Extremal set systems with weakly restricted intersections, Combinatorica 19
Extremal set systems with weakly restricted intersections
Van H.Vu?
September6,2001
Abstract
Theorems on extremal set systems with intersections of restricted cardinality are probably among the most well-known and powerful theorems in extremal combinatorics.
In this paper,we discuss the so-call”weak”version of some theorems of this type,when
the restricted intersection property is weakened by the possible existence of some(maybe
many)intersections having”bad”sizes.In particular,we give a tight upper bound
for the weak version of the”odd town”problem and non-uniform Fisher’s inequality.
The second problem leads to an extremal set theoretic characterization of Hadamard’s
matrices.We also give a tight bound for the weak version of the”even town”problem,
raised by Erd?o s.This bound turns out to be an useful tool to handle systems with
restricted multi-intersections.
1Introduction
Let X be a set of n elemnets,and L a set of non-negative intergers.One of the most interesting and important question in extremal combinatorics is to achieve an upper bound for the cardinality of a family of subsets of X where the cardinality of the intersection of any two members of the family is an integer in L.Such a family is called set system with restricted intersections(with respect to L).Well-known bound concerning these systems are the theorems of odd town([3],chapter1)and even town[4],[9],Fisher inequality [7][5],Erd?o s-de Bruijn inequality[6],Ray-Chaudhury-Wilson theorem[13],Frankl-Wilson theorem[8],etc.These theorems have several important applications in many areas from the theory of codes and designs to geometry.Moreover,all of them can be proved by using the same idea which can be formalized in the following steps:
Step1.Asign vectors of some(properly chosen)linear space to the members of the system. https://www.wendangku.net/doc/b66179746.html,e the restricted intersection property to show some properties of this family of vectors and derive a bound from that.For instance,the most common trick is to show that vectors are independent,so one has a bound by the dimension of the space.In few cases when the space is?nite,an alternative method is to bound the dimension of the subspace spanned by the family of vectors,and obtain a bound by the cardinality of this subspace. Step3.If the bound is not tight,one can sharpen it by adding appropriate vectors to the system,and apply Step2to the new(extended)set of vectors.By this,we gain an additional term(equal to the number of vectors added)on the bound.
?Department of Mathematics,Yale University,10Hillhouse,New Haven,CT-06520,USA.Email: vuha@https://www.wendangku.net/doc/b66179746.html,.
1
Since Step2is crucial and here we de?nitely need to use the restricted intersection property, one may ask now a following natural question:
Can we still prove some bounds if we allow some exceptional intersections of size not in the restricted set?
This relaxation causes some di?culties.Because of the existence of exceptional inter-sections,the statements we can prove(in Step2)under original condition will not hold in general.So the method seems to lose its power.However,if the number of exceptional intersections is relatively small,there are cases when the problem is still solvable using ad-ditional tools(for example from extremal graph theory)combined with the algebraic tool giving tight bounds and interesting optimal constructions.Our purpose in this paper is to discuss some of these possibilities.
Let us now state our goal precisely,First we describe what we understand here by weakly restricted intersection property.Let s be an non negative integer and L a set of non-negative integers,we want to determine m s(n,L),the maximum cardinality of a set system{A1,A2,...,A m}on ground set X of n elements,such that for each index i there are at most s indices j=i satisfying|A i∩A j|/∈L.The case s=0is the original problems with restricted intersections.When s>0we call it a s-weak version.(Sometimes there are further conditions on the size of the members of the family;we will specify these conditions when needed and keep the general notation m s(n,L)).
The paper will be organized as follows:
In Section2,we consider the s-weak version of the so-called”odd town”problem,where all the members of the systems are required to have odd cardinality and L is the set of even integers.We give a tight bound for the case s<2n/4,and an approximately tight bound for s>2n/2.Our methods here is to alternatively apply techniques from extremal graph theory and linear algebra.
Section3stars with the weak version of the”even town”problem,raised by Erd?o s.L is still the set of even integers,but now we do not have any condition on the members of the system.The original problem was solved by Berlekamp[4]and Graver[9].In subsection 3.1we display the bound for the weak version when s<2n/2?log n(Theorem3.1.2).This result is proved in a previous paper[14].
In the rest of Section3,we discuss systems with even multi-intesections(intersections of more than two sets).First,let us take a closer look at the method described in three steps above.One of the reason why linear algebra techniques are powerful for achieving upper-bounds on set systems with restricted intersections is the following.Assume the members of the system is embedded as vectors in some linear space V,with V being properly chosen,it is usually easy to?nd an operator from V2to R(sometimes just the inner product),which(in some sense)indicates the restricted value of the intersection of two vectors(i.e.,two members of the family).A further investigation of this operator actually leads to some desired properties of the family of vectors from which we can obtain strong upper bounds(see Step2).On the other hand,it seems di?cult to apply this approach in case of multi-intersections,since we have not found no”natural”operator which would represent the multi-intersections.In subsection3.2,we will see that the notion of systems with weakly restricted intersections provides an useful tool to handle systems with even multi-intersections succesfully(in fact,we could also allow some odd multi-intersections here,too).Our?rst aim here is to emphasize the use of this notion,so we will sketch the
2
proofs,and refer to[14]for details.
In section4,we consider the1-weak version of the non-uniform Fisher’s inequality. Although Fisher’s inequality is probably the most well-known theorem in the area,the classi?cation of optimal systems(when equality is achieved)is still far from completed. The main conjecture,known as theλ-design conjecture(by Ryser and Woodall[12],[15])is almost thirty-year old,but still open.Rather surprisingly,for the1-weak version,we could ?nd both tight upper-bound and the classi?cation of all optimal systems.This classi?cation leads us to a new extremal set theoretic characterization of Hadamard’s matrices.
In section5,we disaplay some more results concerning weak versions of other classical problems,and close with some open questions and conjectures.
