AN UPPER BOUND FOR THE TRANSVERSAL NUMBERS OF 4-UNIFORM HYPERGRAPHS

Note

An Upper Bound for the Transversal

Numbers

of 4-Uniform

Hypergraphs’

FENG-CHU LAI

Department of Applied Mathematics. National Chiao Tung University, Hsinehu 30050, Taiwan, Republic of China

GERARD J. CHANG

Department of Applied Mathematics. National Chiao Tung University, Hsinchu 30050, Taiwan, Republic of China, and Institute of Informaiion Science, Academia Sinica, Nankang, Taipei 11529, Taiwan, Republic of China

Communicated by the Editors Received April 11, 1988

The main purpose of this paper is to prove that if H is a 4-uniform hypergraph with n vertices and m edges, then the transversal number r(H) <2(m +n)/9. 0 1990 Academic Press, Inc.

All standard terminology of hypergraphs is from [ 11. Suppose

H = (V, E) is a k-uniform hypergraph with n vertices and m edges. Tuza

[2] proposed the problem of finding an upper bound for the transversal

number r(H), of the form t(H) < ck(n + m), where ck depends only on k.

More precisely, we want to determine ck z sup z(H)/(m + n), where H runs

over all k-uniform hypergraphs of n vertices and m edges. It is easy to see

that c1 = 4 and c2 = i. Tuza [2] proved that c3 = $ and asked if ck is 0(1/k).

For any positive integer p we can construct a k-uniform hypergraph H

of n = k + p vertices xi, . . . . x, and m = [n/p1 edges e,, . . . . e,, where

ej= {x, ~~~~r-Yn}-{-xip-p+I~ ...txjp} for l<i,<m-1

’ Supported by the National Science Council of the Republic of China under grant NSC77- 0208-M008-05.

129

00958956190 $3.00 CopyrIght Q 1990 by Academic Press. Inc. All rights of reproductmn I” any form reserved

130 and

LA1 AND CHANG

e, = {Xl) . ..) Xk}.

Then r(H) = 2. To make r(H)/(n + m) large, we only have to find p

such that n + m = k + 1 + p + [k/p] is minimum. It is an easy exercise

to check that n + m achieves minimum at p = LJif]. So ck B bk =

2/(k + 1 + L$j + rk/LkJ). Note that ck = b, = l/(k + 1) for k < 3. The

main purpose of this paper is to prove that cq = b, = $. In fact we prove a

more general theorem.

THEOREM. Suppose H = (V, E) is a 4-uniform hypergraph of n vertices and m edges. Zf H has t end edges, which are edges containing vertices of degree one, then z(H) < (2n + 2m - t)/9.

ProoJ Without loss of generality we may assume that H is connected.

We shall prove the theorem by induction on m. The theorem is trivial for

m=O since z(H)=t=O.

Suppose X= {vi, . . . . tak} c V is a set of k vertices. Denote by H’ the

hypergraph obtained from H by deleting all edges e, , . . . . eP that meet X and ail vertices only in e, u . . . u eP. Then

z(H) d k + z( H’) < (2n’ + 2m’ - t’ + 9k)/9, n’=n-r,

m’=m-p, t’=t-q+s,

where r is the number of vertices only in e, u . . . u eP, q is the number of end edges in e,, . . . . eP, and s is the number of end edges in H’ which are

not end edges in H. Consequently,

T(H) < (2n + 2m - t)/9 - (2p + 2r + s - q - 9k)/9.

So the theorem holds when A* - 2p + 2r + s - q 2 91x1. In the case when

each vertex in X is of degree 2 2, r > q + k and the theorem holds when

A=2p+r+s>81XI.

Case 1. If there is a vertex x of degree p 2 4, then choose X= {x}.

pa4 implies A>81XI.

Case 2. Suppose all vertices of H are of degree 3. Choose a k-cycle C= (x1, el, x2, f3, . . . . xk, ek, x, ) of minimum length. Denote by e; the edge other than e,-i and ei that contains xi.

f? e2 X e2 FIGURE 1 e’ 2 FIGURE 2

132 LA1 AND CHANG

In the case k=2, let X=(x,}. Thenp=3; r>2 when e;=e;; ral and s>l when e;#e;. Thus d>8lXI.

Now assume k 2 3. Choose a vertex x E ek ~ f - { xk _ i, x,}; denote by e’ and e” the edges other than eke, that contain x. By the minimality of the length of C, the edges e,, . . . . ek, e;, . . . . eh, e’, err are distinct. Now consider x= {Xl, x3, x5, . . . . X& 1 } when k is even and X = {x,, x3, . . . . xkm2, x}

when k is odd. Then p = 3k/2 and r, s > /XI = k/2 when k is even (see Fig. 1 for k = 6); and p = 3(k + 1)/2 and r, s 3 1x1 = (k + 1)/2 when k is odd (see Fig. 2 for k = 5). In any case d >, 81X(.

