On the number of cycles of length k in a maximal planar graph

On the Number of Cycles of Length k in a Maximal Planar Graph

S. L. Hakimi

E. F. Schmeichel NORTHWESTERN UNlVERSlN

UNlVERSlN OF SOUTHERN CALlfORNlA

ABSTRACT

Let G be a maximal planar graph with p vertices, and let Ck(G) denote the number of cycles of length k in G. We first present tight bounds for C,(G) and C,(G) in terms of p. We then give bounds for Ck(G) when 5 5 k 5 p , and consider in particular bounds for C,(G), in terms of p. Some conjectures and unsolved problems are stated.

1. INTRODUCTION

We consider only finite undirected graphs without loops or multiple edges. Our terminology and notation will be standard except as indicated. A good reference for undefined terms is Ref. 6.

Let G be a maximal planar graph with p vertices V(G) and q edges E(G). The length of a cycle in G is the number of vertices (or edges) in that cycle. Let Ck(G) denote the number of cycles of length k in G, for 3 5 k ‘p. In Sec. 2, we give tight upper and lower bounds for C,(G) and C,(G). In Sec. 3, we give upper bounds for Ck(G), where 5 5 k s p . In particular, we study C,(G) (i.e., the number of Hamiltonian cycles in G), and explicitly exhibit a family of p-vertex maximal planar graphs in which C,(G) increases exponentially with p. In Sec. 4, we establish a pancyclic behavior in maximal planar graphs.

Journal of Graph Theory, Vol. 3 (1979) 69-86 @ 1979 by John Wiley & Sons, Inc. 0364-9024/79/0003-0069$01 .OO

70 JOURNAL OF GRAPH THEORY

2. BOUNDS ON THE NUMBER OF SHORT CYCLES

We begin with the following theorem:

Theorem 1. Let G be a maximal planar graph with p 2 6 vertices. Then,

The lower bound is attained if and only if G is 4-connected, and the upper bound is attained if and only if G is obtained from K3 by recursively placing a vertex of degree 3 inside a face, and joining this new vertex to the three vertices incident to that face.

Moreover, for every integer r # 3p - 9 such that 2p - 4 I r 5 3p - 8, there exists a p-vertex maximal planar graph G with C3(G) = r. There exists no p-vertex maximal planar graph G with C3(G) = 3p -9.

Proof. A maximal planar graph with p vertices has 3p -6 edges and 2p -4 faces. Since every face of G is bounded by a 3cycle and all such 3-cycles are distinct if G f K3, it follows that C3(G)22p -4 when p r 6 . On the other hand, it can be shown that for each p 2 6 , there exists a 4-connected maximal planar graph on p-vertices (e.g., a (p - 2)-cycle with two additional vertices such that both these vertices are adjacent to every vertex of the (p - 2)-cycle, see Fig. 1). Since a maximal planar graph G is 4-connected precisely if the only 3-cycles in G are those bounding the faces, it follows that C3(G) = 2p -4 if and only if G is 4-connected.

The upper bound is easily seen to be true if p = 6, and hence we proceed by induction on p. We may assume that G is not 4-connected; thus if we embed G in the plane, G will contain a 3-cycle C having some

q P - 3

P

" i

FIGURE 1. A &connected maximal planar graph.

CYCLES IN A MAXIMAL PLANAR GRAPH 71

vertices of G in both its interior and exterior. Hence we can write G = G1 U G,, where Gi will be a maximal planar graph on pi vertices for i = 1, 2, and GI and G, will have precisely the 3-cycle C in common. By the induction hypothesis, C,(Gi) 5 3pi - 8, for i = 1, 2, if pi 2 6 . (It is easy to see, in fact, that C,(Gi) = 3pi -8, if pi = 4 or 5.) So we have

as desired. Also, if C,(G) = 3p -8, we must have C,(Gi) = 3pi -8, for i = 1, 2, and hence by induction G, and G, have the special structure described in Theorem 1. It follows easily then that G must also have this special structure. Moreover, we cannot have C,(G) = 3p - 9, since this would imply that C3(Gi) = 3pi - 9 for i = 1 or 2, which is impossible by the induction hypothesis.