2Weak odd town
Odd town is a village with n inhabitants,and they form m clubs.These clubs should obey the follwing rules:
1.No two clubs have identical membership.
2.Each club should have odd number of members.
3.Every two clubs should share even members in common.
The problem is to maximize m,under these rules.In our terminology,it is equivalent to ?nd m0(n,L),where L is the set of even integers,with an additional condition that all members in the set system have odd cardinality(rule2).We will keep this condition until the end of this section.Consequently,in this Section m s(n,L)means the maximal size of a family of odd subsets of a set of n elements,each subset has at most s odd intersections. The”odd town”Theorem below solves this problem.[3]
Theorem2.1Let L be the set of all even numbers,then m0(n,L)=n,i.e.,if{A1,A2,... ,A m}is a system of subsets of a ground set of n elements such that each|A i|is odd and |A i∩A j|is even then m≤n and the bound is tight.
Let s be a positive integer satisfying log s ≤(n/4)?1and if{log s}=0then{log s}≥log(1+(8/n)).Here log has base2,the other notations are standard.Our main result on the weak version of the problem is:
Theorem2.2If s satis?es the above conditions then m s?1(n,L)=s(n?2 log s ). Proof.Let v i denote the characteristic vector of A i in GF n(2).Consider the simple graph G(V,E)on the set V={1,2,...,m}where(i,j)∈E i?|A i∩A j|≡1(2),i.e.,v i v j=1. The intersecting property of the system A i says that deg i
Consider an independent set D with maximum cardinality|D|=α.It is well known (by Turan)thatα≥m/s,so r=α?m/s≥0.For each i∈D let T i be the set consisting of i and all the vertices of G the only neighbour of which in D is i.Let T=V\∪i∈D T i. Denote by e the number of edges between the two components D and V\D.Since deg i
rs≥|T|.
On the other hand
i∈D
|T i|=m?|T|,one can suppose that1∈D and|T1|=
max i∈D|T i|,hence|T1|≥(m?rs)/α.
Consider the subgraph G1induced by T1.If there were a pair j,j ∈T1where(j,j )/∈E
3
then D∪{j,j }\{1}would be an independent set of cardinality larger thanα.This shows that G1should be a complete graph.Let j,j ,k,k be elements of T1,the de?nition of our graph implies that
(v j+v j )(v k+v k )=v j v k+v j v k +v j v k+v j v k =1+1+1+1=0
(v j+v j )v1=v j v1+v j v1=1+1=0
(v j+v j )v i=v j v i+v j v i=0+0=0
for every i∈D\{1}.(The equalities are in GF(2).)
Denote by V1the subspace of GF n(2)spanned by{v j+v j },j,j ∈T1and D={v i,i∈D}.The equalities above implies that V1?V⊥1and D?V⊥1.Therefore there is a subspace V2?V⊥1such that V⊥1=V1⊕V2.Each vector v i∈D has a unique decomposition v i=v1i+v2i with v1i∈V1,v2i∈V2.Note thatδij=v i v j=(v1i+v2i)(v1j+v2j)=v2i v2j where δij is the Kronecker index for each pair i,j∈D.It follows that the vectors v2i,v i∈D are independent.So
|D|=α=m
s
+r≤dim V2=dim V⊥1?dim V1=n?2dim V1
thus m≤s(n?r?2dim V1).It is clear that V1contains at least|T1|di?erent vectors which yields dim V1≥ log|T1| ≥ log((m?rs)/α)
We complete the proof inderectly.Assume that m>s(n?2 log s ).Furthermore,let f(r)=r+ 2log(m?rs)/α whereα=(m/s)+r.Consider the following two cases:
1.r>(m/s)?4.Since log s ≤(n/4)?1it follows that2 log s ≤(n?4)/2and hence
s(n?2 log s )≥s(n+4)
2
≥
s(α+4)
2
=
s(m
s
+r+4)
2
≥
s(m
s
+m
s
?4+4)
2
=m
a contradiction.
2.r≤(m/s)?4.Suppose r≥2,note that then(m
s ?r)/(m
s
?r+2)≥4/6and
(m s +r?2)/(m
s
+r)≥6/8.Moreover
log
m?rs
(m/s)+r
?log
m?(r?2)s
(m/s)+(r?2)
=log
m?rs
(m/s)+r
.
(m/s)+(r?2)
m?(r?2)s
≥log
4.6
6.8
=?1
which implies that f(r)≥f(r?2).So it su?ces to consider0≤r<2.Observe that
f(r)≥ 2log s (m/s)?r
(m/s)+r
≥ 2log s
n?2 log s ?r
n?2 log s +r
≥2 log s n?2((n/4)?1)?2
n?2((n/4)?1)+2
=2 log
s
1+(8/n)
4
by the inderect assumption and the fact that r ≤2.Now repalce s =2k +δ.The condition on s yields s 1+(8/n )=2k 2δ1+(8/n )
>2k
which shows that f (r )≥2(k +1)=2 log s .Thus we obtain that m ≤s (n ?f (r ))≤s (n ?2 log s ),a contradiction.This proves the upper bound .
Construction 2.3Let p = log s and l =n ?2p
Consider the set X ={a 1,a 2,...,a p ,b 1,b 2,...,b p ,c 1,c 2,...,c l }.Choose s di?erent subsets I 1,I 2,...,I s of the set {1,2,...,p }and set A i,j ={c i }∪{a q ,b q }q ∈I j for every 1≤i ≤l,1≤
j ≤s we obtain a system of s (n ?2p )sets which has the described property.So the bound is tight and our proof is completed.Q.E.D.
Remark.In the Theorem,we insist that the bound is an integer.This costs some technical arguments at the end of the proof.A slightly weaker bound s (n ?2log s )could be proved with less technical details and without the (somewhat arti?cial)condition {log s }≥log(1+(8/n )).However,if we drop this condition,the upper-bound in the Theorem is no longer true.For example,take s =2k +1>n ,one can verify that s (n ?2 log s )<(s ?1)(n ?2 log(s ?1) ).In fact we need this condition to keep the upper-bound described in the Theorem an increasing function on the restricted domain of s .The function s (n ?2log s )is increasing in s for s <2n/2?1,without any further restriction.
