EDGE SUBDIVISIONS OF A GIVEN GEODETIC BLOCK PRESERVING ITS PROPERTY TO BE GEODETIC

1999,19 (1):86-90

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It'lfmJl,m EDGE SUBDIVISIONS OF A GIVEN GEODETIC BLOCK PRESERVING ITS PROPERTY TO BE GEODETIC 1 )

Li Deying ( -t~~ Mao Jingzhong ( ~~ 'f ) Department of Mathematics, Central China Normal University, Wuhan 490079, China Abstract

In this paper authors have obtained a new necessary and sufficient condition

for a graph G to be a geodetic graph. Let 9 : E

~

Z+ be a function from the set of edges

of a graph G to the set of nonnegative integers and let g( G) denote the graph obtained

from G by replacing each edge e E E by a suspended arc (path) P g ( e)+2 of length g( e) + 1. and by using this condition, established a criterion for a function 9 to be a function which can generate a new geodetic block g( G) from a given geodetic block G.

Key words

Geodetic block, subdivision.

1991 MR Subject Classification

1

05C

Introduction In this paper we shall consider only undirected 2-connected simple graphs, i.e, graphs that

are loopless, finite, undirected and without multiple edges. A graph G is said to be geodetic if any pair of points of G are joined by a unique path of shortest length, i.e, a unique distance path[l]. A 2-connected geodetic graph is called a geodetic block. A graph is geodetic iff each of its blocks is geodetic(see Stemple and Watkins[2]). Obviously, odd cycle, tree, complete graph are geodetic graph, we call them the trivial geodetic graph. Now we only consider the nontrivial geodetic graph and 2-connected, i.e, nontrivial geodetic block. An arc(path) P[a, b] connecting a and b of G is called a suspended arc(path) if deg(a) > 2 and deg(b) > 2 and all other points of P[a, b] are degree of 2 in G. Let g : E

-+

Z+ be a function from the set of edges of a graph G to the set of nonnegative

integers and let g(G) denote the graph obtained from G by replacing each edge e E E by a

suspended arc (path) Pg ( e )+2 of length g(e) + 1. Denote by 9 the set of all such fuctions. Then the set 1t(G) of graphs homeomorphic to G can be written as {g(G)lg E Q} Now, let G be a given geodetic block. Discussing possible ways of constructing a new geodetic graph homeomorphic to a given geodetic graph, Parthasarthy and Srinivasan proposed the problem of characterizing the subset of functions g in 9 such that g(G) is also a geodetic block. As far as we know, this problem is still open; however, the case that G is complete graph has been settled[3]. Note that if we want to construct a series of geodetic graphs homeomorphic 1 Received

Mar.3,1997. TIllS is Supported by National Natural Science Foundation of China

No.1

Li & Mao: EDGE SUBDIVISIONS OF A GIVEN GEODETIC BLOCK PRESERVING

87

to a given geodetic graph G, then a complete solution of the problem would enable us to restrict our attention to the case of G having minimum degree b( G) 2: 3 . In this paper we establish a criterion for a graph to be geodetic, and then obtain a partial solution of the problem,

2

A Characterization of Geodetic Graphs

Definition 1 Let Co be an odd cycle with vertices VI, V2, ... , V2k, v2k+l in the order they occur as we proceed around Co. A vertex "i is said to be opposite to Vi if the shortest path from Vi to "i around Co has length k. Similarly, for an even cycle C; of length 2k, the vertex opposite to Vi is defined to be the one located at distance k from Vi (around C e). For example, in the cycle Co the vertices opposite to VI are Vk+l and Vk+2. In the cycle Ce , Vi and vk+i(Inodulo 2k) are opposite ifi:j:. k; Vk is opposite to V2k. Theorem 1 Let G = (V, E) be a geodetic block. Denote by SP = XIX2··· Xk the shortest path connecting two given vertices u and V of G ( Xl = V, Xk = 1/'), and by D a shortest cycle through u. Then {Xio' Xi o+1,···, Xk} ~ V(D), where i o = min{ilxi E V(D)}. Proof Since i o = min{ilxi E V(D)}, it follows that the shortest path connecting u and Xi o is P1 = XioXio+l··· Xk-1U. If the result don't hold, i.e. Pl Cf:. D, then we must have a path Q which is the shortest one from Xi o to u around the cycle D. There are two cases: either (1) IP11= IQI which contradicts the condition that G is a geodetic graph; or (2) IP11 < IQI which implies that P 1 u Q contains another cycle, ,D 1 , containing the vertex u and its length is shorter than D, which contradicts that D is the shortest cycle containing u. Therefore we have P 1 ~ D. Theorem 2 Let G = (V, E) is a geodetic block. Let SP = X1X2 ... Xk denote the shortest path connecting two given vertices u and v, of G ( Xl = V, Xk = u). Let C be the shortest cycle connecting u and v. Fix on some orientation. Suppose that {Xl, X2, ... , Xk} Cf:. V(C). Then there exist integers A and J-L, 1 < A < J-L < k, such that {X1,X2,···,X.x} ~ uCv,{X Il,XJl+1,···,Xk} ~ vCu and {x.x, X.x+1,··:, XIl-l, x ll } n V(C) = 0. Proof Let A = maxfil{x1' X2,···,Xi} ~ V(C)}, It = min{jl{xj,xj+1,···xk} ~ V(C)}. We prove the result in three steps. (1) Firstly we prove that A > .1 and J-L < k . Suppose for a contradiction that A = 1. Then Xz tt: V(C). Therefore we can find a cycle C1 contained in SpuC such that IC11 < ICI and C1 contains the vertices u and v. This, however, contradicts the choice of C. The inequality J-L < k is established similarly. (2) Secondly we prove that {x.x, X.x+1,···, XIJ.-1, xJl} n V(C) = 0. Suppose for a con-