It only remains to show that there exists a p-vertex maximal planar graph G with C,(G) = r for any integer r such that 2p - 4 5 r 5 3p - 10. First construct a 4-connected maximal planar graph G’ on p’= 3p - r - 4 2 6 vertices (such as the graph in Fig. 1). The graph G’ contains exactly (2p’-4) 3-cycles. Then recursively add p -p’ vertices to G’ in the manner described in the statement of the theorem, generating three new 3-cycles with the addition of each new vertex. Call the resulting maximal planar graph G. Then G contains precisely (2gr-4)+3(p-p’)= r 3- cycles. The proof of Theorem 1 is complete. I

Theorem 2. Let G be a maximal planar graph with p L 5 vertices. Then

The lower bound is attained if and only if p = 5 or G is 5-connected. The upper bound is attained if and only if G is the graph in Figure 2.

Proof. If we delete any edge from G, we create a face that is bounded by a 4-cycle. If all the resulting 4-cycles were distinct, then G would contain at least (3p-6) 4-cycles. But the only way that the deletion of two different edges could lead to distinct faces bounded by the same 4-cycle would be if G = K4. Since p 2 5 , we have C,(G)? 3p -6. On the other hand, for any p L 12 except p = 13, there exists a 5-connected maximal planar graph G with p vertices (see Ref. 4). But a maximal planar graph is 5-connected if and only if there are no other 4-cycles besides the ones already described above. Thus C4(G) = 3 p -6 if and only if p = 5 or G is 5-connected. (For 5 <p< 12 or p = 13, there is no maximal planar graph G for which C4(G) = 3p - 6, since each such graph


contains a vertex of degree 5 4 (see Ref. 8). We will say more about this immediately following the proof of this theorem.)

To establish the upper bound C4(G) I;(p’+ 3p - 22), we first prove the following result:

Lemma 1. If G is a 4-connected maximal planar graph with p r 7 vertices, then C4(G)5$(p2-p -2).

Proof of Lemma 1. It is readily verified that there is only one 4-connected maximal planar graph G on 7 vertices with C,(G) = 20, thus we proceed by induction on p. We may assume G is not 5-connected, since otherwise C4(G) = 3p -6<$(p2-p -2) if p 2 7 . Hence if we embed G in the plane, G will contain a 4-cycle C having vertices in both its interior and exterior. (We call such a 4-cycle separating.) Thus we can write G = G1 U G2, where Gi will be a planar graph on pi vertices for i = 1, 2, and G1 and G2 will have precisely the 4-cycle C in common.

Suppose first that pi 27, for i = 1, 2. Then we can add one of the two possible edges, say e,, to Gi to form a maximal planar graph GI which is 4-connected. It follows that C4(G9 I$(p:-pi -2) for i = 1, 2, by the induction hypothesis. There will be precisely four nonseparating 4-cycles in G containing vertices on both sides of C, while the number of separating 4-cycles containing vertices on both sides of C will be at most (pl - 4)(p2 - 4), since p L 7 and G is 4-connected. Also, the edge ei occurs in at least four 4-cycles of G; and so

Combining these observations, we get

C4(G) C4(Gl) -k C4(G2) - 1 +4+ (PI -4>(P2 -4) 5 gp; - p1- 10) + gp; - pz - 10) - 1 + 4 + (pl - 4)(p, - 4)

= $UPl+ P2 - 4Y - (P1+ P2 - 4) - 21 = gpz- p -2),

as desired. If pi = 5, we have C,(Gi) = 5, while if pi = 6 and GI is 4-connected, we

have C4(Gi) = 10. It is then a simple matter to alter the above argument accordingly and conclude that C4( G) ~ ; ( p ’ - p - 2) in these cases also.

This completes the proof of Lemma 1. I We mention in passing that it can be shown that equality holds in

Lemma 1 if and only if G is the graph in Figure 1.