The following theorem gives bounds for the case s takes larger values.
Theorem 2.4
1,If 2n ?2>s >2n/2then m s ?1(n,L )=2s +O (2n/2).
2,If 2n/2≥s ≥2n/4then m s ?1(n,L )
The ?rst bound is approximately tight in the sense that when s 2n/2,the error term is negligible.For s ≥2n ?2,we have m s (n,L )=2n ?1because in this case the family of all
(2n ?1)subsets of odd cardinality satis?es the weak intersection property.
Proof.Consider the graph G where the vertices are all the 2n ?1subsets of odd cardinality of a set of n elements and two subsets form an edge if their intersection is also odd.One
could prove that G has a pseudo-random property that for every large subgraph G on n nodes G has approximately n
2 /2edges.The error term is O (2
n/2n ),hence for subgraphs having more than 2n/2vertices it implies that the average degree is about half of the number of vertices.And so if the maximum degree is smaller than s one concludes that the number of vertices could exceed 2s by a term O (2n/2)only.This implies
m s ?1(n,L )≤2s +O (2n/2)
For general information about pseudo-random graphs,we refer to [2].It is interesting to compare this result with Theorem 2.2and notice the changes in the background of the problem.For smaller s ,the problem is rather algebraic,but when s is su?ciently large,it is related to random structures.
To show that the bound is sharp,let us consider the following construction.
5
Constructions2.5Suppose s=2p1+2p2+...+2p k,where n?1>p1>p2>...>p k≥0. Consider a family of subsets A i,i=1,2...,k,where A1?A2?...?A k and A i has cardinality p i+2for i To prove the second part consider the function h(r)=r+2log(m?rs)/αwhereα= (m/s)+r,we have h (r)=1?4m/s (m/s+r)(m/s?r) . Note that if r<(m/s)?3than h (r)>0,so h(r)is increasing.Hence h(r)≥h(0)= 2log s.Follow the proof of theorem2.2we deduce that m s?1(n,L)≤s(n?2log s)≤sn/2. Suppose r≥(m/s)?3.Note that r=α?(m/s),thusα≥2(m/s)?3.The fact n≥αimplies m≤s(n/2+1.5)which completes the proof.Q.E.D. 3Set systems with(weakly)even(multi-)intersections 3.1Weak even town Erd?o s asked the following question:How many subsets can one choose from a set of n elements so that every intersection is even This problem was answered?rst by Berlekamp[4]and resolved by Grave in a slightly more general form[9]with some applications in Boolean designs.Since then,it has become well-known under the name”even town”problem(see also[3]). Theorem3.1.1(Graver)Let{A1,A2,...,A m}be a system of subsets of a ground set of n elements such that the intersection of any two subsets has even cardinality then m≤2 n/2 + (n),where (n)=1if n is odd and0otherwise.The bound is tight. In our terminology it states that if L is the set of even integers then m0(n,L)=2 n/2 + (n). In[14]we solved the weak version of this problem for all the values of s up to c2n/2?log n with some positive constant c. Theorem3.1.2.[14]Let n0= n/2 . 1.n odd.If s<2n0?2n0?d 22d+1?2where d= log(n+1)?1 2 then m s(n,L)=2n0+s. 2.n is even.If s<2n0?2n0?d?1 22d?1where d= log n+1 2 then m s(n,L)=2n0. In the next section we will apply Theorem3.1.2to?nd sharp upper-bounds for systems with even multi-intersections.We shall also consider a weak version of this,where we allow the existence of some multi-intersection with odd cardinality. 3.2Systems with even multi-intersections The following problem can be considered as a generalization of Erd?o s’s”even town”problem mentioned above. How many subsets can one choose from a set of n elements such that the intersection of any k of them is even. 6 Consider a set of n elements.Partition the set in pairs and consider all the subsets consisting of some pairs.This set system has2 n/2 elements.Moreover,for any?xed k every k members of the system intersect in some pairs so their intersection is always even. As an application of Theorem3.1.2,we show in[14]that2 n/2 is essentially the upper bound for the size of such systems. Theorem3.2.1. Let A1,...A m be subsets of a set X of n elements.Suppose that for every set of k di?erent indices i1,...,i k the intersection∩k j=1A i j has even cardinality.Then there is a function g(k) of order of magnitude O(log k)such that if n≥g(k)then m≤2 n/2 if n is even,and m≤2 n/2 +k?1if n is odd. We shall now sketch the proofs of Theorem3.2.1and its”weak”version,when the system may contain odd multi-intersections(Theorem3.2.3).The proof of3.2.1relies on Lemma3.2.2and Theorem3.1.2.This Lemma also generalizes Theorem2.1. Lemma3.2.2.Let A1,A2,...,A m be odd subsets of a set X of n elements such that for every k di?erent indices i1,...,i k the intersection A i 1∩A i 2 ∩...