On the maximum number of cycles in a planar graph

On the Maximum Numberof Cycles in a PlanarGraph

R. E. L. Aldred1 and Carsten Thomassen2

1DEPARTMENT OF MATHEMATICS AND STATISTICSUNIVERSITY OF OTAGO

P. O. BOX 56, DUNEDIN, NEW ZEALANDE-mail: [email protected]

2DEPARTMENT OF MATHEMATICSTECHNICAL UNIVERSITY OF DENMARK

DK-2800 LYNGBY, DENMARKE-mail: [email protected]

Received November 11, 2006; Revised October 25, 2007

Published online 26 December 2007 in Wiley InterScience(www.interscience.wiley.com).DOI 10.1002/jgt.20290

Abstract: Let G be a graph on p vertices with q edges and let r = q − p +1. We show that G has at most 15

162r cycles. We also show that if G is planar,then G has at most 2r−1 + o(2r−1) cycles. The planar result is best possiblein the sense that any prism, that is, the Cartesian product of a cycle and apath with one edge, has more than 2r−1 cycles. © 2007 Wiley Periodicals, Inc. J Graph

Theory 57: 255–264, 2008

Keywords: graphs; maximum number of cycles

1. INTRODUCTION

It is natural to ask how many cycles a connected graph or multigraph on p verticesmust have or can have. (A multigraph may have loops and multiple edges. A graph

Contract grant sponsor: Danish Research Agency (to R. E. L. A.); Contract grantnumber: 21-03-0486.Journal of Graph Theory© 2007 Wiley Periodicals, Inc.

255

256 JOURNAL OF GRAPH THEORY

has no loops or multiple edges.) The quantity fk(p), defined to be the smallest num-ber of cycles present in a k-connected cubic graph on p vertices, was introducedand investigated by Barefoot, Clark and Entringer in [2] where it was determinedthat f1(p) = (3p + 26)/4 and f2(p) = (p2 + 14p)/8 and the extremal graphs werecharacterized for k = 1, 2. The problem of bounding f3(p) remained open for an-other 10 years. In [1], the present authors determined that 2p0.17 ≤ f3(p) ≤ 2p0.95

,indicating a marked difference between 2-connected and 3-connected cubic graphs.

The related problem of how many cycles can be present was raised by Entringerand Slater [3] in 1981. Let G be a connected graph with p vertices and q edgesand define the parameter r = q − p + 1. The maximum number of cycles in sucha graph is denoted by �(r). (We will use �H (r) (�P (r))to denote the maximumnumber of cycles in a hamiltonian (planar) graph with p vertices and q edges.) Itwas noted in [3] that the dimension of the cycle space of such a graph is known to be2r and consequently, �(r) ≤ 2r − 1. (Of course, this upper bound applies equallywell to multigraphs.) In the same article, the Mobius ladders were used to show�(r) ≥ 2r−1 + r2 − 3r + 3.

In 1994, Shi [6] studied �H (r), where attention is restricted to hamil-tonian graphs, strengthening the lower bound for this class. More re-cently, Rautenbach and Stella [5] improved this lower bound and pro-duced the first improvement on the upper bound for any class of graphswhen they showed 2r−1 + (5/2)(r − 1)2 − (21/2)(r − 1) + 14 ≤ �H (r) ≤ 2r −1 − (r − 1)

( √(r−1)−2

log2(r−1)+2 − 14 log2(r − 1)

)for r ≥ 5.

While it is conjectured in [3] that �(r) ∼ 2r−1, the magnitude of the upperbound of 2r − 1 has stood since 1981. In this article, we reduce the upper boundfor general graphs by a multiplicative constant and determine that for planar graphs2r−1 ≤ �P (r) ≤ 2r−1 + o(2r−1).

The following result from [3] essentially enables us to restrict our attention tocubic graphs.

Theorem 1.1. For every r ≥ 3, there is a cubic graph Gr with p vertices andq = 3p/2 = r + p − 1 edges such that Gr contains precisely �(r) cycles.

In the proof of the above result the construction of Gr from the graph G realizing�(r) preserves planarity so we note the following useful corollary.