We now complete the proof of the theorem by showing that C4(G) 5 i(p2+3p-22), with equality holding precisely when G is the graph of Figure 2. By Lemma 1, we may assume G is not 4-connected, and hence we can choose a 3-cycle C, G, and G, as in the proof of Theorem 1, but with the the additional assumption that p1 is minimum.

Suppose first that p 1 2 7 . Then G, is 4-connected by the minimality of ply and hence C4(Gl)~&p:-p,-2$ by Lemma 1. Also, C4(G,)S $(pq+ 3p, - 22) by the induction hypothesis. Finally, there is at most one vertex in G, adjacent to every vertex in C, and G, contains no separating 3-cycles by the minimality of pl. It follows that the number of 4-cycles in G containing vertices on both sides of C is at most p2 - 1. Combining these observations, we get

as desired, since p 2 2 p l > 6 . If p1 = 6, then G, is 4-connected by the minimality of p,, and hence G,

is the graph in Figure 1. We have at once that C4(G1) = 15. As in the last

FIGURE 2. A planar graph with the maximum number of 4-cycles.


paragraph, we obtain

C4(G)5 C4(G1)+C4(GZ)+(p2- 1> 5 1s +gp; + 3p2- 22) + (pz- 1) < g(p2 + 3)2 + 3(p, + 3) - 221 = i(p’+ 3p - 22)

as desired, since p2 2 p1 = 6.

separating 3-cycle), and hence C,(G,) = 3. It then follows as above that Finally, if p1<S , then in fact p1 must be 4 (otherwise G1 contains a

C4(G) 5 C4(Gi) + C4(G2) + ( ~ 2 - 1) 5 3 +; (p ;+ 3pz-22) +(&- 1) = &[(pz + 1)” + 3(pz + 1) - 221 = gp2+ 3p - 22),

p = 6 , C4 (G1=15 p= 7, C4(G) = 20 p = 8 , C 4 ( G ) = 2 3

p = 9 , C , ( G ) = 2 4 p = 10, C, (GI =26 p= l l , C , ( G ) = 2 9

FIGURE 3. Graphs with possibly the least number of 4-cycles.


as desired. It remains to consider when equality holds. Note that equality holds if and only if C4(G,) = $(p$+ 3p2-22) (in which case G, is the graph in Figure 2 by the induction hypothesis), and there are exactly (p2- 1) 4-cycles in G with vertices on both sides of C. It follows easily now that G must be the graph in Figure 2.

The proof of Theorem 2 is complete. I We note that Theorem 1 is a stronger assertion about C,(G) than

Theorem 2 is about C4(G). For one thing, Theorem 2 does not give a tight lower bound for C4(G), when 5 < p < 12 or p = 13. It is easy to see that C,(G) = 9 when p = 5. The authors believe the following are lower bounds for C4(G) for 5<p<12 or p = 13:

P 6 7 8 9 10 11 13 C,(G)r 15 20 23 24 26 29 34.

In fact, the maximal planar graphs in Figure 3 are believed to have the least number of 4-cycles among the p-vertex maximal planar graphs, for 5<p<12 or p = 13. Finally, in contrast to Theorem 1, there is no assertion in Theorem 2 about the existence of p-vertex maximal planar graphs with a prescribed number of 4-cycles.

3. ON THE NUMBER OF LONGER CYCLES IN MAXIMAL PLANAR GRAPHS

We begin with the following lemma:

Lemma 2. Let G be a maximal planar graph with p 2 8 vertices. Then

There exists a maximal planar graph G for which C,(G) = 6p for every p 2 14.

