The circumference of the square of a connected graph

COMBINATORICABolyai Society – Springer-Verlag

Combinatorica 13pp.

DOI: 10.1007/s00493-014-2899-4

THE CIRCUMFERENCE OF THE SQUAREOF A CONNECTED GRAPH

STEPHAN BRANDT, JANINA MUTTEL, DIETER RAUTENBACH

Received November 11, 2011

The celebrated result of Fleischner states that the square of every 2-connected graph isHamiltonian. We investigate what happens if the graph is just connected. For every n≥3,we determine the smallest length c(n) of a longest cycle in the square of a connected graphof order n and show that c(n) is a logarithmic function in n. Furthermore, for every c≥3,we characterize the connected graphs of largest order whose square contains no cycle oflength at least c.

1. Introduction

Fleischner [2] proved that the square of every 2-connected graph has a Hamil-tonian cycle. We investigate how small the circumference, the length of alongest cycle, of the square of a connected graph can be in terms of the orderof the graph. Since deleting edges cannot increase the circumference of thesquare, the smallest length is always attained by the square of some tree.

We refer to [1] for any undefined notation. The square G2 of a graph Gis the graph with the same vertex set as G where two distinct vertices arejoined by an edge whenever their distance in G is at most two. A tree T isa caterpillar if deleting all its leaves results in a path, which is called thebackbone of T . Note that the trees with one or two vertices are caterpillarswith empty backbone. A rooted tree is a tree in which one vertex is specifiedas its root. It can be considered an oriented tree, where the edges are orientedaway from the root.

Mathematics Subject Classification (2010): 05C38, 05C76, 05C05

http://dx.doi.org/10.1007/s00493-014-2899-4

2 STEPHAN BRANDT, JANINA MUTTEL, DIETER RAUTENBACH

Let ~v = (d0,d1, . . . ,dh) be a vector whose components are non-negativeintegers. A rooted tree of height h is a ~v-tree if each vertex at distance ifrom the root with 0≤ i≤ h has di children. Note that dh = 0 and that a~v-tree is unique up to isomorphism. With this notation the complete k-arytrees are the (k, . . . ,k,0)-trees.

We investigate the function c(n) defined as the minimum circumferenceof the square of a connected graph of order n≥3. Let

g(n) := 6 log3

(2n+ 3

115

)+ 16.

Our central result, which is a direct consequence of a more detailed statementis the following.

Theorem 1. For n≥3,

g(n) ≤ c(n) < g(n) + 1.2.

The lower bound is attained with equality by a unique graph of order nwhenever n is of the form 1

2(115 ·3k−3) for some integer k≥0. This uniquegraph is the ~vk-tree with

~vk = (5, 3, . . . , 3︸︷︷︸k

, 2, 2, 1, 0).

For all other values of n the lower bound is strict.

Fleischner observed that the square G2 of a connected graph G of orderat least 3 contains cycles of all lengths between 3 and the circumferenceof G2 (cf. Theorem 6 in [3]). Therefore, we get the following immediateconsequence of Theorem 1.

Corollary 2. The square of a connected graph of order n≥3 contains cyclesof all lengths between 3 and dg(n)e .

In fact, for every c≥3, we characterize all connected graphs of largest orderwhose square does not contain a cycle of length c. Theorem 1 is a consequenceof this characterization.

The following immediate consequence of Fleischner’s Theorem 6 in [3] isimplicit in [3]. For convenience we give a short proof.

Corollary 3. If G is a connected graph of order n≥3, C is a longest cyclein G2, and H is the subgraph of G induced by the vertices of C, then H2 ispancyclic.

THE CIRCUMFERENCE OF THE SQUARE OF A CONNECTED GRAPH 3

Proof. If xz∈E(C)\E(H2), then G contains the two edges xy and yz forsome vertex y ∈ V (G)\V (H) and inserting y between x and z in C yieldsa cycle of G2 that is longer than C, which is a contradiction. This impliesthat C is a Hamiltonian cycle of H2 and the pancyclicity of H2 follows fromTheorem 6 in [3].

