A Lower Bound on the Independence Number. Vectors. Abstract. This paper proves a lower bound on the independence number of general hypergraphs

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1 SERIE B INFORMATIK A Lower Bound on the Independence Number of General Hypergraphs n Terms of the Degree Vectors Torsten Thele* B 95{2 March 1995 Abstract Ths paper proves a lower bound on the ndependence number of general hypergraphs n terms of the degree vectors. The degree vectorofavertex v s gven by d(v) =(d 1(v) d 2(v) :::)whered m(v) sthenumber of edges of sze m contanng v. We dene a functon f wth the property that any hypergraph H =(V E) satses (H) v2v f(d(v)). Ths lower bound s sharp when H s a matchng. Furthermore ths bound generalzes known bounds of We/Caro and Caro/Tuza for ordnary graphs and unform hypergraphs. *Free Unverstat Berln, Fachberech Mathematk und Informatk, Arnallee 3, Berln, Germany

2 2 1 Introducton We and Caro ndependently dscovered the followng nce lower bound for the ndependence number of a graph n terms of the degrees (see also [G]). Theorem 1 [W,C] Let G =(V E) be agraph wth ndependence number (G). Then where d(v) s the degree of the vertex v. (G) v2v 1 d(v)+1 Ths bound s tght f (and only f) G s the unon of dsjont clques. Ths result rases the queston f a slar lower bound can be found for the ndependence number of hypergraphs. Before statng the results we have to make some dentons. A hypergraph s a par H =(V E) where V s a nte set and E s a collecton of non-empty subsets of V,.e. E 2 V n f g. The rank r of a hypergraph H =(V E) s the maxal sze of an edge n E. Thehypergraph H s k-unform f all edges n E have szek. A set I V s called ndependent f 2 I \ E =,.e. the set I contans no edge of E. The maxal sze of an ndependent set of H s dened as the ndependence number (H). Caro and Tuza proved the followng result, whch s an extenson of Theorem 1. Theorem 2 [CT] Let H =(V E) be ak-unform hypergraph wth k 2. Then (H) v2v f(d(v)) where d(v) s the degree ofv,.e. the number of edges contanng v and the functon f s gven by d 1 f(d) := 1 ; : (k ; 1) + 1 In fact the result of Caro and Tuza s slghtly more general. =1 Remark. ; d+1=(k;1) ;1 The functon f n Theorem 2 can be spled to f(d) = d.thus we may rewrte the result as d(v)+ 1 ;1 k;1 (H) : d(v) v2v For k = 2 (ordnary graphs) ths s the We/Caro bound. In order to generalze ths result to arbtrary (non-unform) hypergraphs we have to generalze the concept of the degree of a vertex. The rst dea maybe sply to dene the degree of a vertex v slarly as the number of edges contanng v. Butwe wll run nto troubles wth ths approach, snce we don't have any nformaton about the szes of the edges contanng v. More useful s the followng approach. Let H =(V E) beahypergraph of rank r. For every vertex v 2 V dene the degree vector d(v) =(d 1 (v) d 2 (v) ::: d r (v)) 2 N r where d m(v) s the number of edges of sze m contanng v for 1 m r. Denton. Let r 1beannteger. Dene the functon f r : N r! R by (;1) f r (d) = (m 2N r m ; 1) +1 : The product and the nner sums are taken over all 1 m r. Note that the outer sum s nte snce all summands are zero unless 2 [ d]:=fj 2 N r : j m d m for all 1 m rg. Now we are n the poston to state our man theorem.

3 Lower Bound on the Independence Number of General Hypergraphs 3 Theorem 3 Let H =(V E) be a hypergraph of rank r. Then (H) v2v f r (d(v)) : Suppose H =(V E) sk-unform, k 2. Let v 2 V be arbtrary, e k denotes the k-th unt vector. Snce H s k-unform f(d(v)) reduces to d k(v) (v) (;1) f(d(v)) = f(d k (v) e k )= (k ; 1) +1 = (v)+ 1 ;1 k;1 d k (v) = (see Concrete Mathematcs [GK] p. 188). Thus the theorem s a generalzaton of the results of We/Caro and Caro/Tuza. Let us also consder the case k = 1,.e. H s 1-unform. Then f(d(v)) reduces to d 1(v) d1 (v) f(d(v)) = f(d 1 (v)) = (;1) 1fd1 (v) = = fd 1 (v) 1 : = Ths s what we expect: The unque maxum ndependent set s gven by the set of vertces of degree. Observaton. Let H =(V E) beamatchng of rank r,.e. H s a hypergraph wth the property e 6= e 2 E ) e \ e =. Then (H) = f r (d(v)): v2v roof. Snce H s a matchng the ndependence number of H s gven by (H) =#vertces of degree vector zero + e2e(jej;1) : On the other hand f r () = 1 and for every edge e 2 E we have v2e f r(d(v)) = jej (1 ; 1=jej) = jej;1. Thus (H) = v2v f r(d(v)) 2 Lemma 4 Let r 2 N, C 1 C 2 ::: C r and C > be gven. The functon g : N r! R gven by s the soluton of the recurrence g(d) = g(d) = m (;1) Cm m + C k C kd k g(d ; e k ) k C kd k + C wth g() = C ;1.Inpartcular g(d) s non-negatve for all d 2 Nr. By ths lemma we nfer that our functon f satses the recurrence f(d) = (k ; 1) f(d ; e k ) (k ; 1) +1 wth f() = 1. In partcular f(d) 1 for all d 2 N r.for later purposes we need the followng equvalent partal derence equaton for f f(d) = m (m ; 1) d m [f(d ; e m ) ; f(d)] (1) for d 6=.

