Graphs with exactly one hamiltonian circuit

JOURNAL OF GRAPH THEORY, VOL. I, 37-43 (1977)

Graphs with Exactly One Hamiltonian Circuit

JOHN SHEEHAN A berdeen University

ABSTRACT

Let h(n ) be the largest integer such that there exists a graph with n vertices having exactly one Hamiltonian circuit and exactly h ( n ) edges. We prove that h ( n ) =[n’/4]+ 1 (n 24) and discuss some related problems.

1. Introduction

Let n be an integer, n 1-4. Let % ( n ) be the set of Hamiltonian graphs with exactly n points. Let Z ( n ) be the set of graphs in %(n) having exactly one Hamiltonian circuit. Let

h(n)=max{q(G) : G E % ( ~ ) } ,

where q ( G ) = IE(G)I. We will prove the following three theorems.

THEOREM 1. h ( n ) = [n2/4]+ 1.

THEOREM 2. There exists exactly one graph H ( n ) in % ( n ) with h ( n ) edges.

J. A. Bondy proved the following useful result.

THEOREM [ l ] . Let G E % ( n ) . Then i f q ( G ) 2 n2 /4 , either G contains all 15 n, or else G is the complete bipartite graph cycles of each length 1, 3

K ( n / 2 , n/2).

We obtain a stronger conclusion from a stronger hypothesis.

THEOREM 3. Let G E & ( n ) . Then i f q ( G ) > [ n 2 / 4 ] + 1 , G contains af least two cycles of each length I , 3 5 1 S n.

2. A lower bound for h(n)

Throughout this paper (unless stated otherwise) integers are taken modulo n. Let H ( n ) (see Figure 1) be the graph with vertices @ 1977 by John Wiley & Sons, Inc. 37

38 SHEEHAN

V.

v7

Figure 1 n (12)

u l , u2, 213, . . , u, and edge-set El U EZ, where

( U Z ~ - ~ , U , ) : 2 i < j < n , i = 1 , 2 , . . - , [ 1 ] } . n - 2

NOTATION. If G E &(n) and C is a Hamiltonian circuit in G, then each edge of G not in C may be regarded as a chord of C. We define the length of u chord of C to be the distance between its end vertices taken around C. Let l be an integer, 1 S IS n/2, and let C(G; I ) denote the set of chords of length 1 contained in G.

LEMMA 1. h (n)Z [nZ/4]+ 1.

Proof. Clearly H(n)E%'(n). Let C be a Hamiltonian circuit in H(n) . We prove by induction that C=(uI, u2, u3, * - * , u,). Let k be a positive integer and P(2k - 1) the proposition:

"C contains no element of ((2s - 1, j ) : 2s < - j < n, s 5 k}."

Since the degrees of u2 and u, are both 2, it follows that (un, u l ) , ( u l , U Z ) E C. Hence P ( l ) is true. Suppose P ( 2 k - 1 ) is true. Then C contains ( 2 k , 2 k + 1 ) and ( 2 k + l , 2 k + 2 ) . Hence P ( 2 k + l ) is true. This completes the induction. Thus C contains no edges in E2, so that C= (VI, u2, u3, * * - , u,) and H ( n ) E X ( n ) . Finally

C(n-2)/21

q(H(n))=IEll+(E21=n+ 1 ( n - 2 i - l ) = i-1

GRAPHS WITH ONE HAMILTONIAN CIRCUIT 39

3. Proof of Theorem 1

Let G E %(n) , and let C= ( u l , u2, * * * , u,) be the Hamiltonian circuit in G.

LEMMA 2 . Let i E (1,2, - , n> and 1 < 1 S 4 2 . The chords (s, ui+{),

Proof. Suppose otherwise. Then G contains (see Figure 2) two Hamil-

In words Lemma 2 states: “NO two chords in C(G; I ) are consecutive.”

(ui+* ,

tonian circuits. I

cannot both belong to C(G; I ) .

LEMMA 3 (C. St.J. A. Nush-Williams).

Proof. Apply Lemma 2. I IC(G; 1)I S n / 2 .

LEMMA 4 (C. St.J. A. Nush-Williams). If n is odd , then h ( n ) S

Proof. By Lemma 3,

[n2 /4]+ I.

LEMMA 5 .

Proof. Apply Lemmas 1 and 4. I

The following theorem appears in a paper by Tutte [ 2 ] , where it is stated that “the theorem, though not this particular proof, is due to Dr C. A. B. Smith.”

If n is odd, then h ( n ) = [n2/4]+ 1.

THEOREM [2]. Let G be a cubic graph. Then i f G contains one Hamil- tonian circuit, it contains at least three.

“ i + , f t ~ “it1

Figure 2

40 SHEEHAN

LEMMA 6 .

Proof.

Let n be even, 1 be odd, 1 < 1 < n/2. Then JC(G; I ) ) < 4 2 .

Suppose JC(G; l ) i= 4 2 . Then, by Lemma 2, G contains 2 cubic subgraph which is Hamiltonian. The lemma now follows from the theorem due to C. A. B. Smith which is stated above. B

LEMMA 7.

Proof. This follows immediately from Lemma 2. F: Let n be even. Then IC(G; n / 2 ) S [n/43.

LEMMA 8.

Proof.

Let n be even. Then h ( n ) = [n’/4]+ 1.

By Lemma I , h(n )2[r z2 /4]+ 1. By definitioE

n::

q ( ~ ) = 1 I C ( G ; I ) / . I = :

Suppose n = 2 (mod 4). Then, from Lemmas 3, 6, 7 and Eq. ( l i ,

k = ?

fn-2)/4

+ k = i 2 I C ( G ; 2 k - l ) ! + l C ( G ; ~ ) ~ 1

n - 2 n 4 i----=-+1.