Harary and Schwenk [5] proved that the square of a tree T is Hamiltonianif and only if T is a caterpillar. Together with Corollary 3 this implies thatthe circumference of the square of a tree T equals the largest order of acaterpillar that is a subtree of T . In the next two sections, we consider theinverse problem asking for the largest order of a tree with no caterpillarsubtree of given order. In the last section we apply these results to cycles inthe square of graphs.

2. Rooted Caterpillars in Rooted Trees

First we consider a rooted version of the problem. For a rooted tree T withroot v, a rooted caterpillar subtree is a subtree C of T such that v ∈ V (C)and deleting all leaves of C that are distinct from v results in a path oneend of which is v. Note that a subtree C of T is a rooted caterpillar subtreeof T if and only if it is a caterpillar such that either v or a child of v is anend of the backbone of C and, in the latter case, v is a leaf of C.

For a positive integer c, we consider the largest order p(c) of a rootedtree that does not contain a rooted caterpillar subtree of order c. Clearly,p(c) is well-defined for every c. We denote the set of rooted trees of orderp(c) that do not contain a rooted caterpillar subtree of order c by T R(c).

Some values of p(c) are given in Table 1, where d denotes the possibledegrees of the root of a tree in T R(c). For c≤ 2, they are obvious and forc ≥ 3, they obey the recursion (1) established in the proof of Theorem 4.Figure 1 shows all trees in T R(c) for 3≤c≤9. We invite the reader to verifythe values of p(c) for small c.

c 1 2 3 4 5 6 7 8 9d 0 1 1 or 2 2 2 or 3 2 3 2p(c) 0 1 2 3 5 7 11 16 23

Table 1: p(c) for 1≤c≤9.

As we shall see below in Theorem 4, for c ≥ 7, the set T R(c) consistsexactly of one (3, . . . ,3︸︷︷︸

k

,2, . . . ,2︸︷︷︸t

,1,0)-tree for integers k and t with k ≥ 0,

1≤ t≤3, and t≡−c (mod 3).


(a) c=3 (b) c=4 (d) c=5

(e) c=6

(i) c=7 (j) c=8 (k) c=9

Figure 1. The trees in T R(c) for 3≤c≤9.

We consider the following functions:

f1(k) =1

2

(23 · 3k − 1

),

f2(k) =1

2

(33 · 3k − 1

),

f3(k) =1

2

(47 · 3k − 1

).

Theorem 4. For every c≥7,

p(c) = fs(r − 2)

where c = 3r+ s for integers r and s with 1 ≤ s ≤ 3. Furthermore, T R(c)consists of a unique element: It is the ~v-tree with

~v =

(3, . . . , 3︸︷︷︸r−2

, 2, 2, 1, 0), if s = 1,

(3, . . . , 3︸︷︷︸r−1

, 2, 1, 0), if s = 2,

(3, . . . , 3︸︷︷︸r−2

, 2, 2, 2, 1, 0), if s = 3.


Proof. We begin with a proof of the following recursion, which is valid forc≥3:

(1) p(c) = max1≤d≤c−2

1 + d · p(c− d).

Indeed, for the value of d for which the maximum is attained, the rootedtree T obtained by taking a root v of degree d each of whose children is theroot of a rooted tree in T R(c−d) has order 1+d·p(c−d) but does not containa rooted caterpillar subtree of order c. This implies that p(c) is at least theright hand side of (1). Conversely, let T be a rooted tree with no rootedcaterpillar subtree of order c. If its root v has degree d, then none of thed rooted subtrees rooted in the children of v can have a rooted caterpillarsubtree of order c−d. Thus none of these d rooted subtrees has more thanp(c−d) vertices. Therefore, p(c) equals the right hand side of (1).

These observations allow to verify the values p(c) given in Table 1 as wellas the sets T R(c) shown in Figure 1. Table 1 also lists the optimum valuesfor d in (1). Observe that d and the optimal rooted trees are unique exceptfor c=4 and c=6.

We complete the proof by induction on c. We already know that thestatement is true for 7 ≤ c ≤ 9. Now let c ≥ 10. We will show that theoptimum value for d in (1) is 3.

