resulting in a configuration forbidden by Lemmas 3.7 and 3.5, respectively. Therefore, we cannot have two consecutive excluded lines.
∎
Excluded lines by definition have slope . To finish the proof pf Theorem 3.3 we extend the notion of excluded line to those lines with even , which do not contain any point of the graph except possibly the point . We call these the orthogonal excluded lines of the sequence . The conclusion of Lemma 3.8 also holds for orthogonal extended lines: there cannot be two consecutive ones. Indeed, is an orthogonal extended line of the avoiding oddeven sequence if and only if is an extended line of the avoiding oddeven sequence , so we can apply Lemma 3.8 for .
Another useful observation is that the line is the fault line of exactly those points that are on the line . Hence if contains a point of the graph (say, it is not excluded), then must be an orthogonal excluded line. Using this observation for one can also obtain that if the orthogonal line contains a point of the graph of (say, it is not excluded), then must be an excluded line for .
Let us now assume that there exists an avoiding sequence and let be a fault line of it for some even . By the above we can make a sequence of conclusions. The lines are not excluded by Lemma 3.8. Then must be orthogonal excluded lines. Then are not excluded by the adaptation of Lemma 3.8 for orthogonal lines. Then must be excluded lines. Again by Lemma 3.8 the lines are not excluded and hence the orthogonal lines must be excluded. This implies that are not orthogonal excluded lines by the adaptation of Lemma 3.8 and are excluded lines.
What can now be the value of ? It must be odd as is even and is an oddeven sequence. The line being excluded shows it cannot be , being excluded shows that it cannot , being excluded shows it cannot be , being excluded shows it cannot be . The line is a fault line of so in principle could be on it. However then, the orthogonal line should also be excluded, meaning that together with they would represent three consecutive orthogonal excluded lines, a contradiction.
∎
To complete the proof of Theorem 1.3 we need the following little proposition.
Proposition 3.9.
Let be a positive even integer. If there is a avoiding oddeven sequence, then there is a avoiding oddeven sequence for every .
Proof.
If is a avoiding oddeven sequence, then define the sequence by
for all . We claim that is a avoiding oddeven sequence.
It is clearly an oddeven sequence as for all . Also, for all . Suppose there were with such that . Then we would have . But this implies , which contradicts the fact that is avoiding. ∎
Proof of Theorem 1.3 (i).
Let be a positive integer. We claim that there is no avoiding oddeven sequence. Indeed, otherwise our previous proposition implied that there is also an avoiding oddeven sequence, which contradicts Theorem 3.3. Now by Proposition 3.2, there is an integer such that every even  tree of order at least contains a leafleaf path of length , which is exactly the statement of part (i) of Theorem 1.3. ∎
4 Characterization of graphs with no subgraphs of minimum degree 3
Let denote the family of graphs with edges and no proper (not necessarily induced) subgraphs with minimum degree . In this section we characterize the members of and deduce Theorem 1.4 as a corollary.
A wheel is an vertex graph with vertices , and with edges and for . The vertex will be called the centre of the wheel and the vertices will be called the outside vertices of . For , Let be the graph on vertices called , , and formed by the edges for , for , , and . We call and the connectors of and the internal vertices of . Note that the roles of the connectors are not symmetric; the letter will always denote one with degree two. See Figure 11 for a picture of the graph .
The next theorem shows that the graphs in must have a very specific structure. See Figure 12 for examples of its members on vertices.
Theorem 4.1.
The family consists of all wheels and those graphs that are formed, for some and , from a copy of with connectors and and a copy of with connectors and by letting and or by letting and .
For the proof we first recall some basic properties of graphs with no induced subgraphs of minimum degree .
Recall from the introduction that the following lemma is easy to prove by induction.
Lemma 4.2.
Every graph on vertices with at least edges contains an induced subgraph with minimum degree .
For degree critical graphs, the induced subgraph of minimum degree (guaranteed by the previous lemma) must be the whole . For these graphs, Erdős et al. [3] presented a special ordering to the vertices. Given an ordering of we let the forward neighbourhood of