Proof. Let u be a vertex of degree >3 in G. Let el(u) and e2(2)) be any two edges incident at u that share a face. Then if we delete el(v) and e2(u) from G, we will form a face bounded by a 5-cycle. It can be seen that all 5-cycles that occur in this way are distinct if p 2 6 , and hence we have d(u) 5-cycles associated with each vertex u of degree >3. This implies that

CAG)2C 4 ~ ) - 3 4 3 ) , U


where o(i) is the number of vertices of degree i in G. Consider now a vertex v of degree 5. We observe that in G - v we have

a face that is bounded by a 5-cycle. Moreover, this 5-cycle is distinct from all 5-cycles counted previously if p 2 8; thus

C5(G)2 1 d ( ~ ) - 3 ~ ( 3 ) + 0 ( 5 ) . u

Now consider a vertex v of degree 4. It can be shown that if p 2 6, then v is adjacent to at most one vertex of degree 3. Note that G - v has a face which is bounded by a 4-cycle. Let this 4-cycle be v1v2v3v4v1. As at most one vertex among vl , v2, v3, v4 is of degree 2 in G - v, there are at least two edges in this cycle such that the removal of either edge from G - v forms a 5-cycle if p 2 8. Therefore, if p 1 8,

C5( G) L 2 d (V ) - 30 (3) + w (5) + 2w (4). u

However, we have

5

~ 3 0 ( 3 ) + 4 ~ ( 4 ) + 5 0 ( 5 ) + 6 ( p - i = 3 C w( i ) )

If we substitute this in Eq. (l), we obtain

C5(G)?6p-6w(3), when p r 8 .

We now observe that if 0(3)=0, Lemma 2 is proved. Hence, we proceed by induction on w(3). Suppose Lemma 2 is correct when w(3) < i. Assume that w(3) = i, and momentarily suppose p 1 9 .

Let v be a vertex of degree 3. Then G - v is maximal planar, and has p 1 8 vertices; by the induction hypotheses, Cs(G - v) 26(p - 1). It can be seen that, as the vertices adjacent to v are of degree 1 4 , there are at least 12 5-cycles in G which pass through v ; thus

C,(G)r Cs(G - v ) + 1216(p - 1)+ 12>6p.

The case when p = 8 can be handled by considering all maximal planar graphs on 8 vertices. Diagrams of all such graphs appear in Ref. 1.


Finally, there exist 5-connected maximal planar graphs having (p - 12) vertices of degree 6 and 12 vertices of degree 5 for every p 2 14 (see, for example, Etourneau [2]). Furthermore, using Etourneau’s construction all resulting graphs have exactly 12 separating 5-cycles. Since every maximal planar graph has exactly 6p - 12 nonseparating 5-cycles, it follows that Etourneau’s graphs have 6p 5-cycles. This completes the proof. I

Lemma 3. Let G be a maximal planar graph with p 2 6 vertices. Then,

Proof. This is easily verified when p= 6. Hence we proceed by induction on p. Let v be a vertex of minimum degree in G ; then 3 5 d(v) 5 5. It can be shown that the number of nonseparating 5-cycles in G that contain v is 125. Using this result, we may write

C5(G) 5 C5(G - V) + 25 + S,(U)

where S,(v) is the number of separating 5-cycles in G containing v. By the induction hypothesis,

Thus the above inequality becomes

C5(G)55(p - 1)2-26(p-l)+25 + S ~ ( V ) .

We observe that if S5(u)510p-56, then C5(G)55pZ-26p as desired. Suppose, therefore, that S,(v) > lop -56 > 1O(p -6). This implies that for some pair of vertices x and y E rv, there exist more than (p - 6) paths of length 3 in G - v joining x and y such that each of these paths together with the path xvy forms a separating 5-cycle in G. Contract vertices x and y in G-v to form a (p-2)-vertex planar graph G’. We observe that in G’ there are more than (p-6) separating 3-cycles containing the vertex x = y. Let GI’ be a (p -2)-vertex maximal planar graph obtained from G’ by adding an edge (if necessary) to G’. Since G” has exactly [2(p -2)-41 nonseparating 3-cycles, it follows that

C3(G”)>[2(p -2) -4]+ (p -6) = 3(p - 2) - 8.

However, this contradicts Theorem 1. The proof of Lemma 3 is complete. I


Lemmas 2 and 3 immediately imply the following theorem:

Theorem 3. Let G be a maximal planar graph with p 2 8 vertices. Then

6p 5 C5(G) I 5p2- 26p.

