Radius, Girth and Minimum Degree

Given a connected graph $G$ on $n$ vertices, with minimum degree $\delta\geq 2$ and girth at least $g \geq 4$, what is the maximum radius $r$ this graph can have? Erd\H{o}s, Pach, Pollack and Tuza established in the triangle-free case ($g=4$) that $r \leq \frac{n-2}{\delta}+12$, and noted that up to the value of the additive constant, this is tight. We determine the exact value for the triangle-free case. For higher $g$ little is known. We settle the order of $r$ for $g=6,8,12$ and prove an upper bound to the order for general even $g$. Finally, we show that proving the corresponding lower bound for general even $g$ is equivalent to the Erd\H{o}s girth conjecture.


Introduction
Consider the following question: given a connected graph G on n vertices, with minimum degree δ ≥ 2 and girth at least g ≥ 4, what is the maximum radius r this graph can have? Denote this maximum value of the radius 1 as r(n, δ, g).
Erdős, Pach, Pollack and Tuza [2] studied this problem for the case g = 4. They established that r(n, δ, 4) ≤ n−2 δ + 12, and noted that up to the value of the additive constant, this is tight.
Our first result settles the triangle-free case fully.
Most importantly, if n ≥ 4δ, then r(n, δ, 4) = n δ − 1 if δ is odd and n = kδ for k odd ⌊ n δ ⌋ otherwise Next we consider the case when the girth g is bigger than 4. We obtain a general upper bound. Theorem 1.2. Let n, δ ≥ 2, and g = 2k such that there exists a connected graph on n vertices with minimum degree δ and girth at least g. Then r(n, δ, g) ≤ nk 2δ(δ − 1) k−2 + 3k.
We also show that in the cases g = 6, 8, 12, this is essentially best possible. Theorem 1.3. Let δ ≥ 2 so that δ − 1 is a prime power.
Note that in the result above, not only do the coefficients of the leading term in δ agree with the upper bound we have from Theorem 1.2, but even the coefficients of the second order terms agree. 1 for those triples n, δ, g with δ ≥ 2 for which there exists a connected graph on n vertices, with minimum degree δ and of girth at least g It would be interesting to see whether the upper bound from Theorem 1.2 is tight, at least up to some constant factor. As our final result, we obtain the following proposition. Proposition 1.4. Let r, c > 0, g = 2k and n ≤ c(r +1)δ k−1 , so that r(n, δ, g) ≥ r. Then there exists a connected graph of girth at least 2k on at most (2k + 1)cδ k−1 vertices with at least 1 2 δ 2 (δ − 1) k−2 edges. This relates the question whether the upper bound from Theorem 1.2 is tight up to some constant factor to the following girth conjecture of Erdős [1]. Conjecture 1.5. (Erdős) For any positive integers l, n, there exists a graph with girth 2l + 1, n vertices and Ω(n 1+ 1 l ) edges. We see that if for some fixed g = 2k and fixed c > 0, we could find graphs G i with δ i → ∞ and n i ≤ c(r(n i , δ i , 2k) + 1)δ k−1 i , then by Proposition 1.4, that would verify the girth conjecture of Erdős for l = k − 1.
The structure of the paper is as follows: in Section 2, we establish a key technical lemma. In the next two sections, we consider separately the triangle-free case and the general case.

Strategy
Throughout the paper, we will use the following lemma as a useful tool. It tells us that if we can find a large collection of vertices in our graph such that any two elements are either neighbours or sufficiently far away from each other, then our graph must in fact have many vertices.
with minimum degree δ. Then for any subset T ⊂ V (G) such that all pairs of non-adjacent vertices in T have distance at least 2k − 1 from each other, we have n ≥ |T |δ(δ − 1) k−2 . Moreover if |T | is odd, we have n ≥ |T |δ(δ − 1) k−2 + 1.
First, we claim that for any distinct v 1 , v 2 ∈ T , the sets S(v 1 ), S(v 2 ) are disjoint. To see that, consider two cases: yielding a contradiction. If d(v 1 , v 2 ) = 1 and for some a ∈ V (G) we have a ∈ S(v 1 ) ∩ S(v 2 ), let v 1 , c 1 , ..., c k−2 , a be a path of length k − 1 from v 1 to a, and let v 2 , d 1 , ..., d k−2 , a be a path of length k − 1 from v 2 to a. First note we can not have c 1 = v 2 , as that would imply d(v 2 , a) ≤ k − 2. Analogously we can not have d 1 = v 1 . But that now implies that the subgraph of G spanned by v 1 , v 2 , a, c 1 , ..., c k−2 , d 1 , ..., d k−2 contains a cycle of length at most 2k − 1, contradicting the assumption that g ≥ 2k.
Next we claim that for any v ∈ T , we have Note that trivially i ≥ 3, as for i = 2 this is just the minimum degree condition. Now, consider any two different a, b ∈ S i−1 (v). Note that every neighbour of a (and analogously of b) must be either in Firstly, note that a and b can not be connected together, as that would contradict the girth assumption. Next, note that a (and analogously b) can be connected to at most one vertex in S i−2 (v), else we would again contradict girth assumption. Finally, again by girth assumption, note that a and b can not share any neighbour in S i (v). Altogether, this implies Finally consider the case when |T | is odd. Note that every vertex in T can be at distance 1 from at most one other vertex of T , else we would get either C 3 in T or a pair of vertices of T at mutual distance 2. Hence, if |T | is odd, some vertex in T is at distance at least 2k − 1 from every other vertex in T . Hence it was not counted in the calculation above, giving n ≥ |T |δ(δ − 1) k−2 + 1.
To find such a large collections of points with restricted mutual distances, we will use several observations. Those can be found in Appendix A.