Note that p(c′ − 1)/p(c′) ≤ 5/7 for 2 ≤ c′ ≤ 7 by Table 1. For c′ with8≤ c′≤ c−1, we obtain, by induction, that p(c′−1)/p(c′) equals one of thefractions f1(k)/f2(k), f2(k)/f3(k), and f3(k)/f1(k+1) for some k≥0. Usingthe definition of the fs, it is easy to check that p(c′−1)/p(c′)≤5/7 holds for2≤c′≤c−1. Since (c′−1)75>c

′ for every c′≥4, this implies

1 + 3p(c− 3) ≥ 1 + 37

5p(c− 4) > 1 + 4p(c− 4)

≥ 1 + 47

5p(c− 5) > 1 + 5p(c− 5)

...

≥ 1 + (c− 4)7

5p(3) > 1 + (c− 3)p(3)

≥ 1 + (c− 3)7

5p(2) > 1 + (c− 2)p(2).

Similarly, using the induction hypothesis and the functions fs, it is easy tocheck that

1 + 1p(c− 1) < 1 + 2p(c− 2) < 1 + 3p(c− 3).


Altogether, we obtain that the optimum value for d in (1) is 3 and hence,by induction,

p(c) = 1 + 3p(c− 3) = 1 + 3fs(r − 3) = fs(r − 2).

Furthermore, T R(c) contains exactly the unique tree T that arises fromthree copies of the unique tree in T R(c−3) by adding the root v of T andadding edges between v and the roots of the three copies. This completesthe proof.

The following observation will be needed below.

Corollary 5. 23<p(c−1)/p(c)< 5

7 for every c≥8.

Proof. This follows from Theorem 4, because the bounds hold for the frac-tions f1(k)/f2(k), f2(k)/f3(k), and f3(k)/f1(k+1) for k≥0.

3. Caterpillars in Trees

Now we consider the unrooted version of the problem. For a positive inte-ger, let T (c) denote the set of trees of largest order that do not contain acaterpillar subtree of order c.

Define the function q(c) as q(c)=c−1 for 1≤c≤6 and

(2) q(c) = 1 + max d · p(c− d

2+ 1

)for c≥7, where the maximum extends over all d with the same parity modulo2 as c and 1≤ d≤ c−2. As we shall see later, q(c) is the order of the treesin T (c). As the reader can easily verify, for the small values of c, we get thefollowing table:

c 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16d 3 4 5 4 or 6 3 or 5 4 5 6 5 or 7 4q(c) 0 1 2 3 4 5 7 9 11 13 16 21 26 31 36 45

Table 2: q(c) and the optimum values for d for 1≤c≤16.

Lemma 6. For c≥17, the unique maximum in (2) is attained for d=5 if cis odd and for d=6 if c is even.


Proof. Let c ≥ 17. The first inequality of Corollary 5 implies p(c+12

)<

3p(c−12

)< 5p

(c−32

)for odd c and 2p

(c2

)< 4p

(c−22

)< 6p

(c−42

)for even c.

Similarly, Table 1 and the second inequality of Corollary 5 imply

5p

(c− 3

2

)> 7p

(c− 5

2

)≥ 9p

(c− 7

2

)≥ . . . ≥ (c− 4)p(3) ≥ (c− 2)p(2)

for odd c and

6p

(c− 4

2

)> 8p

(c− 6

2

)≥ 10p

(c− 8

2

)≥ . . . ≥ (c− 4)p(3) ≥ (c− 2)p(2)

for even c. Therefore, the maximum in (2) is attained exactly if d = 5 ord=6.

Table 1, Theorem 4, (2), and Lemma 6 together easily imply that

q(c) =

3(23 · 3k − 1

)+ 1, if c ≡ 0 (mod 6),

52

(33 · 3k − 1

)+ 1, if c ≡ 1 (mod 6),

3(33 · 3k − 1

)+ 1, if c ≡ 2 (mod 6),

52

(47 · 3k − 1

)+ 1, if c ≡ 3 (mod 6),

3(47 · 3k − 1

)+ 1, if c ≡ 4 (mod 6),

52

(23 · 3k − 1

)+ 1, if c ≡ 5 (mod 6),

for every c≥17 and k=⌊c−176

⌋.

Our next aim is to characterize the trees in T (c). We start with a basicobservation on a tree and its largest caterpillar subtrees.