∩A i k has even cardinality then m≤(k?1)(n?log(k?1)) Proof.Let d denotes the dimension of the space spanned by the characteristic vectors of the members of the system in GF n(2).If d≤log(k?1)then m≤(k?1)/2(note that at least a half of the vectors of any subspace of GF n(2)must have even weight).If d>log(k?1),we will prove a stronger inequality that m≤(k?1)(d?log(k?1)). We use induction on d.To start let us note that if d=log(k?1)+1then m≤k?1 by the above reasoning. Let t be the largest number such that there are t members of the family with their intersection having odd cardinality.Suppose A ij,j=1,2,..,t are such members,so A=∩t j=1A i j has odd cardinality.Since the intersection of any k members has even cardinality, we have t≤k?1.Let v i be the characteristic vector of A i and v be that of A.Note v i 1.v=|A i 1 ∩A|(mod2)=|A|(mod2)=1 and v q.v=|A q∩A|(mod2)=|A q∩∩k?1 j=1A i j |(mod2)=0 for every q=i j,j=1,2,...,t.So it is clear that the vector v i 1 is independent from the set {v q,q=i j}.So if one remove these t≤k?1vectors v i j from the system then the dimension of the space spanned by the v i decreases by at least one.We complete the proof by the induction hypothesis.Q.E.D To prove Theorem3.2.1,one can?rst bound the number of odd intersections of any member in the family using the previous Lemma,then apply Theorem3.1.2.For the detailed proof,we refer to[14](few technical arguments are needed to obtain the term k?1 in the case n odd). Let us now consider the s-weak version of Theorem3.1.2.The problem is as follows; Let A1,...,A m be subsets of a set of n elements such that for every set of k?1di?erent indices i1,i2,...,i k?1there are at most s index i k such that|∩k j=1A i j |is odd.One wants an upper bound on m in term of s,k and n. 7 Again,we are able to prove a sharp bound,in case s and k are not too large: Theorem3.2.3.Given s and k there is a function g(s,k)of order of magnitude O(log k+ log s)such that if n>g(s,k)then m≤2 n/2 if n is even and m≤2 n/2 +k+s?1if n is odd. The proof is similar to that of3.2.1.All one needs to do is to consider the s-weak version of Lemma3.2.2. Lemma3.2.4.Let A1,A2,...A m be odd subsets of a ground set of n elements such that for every k?1di?erent indices i1,i2,...i k?1there are at most s index i k(i k=i j j that|∩k j 1A i j |is odd,then m<(k?1+s)(n?log(k?1)). https://www.wendangku.net/doc/b66179746.html,e the same notation as in3.2.2.If t odd(multi)intersection;A=∩k?1 l=1A il,|A|is odd.Let A j1,...,A js be the members of the system which have odd intersection with A.It is clear that s ≤https://www.wendangku.net/doc/b66179746.html,e similar method we can show that if we remove k?1+s vectors A i1,...,A i(k?1),A j1,...,A js from the system then the dimension decreases by at least one.Now we can?nish by the same argument as in3.2.2.Q.E.D The bound2n/2for the case n even follows directly from Theorem3.1.2.To get the bound in case n odd,one needs to repeat the detailed proof of3.2.1in[14]with minimal modi?cation. To see that the bounds in3.2.3are sharp,consider the following construction.Assume n is odd,partition a subset of n?1elements into n?1/2pairs.Let B be the family consting of the unions of pairs(|B|=2n?1/2).Choose k+s?1members of B and add to each of them the remaining element of the ground set.The union of B and these k+s?1subsets forms a family with achieving the upper-bound.For n even,simply take the all possible unions of n/2pairs. 4Non-uniform Fisher inequality under weak condition Let L={λ},λis a non-negative integer,the nonuniform Fisher’s inequality[5][10]states that m0(n,L)≤n.Althout it is probably the?rst and most famous result in this area,it is still di?cult to characterize all optimal systems(when equality is achieved).There are characterizations for some special cases;the most well-known is probably the theorem of Erd?o s and de-Bruijn for the case L={1},when the optimal system is either a projective plane or a”pencil”.If the system is uniform(meaning all members have the same cardinal-ity),it is known that an optimal system should be a symmetric design[11].For the general case,Ryser and Woodall(independently)made the so-callλ-design conjecture([12],[15]). Although almost thrity year-old,this conjecture is still open. In this section we are going to consider the1-weak version of non-uniform Fisher’s inequality.First,we are able to obtain a tight bound and surprisingly all the optimal constructions are well characterized by Hadamard’s matrices.So in the same time we obtain a new extremal set theoretic characterization of Hadamard’s matrices. Theorem4.1If n>2then m1(n,L)≤2(n?1).Moreover,except the case n=3the equality holds i?n=4λand a Hadamar’s matrix of order n exists. 8 It is clear that m 1(1,L )=2and m 1(2,L )=3,so the bound does not hold in these cases.In order to prove the Theorem ?rst we need the following Lemma. Lemma 4.2.Let M be a n by m matrix of rank r .