We expect that the upper bound for C5(G) given in Theorem 3 is weak. We conjecture that C5(G)s2p2- lop -t 12. In fact for the graph shown in Figure 1, the conjectured upper bound is achieved.

Before considering an upper bound for Ck(G) for an arbitrary k, we need the following definitions.

Let f and g be two given real valued functions defined on the positive integers. We write f(n) E O(g(n) ) if there exists a positive constant Co such that f (n)sCog(n) for all n>O; otherwise we write f(n)& O(g(n)). On the other hand if there exist a positive constant co and an integer no such that f (n) rcog(n) for all n 2 no, then we write f(n) E O ( g ( n ) ) .

Let ck(p) = max, Ck(G), the maximum being taken over all maximal planar graphs G with p vertices.

Theorem 4. Let G be a maximal planar graph with p vertices. Then,

Ck(G) and C,(p) E O(pLknJ), for k = 3,4, . . . , p.

Proof. We need only to show that ck(p) E O(pLk’2J), and we do this by induction on k. We have seen that the result is true for k =3, 4, 5. Suppose that it is true for all integers < k. To prove it is also true for k, it suffices to show that

because

and hence we would have C,(p) E O(pLknl). Otherwise for any constant C,, c k ( P ) - C k ( p - l ) > C o p L k n ’ - l , for some, integer p. Let G be a p- vertex maximal planar graph with ck(p> k-cycles. Let v be a vertex of minimum degree in G, so that d ( u ) 5 5 . Then,

therefore, v must occur in more than Cop Lk’zJ-l k-cycles in G. Hence, for


some pair of vertices ‘ul and ‘u2 E rv, there are more than Cop LknJ-l/($)

paths between ‘ul and u2 of length k - 2 in G - ‘u. Form the (p - 2)-vertex planar graph G’ by contracting ‘ul and u2 in G - ‘u. Then

As C, was any constant, this implies that Ck-2(G’) 4 O(p lknJP1), which violates the induction hypothesis. We conclude that c k ( p ) E O(pLk”’) as asserted. I

To study the quality (tightness) of the upper bound for Ck(G) given in Theorem 4, we may ask the following question: are there p-vertex maximal planar graphs for which Ck(G) E O(pLkn’) for k = 3, 4, . . . , p? We have already shown that the answer is positive for k = 3,4,5. We will show that for k > 5, the answer is again positive with some qualifications.

First suppose that p I q k . Define i = LikJ, so that (p - i)/i 12. By the Euclidean division algorithm, p - i = bi + r, where 0 5 r < i. Consider the planar graph G shown in Figure 4, where the set of vertices Aj has

Ai

A 3

FIGURE 4. Graph with exponentially many k-cycles.


cardinality b, for j = 1, 2, . . . , i. We note that

V(G)={ (J Ai]U{vl, v2,. . . , vi}U{vl,,vl,, . . . , v:}. j = l

It is easy to see that IV(G)l= p. The edge set of G includes all the edges indicated in Figure 4 together with any others that do not destroy planarity. It is easy to verify that

is a cycle of length 2i in G. Moreover, observe that ~ , E C can be replaced by any other vertex in Ai and we will still have a cycle of length 2i. Also this is true for any j , 1 5 j 5 i. Thus there are at least b' cycles of length 2i in G.

Suppose k is even; consequently k = 2i. Then for the graph G of Figure 4, C k ( G ) 2 b'"'. Now assume k is odd; then k = 2i + 1. A typical cycle of length 2i + 1 in G is

Since there are, as before, b choices for each of the vertices a,,, aZ1, u ~ ~ , . . . , u+l)l, and b- 1 choices for ail, there are at least bipl(b - 1) k-cycles in G. Moreover,

is a typical k-cycle in G distinct from all the previous ones. This in turn implies that there are at least [2bi-'(b - l)] k-cycles in G. As b 2 2, this in fact implies that Ck(G) 2 b', or again Ck(G) 2 b Lk'2J. We note that

and thus if p ?$k, then

Therefore we have that for the graph G of Figure 4 and for a fixed value of k (i.e., k is fixed but p is sufficiently large), C k ( 6 ) E O(pLk/2').