Corollary 1.2. For every r ≥ 3, there is a planar cubic graph Gr with p verticesand q = 3p/2 = r + p − 1 edges such that Gr contains precisely �P (r) cycles.

2. GRAPHS WITH MANY CYCLES

Definition 2.1. We define the ladders L2n as follows:V (L2n) = {xi, yi : 0 ≤ i ≤ n − 1}. For n = 1, L2 is simply the double bond, thatis, two edges joining x0 to y0, while for n ≥ 2E(L2n) = {xixi+1, yiyi+1, xiyi : 1 ≤ i ≤ n − 2} ∪ {x0x1, x0y1, xn−1y0, yn−1y0,

xn−1yn−1}.Journal of Graph Theory DOI 10.1002/jgt

MAXIMUM NUMBER OF CYCLES IN PLANAR GRAPHS 257

The cross-ladders CL2n are obtained by taking a ladder L2n and selecting someset S (possibly empty) of disjoint pairs (j, j + 1) with 1 ≤ j ≤ n − 2. For each pair(j, j + 1) ∈ S, delete the edges xjyj, xj+1yj+1 and add the edges xjyj+1, xj+1yj. Aladder string LS2n is a collection graph, �2n1, . . . , �2nk

joined by edges yi0x

i+10 , 1 ≤

i ≤ k − 1, where �2niis either a ladder L2ni

or a cross-ladder CL2ni,∑k

i=1 ni = n,and �2ni

has degree 2 vertices xi0 and yi

0. We denote F as the set of all ladderstrings.

Note: In LS2n, there are 2n vertices and precisely 2n paths between the twovertices of degree 2.

Theorem 2.2. Let G be a multigraph on 2n vertices such that two vertices, x, y

have degree 2 and all others have degree 3. Then either G ∈ F or the number ofpaths between x and y, Px,y, is at most 15

16 2n.

Proof. We may assume that G is connected as otherwise paths between x andy would be restricted to that component of G containing both x and y and deletingany other components would reduce n while not affecting Px,y. We proceed byinduction on n. When n = 1, G = L2 which clearly belongs to F. Assume that theresult is true for all 1 ≤ n ≤ k and consider such a graph on 2k + 2 vertices. Letu, v be the neighbors of x in G. If u = v, let w �= x be the remaining neighbor ofv. Then the number of paths from x to y in G is twice the number of paths fromw to y in G − x − v. By induction, either G − x − v is in F and thus so is G, orG − x − v has at most 15

16 2k paths from w to y so that there are at most 1516 2k+1 paths

from x to y in G. So assume that u �= v and form a new graph G′ by deleting x andsuppressing u. Thus G′ is a graph of the desired type on 2k vertices. By our inductivehypothesis, either Pv,y ≤ 15

16 2k or G′ ∈ F. Similarly, if G′′ is obtained from G bydeleting x and suppressing v, then either Pu,y ≤ 15

16 2k or G′′ ∈ F. Consequently, Ghas Px,y = Pu,y + Pv,y ≤ 15

16 2k + 1516 2k = 15

16 2k+1 paths between x and y unless oneof G′ and G′′ belongs to F. Without loss of generality, assume that G′ ∈ F.

First consider G′ to be either a ladder (L2k, say) or cross-ladder (CL2k, say)labeled according to the definition so that v = x0 and y = y0. Furthermore, wethink of G′ being drawn so that x0 lies to the far left of our picture and y0 to the farright. We will also think of the xis being drawn at the top of our picture and the yisat the bottom. Reconstruct G from G′ by replacing x adjacent to v and then considerthe position of u. If u subdivides any of the edges vx1, x1y1, vy1, then G ∈ F. Thus,by symmetry, we have three cases to consider:

Case (i). u subdivides xk−1y or xjxj+1 for some 1 ≤ j ≤ k − 1 and xjyj+1 �∈E(G′);

Case (ii). u subdivides xjxj+1 for some 1 ≤ j ≤ k − 2 and xjyj+1 ∈ E(G′);Case (iii). u subdivides xjyj for some 2 ≤ j ≤ k − 1.

