Some localization theorems on hamiltonian circuits

 

 

Some localization theorems on hamiltonian 

circuits 

Armen S Hasratian and N. K Khachatrian

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Hasratian, A. S, Khachatrian, N. K, (1990), Some localization theorems on hamiltonian circuits, Journal of combinatorial theory. Series B (Print), 49(2), 287-294. https://doi.org/10.1016/0095-8956(90)90032-U

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Some Localization Theorems on Hamiltonian Circuits

A. S. HASRATIAN

Department oj Applied Mathematics, University af Yerevan, Yerevan, 375049, USSR

AND

N. K. KHACHATRIAN

Computing Center, Academy af Sciences af the Armenian SSR, Yerevan, 375014, USSR

Communicated by the Editors Received April 13, 1987

Theorems on the localization of the conditions of G. A. Dirac (Proc. London Math. Soc. (3) 2, 1952, 69-81), 0. Ore (Amer. Math. Monthly 67, 1960, 55), and Geng-hua Fan (J. Combin. Theory Ser. B 37, 1984, 221-227) for a graph to be hamiltonian are obtained. It is proved, in particular, that a connected graph _{G on} p;;;, 3 vertices is hamiltonian if d(u);;;, [M3_{(u)l/2 for each vertex} u _in G, _where M3(u) is the set of vertices v in G that are a distance at most three from u. i'.; 1990 Academic

Press. Inc.

1. INTRODUCTION

Our notation and terminology follows Harary [ 4]. Let k be a positive integer. For each vertex u of a graph G = ( V, X) we will denote by Mk_(u) and N(u) the sets of all vE V with d(u, v):::;;k and d(u, v)= 1, respectively. The subgraph of G induced by Mk_{(u) is denoted by G}

k(u). The degree in Gk(u) ofa vertex vEMk(u) is denoted by dGkCui(v).

The closure C( G) of G is the graph obtained from G by recursively joining pairs of nonadjacent vertices whose degree-sum is at least

I VI,

until

no such pair remains.

The following results are known. A graph G

= (

V, X) on p � 3 vertices is hamiltonian if:

d(v)�p/2 for each v EV (Dirac [2]). uv rt X� d( u) + d( v) � p ( Ore [ 6] ). d( u) = k < ( p - 1 )/2 =:>

I {

v E V/ d( v) ::::; k}

I

< k

287

(Posa [7]). ( 1.1) ( 1.2) (1.3)

288 HASRATIAN AND KHACHATRIAN and

d(u)

=

(p - 1 )/2:::;. I { v E V/d( v) ~ (p 1 )/2} I ~ (p - 1 )/2.

C( G) is a complete graph (Bondy and Chvatal [ 1] ). ( 1.4) G is 2-connected and d(v) < p/2, d(u, v)

=

2:::;. d(u) ~ p/2

(Geng-hua Fan [3]). (1.5)

In [ 5] the following theorem on a localization of condition ( 1.3) is

proved:

THEOREM. A connected graph G on p ~ 3 vertices is hamiltonian if d(u)

=

k < (p 1 )/2::;. I { v E M2(u)/d(v) ~ k} I

<

k

and

d(u) (p 1)/2:::;.l{vEM2(u)/d(v)~(p l)/2}1~(p-1)/2.

In this paper we obtain the theorems on localizations of conditions (l.1 ), ( 1.2 ), and ( 1.5 ).

2. RESULTS

LEMMA. Let G be a graph with d(u,v)=2, weN(u)nN(v), and d(u)

+

d(v) ~ IN(u) u N(v) u N(w)I. Then IN(w)\(N(u) u N(v))I ~

IN(u)nN(v)I. Proof

IN(w)\(N(u) u N(v))I

= IN(w)I - IN(w) n (N(u) u N(v))I

=

IN(w)I - (IN(w)I

+

IN(u) u N(v)I IN(u) i...LN(v) u N(w)I)

=

IN(u) u N(v) u N(w)I - (IN(u)I + IN(v)I - IN(u) n N(v)I)

IN(u) n N(v )I (d(u)

+

d(v) IN(u) u N(v) u N(w)I)

~ IN(u) n N(v)I.

THEOREM 1. Let G

= (

V, X) be a connected graph with at least three

vertices. If

d(u)

+

d(v) ~ IN(u) u N(v) u N(w)I

for each triple of vertices u, v, w with d( u, v)

=

2 and w E N( u) n N( v ), then G is hamiltonian.

SOME LOCALIZATION THEOREMS

289

Proof Let G satisfy the hypothesis of Theorem 1. Clearly, G contains a circuit; let C be the largest one. If G has no hamiltonian circuit, then there is a vertex u outside of C that is adjacent to at least one vertex in C.