However, we may wish to examine Ck(G) when k is a function of p ; for example k = $ p or k = p . If k = i p , we note that the relation p >$k is


satisfied; therefore for the graph G of Figure 4,

Cp,(G) 2 LP - LaPJ lip1 1 lPI4j.

But this, implies that Cp,(G) E 0(3Lp/4J). Using the same reasoning, in general, we have that if k is a linear function of p but p’qk, then C,(G) E O(aLk”’) where a 22. This is a weaker result than the previous case, nevertheless it indicates the existence of planar graphs with exponentially many k-cycles in this case.

We will now show that there exists a p-vertex planar graph G with Ck(G) E d(2Lp/3’) if p <?k. 2 Let i = Lip], or equivalently p = 3 i - t r with 0s r < 3. This implies that L(p - i ) / i J = 2. We again refer to the graph G of Figure 4 where now, of course, b = 2 and 0 5 r < 3. We will first examine the case when k = p. To do this, we first split the vertex u1 in the graph G of Figure 4 into two vertices q1 and u12, where oll (respectively, u12) is adjacent to the vertices in A, and u: (respectively, Ai and ui ) . We call the resulting graph G’. We observe that now in the graph G’ of Figure 4 (or really a slight modification of that graph), we have the path

12 u r . . . u:ull. (If I = 0, this modification of G is not necessary.) Then, a typical p-cycle (Hamiltonian cycle) in G’ would be

We observe that in the above Hamiltonian cycle for each j , 1 sj I i, we may change the order of vertices aI1a,, to aI2ql; therefore, G contains at least 2’ Hamiltonian cycles. This implies that, the graph G’ of Figure 4, C,(G’)E O(2Lp/3’). We finally note that, if in the above typical cycle we replace allalz by a,, (or a12), then we will have a typical (p-1)-cycle in G. It is again clear that we have at least 2‘ ( p - 1)-cycles. Repeated use of this process shows that in fact for the graph G’ of Figure 6, Ck(G‘)E O(21p13’) for all k satisfying $ p < k I p .

Since Hamiltonian cycles are of special interest, we now give a recursive construction leading to maximal planar graphs with a much larger number of Hamiltonian cycles. We state the result as a theorem.

Theorem 5. There exists a maximal planar graph G with p 3 4 vertices for which

3 * 6(p-4)/3, if p = 1 (mod 3),

10 * 6‘p-6”3, if p = O (mod 3). 6 6(p-5)/3, if p = 2 (mod 3),


Proof. It is readily verified that for p =4, 5, o r 6, the required graphs are given in Figure 2. Note that each of these graphs contains a vertex of degree 3.

Suppose then that we have constructed a p-vertex maximal planar graph G with the desired number of Hamiltonian cycles that contains a vertex v of degree 3. To construct a p + 3 vertex maximal planar graph G’ with the desired properties, proceed as follows: Delete u from G, and in the face of G - v which contained v, place four new vertices v’, v:, v:, u: with adjacencies as shown in Figure 5. Note that G‘ is maximal planar and has a vertex v‘ of degree 3.

To prove Theorem 5, clearly it s f i c e s to show that Cp+3(G’)z6Cp(G). Consider any Hamiltonian cycle C in G, say C = (. . . v1uv2 . . . ). We can obtain six distinct Hamiltonian cycles in G’ by replacing the sequence of vertices v lvv2 in C by any one of the following six sequences in G’:

v ,v;v:vrv:u2, u 1 v;lJ’u:V;u,, v,u:v’v;v;v,,

v,u;v;v’v;u,, ,til u~v;v’v;u2, or vIv;v’v; v:v2.