Before considering these cases further, we introduce the following notation andobservation.

Journal of Graph Theory DOI 10.1002/jgt


Observation. Let α ≥ 0 be the number of edges xiyi (in G), 1 ≤ i ≤ j (called“rungs”) and β ≥ 0 be the number of “crosses” (i.e., pairs of edges xiyi+1, xi+1yi,

1 ≤ i ≤ j − 1) in the graph G′. Then α + 2β ≤ j ≤ α + 2β + 1.The graph G′ − xj+1 − yj+1 has exactly two components. Call that component

of G′ − xj+1 − yj+1 containing xj, LG′j and the other, containing xj+2, RG′

j+2.The number of paths, Pxj,yj

, from xj to yj in LG′j is maximized when all β crosses

come after the α rungs. This is because a path in LG′j from xj to yj either uses

the path segment x1x0y1, in some order, exactly one rung xiyi, or exactly one ofthe edges xiyi+1, xi+1yi from some cross. Once one of these elements is chosen, nopart of LG′

j to the left of this element can be used in the path but any selection ofcrosses to the right of this element can be combined with it to form a path from xj

to yj. Hence the maximum number of paths from x to y in LG′j is achieved with all

rungs to the left of all crosses. Furthermore, Pxj,yjis maximized when β is as large

as possible. Note, this means that Pxj,yjis maximized when α = 0, β = j/2 for j

even and when α = 1, β = (j − 1)/2 for j odd. Thus

Pxj,yj≤ (α + 1)

β∑i=0

(β

i

)+ 2

β∑i=1

(β

i

)= (α + 3)2β − 2.

We now return to consider the three cases enumerated above. In each case, thereare 2k paths from x to y not using the edge xu. The number of paths using theedge xu is dependent upon the location of u. Generally speaking, u subdivides anedge incident with xj for some j. We can either enter RG′

j+1 without first passingthrough LG′

j−1 or via some xj−1 to yj−1 path in LG′j−1 with slight variations in the

subscripts here depending upon whether u lies in a cross. In each of the followingcases we count the number of paths as indicated above.

Case (i). If u subdivides either xk−1y or an edge xjxj+1 for some 1 ≤ j ≤ k − 1and xjyj+1 �∈ E(G′), then,

Px,y = 2k + 2(k−1−j)(1 + Pxj,yj) ≤ 7

82k+1.

Case (ii). If u subdivides an edge xjxj+1 for some 1 ≤ j ≤ k − 2 and xjyj+1 ∈E(G′), then, either j = 1, α = β = 0 and

Px,y = 2k + 2(k−3)(4 + 3Px1,y1 ) = 15

162k+1,

or j ≥ 2 and

Px,y = 2k + 2(k−2−j)(4 + 3Pxj−1,yj−1 ) ≤ 13

162k+1.



Case (iii). If u subdivides an edge xjyj for some 2 ≤ j ≤ k − 1, then, eitherj = 2, x1y1 ∈ E(G′) and

Px,y = 2k + 2(k−3)(2 + 2Px1,y1 ) = 7

82k+1,

or j ≥ 3 and

Px,y = 2k + 2(k−1−j)(2 + 2Pxj−1,yj−1 ) ≤ 13

162k+1.

So the result holds if G′ is either a ladder or cross-ladder.Thus we may assume that G′ is a non-trivial ladder string with blocks

�′2k1

, . . . , �′2kj

,∑j

i=1 ki = k. Now we use the above analysis of the single lad-der or cross-ladder case to determine that if u subdivides any of the edges in �′

2k1,

then either G ∈ F or there are at most 1516 2k1+1 paths from x to y1

0 and hence at most1516 2k1+1�

j

i=22ki = 1516 2k+1 paths from x to y in G. If u subdivides an edge which is

either a bridge joining �′2ki

to �′2ki+1

or belongs to �′2ki+1

for some i ≥ 1, then the

number of paths from x to y is at most 2k + �j

m=i+12km ≤ 2k + 2k−1 ≤ 34 2k+1. Con-

sequently, the result holds for n = k + 1 and hence, by induction, for all integersn ≥ 2. �

It should be noted that by iteratively expanding the classF to include worst casegraphs (such as those graphs G obtained in the Case (ii) of the proof above withj = 1 in the cross-ladder G′) the constant c such that G contains at most c2n pathsfrom x to y can be reduced from 15/16 (to 29/32 in the first iteration). However,this process leads to a rapid growth in the number of cases to be considered and isunlikely to produce a value near 1/2, as conjectured by Entringer and Slater [3].