Let { w 1 , ... , w n} be the set of vertices in C that are adjacent to u, and for

each i = I, ... , n let v; be the successor of w; in a fixed cyclic ordering of C.

Note, that if 1 '!(_i<j'!(_n then v;vjrf=X. Otherwise delete the edges W;V;,

w₁v₁ from C and add the edges v;vJ, w;u, uw_1. In this way we obtain a circuit longer than C, which is a contradiction.

For each i=l, ... ,n let F; and T; denote the sets N(u)nN(v;) and N(w;)\(N(u)uN(v;)), respectively. Since C is the longest circuit, uv;rf;X, i = l, ... , n. Then d(u, v;) = 2 and from the Lemma we have I T;I '!(_ IF;I for each i = 1, ... , n.

We shall show that there is a vertex u' such that u' rf= C and u' E F; for some i, l '!(_ i '!(_ n.

Consider the following iterated algorithm.

Stepl. k:=l, m:=1, and Zf:={u,v;}, YJ:={w;} for each

i= 1, ... , n.

Step 2. If the set

F

m \

Y!

contains a vertex u'

ff

C,

then stop. Otherwise, choose an arbitrary vertex w in F m \

Y!.

Clearly, w

=

w, for some r, 1 '!(_ r '!(_ n.

Set

Yk+ 1 ·= yk U {w } . zk+ 1 ·= zk.

m · m r ' m · m'

for i =I= m, r and 1 '!(_ i '!(_ n.

Step 3. k :=k+ l, m :=rand go to Step 2.

It is not difficult to see that before the kth iteration of the algorithm we have

(a) Zjc;;T;, Yjc;;F;, IZjl?:IYjl for each i, l'!(_i'!(_n.

(b) F m \

Y!

=I=

0,

because IT ml'!(_

IF

ml and IZ!I > I Y!I. (c) Yjc;; {w1 , ... , wn} for each i, l '!(_i'!(_n.

(d) IY!l=IY!-11+1ifk?:2.

From (b), (c), (d) it follows that if k?:2 then

L7-i

IY;-1 1

<

L7-

1 I Yjl '!(_ n

2

. Hence there exists k such that 1 '!(_ k '!(_ n2 and the set F m \ Y!

contains a vertex u'

ff

C. Delete the edge wmvm from C and add the edges wmu, uu', u'vm. In this way we obtain a circuit longer than C, which is a contradiction. The proof is complete.

290 HASRA TIAN AND KHA CHA TRIAN

Note that for every t ~ 5 there exists a graph G,

= (

V" X,) with

V, = {v1 , v2 , ••• , v2, } and t-2

X,=

U

{v;v/2k+ 1 ~i<j~2k+4}, k-0

which fulfills the condition in Theorem 1 and does not fulfill conditions (l.1}-(1.5). Clearly, C(G,)=G" IV,\=2t, and IX1\=5t-4=(5/2)·

IV,1-4.

COROLLARY 1. Let G be a connected graph on p ~ 3 vertices.

If

d(u)

+

dGz(uJ(v) ~ IM2

(u)I for each pair of vertices u, v with d(u, v) = 2, then G is hamiltonian.

Proof Clearly,

d(u) + d(v)- IN(v)\M2(u)I

=

d(u) + da2(u)(v) ~ IM

2

(u)I.

Then

d(u) + d(v) ~ \M2(u)I + IN(v)\M2_{(u)I ~}

\N(u) u N(v) u N(w)I

for each vertex w EN( u) n N( v ). Hence, Corollary 1 follows from Theorem 1. COROLLARY 2. Let G be a connected graph on p ~ 3 vertices. If d(u)

+

d(v) ~ \M3_{(u)I for each pair of vertices u, v with d(u,}

v) = 2, then G is hamiltonian.

Corollary 2 follows from Theorem 1 because \M3_{(u)I ~ IN(u)}

u N( v) u N( w) I for each vertex w E N( u) n N( v ).

COROLLARY 3. Let G be a connected graph on p ~ 3 vertices.

If

d(u) ~ \M3(u)l/2for every vertex u in G then G is hamiltonian.

Proof Let Gi'Kp, d(u,v)=2, and d(u)~d(v). Since d(u)~\M3_(u)\/2,

then d(u)+d(v)~IM3_{(u)l;;:31N(u)uN(v)uN(w)I}

for each vertex WE

N( u) n N( v ). Therefore Corollary 3 follows from Theorem l. COROLLARY 4. Let G be a connected graph on p ~ 3 vertices.

If

or

d(u)+d(v)~

\M

2

(w)I

for each triple of vertices u, v, w with d(u, v)

=

2 and w E N(u) n N(v ), then G is hamiltonian.

SOME LOCALIZATION THEOREMS 291 Proof Let d( u, v) = 2 and w E N( u) n N( v ).