4 4 2 roof of the Man Theorem For convenence let us dene the functon F (H) := v2v f(d(v)) for every hypergraph H =(V E), where f = f r and r = rank(h). Suppose x savertex of H. Let H n x denote the resultng hypergraph after removng x together wth all ncdent edges from H. The key to the proof of our man theorem s Lemma 5 Let H =(V E) be ahypergraph wth E 6=. Then there exsts a vertex x 2 V F (H n x) F (H). wth The man work wll be the proof of ths lemma. roof of Theorem 3. Lemma 5 enables us to use the followng algorthm to nd an ndependent set I n H. WHILE E(H) 6= DO Choose x 2 V (H) wth F (H n x) F (H) H := H n x END Output ndependent set I = V (H). Snce f()=1we know thatf (I) =jij. On the other hand the value of F never decreases by the choce of the deleted vertces. Thus F (H) F (I) =jij (H). 2 We remark that the proof ples a polynomal algorthm that computes an ndependent set of sze at least F (H) n an arbtrary hypergraph H of constant rank. In partcular, for unform hypergraph, ths s the so-called max-algorthm (see also [CT, G]): Successvely remove vertces of maxum degree wth all ncdent edges untl no edges are left. It s easy to see that a vertex x wth maxum degree n a unform hypergraph has always the property F (H n x) F (H). 3 roofs of Lemmas For the proof of Lemma 5 we need Lemma 6 Let r 2 N, d 2 N r and 2 [ d] be gven. Then f(d ; ) ; f(d) r m=1 m [f(d ; e m ) ; f(d)] : roof of Lemma 5. Let H = (V E) beahypergraph of rank r wth E 6=. Dene V to be the set of all non-solated vertces,.e. vertces x wth d(x) 6=. By assumpton, V 6=. Furthermore for two dstnct vertces x w 2 V the co-degree vector s gven by d(x w) = (d 1 (x w) d 2 (x w) ::: d r (x w)) 2 N r,whered m(x w) s the number of edges of sze m contanng both x and w. Set d(w w):=. Now let x 2 V be arbtrary, then F (H n x) ; F (H) = w2v [f(d(w) ; d(x w)) ; f(d(w))] ; f(d(x)) : Consder one summand. Lemma 6 ples [f(d(w) ; d(x w)) ; f(d(w))] m d m (x w) [f(d(w) ; e m ) ; f(d(w))] :

5 Lower Bound on the Independence Number of General Hypergraphs 5 Thus F (H n x) ; F (H) w2v m We sum these derences up over all x 2 V : [F (H n x) ; F (H)] x2v = m x2v w2v m x2v d m (x w) [f(d(w) ; e m ) ; f(d(w))] ; f(d(x)) : d m (x w) [f(d(w) ; e m ) ; f(d(w))] ; f(d(x)) x2v d m (x w) [f(d(w) ; e m ) ; f(d(w))] ; f(d(x)) w2v x2v! = d m (x w) [f(d(w) ; e m ) ; f(d(w))] ; f(d(x)) m w2v x2v x2v = (m ; 1) d m (w)[f(d(w) ; e m ) ; f(d(w))] ; f(d(x)) m w2v x2v! = (m ; 1) d m (x)[f(d(x) ; e m ) ; f(d(x))] ; f(d(x)) x2v = : There we made use of the followng observaton m x2v d m (x w) =(m ; 1) d m (w) and the fact that f(d) satses the partal derence equaton (1) for d 6=. By denton, d(x) 6= for all x 2 V. We nfer that for a random x 2 V the expectaton of F (H n x) ; F (H) s non-negatve. Thus there exsts an x 2 V V wth F (H n x) F (H). 2 Lemma 7 For r 2 N, 1 k l r and d 2 N r wth d k 1 we have roof. We wllshowthat Consder the case k 6= l rst. f(d ; e k ) ; f(d) f((d + e l ) ; e k ) ; f(d + e l ) : [f(d ; e k ) ; f(d)] ; [f((d + e l ) ; e k ) ; f(d + e l )] : f(d ; e k ) ; f(d) = ; = ; ; 1 k k ; 1 k ; 1 (;1) (m m ; 1) +1 (;1) (m m ; 1) +1 (;1) m (m ; 1) +1 : Slarly f((d + e l ) ; e k ) ; f(d + e l )=; ; 1 dl +1 k ; 1 l (;1) (m m l ; 1) +1 :