4 4

Suppose n S O (mod 4). Then, fron Lemmas 3, 6, 7 and Eq. (l),

r, n L 4 4

+-=-+ l .

Therefore, from Eqs. (2) and (3), q ( G ) S [n2/4]+ 1. This completes the proof. I


THEOREM 1.

Proof. Apply Lemmas 5 and 8. I

In the interests of brevity most of the details of the proof of the next

h (n) = [ n2/4] + 1.

theorem are omitted.

THEOREM 2. H ( n ) is the only element of X ( n ) wifh [n2/4]+ 1 edges.

Proof. Let G E Z( n) and q ( G) = [ n2/4] + 1. By definition

[n/Zl E ( G ) = U C(G; l ) .

1-1

Suppose n is odd. Then by Lemma 3 and the proof of Lemma 4,

n - 1 n 1<1<-.

2 1C(G; Z)l=-

2 ’

Suppose n is even. Then if l is odd, from Lemmas 3, 6 and the proof of Lemma 8,

n - 2 n 1<1<-.

2 IC(G; I ) / = -

2 ’

If I is even, then by Lemma 3 and the proof of Lemma 8,

Finally, from Lemma 7 and the proof of Lemma 8,

(3)

Using Lemma 2 and Eqs. ( l ) , (2), (3), (4), and (3, the theorem follows after very lengthy but straightforward calculations. I

4. Multiplicity of cycles

a Hamiltonian circuit in G. If u E V(G), then d(u) is the degree of V. In this section we assume that G E %(n) and that C= (UI, UZ, * * , Un) is

42 SHEEHAN

LEMMA 9. Ler G E % ( ~ ) . Suppose d ( u , ) + d ( ~ ~ + ~ ) S n + l , i = 1,2, . . , n - 1. Then i f q(G)>[n2/4]+ 1, there exist distinct integers i , j , k ~ { 1 , 2 ; . . , n } such that d (u i )+d(u i+ l )=n+l , d ( ~ ~ ) + d ( q + ~ ) =

Suppose there do not exist distinct integers i, j , k E {1,2, * * * , n }

n + I , and d ( U k ) + d ( U k + l ) = n + I .

Proof. satisfying the conditions of the lemma. Then if n is even,

If n is odd, there exists at least one vertex with degree S ( n / 2 ) , and therefore

Therefore, from (1) and (2) we obtain contradictions. I In the proof of the next theorem we use an idea due to Bondy [l].

THEOREM 3. Let GE %(n). Then if q(G)>[n2/4]+1, G contains at least two cycles of each length 1, 3 5 1 S n .

Proof. By Theorem 1, G contains 2 cycles of length n. Let 3 S I < n. Suppose d(ui) + ~ ( U ~ + ~ ) S n + 1, i = 1,2, - * , n - 1. Choose distinct integers i, j , k E {1,2, - - , n } such that ui, q, and u k satisfy the conditions of Lemma 9. For the adjacent vertices ui, ui+l on C consider the following pairing of chords of C:

for i + 1 - 1 S s S i - 1 pair (ui7 us) with (ui+l , u,-l+3),

for i + 2 Ss 5 i + 1 - 2 pair (ui, u s ) with ( u ~ + ~ ,

If at most one chord of each pair is an edge of G, then d ( ~ ~ ) + d ( u i + ~ ) S n , which is false. Therefore G contains two chords from one pair. Hence G contains (see Figure 3) a cycle C(ui ; I ) of length I. This cycle consists of part of C and exactly two “internal” chords of C. One of these chords is incident to ui and the other to u ~ + ~ . In exactly the same way G contains cycles C(vj; [).and C(u,; 1) of length 1. Since i, j, and k are distinct, it is clear from their definitions that at least two of the cycles C(ui; I ) , C(ui; I ) , and C(uk; I ) are distinct. Hence G contains at least two cycles of each length I , 3 S I < n.


Figure 3

Finally, if there exists i E {1,2, - - , n - 1) such that d(ui) + d(ui+l) > n + 1, then the result follows immediately by the above argument.

(1) An analogous result to Theorem 1 can easily be obtained when instead of Hamiltonian circuits we consider Hamiltonian paths. In this case h(n ) is replaced by [n2/4]-[n/2]+ 1, and the extremal graph is again unique.

(2) Theorem 3 is the best possible in the sense that “contains at least two cycles of each length. I , 3 S 1 S n” in the statement of Theorem 3 cannot be replaced by “contains at least k cycles of each length 1, 3 5 1 S n” for some k h 3. It is easy to see that by adding one edge to H ( n ) in a suitable way we obtain a graph with [n2/4]+2 edges and having exactly two Hamiltonian circuits.

(3) An interesting conjecture has been suggested to me by Dr. A. J. W. Hilton, viz.,

I

CONJECTURE. Let G E &(n). Then if q(G)>[n2/4]+ 1, G contains at least n - 1 + 2 cycles of each length I , 3 S 15 n. (This conjecture is true when 1 = n by Theorem 3, and by a result of P. Erdos [3] it is true also when I = 3.)

However although this conjecture may well be true, it is for some values of 1 very weak indeed.

The author is indebted to Professor C. St.J. A. Nash-Williams not only for the specific lemmas due to him in this paper but also for several extremely useful conversations during its preparation.

REFERENCES 1. J. A. Bondy, Pancyclic graphs. J. Comb. Theory. (B) 1 1 (1971) 80-84. 2. W. T. Tutte, On Hamiltonian circuits. J. London Math. Soc. 21/22 (1946) 98-101. 3. P. Erdos, On a theorem of Rademacher-Turin. Ill. J . Math. 6 (1952) 122-127.

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Graphs with exactly one hamiltonian circuit