Triangle-Free Graphs
To prove Theorem 1.1, we will establish the following three propositions.
Proposition 3.1. Fix δ ≥ 2 and n. Then every connected triangle-free graph on n vertices with radius r satisfies r ≥ 2 and n ≥ 2δ. Moreover, if r = 3, we have n ≥ 2δ + 2.
Further, for any n ≥ 2δ, we have a connected triangle-free graph on n vertices with radius 2. And for any n ≥ 2δ + 2, we have a connected triangle-free graph on n vertices with radius 3.
Proposition 3.2. Let r ≥ 4, δ ≥ 2, c ≥ 0 be integers. Then there exists a connected triangle-free graph with n = 2⌈ rδ 2 ⌉ + c vertices, minimum degree δ and radius r. Proposition 3.3. If G is a connected triangle-free graph on n vertices with minimum degree δ ≥ 2 and radius r ≥ 4, then we have n ≥ 2⌈ rδ 2 ⌉. Let us first see how Theorem 1.1 follows from these.
First consider the case when δ is odd and n = kδ for k odd. Using Proposition 3.2 with r = n δ − 1 and c = δ, we conclude there exists a connected trianglefree graph with n vertices, minimum degree δ and radius n δ − 1, and hence that r(n, δ, 4) ≥ n δ − 1. Also, for any connected triangle-free graph with n vertices and minimum degree δ, it follows in this case from Proposition 3.3 that its radius r satisfies r < n δ . Since r is integer, that implies r ≤ n δ − 1. Hence, we conclude r(n, δ, 4) = n δ − 1. Next consider the other case. Using Proposition 3.2 with r = ⌊ n δ ⌋ and c = n − 2⌈ rδ 2 ⌉, we conclude there exists a connected triangle-free graph with n vertices, minimum degree δ and radius ⌊ n δ ⌋, and hence that r(n, δ, 4) ≥ ⌊ n δ ⌋. Also, for any connected triangle-free graph with n vertices and minimum degree δ, it follows from Proposition 3.3 that its radius r satisfies r ≤ n δ . Since r is integer, that implies r ≤ ⌊ n δ ⌋, and hence we conclude r(n, δ, 4) = ⌊ n δ ⌋. In the rest of the section, we will prove Propositions 3.1, 3.2 and 3.3 and thus prove Theorem 1.1. The section will be divided into four subsections -in the first subsection we prove Proposition 3.2; in the second subsection we prove a technical lemma we will need to prove Proposition 3.3; in the third subsection we prove Proposition 3.1; in the fourth subsection we prove Proposition 3.3 when r = 4k, r = 4k + 1 or r = 4k + 2; and in the final subsection we prove Proposition 3.3 when r = 4k + 3.

Proof of Proposition 3.2.
To prove Proposition 3.2, it is enough to consider the following simple example.