If d(u)

+

d(v) ~ IM2

(w)I, then d(u)

+

d(v) ~ IN(u) u N(v) u N(w)I because IM2(w)I ~ IN(u) u N(v) u N(w)I.

Suppose that dGi(w)(u) + dGi<w>(v) ~ IM 1(w)I. Clearly, dG,(wl(u) = d(u)-IN(u)\M1(w)I and dGi<w>(v) = d(v)- IN(v)\M1

(w)I. Hence d(u)

+

d(v) ~ IM 1(w)I

+

IN(u)\M1

(w)I

+

IN(v)\M1(w)I

~ IN(u)uN(v)uN(w)I and Corollary 4 follows from Theorem 1.

COROLLARY 5. Let G be a connected graph on p ~ 3 vertices. If for each vertex u in G at least one of the graphs G₁(u) or Gi(u) satisfies Ore's condition, then G is hamiltonian.

Corollary 5 follows from Corollary 4.

THEOREM 2. Let G

= (

V, X) be a 2-connected graph on p ~ 3 vertices and let v and u be distinct vertices of G. If

d(u)

<

p/2, d(u, v)

=

2 ==> d(v) ~ IM3(u)l/2, (2.1) then G is hamiltonian.

Proof Let A = { P 1

, ••• , ph} be the set of all longest paths in G. For each

i = 1, ... , h let pi= v~v; · · · v~ and /(Pi) be the smallest r from {O, 1, ... , m - 1} such that v~ v~ EX. We denote by A 1 the set of all pi EA

with d(v~)=max1,;;J.;;hd(vi).

Suppose that G is a graph satisfying the condition of Theorem 2 and that

G has no hamiltonian circuit. We shall arrive at a contradiction.

Let P

=

v0v, · · · vm be some longest path in G of length m, chosen so that

f(P)

= minP,eA,

f(P1

). Clearly, d(v_{0 )} ~ d(vm). If d(v_{0 )}

+

d(vm) ~ p then there are at least two consecutive vertices on P, vi, and v1+ i , such that V;Vm EX and v,. + 1 v0 EX, and so we obtain a circuit oflength m + 1. By the

connected-ness of G, we have either a hamiltonian circuit or a path of length m

+

1. Each leads to contradictions. Consequently d(v0 )

+

d(vm) < p. Since

d(vo) ~ d(vm), d(vm) <p/2. I

From the proof above we can also suppose that (a) G has no circuit of length m + 1.

Since G is 2-connected, d(vm)~2. Let N(vm)= {v_{11 , ... ,} v₁,} and

J,

< · · · <},. Clearly, }1 ~ 1, otherwise G has a circuit of length m

+

1, which is contrary to (a). We show now that

292 HASRATIAN AND KHACHATRIAN

(b) if vm V; EX and j 1

<

i

<

m - 1 then N( v1 +;) <;; { vh, ... , vm},

( c) v m v; rf X for some i, j ₁ < i < m,

( d) vJ, _ 1 vJ; + 1 rt X for every i, 1

<

i

<

t.

Proof (b) Clearly, we have N(v1+;)c;; {v0 , v1 , •.• , vm} otherwise G has a

path of length m

+ 1.

Suppose that there is s such that 1

<

s < j I and v s v 1 + ; E X. Then

is the longest path in G with f(P') <f(P). This contradicts the choice of P.

Therefore N(v1+;)S {vJ;• Vi+J,, ... , vm}.

(c) If vmv;EX for every i, }1<i<m, then

U;"=1

/N(v1+;)S

{v1i, vi+J,• ... , vm}· This contradicts the 2-connectedness of G.

( d) It is obvious that ( d) follows from (b ).

From (c) it follows that there is a k such that }1 <k<m-1, vmvkrf;X,

and vm V; EX for every i, }₁

<

i

<

k-1. Thus we have

( e) there is no i such that j 1 < i

<

m - 1, vh _ 1 v; E X, and v k v 1 + ; E X.

Indeed, if V;V1i-i EX and vkv1 +;EX then from (d) it follows that i > k.

Then G has the longest path P',

with f(P') <f(P). This contradicts the choice of P.

Clearly, d(vbvm)=2 and d(v1i_1,vm)=2. Since d(vm)<p/2, it follows

from (2.1) that d(vJ,-d~ IM3(vm)l/2 and d(vd~ IM3(vm)l/2.

Since vkvh_1rtXand the degree-sum of vertices vk and v1i_1 in G3(vm) is at least IM3(vm)I, d(vk> V1i_ 1)=2.

From (d) it follows that d(v1i_1)< IM3(vm)l-d(vm). Since d(vJ,-i)~

IM3

(vm)l/2, then d(vm) < IM3(vm)l/2. Therefore d(vh-i) > d(vm) and

d(vk) > d(vm).