tradiction that {x.x+l,···,XJL-l} n V(C) # 0. Say, Xj E {x.x+l,···,XJL-l} n V(C) ( where j = llnn{ili > A, Xj E V(C)}). Obviously, j # A + 1 ( otherwise it would contradict the choice of A) and j :j:. J-L - 1 (otherwise it would contradict the choice of J-L). Suppose we have

{x.x,Xj} C uCv (or vCu) in SOIne orientation of C. We replace the arc of C from X.x to Xj by the part of SP from X.x to Xj. We thus get a cycle containing 1t and v and its length is shorter than C. This contradicts the choice of C. ( The length of X.x CXj is different from the length of x.xSPXj, for otherwise we would ·C0111e to a contradiction with the condition that G is a geodetic graph). The case {Xj,xJl} ~ uCv (or vCu) is settled similarly, (3) Finally, we prove that if X.x E uCv, then x ll E vCu (Note that, reversing the orientation of G, we get X.x E vCu, x ll E uCv). In fact, if X.x and xJl both belong to the arc uCv (or both

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vCU ),

Vol. 1 9

then we can get the same contradiction as in (2).

A pair u and v of vertices of G are said to be balanced if there is cycle C of G ~ satisfy the properties: (1) C is a shortest cycle connecting u and v, (2) C is even, (3') 'U and v are opposite on C. Then C is said to be a balanced cycle by u and v.

=

Theorem 3 Let G (V, E) be a 2-connected graph. Assume that for any pair balanced vertices u and v of G, and any balanced cycle C= VX1X2 ... Xm-1 UY1Y2 ... Ym-1'V, there is path x>tPYI-t (1 S; A, J.L S; m - 1) satisfing V(P) n V(C) = 0 and its length is less than IA + J.L - 'lnl. Then G is a geodetic block. Proof

Under the assumptions of the theorem we prove that for any two vertices

u, 'v

EV

there is exactly one shortest path connecting 'U and v. Suppose for a contradiction that there are two vertices u and v with dCu, v) = k such that there, are at least two shortest paths connecting u and v. Take any two of them and denote them by P 1 = VX1X2'" Xk-1 U and P2

= UYlY2 · · · Yk-l v.

We shall first consider two cases.

Case 1 {Xl, X2, ... ,Xk-l}n {Yl' Y2, · · . ,Yk-l} = 0. Then PI UP2 form an even cycle with length 2k. According to the assumption, there is a path X>t QYI-t such that the length of X>t QY/-L

is less than IA + J.L - kl, therefore either the length of VXl ... X>t QYI-t ... Y1'n is A+ J.L+ IX>t QYJL I < k, or the length of VYk-1 ... Yl-tQx>t ... Xk-l U is (k - J.L) + (k - A) + IX>t QYI-t I < k. This contradicts the assumption that d( u, v) = k.