Case 3. Suppose all vertices are of degree at most 3 and there is a vertex x of degree 3 contained in e,, e,, e3 ; but e, has at least one vertex of degree 6 2.

If e, is an end edge, then consider X= {x}. p = 3 and r > 2 imply A > 81x1. So we can assume that e, is not an end edge. Suppose y is a ver- tex of degree 2 in e,, say yee, ne. If e=e2 or e3, then consider X= {x}. Againp=3 and r>2 imply A>81XI. So assume e#e, and e#e,. If e is not an end edge, then consider X= ix}. p = 3, r 2 1 and s B 1 imply A 2 81x1.

Now we can assume e = (y, y,, 112, y3} with 1 = deg(y,) ddeg(y,) 6 deg(y,)<3. If deg(y,)=3, then choose X= {y3} to get A>8lXI. So assume deg( u3) = 2, say y, E e n e’. In the case of deg( yz) = 1 or e’ is an end edge we consider X= { y3}, then p = 2, r > 3, and s > 1; in the case of deg(y,) = 2 and e’ is not an end edge we consider X= { y2}, then p = 2, r>2, and ~22. In any case A~8lXl.

Case 4. Suppose all vertices of H are of degree at most 2. Suppose there is an end edge e containing at least two vertices of degree one. Choose X= {x}, where x is a vertex in e of maximum degree p. Then either p=q=l and r=4, which imply d*>9=9lXI; or else p=2 and r>q+233 which imply A*=2p+2r-q+s>2p+2q+4-q+sa

9 = 91x1. So we can assume that every and edge has exactly one vertex of degree one. Count the degrees of all vertices. There are t vertices of degree one and n - t of degree 2; then we have 4m = t + 2(n - t) = 2n - t.

Suppose there are two distinct edges e and

f

such that le n

f 1 =

3, say

e={x,,x,,x,,x,}

andf=( x1, x2,

x3, x5}. If deg(x,) = 1 or deg(x,) = 1, then we choose X= (xi}. In this case p = 2 and r 2 4, which imply Aa8lX(. So we may assume deg(x,)=deg(x,)=2, say x4eene’ and x5 E f n f ‘. If e’ or

f'

is not an edge, then we choose X= (xi }. In this case p = 2, r = 3, and s 2 1, which imply A 3 8 [XI. So we may assume that e’ and f’ are end edges, say e’= (x4,~,,y2,y3} with deg(y,)= 1 and deg(y,)=

deg(y,) = 2 (since every end edge has exactly one vertex of degree one). Let y1 E e’ n e, and y, E e’ n e2. If e, or e2 is an end edge, we can without loss of generality assume that e, is an end edge. Now choose X= { JJ, }. Then

either p = 2, r = 3, and s > 1 (when e, is an end edge), or else p = 2, r = 2,

and s3 2 (when e, is not an end edge). In any case A > 8. So we can

assume that Je n fl d 2 for any two distinct edges e and J:

Choose maximum number of vertices xr , . . . . xk of degree 2, say xi E ej n f,

for 1 d id k, such that e,, fr, e,,

fi,

. . . . ek,

fk

are distinct. Let g1 , . . . . g,- Zk

be the remaining edges of E. By the maximality of k, the edges gj are

pairwise disjoint. Each edge gj has at least three vertices of degree 2, which

are also in some e,. or

f,;

call such vertices common vertices. Note that there

are at least 3(m - 2k) common vertices. On the other hand, e, u

fj

has at

most three common vertices for any i. Otherwise the fact that any two

distinct edges intersect at no more than two vertices implies that either lej n f, 1 = lei n gj 1 = (f, n gr 1 = 2 for some gj or else there are common ver- tices x7 E e, A gj and x,* *

Ef,

n gj, for distinct j and j’. For the former case, H has exactly 3 edges and 6 vertices and t(H) = 2, so the theorem holds.

For the latter case, we can replace xi by XT and x,+* to get k + 1 vertices

whose containing edges are distinct, in contradiction to the maximality of

k. Hence 3(m - 2k) < 3k, i.e., m d 3k. Since x1, . . . . xk together with a vertex in gj for 1 < j < m - 2k form a transversal,

r(H)<k+m-2k d 2mJ3 = (2m + 2n - t)/9 (since m < 3k) (since 4m = 2n - t). Q.E.D. REFERENCES

1. C. BERGE, “Graphs and Hypergraphs”, North-Holland, Amsterdam, 1973. 2. Z. TUZA, Covering all cliques of a graph, preprint.