(r (M )denotes the rank of M ).Denote by M +u the matrix obtained from M by adding to every column a non-zero vector u .Then r (M +u )≥r (M )?1and r (M +u )=r (M )?1i?for every choice of r independent columns of M the vector ?u and every column of M can be seen as an a?ne combination of the chosen columns. Proof.The inequality r (M +u )≥r ?1is trivial by the subadditivity of the rank of matrices.Assume that r (M +u )=r ?1.Choose r independent columns v 1,v 2,...,v r of M .The n by r matrix with columns v i +u (i =1,2,...r )has rank r ?1,so we can suppose that v i +u,i ≤r ?1are independent.One can ?nd r ?1coe?cients α1,α2,...,αr ?1satisfying that:v r +u = r ?1i =1αi (v i +u )hence ?u ( r ?1i =1αi ?1)=?v r + r ?1i =1αi v i .Devide both sides by ( r ?1i =1αi ?1)we have ?u as an a?ne expression of the v i . Consider a column vector v l of M .The vector v l +u can be written as a linear combinations of the v i +u,i =1,2...,r ?1:v l +u = r ?1i =1γi (v i +u ).Thus we have v l = r ?1i =1γi v i +( r ?1i =1γi ?1)u .Replace u = r i =1βi v i we can express v l as a linear combination of the v i ,i =1,2,...,r ?1.The sum of the coe?cients in this combination is r ?1i =1γi +( r ?1i =1γi ?1)( r i =1βi )=1,since r i =1βi =?1.This completes the ?rst implication of the statement.The second implication is left to readers.Q.E.D Remark.Let λby a non-negative number.Consider a square matrix A where all diagonal entries of are larger than λand all o?-diagonal entries are equal to λ.The ?nishing step of all algebraic proofs of non-uniform Fisher inequality consists in showing such A matrix must have full rank.The usual approach is either to compute the determinant of A or to show that A is positive de?nite.Lemma 4.2o?ers a new proof.Let u be a vector with every coordinate equals to λ.Note that r (A ?u )=n .We need only show that r (A )=r (A ?u )?1.Since A ?u is diagonal with positive entries,no a?ne combination of its columns would give the vector ?u .So A has full rank by the criteria provided in the Lemma. Proof of Theorem 4.1It is easy to show that m 1(3,L )=4.So from here we can assume n >3and m >4.We call a subset A i normal if all of its intersections with the others have λelements.If |A p ∩A q |=λwe say the pair (A p ,A q )is irregular.Apparently A can be partitioned into irregular pairs and a set of normal members A ={A 1,A 1 }∪{A 2,A 2 }∪...∪{A k ,A k }∪{A k +1,A k +2,...,A k +l } where A i ,A i are the irregular pairs and A k +j are the normal members. If there is a member of A (say A p )having exactly λelements,then the sets (A i ∪A i )\A p ,i =p and A k +j \A p ,k +j =p are all disjoined.Moreover,(A i ∪A i )\A p ,i =p has at least two elements.Thus it is clear that n ?λ≥m ?2,that is,m ≤n +2?λ<2(n ?1). So from here we can suppose that every set in the family has more than λelements.Let v i ,v i and u j be the characteristic vectors of A i ,A i and A k +j ,respectively.By the property of the family we obtain that v i v i =λand every other pair of vectors has product λ(?).Let M be the n by m matrix with columns (v 1,v 1 ,...,v k ,v k ,u 1,...,u l ).Denote by J t the all-one t by t matrix and let 1t denote its column vector.Furthermore,let M 1=M T M and M 2=M 1?λJ m .By (?)it is clearl that M 2is the direct sum of k 2by 2matrices 9 corresponding to the irregular pairs and l1by1matrices corresponding to the normal members of the family A M2=V1⊕V2⊕...⊕V k⊕U1⊕U2⊕...⊕U l Note that all of the U j are non-zero since|A k+j|>λ,so we have that r(M2)= r(V i)+l. Moreover,n≥r(M1)≥r(M2)?1.Observe that if r(M2)≥m+5 2,then the above inequality implies that m≤2n?3,so in the followings we may assume r(M2)≤m+4 2. Since m+4 2 entries of V1are|A1|?λ,|A1∩A1 |?λ,|A1∩A1 |?λ,|A1 |?λthis matrix has rank one i?|A1∩A1 |?λ=?(|A1|?λ)=?(|A1 ?λ)(??)which implies|A1|=|A1 |.In this case it is obvious that1m is not contained in the subspace spanned by the column vectors of M2, thus r(M1)=r(M2)by Lemma4.1,and we have that n≥r(M2).So if r(M2)>m+2 2again we obtain that m<2(n?1).The remainning cases can be treated as follows. 1.r(M2)=m+2 2.We show in this case m<2(n?1).This case could occur i?either k=m/2or k=m/2?1and exactly m/2?1matrices among the V i have rank one.Consider, let’s say the?rst case and suppose that r(V i)=1for every i So we obtain that m 2?1=dim(x1,x2,...,x m 2 ?1 )≤n?dim(v k,v k ,v1+v1 ,1n)=n?p. Note that p is at least two.If p>2we obtain that m<2(n?1).We will show p=2 cannot occur.Assume that p=2,hence1n and v1+v1 are linear combinations of v k and v k .Since v k and v k are(0,1)vectors as well as v1and v1 the only possible combination is v k+v k =1n=v1+v1 .But then v k1n=v k 1n=v k(v1+v1 )=v k (v1+v1 )=2λwhich implies that|A k|=|A k |=2λ.On the other hand,these set are disjoined,so they satisfy(??)and therefore r(V k)=1,a contradiction.Readers could treat the other case (k=m/2?1)similarly. 