Transversals of circuits in the lexicographic product of directed graphs

M ATHÉMATIQUES ET SCIENCES HUMAINES

C ^ARSTEN T ^HOMASSEN

Transversals of circuits in the lexicographic product of directed graphs

Mathématiques et sciences humaines, tome 51 (1975), p. 43-45

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43

TRANSVERSALS OF CIRCUITS IN THE LEXICOGRAPHIC PRODUCT OF DIRECTED GRAPHS

Carsten THOMASSEN

A directed

graph

^{G consists of a set}

^V(G)

^{of vertices and a set}

^E(G)

çonsisting

of ordered

pairs (x,y) (called edges)

^of distinct vertices _x,y.

The directed circuit of

length

ⁿ

^{(n ~} 2)

^{is the directed}

graph consisting

^of

the vertices

x2,’

^...,

xn

^{and the}

^edges (x., 1 x. 1+ 1),

^{1 ~ i ~}

^n-1,

^and

(xn x1).

^Let

Sq

^{dénote the}

^graph

^{with q vertices and no}

^edges.

The

lexicographie product

^{G 0 H of the directed}

^{graphs G,}

^{H can be defined}

as follows :

For each vertex x of

G,

^{let H}

x

be a copy of H such that _x,y E

V(G), ^{x 0}

y

implies Hx

ⁿ

Hy ^{= 0.}

^{Then add to the}

^graph

_x ^U

^H

^ail

^edges

^such

that

x1

^E

Hxo, ^Yi

^E

Hy

^and

^{(x,y)~ E(G).}

x E

V(G)

G

being

^{a directed}

graph,

^{we define}

T (G)

^{= min}

{ml

^{3A g}

^{E (G) :} 1 AI =m,

^G-A

contains no directed

circuit}.

We also define

Tk (G)

^{= min}

^{ml

^{3A ç}

^{E(G) : lAI =m,}

^{G-A contains no directed}

circuit of

length

^{less than}

^k}.

J.C.Bermond showed that

T (G 0 11) IV (G) 1 ^{T (H)} +IV(H)12 ^T(G)

^and

conjectured [2, Conjecture ^2]

that

equality

^holds.

^(The

^weaker

conjecture

^for tournaments was made in

[11).

We shall hère prove this

conjecture.

THEOREM 1

Let G and H be directed

graphs

^{with p and q}

^vertices, respectively.

^Then

T(G 0 H)

⁼

PT(H)

⁺

q2T(G).

Clearly T(G)

⁼

Ti.(G)

^{when k}

&#x3E;IV(G)I,

^{so Theorem 1 follows} from Theorem 2 below.

* Dept of Combinatorics and Optimization, Faculty ^of Mathematics, University ^of Waterloo, ^Canada

44

THEOREM 2

Let G and H be directed

graphs

^{with p and q}

vertices, respectively.

^Then

for every

^{k ~}

2, Tk (G 0 ^H)

⁼

pT(H)

⁺

Proof : Let A ç

E(G),

^B

^{9 E (H)}

such that

1 AI= T k (G), ^IBI= TH&#x3E;

^and

G -

A, respectively

^{H -}

B,

^{contains no circuit of}

length

less than k. From each of the

graphs

.

H(x

^E

^V(G))

^we

delete

^the

^edges

of B and for each

edge

of A we delete the

corresponding q ^edges

^{of G 0 H. We} have then deleted

pTk(H)

⁺

qT k (G) ^edges

^{from G 0 H and the}

^resulting

^directed

^graph

^contains

no directed circuit of

length

less than k.

This proves the

inequality Lk(G

⁰

^H)

^{The reverse}

inequality

^{follows from} Theorem 3 below.

If 5

is a

family

^{of directed}

graphs

^{and G is a directed}

graph

^{then an}

5-tranversal

^{of G is a subset A of}

E(G)

^such that every

subgraph

^{of G}

which is

isomorphic

^{to one of the}

graphs of "9

contains an

edge

^{of A. We}

define as the minimum number of éléments in an

5-transversal

^{of G.}

If,

ⁱⁿ

particular, 57

^{consists of the} directed circuits of

length

^{less than}

k then

clearly f(G,

⁼

Tk(G) .

THEOREM 3

Let G and H be directed

graphs with p

^and q

vertices, respectively,

^{and let}

-

~

^{be a}

_family

^{of directed}

_graphs.

^Then

f(G

⁰

H, -Ç ) &#x3E;- pf(H, 5 )

⁺

q 2f(G,5).

Proof : Since G 0 H is the

edge-dis joint

^{union of G 0}

S

^{and the}

^graphs

H ,

^{x x ~}

^V(G),we

^have

^f(G

⁰

H, 5 ) &#x3E;-

_/

^{pf (H,}

₂

⁵¹⁾

_r ⁺

^f(G

⁰

sq, 5 ) t

^{q so it is}

sufficient ^to show that

f(G 0 S , % ) &#x3E;, q 2f(G, 5 ).

^{Let A be any}

F-tranversal

^{of G 0}

S.

q ^{Sélect one vertex} from each of the

graphs (S ) ,

q ^x

x E

V(G),

^{and consider the}

subgraph

^{of G 0}

S q ^{spanned by}

^{thèse vertices.}

This

subgraph

^is

isomorphic

^to

G,

and there are

qp

^such

subgraphs,

^say

GII,G 2" ...1, G p.

^{For each}

^i,

^{1 - 1 * A n}

^{E(G i)}

^{is an}

~-tranversal

^of

G. so 1

Moreover,

^each

edge

^{of A} is contained in

precisely

^{of the}

^graphs G.

¹

1 Hence

45

Combining ^(J)

^and

⁽²⁾

^{we obtain the}

inequality

. Since A is _any

st-transversal,

REFERENCES

[1]

BERMOND

J.C.,

"Ordres à distance minimum d’un tournoi et

graphes partiels

^{sans circuits}

maximaux",

Math. Sci.

hum.,

³⁷

(1972),

^5-25.

[2]

^BERMOND

J.C.,

"The circuit

hypergraph

^{of a}

tournament",

^Proc.

Colloquia

Mathematica Societatis

János Bolyai, Infinite

and

finite sets, Keszthely (Hungary),(1973),

^165-180.