=

=

Case 2 {Xl' X2,"', Xk-l} n {Yl' Y2,"', Yk-l} {x>t} {YJi'}' i.e., the paths PI and P2 meet at exactly one vertex besides u and v. Then there are two possibilities. (1) The vertices v, Xl, X2, .• · ,X>t(= YJL)' YI-t+1, ... ,Yk-1,and v form a cycle, C1 ,with length k; and the vertices u, Y1,Y2,"', YI-t(= x>t), X>t+1,"', Xk-1, U form another cycle, C 2, with length k. If k is odd , then there is a path from v to YJl around C1 with length S; l ~ J, and there is

another path from YI-t to u around C 2 with length S; l ~ J. Therefore there is a path from v to u with length at most k - 1, which contradicts d( u, v) = k. If k is even, observe first that C1 is balanced by v and X>t (=YI-t), and C2 by X>t and v. Then apply the argument of Case 1 to both C l and C 2 • Finally obtain a desired shorter (u, v)-path by using appropriate arcs of C1 and C2 , and some two paths (one of C 1 and the other of C 2 ) . we can still get a contradiction. (2) The vertices v, Xl, X2, ... , X>t (= YJi')' YJi.+1, ... ,Yk-1, V form a cycle, C 1, with length /1, and the vertices u, Y1,Y2, .. " YJL( = x>t), X>t+1," " Xk-1, U form another cycle, C 2 , with length 12 • Since II + 12 = 2k, 11 and 12 must have the same parity. If /1 and 12 is even, we can derive the contradiction as in (1). If /1 and 12 is odd, then take the shorter (v, x.x)-path among

-c»,

and x)..Clv, and the shorter (x).., u)-path among x>..C2u and uC2x).., we can find a shorter (v u )-path. a contradiction.

The same method can be used in the case I{X1, X2,"', Xk-1} n {Y1' Y2,"', Yk-1}1 2:: 2. In fact, the path P1 = VX1X2" 'Xk-1U and P 2 = VYk-1Yk-2' "Y1U meet at w frist time, therefore, vP1wP2v is a cycle, by C. If C is odd, then take the shorter (v w)-path among »c-» and wCv, therefore we can find a shorter (v u)-path, a contradiction. If C is even, we observe first that

C is balanced by v and w, then apply the argument of Case 1 to C, we can obtain a shorter (v u )-path, we can still get a contradiction.

Therefore there is only one shortest path connecting

'U

and v, so that G is geodetic graph.

From Theorems 2 and 3 it follows that the condition of Theorem 3 is necessary and sufficient for G to be a geodetic graph.

Li & Mao: EDGE SUBDIVISIONS OF A GIVEN GEODETIC BLOCK PRESERVING

No.1

Corollary

Let G

89

= (V, E) be a 2-connected graph, and let g : E ~ N be a lllapping(N

is a set of positive integer). If the graph G* obtained from G by subdividing each edge e E E by the g(e) new point, satisfies the condition of Theorem 3, then G* is a geodetic block.

3

Application to the Petersen Graph In this section, let G be the familiar Petersen graph, see Figure 1. Denote by E 1 the set of

the edges of the inner cycle: cycle:

'V1'UZ,'UZ'V3~'U3'V4,'V4'U5~'U5'V1~

'U1 'Uz, 'UZ'l·L3, 'U3'U4, 'U4'lL5, 'U5'U1;

and by

E 3

by

E 2

the set of the edges of the outer

the set of the edges joining the inner cycle with

the outer cycle: 'U1 'U1, 'U2V3, 'U3'V5, 'l.L4'U2, 'U5'U4. Consider a mapping g from the set E = E 1 UEzUE3 to N, the set of positive integers, such that the restriction of g to any of EiCi = 1,2,3) is a

constant function.

Uz

Us

U3

Figure 1 Suppose g(e) = x if e EEl, g(e) = y if e E E z, g(e) = z if e E E 3 • Now we establish sufficient conditions for g ensuring that g( G) is a geodetic graph. We shall call the vertices 'lli and 'ui(i = 1"",5) basic vertices, and the members of E basic edges. First we consider any shortest odd cycles of G. They are of four kinds. (1) the one consisting of the edges of the outer cycle. Its length after subdivision by adding

g(e) vertices in each edge is 5x

+ 5.

(2) the one consisting of the edges of the inner cycle of G. Its length after subdivision is

5y+ 5. (3) those containing only one edge of the inner cycle.

For example,

'Ul'Ul'U5'U3'll,Z'Ul;

its

(4) those containing only one edge of the outer cycle. For example,

'U1 '01 'UZ'U3'll,Z'll,1;

its

length after subdivision is x

+ 2y + 2z + 5.

length after subdivision is 2x

+ y + 2z + 5.