2.r(M2)=m+1 2.This case occurs i?k=m?1 2 and each of the V i,i≤k has rank one. De?ne the vectors x i,i≤k the same way.A similar reasoning shows that these vectors are independent.Moreover,each x i is orthogonal to u1and1n.Since u1and1n are independent it is clear that m?1 2 ≤n?2thus m≤2n?3. 3.r(M2)=m 2.This case occurs i?k=m 2 and each V i,i≤k has rank one.The vectors x i and1n are pairwise orthogonal.The?rst consequence of this fact is that m 2≤n?1,and hence m≤2(n?1).This gives us the desired upper bound.To characterize the optimal constructions note that the x i are orthogonal to v j+v j for every j≤k,so in the optimal system v j+v j must equal to1n,namely every irregular pair should form a partition of X.It yields that each x i is an(1,?1)vector.From this we can conclude that the optimal construction exists i?we could?nd an orthogonal basis of R n consisting of1n and some (1,?1)vectors.This is equivalent to the problem of?nding a Hadamard’s matrix of order n. Remark.From any Hadamard’s matrix{h ij}of order n,one can construct an optimal system as follows.First we can assume that the?rst row of the matrix is all-one.Hence each of the remaining n?1rows consists of(n/2)1’s and(n/2)(?1)’s.Let X={1,2,...,n},and take A+i={j|h ij=1},A?i={j|h ij=?1},for i=2,3,...,n.Thus we obtain a family of 2n?2subsets of X.Moreover,|A+i∩A?i|=0and|A+i∩A?j|=|A+i∩A+i|=|A?i∩A?i|=n/4, for all i and j. 10 5 Remarks and open problems 5.1Remarks Suppose there is a system with m members having the weak intersection property with parameter s .Using an obvious greedy algorithm,we can select a subsystem with at least m/(s +1)members,which contains no exceptional intersection.Thus,we obtain a trivial upper-bound: m s (n,L )≤(s +1)m 0(n,L ) Weak version of Frankl-Wilson Theorem Let us ?rst state the (uniform)Frank-Wilson Theorem,which is one of the strongest result among Theorems on Systems with restricted intersections.Also note that Theorem 2.1is a subcase of this Theorem when p =2. Theorem 5.1.1(Frankl-Wilson)Let p be a prime number and H a set of l (i)|E |=k (modp )for every member E of F (ii)|E ∩F |∈L (modp )for every E and F in F ,E =F .Then |F|≤ n l In our terminology,it says m 0(n,L )≤ n l ,under condition (i).Consider now its weak ver- sion,which we obtain by keeping condition (i)and replacing (ii)by the following weakened condition: (ii’)For each member E ,there are at most s member F so that |E ∩F |/∈L (modp ).By the upper bound shown above,we have: m s (n,L )≤(s +1)m 0(n,L )≤(s +1) n l The following construction suggests that this bound may be approximately tight.Construction 5.1.2Consider the following partition of the ground set X (of n elements): X =A 0∪ log(s +1) i =1B i ,where |B i |=p for every i ;|A 0|=n ?p log(s +1) .De?ne an optimal system A 0with restricted intersections on A ;|A 0|=m 0(n ?p log(s +1) ).We can select s +1di?erent subsets of cardinality divisible by p ,each is the union of some B i .Call this system B .De?ne A ={A |A =A 0∪B,A 0∈A 0,B ∈B}.The family A has (s +1)m 0(n ?p log(s +1) ,L )members and each member has s intersections of non-restricted size.This yields a lower bound: (s +1)m 0(n ?p log(s +1) ,L )≤m s (n,L ) Together with the upper bound we have: Fact 5.1.3 (s +1)m 0(n ?p log(s +1) ,L )≤m s (n,L )≤(s +1)m 0(n.L ) Assume p,s,L are ?xed,if we can show: 11 lim m0(n?p log(s+1) ,L)/m0(n,L)=1 n→∞ then m s(n,L)is asymptotically(s+1)m0(n,L).In fact,it su?ces to show lim n→∞m0(n?1,L)/m0(n,L)=1.Though true it seems,we do not see how to prove this. Weak version of Erd?o s-de Bruijn Theorem When L={1},we deal with the weak version of Erd?o s-de Bruijn’s theorem.If s=1 we can show that m1(n,L)=n+1.The only optimal construction is the pencil with its top as the only additional set.In the case s=2the tight upper bound is roughly1.5n.This bound can be achieved within a constant by considering the system{0,1},{0,2},...{0,n?1},{0,1,2}{0,3,4},...,{0,2 (n?1)/2 ?1,2 (n?1)/2 }on the ground set{0,1,..,n?1}. The proof is rather technical and somewhat similar to that of4.1,so we omit this here. Weak version of Fisher Inequality For the weak version of Fisher’s inequality,the trivial upper bound is(s+1)n.Except the case s=1discussed in the previous section,we have not found any construction which gives asymptotically(s+1)n.The best construction we have gives sn/4and is described below: Construction5.1.4Consider two separate projective planes P1and P2of rank p.Let l i1,l i2,...l i q denote the lines of P i(q=p2+p+1).Let A i,k=l1i∪l2i+k for k=0,1,..,t?1 where l2q+k=l2k.The system has qt members.The ground set has n=2q points.A regular intersection has size2,and an exceptional intersection has size p+2.This occurs whenever the two members contain the same line in P1or P2.So each member in the system has exactly2t exceptional intersections. Twin versions of results in Section2and3 Each problem considered in Section2and3has its”twin”version obtain by changing the parity of the intersections(for example:consider a setsystem when every k members has odd intersection)and the parity of the size of the members in the system(if such condition is needed like in Theorem2.2,Lemma3.2.2etc).To handle the”twin”problems,we should modify the original proofs presented here(in a very natural way to?t the change of parity) and use some additional(but simple)arguments.For instance we refer to[14],where the ”twin”version of Theorem3.1.2is discussed. 