Corollary 2.3. Let G be a cubic multigraph on 2n vertices. Then G contains atmost 15

16 2n+1 cycles.

Proof. The result is easily seen to be true for n ≤ 3. Let G be a smallest cubicmultigraph such that G has 2n vertices and contains more than 15

16 2n+1 cycles.Choose an edge xy ∈ E(G) so that G − xy �∈ F. Note it is always possible to findsuch an edge since, if for some edge x0y0, G − x0y0 ∈ F, then either of the other twoedges incident with x0 will work when n ≥ 4. By Theorem 2.2, for the edge xy inG, there are at most 15

16 2n paths from x to y in G − xy, and hence at most 1516 2n cycles

in G through the edge xy. By the minimality of G, if Gxy is obtained from G byremoving the edge xy and suppressing the resulting degree two vertices x and y, thenGxy is cubic on 2(n − 1) vertices and has at most 15

16 2n cycles (which correspond tocycles in G avoiding the edge xy). Thus, G contains at most 15

16 2n + 1516 2n = 15

16 2n+1

cycles. This contradiction establishes the result. �



Note that Corollary 2.3 is sharp for K3,3.

Corollary 2.4. �(r) ≤ 1516 2r.

Proof. By Theorem 1.1, there is a cubic graph G realizing �(r). Thus, byCorollary 2.3, G contains at most 15

16 2n+1 = 1516 2r cycles. �

Finally, we observe that Theorem 2.2 reduces the conjecture of Entringer andSlater [3] to the 3-connected case, a fact we shall use for our main result in the nextsection.

Corollary 2.5. Let G be a cubic multigraph of connectivity at most 2 on 2n

vertices. Then G has at most 2n + o(2n) cycles.

Proof. If G is disconnected or has cut edges, then we delete all cut edges andapply the trivial upper bound on the number of cycles based on the dimension ofthe cycle space to each component.

On the other hand, if G is 2-connected and contains two edges e1 = x1x2 ande2 = y1y2 such that G − e1 − e2 has two components G1, G2 containing x1, y1 andx2, y2, respectively, then we apply Theorem 2.2 to each of G1 and G2. Thus thereare at most 2n cycles in G containing e1 and e2. Applying the trivial upper bound onthe number of cycles to each of G1 and G2 shows the number of additional cyclesis o(2n). �

The proof of Corollary 2.5 shows more, namely that the number of cycles canbe reduced by the multiplicative factor 15/16 unless the multigraph G has a veryspecial structure.

3. PLANAR GRAPHS WITH MANY CYCLES

For the remainder of the article, we will focus on planar graphs.

Theorem 3.1. Let G be a cubic planar multigraph on 2n vertices. Then G has atmost 2n + o(2n) cycles.

Proof. Let G be chosen to be a cubic planar multigraph of order 2n with asmany cycles as possible. By Corollary 2.5, we may assume that G is 3-connected(and hence a graph).

Consider a plane embedding of G and a set of face-bounding cycles � ={C1, C2, . . . , Ck} chosen as follows. We say that two face-bounding cycles Ci andCj are suitably separated if in the planar dual, G∗, of G, vi is the vertex correspond-ing to Ci and vj is the vertex corresponding to Cj and the distance from vi to vj isat least 5. Choose C1 to be a largest face-bounding cycle. As we may assume thatn > 12, we may assume that C1 has length at least 6. Choose C2 to be a largestface-bounding cycle suitably separated from C1. Continuing, we choose each Ci

to be a largest face-bounding cycle suitably separated from all those chosen sofar. Stop when all remaining suitably separated face-bounding cycles (if any) havelength at most 5. For each Ci ∈ �, the length of Ci is denoted by mi.