6 6 uttng ths together yelds [f(d ; e k ) ; f(d)] ; [f((d + e l ) ; e k ) ; f(d + e l )] ; 1 dl (;1) = k ; 1 l ; 1 m (m l ; 1) +1 ; 1 (;1) = k m = g(d ; e k ) where g s gven by the recurrence g(d) = (m ; 1) g(d ; e m ) (m ; 1) + C (m ; 1) +[(k ; 1)+(l ; 1) + 1] {z } =:C > wth g() = C ;1 > accordng to Lemma 4. In partcular g(d ; e k ) s non-negatve whch proofs the cla for k 6= l. Now let k = l. Wehave toprove that [f(d ; e k ) ; f(d)] ; [f(d) ; f(d + e k )]. Consder agan ; 1 (;1) f(d ; e k ) ; f(d) =; (m m ; 1) +1 k ; 1 and slarly We nfer that f(d) ; f(d + e k )=; k ; 1 (;1) m (m ; 1) +1 : [f(d ; e k ) ; f(d)] ; [f(d) ; f(d + e k )] ; 1 (;1) = (m k ; 2 m ; 1) +1 ; 1 (;1) = k m = g(d ; e k ) where g s agan gven by the recurrence g(d) = (m ; 1) g(d ; e m ) (m ; 1) + C (m ; 1) +[2(k ; 1) + 1] {z } =:C > wth g() = C ;1 > accordng to Lemma 4. In partcular g(d ; e k ) s non-negatve and the cla follows also for k = l. 2 Remark. Lemma 7 tells us that for any d and k the derence f(d;e k );f(d) decreases whenever we ncrease any componentofd. Ths s essental for the proof of Lemma 6 roof of Lemma 6. Let r 2 N, d 2 N r and 2 [ d]begven. Consder the ponts (d;) and d on the N r grd. A monotoncal path between these ponts s a sequence of grd ponts startng wth (d;) and termnatng wth d where two neghborng ponts are of the form (d ; e m ), d for some

7 Lower Bound on the Independence Number of General Hypergraphs 7 1 m r. Each monotoncal path between (d ; ) and d has length := m and the number ; m of such paths s gven by the multnomal coecent 1 ::: r.nowlet = p p 1 ::: p be such a monotoncal path, p = d ; and p = d. Accordng to ths path we rewrte f(d ; ) ; f(d) as the telescopng sum f(d ; ) ; f(d) = j=1 [f(p j;1 ) ; f(p j )] : Note that all derences havetheformf(d ;e m );f(d ) for some 1 m r and d 2 [d;+e m d]. For each 1 m r there are exactly m derences of the form f(d ; e m ) ; f(d )nthe telescopng sum snce s monotonc. By Lemma 7 we see that each such derence satses Thus we can estate f(d ; e m ) ; f(d ) f(d ; e m ) ; f(d) : f(d ; ) ; f(d) m m [f(d ; e m ) ; f(d)] : 2 It remans to proof Lemma 4. roof of Lemma 4. Let r 2 N, C 1 C 2 ::: C r and C > begven. We have toshowthat the functon g : N r! R gven by g(d) = m (;1) Cm m + C satses the recurrence g(d) = Ck d k g(d ; e k ) Ck d k + C wth g() = C ;1.Itseasytocheck thatg() = C;1 holds. Let us rewrte the recurrence as a partal derence equaton C g(d) = k C k d k [g(d ; e k ) ; g(d)] for d 6=. Suppose d k > thenwehave ; 1 (;1) g(d ; e k ) ; g(d) = ; k ; 1 m Cm m + C = ; 1 d m (;1) k : d k m Cm m + C Hence, C k d k [g(d ; e k ) ; g(d)] = ; d m (;1) C k k m Cm m + C and therefore k C k d k [g(d ; e k ) ; g(d)] = ; = ; d m d m (;1) k C k k m Cm m + C (;1) 1 ; m C Cm m + C

8 8 = ; d m (;1) m {z } + C = for d6= d m (;1) m Cm m + C {z } =g(d) = C g(d) as desred. 2 References [C]. Caro, New Results on the Independence Number, Tech. Report, Tel-Avv Unversty (1979). [CT]. Caro, Z. Tuza, Improved Lower Bounds on k-independence, J. of Graph Theory, (1991), Vol. 15, p [G] J. Grggs, Lower Bounds on the Independence Number n Terms of the Degrees, J. of Comb. Theory, Ser. B 34 (1983), p [GK] R. L. Graham, D. E. Knuth, O. atashnk, Concrete Mathematcs, Addson-Wesley, (1992), Eght prntng. [W] V. K. We, ALower Bound on the Stablty Number of a Sple Graph, Bell Lab. Tech. Memo. No (1981).