Case 1. d(vd<p/2. Since d(vk>vm)=d(vk>vh_1)=2, it follows from

(2.1) that d(vm)~IM3

(vk)//2 and d(v1i-d~IM3

(vdl/2. Together with

d(vk) > d(vm) this implies that

(2.2) From ( d) it follows that v; vh _ 1 rt X for each i, 1

+

j 1

<

i < k. From ( e) it

follows that V;V1i _, rt X for every i such that i > }1 and vkv;+ 1 EX. Besides, vmvJi-I rt X and vm, vh-I E M3(vk). Thus d(vh- i)

<

IM3(vdl - d(vk)- 1.

This contradicts (2.2).

SOME LOCALIZATION THEOREMS 293

d(v1i-1 , vd

=

2 imply that d(vm)"~ IM 3

(vJi-l )1/2 and d(vd ~ IM3(v1i _ ill/2.

Since d( vii_ i) > d( vm ), we have

• 3

d(vh i)+d(vk)~ IM (v11 1)1. (2.3) If d(v1i i)~p/2 then d(v₁₁_i)+d(vk)~p~IM3(vil

di,

so (2.3) holds again.

From (b) it follows that vk v₁₁ 1

1

X and vk is not adjacent to every

vertex veN(vil_i)\{v_{11 ,} v1+1,, ... , vm}.

From (e) it follows that vkv1+d:,X for every i such that viEN(v11 i)n {v1+11,v2+Ji, ... ,vm}. Besides, we have v11 i,vkeM3(v11_i). Therefore d(vk)~IM3_{(vh_i)I}

d(vh_i) 1.

This contradicts (2.3 ). The proof is complete.

Note that for every r ~ 2 there exists a graph G,

= (

V,, X,) with

V,=

{w

1 ,

w

2 } u {v1 , ... , v3

,_i}

u {u1 , ... , u3, _ 1 } and X,= {w1v;, w2v;/i=

1, ... , 2r} u { v;v_1, u;ujl ~ i < j ~ 3r - 1} u { v;ujl

+

2r ~ i, j ~ 3r - 1} that satisfies the condition of Theorem 2 and does not satisfy the condition

( 1.5 ).

Besides, for every n ~ 5 there exists a graph G n

= (

Vn, Xn) with

Vn {v1 , ... ,vn} and Xn={v;vjl~i<j~n 2}u{vn 1V1,Vn-1V2}U

{ v,, v Ji= 2, 3, ... , n - 2} that satisfies the condition of Theorem 2 and does not satisfy the condition of Theorem 1.

Let G

= (

V, X). It is shown in [ 1 ] ( by paraphrasing Ore's proof [ 6

J )

that if G

+

uv is hamiltonian and d(u)

+

d(v) ~

IVI

then G itself is hamiltonian.

THEOREM 3. If G

+

UV is hamiltonian, d( u, v)

=

2, and

(2.4)

then G itself is hamiltonian.

Proof Suppose G

+

uv is hamiltonian but G is not. Then G has a hamiltonian path u1 , u2 , ... , uP with u1

=

v and uP

=

u. Let N(u)

=

{u;

1, ...

,u;,}.

If vu1+₉

1X

for every j, 1~j~t, then dG2(uJ(v)<IM

2

(u)l-d(u ). This contradicts (2.4 ). Hence there is m such that 1 ~ m ~ t - 1,

VU1+;mEX, and UU;mEX.

But then G has the hamiltonian circuit

This contradicts the hypothesis.

COROLLARY 6. If G

+

UV is hamiltonian, d( u, v)

=

2, and d( u)

+

d( v) ~

IM3(v)I, then G itself is hamiltonian.

294 HASRA TIAN AND KHA CHA TRIAN REFERENCES

1. J. A. BONDY AND V. CHVATAL, A method in graph theory, Discrete Math. 15 (1976), 111-135.

2. G. A. DIRAC, Some theorems on abstract graphs, Proc. London Math. Soc. ( 3) 2 ( 1952 ), 69-81.

3. GENG-HUA FAN, New sufficient conditions for cycles in graphs, J. Combin. Theory Ser. B

37 (1984), 221-227.

4. F. HARARY, "Graph Theory," Addison-Wesley, Reading, MA, 1969.

5. A. s. HASRATIAN AND N. K. KHACHATRIAN, A two theorems on a hamiltonian graphs (Russian), Math. Zametki 35, No. 1 (1984), 55-61.

6. 0. ORE, Note on Hamiltonian circuits, Amer. Math. Monthly 61 (1960), 55.

7. L. POSA, A theorem concerning Hamilton lines, Magyar Tud. Acad. Mat. Kutato /nt. Kazi.