It is easy to see that all Hamiltonian cycles in G’ generated in this way are distinct. Hence C,+,(G’)r6Cp(G) as required. The proof is complete. I

In general, the above construction will not yield a p-vertex maximal planar graph with the maximum number of Hamiltonian cycles. For example, for the graph GI of Figure 6, C,(G,) = 16, while for the graph G with 6 vertices constructed earlier, we had only C,(G) = 10. The question o f what C,(p) is in general remains open, and appears to be difficult. However, as the rank of a cycle basis in a p-vertex maximal planar graph is 2p-5, certainly Cp(G)<22p-’. The authors conjecture that, in fact, C p ( G ) 52”.

G G‘

FIGURE 5. Transformation in the proof of Theorem 5.


FIGURE 6. Graph with 16 Harniltonian cycles.

We have no interesting lower bound for C,(G) in a maximal planar graph for k 2 6 . As there are maximal planar graphs that are not Hamiltonian, there are maximal planar graphs for which C, ( G ) = 0. But an interesting question is the following: among the family of p-vertex Hamiltonian maximal planar graphs G, what is a lower bound for C,(G)? We will treat this question in a forthcoming paper [5].

4. PANCYCUC BEHAVIOR IN MAXIMAL PLANAR GRAPHS

A planar graph G is called outerplanar if G has a plane embedding such that all vertices of G belong to a cycle bounding a single face of G (which is usually taken to be the exterior face). Clearly an outerplanar graph is Hamiltonian. An outerplanar graph G is called maximal outerplanar if no edges can be added to G without destroying outerplanarity. A good reference for outerplanar graphs is Ref. 6.

A graph G with p-vertices is called pancyclic if C,(G)>O for k = 3, 4, . . . , p. We begin with the following result:

Lemma 4. A maximal outerplanar graph G is pancyclic.

Proof. It is easy to see that G has at least two vertices of degree 2 (see Ref. 6, p. 107). Let u be a vertex of degree 2 in G, and let ul, v2 be the two vertices in G adjacent to u. Since G is maximal outerplanar, the edge (q, u2) must belong to G, and hence G has a p - 1 cycle.

Now consider G-u. Since G - u is maximal outerplanar with p - 1 vertices, the argument in the last paragraph implies that G - u (and hence G ) contains a p - 2 cycle. A recursive use of this argument shows that G is pancyclic, and proves Lemma 4. I


Lemma 3 immediately implies the following result:

Corollary 1. A Hamiltonian maximal planar graph is pancyclic.

As not all maximal planar graphs are Hamiltonian, we seek a stronger result. A graph G is inner triangulated if G is planar and it has exactly one face bounded by a cycle C which is not a triangle (a 3-cycle). We assume that C bounds the exterior face in a planar representation of G. Let C = ~11212113 - * * up1.

Lemma 5. An inner triangulated graph has a cycle of length k for every k satisfying 3 i k 5 r.

Proof. Let G be the least order graph for which Lemma 5 does not hold. Assume G has p vertices. Let j be the number of vertices of G in the interior of C, so that p = j + r. We observe that j # 0; because if j = 0, G is an outerplanar graph, and by Lemma 4, is pancyclic. Suppose j = s > 0 is the least integer for which Lemma 5 does not hold. Pick any edge, say (vl, u,) on C. This edge is in a 3-cycle bounding a face, say v,u,v,vl. IT v, is in the interior of C, then consider the cycle C’= V , U , V ~ V ~ * - vrul in the graph G’ = G - (vl, u,). We note that G‘ is an inner triangulated graph with p vertices and s - 1 interior vertices; thus by minimality of s, G’ has cycles of length k for k i r + 1. This implies G has the desired cycles. Therefore, we must assume u, is on C. If v, = v3, then d(v,)=2 and G’= G-v, is an inner triangulated graph with (p-1) vertices and an ( r - 1) face C‘= u1u3u4 - - - urul. By minimality of p, G’ has cycles of length k for all k i r - 1. This in turn implies that G has a cycle of every length i r . We must therefore assume x > 3.