5.2Open questions. Question1.Theorem2.2and2.4give tight and approximately tight bounds for the weak odd town problem,when s<2n/4and s>2n/2,respectively.It remains an open question to determine the tight upper-bound for the case of s is between2n/4and2n/2.We think the bound s(n?2log s)is still valid for most value of s in this interval.Except the case when n?2log s is very small(say1),we have not found any construction violating this bound. Question 2.We do not know if the bound in Lemma3.2.2is sharp,except the case k=2.It is shown that there is a system with the described property,having cardinality (k?1)(n?2log(k?1))[14]. 12 Question 3.Find the sharp bound in Lemma 3.2.4.It would generalize Theorem 2.2.Question 4.The optimal systems in Theorem 3.1.1and 3.1.2are characterized.([14])It would be interesting to characterize the optimal systems in Theorem 3.2.1and 3.2.3.The construction described at the end of Section 3is the only optimal construction we have found.Is it the only one ? Question 5.We wonder if one could prove results similar to Theorem 3.2.1for (prime)moduli other than 2.The proof of Theorem 3.2.1based on Lemma 3.2.2and Theorem 3.1.2.Lemma 3.2.2is easy to generalize for other moduli.So what we need is an extension of Theorem 3.1.2which determines m s (n,L )where L is the set of intergers divisible by a (prime)number p (p =2).We conjecture that m s (n,L )is 2n/p when n is su?ciently large and divisible by p and s ralatively small (in sense of Theorem 3.1.2).Even the case p =3and s =0is open. Question 6.Consider Fact 5.1.3.To prove that m s (n.L )is approximately (s +1)m o (n,L )it su?ces to prove the following: lim n →∞m 0(n ?1.L )/m 0(n,L )=1 Let us mention here that the original Frankl-Wilson bound is sharp for certain pairs of set L and number k (i.e,for certain L and k :m 0(n,L )= n |L | ),so in these cases the limit is really 1.However,there is a chance that Frankl-Wilson bound is not (approximately)tight for some L and k .It would be interesting to prove or disprove this. Question 7.The most challenging question may be:Find any bound for the weak version of Ray-Chaudhury-Wilson theorem.Precisely speaking,the problem is the following: What is m s (n,L )with L being a ?nite set of non-negative integers ? Except Theorem 4.1,which is a very special case,we do not know anything about this prob-lem.This seems already di?cult to estimate the ratio m s (n,L )/m 0(n,L ).We conjecture that,under general circumstances (e.g.|L |and n are large,s is su?ciently small,etc),this ratio is close to 1. 6References References [1]N.Alon,L.Babai and H.Suzuki,Miltilinear polynomials and Frankl-Ray-Chaudhuri- Wilson type interestion theorems,https://www.wendangku.net/doc/b66179746.html,bin.Th.A 58(1991),165-180. [2]N.Alon and J.Spencer,”The Probabilistic Methods,”Wiley,NewYork,1992. [3]L.Babai-P.Frankl,”Linear algebraic methods in combinatorics”,preprint [4]E.R.Berlekamp,On subsets with intersection of even cardinality,Canad.Math.Bull. 12(1969),363-366 [5]Bose,A note on Fishre’s inequality for balanced incomplete block designs,Ann.Math. Stat.20(1949)389-419. 13 [6]P.Er?o s-N.G.de Bruijn,On a combinatorial problem,Proc.Konink.Nederl.Akad. Wetensch.SerA.51(1948),1277-1279. [7]R.A.Fisher,An examination of the di?erent possible solutions of a problem in incom- plete blocks,Annals of Eugenics(London)10(1940),52-75. [8]P.Frankl and R.Wilson,Intersection theorems with geometric consequences,Com- binatorica1,(1981)357-368. [9]J.E.Graver,Boolean designs and self-dual matroids,Lin.Alg.Appl.10(1975), 111-128. [10]K.M.Majumdar,On some theorems in combinatorics relating to incomplete block designs,Ann.Math.Statist.,24(1953),377-389. [11]H.J.Ryser,A note on a combinatorial problem,Proc.A.M.S.(1950),422-424. [12]H.J.Ryser,An extension of a theorem of de-Bruijn and Erd?o s on combinatorial de- signs,J.Algebra10(1968),246-261. [13]D.K.Ray-Chaudhury and R.Wilson,On t-designs,Osaka,J.Math.12(1975)737-744. [14]Van H.Vu,Extremal systems with upper bounded odd intersections,to appear in Graphs and Combinatorics. [15]D.R.Woodall,Squareλ-link designs,Proc.London Math.Soc.(3)20(1970),669- 687. 14 with复合结构专项练习（二） 一请选择最佳答案 1）With nothing_______to burn，the fire became weak and finally died out. A.leaving B.left C.leave D.to leave 2）The girl sat there quite silent and still with her eyes_______on the wall. A.fixing B.fixed C.to be fixing D.to be fixed 3）I live in the house with its door_________to the south.