Proof of Proposition 3.2. Consider 2r boxes labelled
If c is such that at this point we have less than 2⌈ rδ 2 ⌉ + c vertices, put the remaining vertices into any of these boxes arbitrarily. Then connect all vertices in B i to all vertices in B j whenever i − j ≡ ±1 mod 2r. It is now easy to see that this graph has required properties for any choice of parameters r, δ, c in our range.
3.2. Technical lemma. First, recall Lemma 2.1 which implies the following result for triangle-free graphs.
Lemma 3.4. Let G be a triangle-free graph on n vertices and with minimum degree δ. Then for any subset T ⊂ V (G) such that no two vertices of T are at mutual distance 2, we have n ≥ 2 δ|T | 2 .
We will also need another lemma of similar flavour here.
Further, no vertex can be contained both in some set of the form N (c i ) and in some set of the form N (d j ). This is true because no c i and d j are at mutual distance 2, and hence by the triangle-free condition can not share a neighbour.
Let B be the set of vertices of G contained in at least one set of the form N (c i ) (and hence no set of the form N (d j )). By above, we find Since |B| is integer, we have |B| ≥ ⌈ rδ 2 ⌉. Let B ′ be the set of vertices of G contained in at least one set of the form N (d j ). By analogous argument, we get |B ′ | ≥ ⌈ rδ 2 ⌉. Using that B, B ′ are disjoint, we obtain n ≥ 2⌈ rδ 2 ⌉. 3.3. Proof of Proposition 3.1. Here we handle the small radius cases.
Proof of Proposition 3.1. Consider a connected triangle-free graph G on n vertices of radius r and minimum degree δ ≥ 2. We must have r ≥ 2, since the only connected triangle-free graphs of radius 1 are star graphs, but those have minimum degree 1. Now consider any two adjacent vertices a, b ∈ V (G). It follows from Lemma 3.4 applied to T = {a, b} that n ≥ 2δ.
If r = 3, then we can take a, b which instead satisfy d(a, b) ≥ 3. But then even their closed neighbourhoods are disjoint, which implies Now if n ≥ 2δ, we can easily check that K δ,n−δ is a connected triangle-free graph on n vertices of radius 2 and minimum degree δ.
3.4. Proof of Proposition 3.3 for r = 4k, 4k + 1, 4k + 2. In this subsection, we prove the following. Proposition 3.6. If G is a connected triangle-free graph on n vertices with minimum degree δ ≥ 2 and radius r ≥ 4 such that r = 4k + i for some k and some i ∈ {0, 1, 2}, then we have n ≥ 2⌈ rδ 2 ⌉. Proof. Let v 0 be a center of our graph G, which moreover has the property that every other center of G is at distance r from at least as many vertices as v 0 is.
It follows from Observation A.1 that t ≤ 3. Claim 3.7. Assume that either r = 4k (and t is any), or r = 4k + 2 (and t is any), or r = 4k + 1 and 0 ≤ t ≤ 2. Then we have n ≥ 2⌈ rδ 2 ⌉. Proof of Claim 3.7. We will show that in each of these cases, we can find a collection C of r vertices in G such that no two are at mutual distance 2. The result then follows from Lemma 3.4.
Depending on the values of r and t, choose C to be the following collection.
We need to check two things; that C genuinely consists of r distinct vertices, and that no two vertices of C have mutual distance 2.
None of the collections above contains both v 1 and v ′ 1 . For all other pairs v i , v ′ j , it follows in our particular case from Observation A.3 that v i = v ′ j . So C consists of r distinct vertices.
Note that v 0 , ..., v r is a path of length r and v ′ 0 , ..., v ′ r−t is a path of length r −t. Hence we can trivially check that C contains no two vertices of the form v i , v j such that d(v i , v j ) = 2 and no two vertices of the form v First consider the case i ≥ 3. Note that by our choice of v 0 , v ′ r−t , we always

follows from Observation A.2 under
additional assumption that d(v 3 , v ′ r−t ) + t ≥ r + 2. Hence, for i ≥ 3, it is enough if C does not contain both v i and v ′ i−2 in the case when we have d(v 3 , v ′ r−t )+t ≤ r+1. Next, consider the case i = 2. It follows from Observation A.2 that it suffices to ensure that if our collection contains v 2 , then it does not contain: • Finally, consider the case i = 1. It follows from Observation A.2 that it suffices to ensure that if our collection contains v 1 , then it does not contain: • Hence, it can be checked trivially going through the discussion above that in each of the cases, no two vertices in C are at mutual distance 2. The result follows.
Assume for a contradiction two vertices of T have mutual distance 2. It follows from Observation A.2 that one of them has to be v ′′ r−s . Since for any v, w ∈ V (G), Next, consider the case