A generalized cycle is an element of the cycle space, that is a collection of pairwisedisjoint cycles. Let C denote the set of generalized cycles in G, C0 = {C ∈ C : C isa single cycle}, C1 = {C ∈ C0 : C passes through each cycle in � exactly once} andC2 = {C ∈ C0 : there is some cycle in � through which C passes either not at all orat least twice}. We define each cycle in � to be in C2.

Let Cg = C\C0. Then clearly C = C1 ∪ C2 ∪ Cg. Thus, if we let �i = |Ci| for i =0, 1, 2, we have

�1 + �2 + �g = 2n+1. (1)

We define a mapping f on the cycles in C2 as follows. If C is one of the cyclesin �, then f (C) is the sum of all other Ci in �. For all other cycles C in C2,f (C) = C + Cj, where the addition is the symmetric difference, and Cj is the firstcycle in � qualifying C to be in C2.

Now, if k = 1, then f (C1) is empty, and if k = 2, then f (C1) = C2 and f (C2) =C1. (In either of these cases, for all cycles C ∈ C2 − �, f (C) ∈ Cg and f (C) = f (C′)if and only if C = C′.)

For k > 2, we note that, since each cycle in � is face-bounding, for each C ∈ C2,f (C) consists of at least two cycles, hence f (C) ∈ Cg. Moreover, f : C2 −→ Cgis an injective map. To see this, suppose C and C′ are distinct cycles qualified tobe in C2 by the same cycle Cj ∈ �. Then clearly f (C) and f (C′) are also distinct.On the other hand, if f (C) = C + Cj and f (C′) = C′ + Cj′ , and j < j′, then theintersection of f (C) with Cj either equals Cj or has at least two components,whereas the intersection of f (C′) with Cj is a path. So f (C) and f (C′) are distinctin this case.

In the calculations below, we assume for notational convenience that k > 2.From the above discussion, we have

�2 ≤ �g.

Thus, from (1) we have

�1 + 2�2 ≤ �1 + �2 + �g = 2n+1

giving

�2 ≤ 2n − 1

2�1.

So

�0 = �1 + �2 ≤ 2n + 1

2�1. (2)

We focus on �1.Consider a cycle C ∈ C1. For each cycle Ci ∈ �, we have a segment of

C, . . . uivi . . . wixi . . ., where ui, xi are not on Ci and vi . . . wi represents an arcof Ci. We refer to vi, wi ∈ V (Ci) as the vertices of egress in Ci. Clearly there are



two such vi . . . wi arcs available in Ci for each pair of vertices of egress. For eachC ∈ C1, form a new cubic planar multigraph G′ from G by successively deletingthe vertices in � and successively suppressing all vertices neighboring vertices in� apart from ∪k

i=1{ui, xi}. Finally, add the edges {ui, xi : 1 ≤ i ≤ k}.We claim that the multigraph G′ is connected and has 2n −∑k

i=1(2mi − 2) ver-tices. As G is 3-connected, the deletion of any fixed cycle in � leaves a connectedgraph. As the cycles in � are suitably separated, the deletion of all of them leaves aconnected graph. Also, no vertex can be adjacent to two distinct cycles in �. How-ever, some vertex outside � may have two neighbors in the same cycle in �. Thisis why it is convenient to delete the vertices in the cycles in � one by one becauseafter deleting a vertex in a cycle in � we suppress its neighbor outside � and thus“remove” two vertices altogether. At each stage the current multigraph is connected(again because every cycle in � is a face boundary and therefore non-separating).Therefore we never encounter a multigraph having a vertex adjacent to three ver-tices in a cycle in �. This proves the claim that G′ has 2n −∑k

i=1(2mi − 2) vertices.Such a multigraph G′ can be formed for each selection of one pair of vertices ofegress in each cycle in �.

So the number of cycles in C1 can be bounded as follows:

�1 ≤(

k∏i=1

[(mi

2

)× 2

])× 2

n−k∑

i=1(mi−1)+1

from which it follows that

�1 ≤ 2

(k∏

i=1

[mi(mi − 1)

2mi−1

])× 2n ≤ 2

(30

32

)k

2n.