Consider edge (v,, u,). This edge belongs to a 3-cycle ~ 2 t 1 , ~ 3 ~ 2 . With identical reasoning as the first half of this proof, we can show that vy is not an interior vertex and if v, lies on C then 4 < y i x. We may proceed to consider edge (u3, u4) and so forth, say (q, Q+~) . At every stage either we have a contradiction or the length of the cycle ui+1vi+2 - u,vi+l becomes smaller. As this length eventually becomes three, at that stage d(vi+,) = 2. However, if d(ui+,) = 2, then G’ = G - vi+, is inner triangulated with (p - 1) vertices and face C‘= u1u2 - - - ~ i + 1 ~ i + 3 * * * urul. But by minimality of G, G’ must have cycles of all length < r - 1. This would imply G has all cycles of length 5 r . Thus Lemma 5 is proved. I

Theorem 6. Let G be a maximal planar graph. Let r be the length of a longest cycle in G. Then C,(G) > 0 for all k satisfying 3 I k i r.

Proof. Let C be a longest cycle in G. Consider a representation of G


on the plane. In this representation of G, certain vertices and edges of G lie in the exterior of the cycle C. Let G’ be obtained from G by deleting all edges and vertices of G that lie in the exterior of cycle C. It is clear that G’ is an inner triangulated graph and by Lemma 5 has cycles of all length k, 3 I k 5 r. This, of course, implies that for graph G, Ck(G)> 0 for all k, 3 5 k ~ r . I

In a forthcoming paper, the length of the longest cycle in a maximal planar graph will be studied [5 ] .

It appears that a result stronger than Theorem 6 is possible. In fact, we conjecture that for a maximal planar graph G, Ck(G)>min{Ck-,(G), Ck+l(G)}, for 3 1 k < r , where r is the length of the longest cycle. It is worth noting that the same behavior has been conjectured about the absolute values of the coefficients of the chromatic polynomial of a graph (see Ref. 7). This type of phenomenon occurs in another graph theoretic problem as well. Let G be a labeled graph and g a subgraph of G. Let N(k) denote the number of (labeled) spanning trees of G each of which contains exactly k edges of g. Let G - g be a graph obtained from G by deleting all edges in g from G. Then N ( k ) > min {N(k - l), N(k + l)}, for all k such that R ( G ) - R ( G - g ) r k < R ( G ) , where R(G’) denotes the rank of G’ (i.e., the number of edges in a maximal circuitless subgraph of G’) (see Ref. 3).

ACKNOWLEDGMENTS

The authors wish to express their gratitude to Carsten Thomassen for generously providing shorter proofs for Theorems 1 and 2 as well as other helpful suggestions. This work was supported in part by the U.S. Air Force Office of Scientific Research, Systems Command, under grant No. AFOSR-76-3017.

References

[l] V. Chvatal, Planarity of graphs with given degrees of vertices. Nieuw Arch. Wisk. 17 (1969) 47-60.

[2] E. Etourneau, Existence and connectivity of planar graphs having 12 vertices of degree 5 and n-12 vertices of degree 6. Infinite and Finite Sets. North-Holland, Amsterdam (1975).

[3] S. L. Hakimi, Graphs with two kinds of elements. J. Franklin Inst. 270 (1960) 451-467.


[4] S. L. Hakimi and E. F. Schmeichel, On the connectivity of maximal planar graphs. J. Graph Theory 2 (1978) 307-314.

[S] S. L. Hakimi, E. F. Schmeichel, and C. Thomassen, Two problems concerning the longest cycle in a maximal planar graph. J. Graph Theory. To be submitted.

161 F. Harary, Graph Theory. Addison-Wesley, Reading, Mass. (1969). [7] R. C. Read, An introduction to chromatic polynomials. J. Com-

[8] E. F. Schmeichel and S. L. Hakimi, On planar graphical degree binatoriuf Theory 4 (1968) 52-71.

sequences. S I A M J . A p p l . Math. 32 (1977) 598-609.

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On the number of cycles of length k in a maximal planar graph