Lemma 7. If T is a tree of order at least three, then the intersection of thebackbones of all largest caterpillar subtrees of T is a non-empty path in T .

Proof. Note that every largest caterpillar subtree of T has order at leastthree and hence has a non-empty backbone. Let S be a caterpillar of ordern≥3 and let v be a vertex of its backbone. Then S contains two caterpillarsubtrees S′ and S′′ whose backbones both end in v such that S=S′∪S′′ andS′ and S′′ share at least three vertices. Hence, one of S′ and S′′ has orderat least n+3

2 .For a contradiction, we assume that S1 and S2 are two caterpillar subtrees

of T of largest order n such that the backbones of S1 and S2 do not intersect.Let P be a shortest path in T between a backbone vertex, say v1, of S1 anda backbone vertex, say v2, of S2. If S′1 and S′2 denote caterpillar subtreesof order at least n+3

2 of S1 and S2, respectively, such that the backbone ofS′i ends in vi, then S′1∪P ∪S′2 is a caterpillar subtree of T of order at least


n+32 +n+3

2 −2>n, which is a contradiction. Hence, the backbones of every twolargest caterpillar subtrees of T intersect. Therefore, by the Helly propertyof subtrees in a tree (see e.g. Golumbic [4, p. 80] for a proof) applied tothe backbones of the largest caterpillar subtrees, all of them have a commonintersection and the vertices in the common intersection induce a path in T .

Since every tree of order at most six is a caterpillar, we obtain for c≤6,that every tree of order q(c) belongs to T (c). For c ≥ 7, we obtain thefollowing characterization of the trees in T (c).

Theorem 8. For c ≥ 7, every tree in T (c) has order q(c). Moreover, ev-ery tree in T (c) is obtained from d disjoint rooted trees T1,T2, . . . ,Td ∈T R( c−d2 +1) by adding a vertex v and adding edges between v and the rootsof the trees T1,T2, . . . ,Td, where d is such that the maximum in (2) is ob-tained for d.

Note that the optimum values of d are displayed in Table 2 for c≤ 16. Forc≥ 17, Lemma 6 implies that d∈ {5,6} and d≡ c (mod 2). Therefore, d isuniquely determined except for c∈{10,11,15}.

Proof. It is easy to verify that a tree as described in the statement does nothave a caterpillar subtree of order c. Hence, the trees in T (c) have order atleast q(c).

We prove the statement by induction on c. For c= 7, the statement istrue with T (c) consisting of exactly one tree, being a claw K1,3 with eachedge subdivided once (see the last tree in Figure 1 (e)). Now let c≥8. Let Tbe a tree in T (c). Let B be the intersection of the backbones of all largestcaterpillar subtrees of T . By Lemma 7, B is a non-empty path in T .

If B contains an edge e, then contracting e in T results in a tree T ′ oforder |V (T )|−1 containing no caterpillar subtree of order c−1. By induction,the order of T ′ is at most q(c−1) and therefore the order of T is at mostq(c−1)+1. In view of the explicit values and formula for q(c) above, we haveq(c−1)+1<q(c). Since the trees in T (c) have order at least q(c), this is acontradiction.

Therefore, B consists of a single vertex v ∈ V (T ) and we root T in v.Deleting v from the rooted tree T , results in a collection of subtrees T1, . . . ,Tdrooted in the children w1, . . . ,wd of v, where d is the degree of v in T . LetS be a largest caterpillar subtree of T . Since v belongs to the backboneof S, all the vertices w1, . . . ,wd belong to S. The intersection of S withTi is a rooted caterpillar subtree Si of Ti, rooted in wi. Let ci− 1 denotethe order of a largest rooted caterpillar subtree of Ti and assume that thenumbering of the trees is such that c1 ≥ c2 ≥ . . . ≥ cd. Since T is not a


caterpillar, we have c2≥3 or c1≥5. By the maximality of S, the order of Sis c−1=(c1−1)+(c2−1)+(d−1). Moreover, c1=c2, otherwise every largestcaterpillar subtree of T contains the vertex w1, which is a contradiction toB consisting of a single vertex. Hence, c1 = c−d

2 + 1. By the maximality of

T , both T1 and T2 are rooted trees in T R(c1). Similarly, Ti ∈ T R(c1) for3≤ i≤d. Therefore, the order of T is at most q(c).