Also, looking at just m1 = m

�1 ≤ 2m(m − 1)2n−m+1 ≤ m222−m2n. (3)

Recall that each cycle in � was chosen to have length at least 6. We considerG∗, the planar dual of G. Since G is cubic and planar on 2n vertices it has 3n

edges and n + 2 faces. Thus G∗ is a planar triangulation with n + 2 vertices and3(n + 2) − 6 = 3n edges. If we denote by ni the number of vertices of degree i inG∗ we have

m1∑i=3

ini =m1∑i=6

ini +5∑

i=3

ini = 6n.

Consequently,

m1

m1∑i=6

ni + 55∑

i=3

ni ≥ 6n.



Let n≥6 = ∑m1i=6 ni and n≤5 = ∑5

i=3 ni, then n≥6 + n≤5 = n + 2 and we have

m1n≥6 + 5n≤5 = m1n≥6 + 5(n + 2 − n≥6) ≥ 6n.

So

n≥6 ≥ n − 10

(m1 − 5).

We remain in G∗ to find a bound on k. To ensure that the cycles in � are suitablyseparated, if N2(u) denotes the set of vertices distance at most 2 from u, then wemay take u and v to be vertices in V (G∗) corresponding to distinct cycles in � anddemand that N2(u) ∩ N2(v) = ∅. Thus

k ≥ (n − 10)/(m1 − 5)

1 + m21

≥ n − 10

m31

>n

2m3, if n > 20.

So

�1 ≤ 2

(30

32

) n

2m3

2n.

Since (3) also holds, we have

�1 ≤ 2

(30

32

)n14

2n

for n sufficiently large. To see this, suppose m3 ≤ n34 /2. Then

�1 ≤ 2

(30

32

) n

2m3

2n ≤ 2

(30

32

)n14

2n.

So we may assume that m3 ≥ n34 /2, that is, m ≥ n

14 /2

13 . Then from (3), we have

�1 ≤ m222−m2n ≤ 4m32−m2n ≤ 2n34 2(−n

14

)/2

13 2n

since m32−m is decreasing for large m.

It remains to show that n34 2(−n

14

)/2

13 ≤ (

3032

)n 14. Taking logs we get

3

4log n ≤ n

14

(log

(30

32

)+ (log 2)/2

13

)

which is certainly true for n sufficiently large.Thus, from (2),

�0 ≤ 2n + 1

2�1 ≤

1 +

(30

32

)n14

2n

and the result follows. �Journal of Graph Theory DOI 10.1002/jgt


We note that Theorem 3.1 is best possible in the sense that the n-gonal prism,Cn × K2 is cubic, planar and has 2n + 2

(n

2

)cycles.

Corollary 3.2. �P (r) ≤ 2r−1 + o(2r−1).

Proof. The proof follows from Corollary 1.2 and Theorem 3.1. �

ACKNOWLEDGMENT

The research work of R. E. L. A. was supported by Danish Research Agency Grantno. 21-03-0486.

REFERENCES

[1] R. E. L. Aldred and C. Thomassen, Counting cycles in cubic graphs, J CombTheory B 71 (1997), 79–84.

[2] C. A. Barefoot, L. Clark, and R. C. Entringer, Cubic graphs with the minimumnumber of cycles, Congr Numer 53 (1986), 49–62.

[3] R. C. Entringer and P. J. Slater, On the maximum number of cycles in a graph,Ars Combin 11 (1981), 289–294.

[4] D. R. Guichard, The maximum number of cycles in graphs, Proceedings ofthe Twenty-seventh Southeastern International Conference on Combinatorics,Graph Theory and Computing (Baton Rouge, LA, 1996). Congr Numer 121(1996), 211–215.

[5] D. Rautenbach and I. Stella, On the maximum number of cycles in a Hamiltoniangraph, Discrete Math 304 (2005), 101–107.

[6] Y. Shi, The number of cycles in a Hamilton graph, Discrete Math 133 (1994),249–257.


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On the maximum number of cycles in a planar graph