From the necessery and sufficient condition for g( G) to be a geodetic graph obtained in the preceding section we know that any the shortest odd cycle of G must be still an odd cycle after subdivision by adding g(e) vertices in each edge. This implies that x

== y == O(11l0d 2).

Now we consider any shortest even cycles of G. They are of two kinds. (1) those consisting of one edge of

C1

= 'Ul 'U1 'Uz'U4 'U3 'Uz 'Ul ;

E1~

(2) those consisting of three edges of

Cz

=

three edges of

E z,

and two edges of

E3~

for example,

one edge of

E z,

and two edges of

E 3;

for example,

E 1,

'01 'UZ'U3'UZ u3'U5'U1'

By symmetery, we can only consider one even cycle, say C1 . Consider the vertex p which

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ACTA MATHEMATICA SCIENTIA

Vol.19

is the new vertex after subdivising the basic edge VI V2 such that the distance between VI and p is ~, and the vertex q which is the new vertex resulting from subdividing the basic edge 'U2U3 such that p and q are opposite vertices on the cycle GI . Because their are only three suspended arcs of the cycle GI , its basic vertices are VI, V5, U3; V2, V3, U2 and 't/'l, 1/'5, u4, respectively. If q is on the basic edge UI U2 (or U3U4) and the shortest cycle containing q and p is not an odd cycle UIVIV2V3U2UI, then, obviously, g(G) cannot be a geodetic graph. Therefore q must be on the basic edge U2U3. There are three suspended arcs, VIV5U3, V2V3U2 and UI U5U4, connecting the two arcs of GI divided by p and q. Notice that IGII x + 3y + 2z + 6 . Obviously, the shortest path connecting p and q cannot be PVIUIUSU4U3q (because the length of the path from UI passing through U2 to q is less than the length of the path from UI passing through Us, 'tL4, 'U3 to q ). Therefore we only need to investigate the arc from p passing .through VI, the arc contained in the odd cycle VIVSU3U2UI, the arc fromp passing through V2, and the arc contained in the odd cycle V2V3U2U3U4. From the assumption d(VI'P) = ~ < d(V2'P) it follows that

=

1

z

2" + 2" [x + 2y + 2z + 5 -

1 1] < 2" [x + 3y + 2z

+ 6].

Therefore ~ < ~ + 1, so that z ::s y. Since the inner cycle and the outer cycle in the Petersen graph are symmetric, we can obtain y ::s x by a similar argument, Thus z = y. Since the shortest cycle containig Ul and U4 is necessarily an odd cycle, it follows that the length of the odd cycle containing Ul and U4 is less than the length of the even cycle

+ 2y + 2z + 5 < x + 3y + 2z + 6, this 5y + 5 < x + 3y + 2z + 6, which irnplies 2y < x + 2z + 1. Since x = y we have x = y < 2z + 1.

containing

'Ul

and

U4.

Thus we have: z

obviously holds,

Combining all these discussions, we obtain that the mapping g is as follows: O(mod 2) if e E E I U E 2 , g(e) z if e E E 3 , where x and z satisfy the x < 2z + 1, It is easy to see that the graphs obtained by virtue of mapping of this kind the graphs described in [4]. For example, denoting a geodetic block with diameter

=

=

g( e)

=x

==

I.

i.e., z ~ are exactly d and girth

= O,z 1, then we get a (3,5)-graph; if z = O,z = 2, then we get a (4,5)-graph; if x = 0, Z = 3, then we get a (5,5)-graph; if x = 2, Z = 1, then we get a (7,13)-graph; if x = 2, Z = 2, then we get a (8,15)-graph; if z = 2, z = 3, then we get a (9,15)-graph; if x = 4, Z = 2, then we get a (12,21)-graph; if x = 4, Z = 3, then we get a g by a (d,g)-graph, we have: if x

=

=

(13,23)-graph; if x 6,z 3, then we get a (17,29)-graph. However, it is a still an open problem to construct a (6, 11)-graph. Acknowledgement

The authors thank Serge Lawrencenko for his help when preparing

this paper. References 1 Ore O. Theory of graphs. American Mathematics Society Providence,R 1,1962 2 Stemple J G, Watkins M E. On planar geodetic graphs. J Combin Theory, 1968,B4: 101-117 3 Parthasarathy K.R, Srinivasan N. Some General Constructions of Geodetic Blocks. Journal of Combinetorial Theory, 1982,B33: 121-136 4 Mao Jingzhong. On the Construction of Geodetic Block with 9 ~ (2d + 1). Kexue Tongbao, 1988,33: 227-230