Summarizing the above, we obtain that all trees in T (c) have order q(c),and that their structure is as described in the statement of the theorem.

Combining Theorems 4 and 8, Lemma 6, Tables 1 and 2, and Figure 1implies the following observations: The trees in T (c) are unique for 7≤c≤9,12 ≤ c ≤ 14, and c ≥ 16. For 7 ≤ c ≤ 16, the trees in T (c) are exactly the~v-trees with

~v =

(3, 1, 0), if c = 7,

(4, 1, 0), if c = 8,

(5, 1, 0), if c = 9,

(6, 1, 0), if c = 10,

(3, 2, 1, 0), if c = 11,

(4, 2, 1, 0), if c = 12,

(5, 2, 1, 0), if c = 13,

(6, 2, 1, 0), if c = 14,

(7, 2, 1, 0) and (5, 3, 1, 0), if c = 15,

(4, 2, 2, 1, 0), if c = 16,

as well as all spanning trees of the graphs Hc,i displayed in Figure 2 belowfor c∈{10,11,15}.

The graphs Hc,i are a convenient way of summarizing the different pos-sibilities. For c=10 for instance, Table 2 gives the two possible values 4 and6 for d. For d= 6, the tree in T (10) is formed using 6 copies of the uniquerooted tree in Figure 1 (a), which leads to a (6,1,0)-tree. For d=4, the treesin T (10) are formed using 4 copies of the two rooted trees in Figure 1 (b).This leads to five non-isomorphic different trees, which all arise by deletingone edge in each of the four triangles of H10,0. Similar comments apply toc=11 and c=15.


For c≥17, the trees in T (c) are exactly the ~v-trees with

~v =

(6, 3, . . . , 3︸︷︷︸k

, 2, 2, 1, 0), if c ≡ 0 (mod 6),

(5, 3, . . . , 3︸︷︷︸k+1

, 2, 1, 0), if c ≡ 1 (mod 6),

(6, 3, . . . , 3︸︷︷︸k+1

, 2, 1, 0), if c ≡ 2 (mod 6),

(5, 3, . . . , 3︸︷︷︸k

, 2, 2, 2, 1, 0), if c ≡ 3 (mod 6),

(6, 3, . . . , 3︸︷︷︸k

, 2, 2, 2, 1, 0), if c ≡ 4 (mod 6),

(5, 3, . . . , 3︸︷︷︸k

, 2, 2, 1, 0), if c ≡ 5 (mod 6),

where k=b c−176 c. Note that for c≥17 with c≡5 (mod 6), these trees are the~vk-trees. Note furthermore that for c≥ 7, each of these trees has a uniquevertex of maximum degree.

4. Cycles in Squares of Connected Graphs

By Fleischner’s Theorem 6 in [3], for every c≥3, the square of some graphhas a cycle of length at least c if and only if it has a cycle of length exactlyc.

Since the circumference of the square of a connected graph G is at leastthe circumference of the square of each spanning tree of G, the largest orderof connected graphs whose square does not contain a cycle of length c isexactly q(c). Our next result characterizes all these graphs.

Theorem 9. The connected graphs of largest order whose square does notcontain a cycle of length c have order q(c) and are exactly

• the connected graphs of order c−1 for 3≤c≤6,• the trees in T (c) for c≥7 with c 6∈{10,11,15},• the trees in T (c) as well as all connected spanning subgraphs of thegraphs Hc,i shown in Figure 2 for c∈{10,11,15}.