by assumption, it further follows that the other vertex would have to be
Hence there exists a vertex a, such that a is neighbour of both v ′′ r−1 and v ′ 4k−2 . Moreover, clearly d(a, v 0 ) = r − 2. Consider two cases. If d(a, v 4k+1 ) ≥ 3, take a Assume for a contradiction two vertices of T have mutual distance 2. It follows from Observation A.2 one of them has to be a. Since for any v, w in G, we have Hence, no two vertices of T have mutual distance 2 and |T | = r. The result follows from Lemma 3.4.
Next, consider the case d(a, v 4k+1 ) < 3. By the triangle inequality, we have d(a, v 4k+1 ) ≥ |d(a, v 0 ) − d(v 0 , v 4k+1 )| = 2, so that d(a, v 4k+1 ) = 2. Hence, there exists a vertex b such that b is neighbour of both a and v 4k+1 . Consider We have |U | = 8k + 2 = 2r. Consider auxiliary graph H on V (H) = U in which we connect two vertices if their distance in G is precisely 2. H is union of two disjoint cycles of length r, first being v 0 , v 2 , ..., v 4k , b, v ′ 4k−2 , ..., v ′ 2 , and second being v 1 , v 3 , ..., v 4k+1 , a, v ′ 4k−3 , ..., v ′ 1 . The result then follows from Lemma 3.5. The only non trivial relationships needed to prove that H is union of two disjoint cycles of length r are If any of these distances was at most 2, we could find a path of length at most r − 1 from v 3 to v ′ 4k−2 . That would be a contradiction. Putting Claim 3.7 and Claim 3.8 together now finishes the proof of Proposition 3.6.
3.5. Proof of Proposition 3.3 for r = 4k + 3. In this subsection, we prove the following.
Proposition 3.9. If G is a connected triangle-free graph on n vertices with minimum degree δ ≥ 2 and radius r ≥ 4 such that r = 4k + 3 for some k, then we have n ≥ 2⌈ rδ 2 ⌉. We use a slightly weaker and more general set-up than we did in the proof of Proposition 3.6. This will have the advantage that we have more freedom in our choice of a center v 0 as well as in the choice of v ′ r−t . Proof. Take v 0 to be any center of our graph G. Let v r be any vertex such that be a path of length r − s form v 0 to v ′′ r−s . By Observation A.1, we have s ≤ 4. We will consider four cases depending on the value of t. Consider the case t = s = 0. Let By Observation A.2, no two elements of T have mutual distance 2. The result follows from Lemma 3.4.
Next consider the case t = 0, 1 ≤ s ≤ 4. We claim that we can find four vertices z 1 , z 2 , z 3 , z 4 such that no two out of z 1 , z 2 , z 3 , z 4 have mutual distance 2, and r − 3 ≥ d(v 0 , z i ) ≥ r − 4 for i = 1, 2, 3, 4. Set Hence, we can always find suitable z 1 , z 2 , z 3 , z 4 . Let It follows from Observation A.2 that no two elements of T have mutual distance 2. The result follows from Lemma 3.4.
Consider the case t = 3. Let w r−u be so that d(v 1 , w r−u ) ≥ r and d(v 0 , w r−u ) = r − u for some 0 ≤ u ≤ 1.
First, consider the case when d( , w r−u Assume for a contradiction that two vertices of T have mutual distance 2. It follows from Observation A.2 that one of them has to be w r−u . Since for any v, w Hence, no two vertices of T have mutual distance 2, and |T | = r. The result follows from Lemma 3.4.
Next, consider the case d( Hence there exists a vertex a such that a is neighbour of both w r−1 and v ′ 4k . Moreover, clearly d(a, v 0 ) = r − 2.
Consider two cases. If d(a, v 4k+3 ) ≥ 3, let Assume for a contradiction that two vertices of T have mutual distance 2. It follows from Observation A.2 that one of them has to be a. Since for any v, w in Hence, no two vertices of T have mutual distance 2 while |T | = r. The result follows from Lemma 3.4.
Next, consider the case d(a, v 4k+3 ) < 3. By the triangle inequality, we have d(a, v 4k+3 ) ≥ 2, so that d(a, v 4k+3 ) = 2. Hence, there exists a vertex b, such that b is neighbour of both a and v r . Consider We have |U | = 8k + 6 = 2r. Consider the auxiliary graph H on V (H) = U in which two vertices are connected if their distance in G is precisely 2. H is union of two disjoint cycles of length r, first being v 0 , v 2 , ..., v 4k+2 , b, v ′ 4k , ..., v ′ 2 , and second being v 1 , v 3 , ..., v 4k+3 , a, v ′ 4k−1 , ..., v ′ 1 . Indeed, the only non trivial relationships needed to prove that H is union of two disjoint cycles of length r are If any of these distances was at most 2, we could find a path of length at most r − 1 from v 3 to v ′ 4k . The result follows from Lemma 3.5.
Finally, consider the case t = 1.
Proof of Proposition 1.4. Let v 0 be a center of our graph, v r a vertex with d(v 0 , v r ) = r and v 0 , ..., v r a path of length r.
Consider the sets Q(v 0 ), ..., Q(v r ), defined for each v i on our geodesic to be the points at distance at most k from v i . Every vertex in our graph is in at most 2k + 1 of these sets, so in particular some of these sets contains no more than (2k + 1)cδ k−1 vertices.
Also, it follows same as in the proof of Lemma 2.1 that each vertex has at least δ(δ − 1) k−2 vertices at distance at most k − 1 from it. Hence, as all edges from these vertices are included in Q(v i ), we get that for every i the subgraph induced by Q(v i ) has at least 1 2 δ 2 (δ − 1) k−2 edges. Putting this together, we get a connected graph of girth at least 2k on at most (2k + 1)cδ k−1 vertices with at least 1 2 δ 2 (δ − 1) k−2 edges.