Proof. Since for a connected graph of order at most 6, every spanning treeis a caterpillar, the statement is obvious for c ≤ 6. Let c ≥ 7 and G be aconnected graph of largest order whose square does not contain a cycle of


(a) c=10:H10,0 (b) c=11:H11,0

(c) c=15:H15,0

(d) c=15:H15,1

(e) c=15:H15,4

Figure 2. The graphs Hc,i for c=10,11,15.


length c. As we have noted in the second sentence of this section, the orderof G is q(c). Furthermore, every spanning tree of G must belong to T (c). LetT be a spanning tree of G and let v denote its unique vertex of maximumdegree d. We may assume that there is some edge e in E(G)\E(T ). If either eis incident with v or e joins two vertices in distinct components of T−v, thenG has a spanning tree T ′ with maximum degree either d+1 or d−1, whichare both of a different parity modulo 2 than c. Hence, T ′ cannot belong toT (c), which is a contradiction. Therefore, e joins two vertices, say u and w,in the same component of T − v. Let the distance between v and w be atleast as large as the distance between v and u. Let the tree T ′ arise fromT by adding e and deleting the edge between w and its parent in T . Notethat the distance between v and u remains unchanged but the degree of uin T ′ is larger than the degree of u in T . Since d>1, not both trees T andT ′ can be ~v-trees. In view of T (c), this implies c∈{10,11,15}. In view of thecomplete description of the trees in T (c) for c∈{10,11,15} given at the endof Section 3, both trees T and T ′ are spanning trees of one of the graphs Hc,i

for some fixed i. Therefore, e belongs to one of the triangles contained inthese graphs. Since e was an arbitrary edge from E(G)\E(T ), the graph Gmust be a connected spanning subgraph of Hc,i, which completes the proof.

We return to our initial question concerning the minimum circumferencec(n) of the square of a connected graph of order n. In fact, the trees in T (c)are graphs of largest order such that the circumference of their square isc−1. This implies that for n∗≥3, the value of c(n∗) is the unique integer cwith

(3) q(c) + 1 ≤ n∗ ≤ q(c+ 1).

Expressing c−1 as a function of n=q(c) for c≥17 with c≡5 (mod 6) yieldsc(n)=c−1=g(n) where g(n) is as defined just before Theorem 1. For valuesof c in one of the other five residue classes modulo 6 as well as for c≤ 16,the squares of the trees in T (c) have slightly larger circumference than g(n).Since the ~vk-tree for k= c−17

6 is the unique element of T (c) for c≥ 17 withc≡5 (mod 6), it follows that

c(n) ≥ g(n)

for every n≥3 with equality if and only if n is the order of some ~vk-tree.Now we consider c(n)−g(n). Since g(n) is monotonously increasing, (3)

implies that this difference is at most the supremum of c− g(q(c) + 1) forc≥3. Using the specific values of q(c) for c≤16 and the general expressionsfor c≥17, it follows easily that

c(n)− g(n) ≤ supc≥3

(c− g(q(c) + 1)) = 11− g(q(11) + 1) < 1.19335


for every n≥3, and Theorem 1 follows.Applying the same methods, we can obtain the following more refined

and slightly stronger statement, where

γ := 6 log3115

94< 1.10123.

Theorem 10. For n 6=6,14,17,37,

g(n) ≤ c(n) < g(n) + γ,

where c(n) attains the lower bound if n=q(6k+17) for some k≥0, and forthe sequence nk =q(6k+16)+1, k≥0, we have

limk→∞

(c(nk)− g(nk)) = γ.

Acknowledgement. We were supported by the DFG project “Cycle Spec-tra of Graphs” RA873/5–1.

References

[1] R. Diestel: Graph Theory, Springer, 2010.[2] H. Fleischner: The square of every two-connected graph is Hamiltonian, J. Comb.

Theory, Ser. B 16 (1974), 29–34.[3] H. Fleischner: In the square of graphs, Hamiltonicity and pancyclicity, Hamilto-

nian connectedness and panconnectedness are equivalent concepts, Monatsh. Math.82 (1976), 125–149.

[4] M. Ch. Golumbic: Algorithmic graph theory and perfect graphs, Elsevier, Amsterdam,2004.

[5] F. Harary and A. Schwenk: Trees with Hamiltonian square, Mathematika, Lond.18 (1971), 138–140.

Stephan Brandt

Department of Mathematics and

Computer Science (IMADA)

University of Southern Denmark

Odense, Denmark

[email protected]

Janina Muttel, Dieter Rautenbach

Institut fur Optimierung und Operations Research

Universitat Ulm

Ulm, Germany

janina.muettel,[email protected]

mailto:[email protected]




Documents

The circumference of the square of a connected graph