（这里with结构作定语） A.facing B.faces C.faced D.being faced 4）They pretended to be working hard all night with their lights____. A.burn B.burnt C.burning D.to burn 二：用with复合结构完成下列句子 1）_____________（有很多工作要做），I couldn't go to see the doctor. 2）She sat__________（低着头）。 3）The day was bright_____.（微风吹拂） 4）_________________________，（心存梦想）he went to Hollywood. 三把下列句子中的划线部分改写成with复合结构。 1）Because our lessons were over，we went to play football. _____________________________. 2）The children came running towards us and held some flowers in their hands. _____________________________. 3）My mother is ill，so I won't be able to go on holiday. _____________________________. 4）An exam will be held tomorrow，so I couldn't go to the cinema tonight. _____________________________. With的用法全解 with结构是许多英语复合结构中最常用的一种。学好它对学好复合宾语结构、不定式复合结构、动名词复合结构和独立主格结构均能起很重要的作用。本文就此的构成、特点及用法等作一较全面阐述，以帮助同学们掌握这一重要的语法知识。 一、 with结构的构成 它是由介词with或without+复合结构构成，复合结构作介词with或without的复合宾语，复合宾语中第一部分宾语由名词或代词充当，第二部分补足语由形容词、副词、介词短语、动词不定式或分词充当，分词可以是现在分词，也可以是过去分词。With结构构成方式如下： 1. with或without-名词/代词+形容词； 2. with或without-名词/代词+副词； 3. with或without-名词/代词+介词短语； 4. with或without-名词/代词 +动词不定式； 5. with或without-名词/代词 +分词。 下面分别举例： 1、 She came into the room，with her nose red because of cold.（with+名词+形容词，作伴随状语） 2、 With the meal over ， we all went home.（with+名词+副词，作时间状语） 3、The master was walking up and down with the ruler under his arm。（with+名词+介词短语，作伴随状语。） The teacher entered the classroom with a book in his hand. 4、He lay in the dark empty house，with not a man ，woman or child to say he was kind to me.（with+名词+不定式，作伴随状语）He could not finish it without me to help him.（without+代词 +不定式，作条件状语） 5、She fell asleep with the light burning.（with+名词+现在分词，作伴随状语） Without anything left in the with结构是许多英 语复合结构中最常用的一种。学好它对学好复合宾语结构、不定式复合结构、动名词复合结构和独立主格结构均能起很重要的作用。本文就此的构成、特点及用法等作一较全面阐述，以帮助同学们掌握这一重要的语法知识。 二、with结构的用法 with是介词，其意义颇多，一时难掌握。为帮助大家理清头绪，以教材中的句子为例，进行分类，并配以简单的解释。在句子中with结构多数充当状语，表示行为方式，伴随情况、时间、原因或条件（详见上述例句）。 1.带着，牵着…… (表动作特征)。如： Run with the kite like this. 精神分裂症的病因及发病机理 精神分裂症病因：尚未明，近百年来的研究结果也仅发现一些可能的致病因素。（一）生物学因素1．遗传遗传因素是精神分裂症最可能的一种素质因素。国内家系调查资料表明：精神分裂症患者亲属中的患病率比一般居民高6.2倍，血缘关系愈近，患病率也愈高。双生子研究表明：遗传信息几乎相同的单卵双生子的同病率远较遗传信息不完全相同 的双卵双生子为高，综合近年来11项研究资料：单卵双生子同病率（56.7%），是双卵双生子同病率（12.7%）的4.5倍，是一般人口患难与共病率的35-60倍。说明遗传因素在本病发生中具有重要作用，寄养子研究也证明遗传因素是本症发病的主要因素，而环境因素的重要性较小。以往的研究证明疾病并不按类型进行遗传，目前认为多基因遗传方式的可能性最大，也有人认为是常染色体单基因遗传或多源性遗传。Shields发现病情愈轻，病因愈复杂，愈属多源性遗传。高发家系的前瞻性研究与分子遗传的研究相结合，可能阐明一些问题。国内有报道用人类原癌基因Ha-ras-1为探针，对精神病患者基因组进行限止性片段长度多态性的分析，结果提示11号染色体上可能存在着精神分裂症与双相情感性精神病有关的DNA序列。2．性格特征：约40%患者的病前性格具有孤僻、冷淡、敏感、多疑、富于幻想等特征，即内向 型性格。3．其它：精神分裂症发病与年龄有一定关系，多发生于青壮年，约1/2患者于20～30岁发病。发病年龄与临床类型有关，偏执型发病较晚，有资料提示偏执型平均发病年龄为35岁，其它型为23岁。80年代国内12地区调查资料：女性总患病率（7.07%。）与时点患病率（5.91%。）明显高于男性（4.33%。与3.68%。）。Kretschmer在描述性格与精神分裂症关系时指出：61%患者为瘦长型和运动家型，12.8%为肥胖型，11.3%发育不良型。在躯体疾病或分娩之后发生精神分裂症是很常见的现象，可能是心理性生理性应激的非特异性影响。部分患者在脑外伤后或感染性疾病后发病；有报告在精神分裂症患者的脑脊液中发现病毒性物质；月经期内病情加重等躯体因素都可能是诱发因素，但在精神分裂症发病机理中的价值有待进一步证实。（二）心理社会因素1．环境因素①家庭中父母的性格，言行、举止和教育方式（如放纵、溺爱、过严）等都会影响子女的心身健康或导致个性偏离常态。②家庭成员间的关系及其精神交流的紊乱。③生活不安定、居住拥挤、职业不固定、人际关系不良、噪音干扰、环境污染等均对发病有一定作用。农村精神分裂症发病率明显低于城市。2．心理因素一般认为生活事件可发诱发精神分裂症。诸如失学、失恋、学习紧张、家庭纠纷、夫妻不和、意处事故等均对发病有一定影响，但这些事件的性质均无特殊性。因此，心理因素也仅属诱发因 with用法归纳 （1）“用……”表示使用工具，手段等。例如： ①We can walk with our legs and feet. 我们用腿脚行走。 ②He writes with a pencil. 他用铅笔写。 （2）“和……在一起”，表示伴随。例如： ①Can you go to a movie with me? 你能和我一起去看电影'>电影吗？ ②He often goes to the library with Jenny. 他常和詹妮一起去图书馆。 （3）“与……”。例如： I’d like to have a talk with you. 我很想和你说句话。 （4）“关于，对于”，表示一种关系或适应范围。例如： What’s wrong with your watch? 你的手表怎么了？ （5）“带有，具有”。例如： ①He’s a tall kid with short hair. 他是个长着一头短发的高个子小孩。 ②They have no money with them. 他们没带钱。 （6）“在……方面”。例如： Kate helps me with my English. 凯特帮我学英语。 （7）“随着，与……同时”。例如： With these words, he left the room. 说完这些话，他离开了房间。 [解题过程] with结构也称为with复合结构。是由with+复合宾语组成。常在句中做状语，表示谓语动作发生的伴随情况、时间、原因、方式等。其构成有下列几种情形： 1.with+名词(或代词)+现在分词 此时，现在分词和前面的名词或代词是逻辑上的主谓关系。 例如:1)With prices going up so fast, we can't afford luxuries. 由于物价上涨很快，我们买不起高档商品。（原因状语） 2)With the crowds cheering, they drove to the palace. 在人群的欢呼声中，他们驱车来到皇宫。（伴随情况） 2.with+名词(或代词)+过去分词 此时，过去分词和前面的名词或代词是逻辑上的动宾关系。 独立主格篇 独立主格，首先它是一个“格”，而不是一个“句子”。在英语中任何一个句子都要有主谓结构，而在这个结构中，没有真正的主语和谓语动词，但又在逻辑上构成主谓或主表关系。独立主格结构主要用于描绘性文字中，其作用相当于一个状语从句，常用来表示时间、原因、条件、行为方式或伴随情况等。除名词/代词+名词、形容词、副词、非谓语动词及介词短语外，另有with或without短语可做独立主格，其中with可省略而without不可以。*注：独立主格结构一般放在句首，表示原因时还可放在句末；表伴随状况或补充说明时，相当于一个并列句，通常放于句末。 一、独立主格结构： 1. 名词/代词+形容词 He sat in the front row, his mouth half open. Close to the bank I saw deep pools, the water blue like the sky. 靠近岸时，我看见几汪深池塘，池水碧似蓝天。 2. 名词/代词+现在分词 Winter coming, it gets colder and colder. The rain having stopped, he went out for a walk. The question having been settled, we wound up the meeting. 也可以The question settled, we wound up the meeting. 但含义稍有差异。前者强调了动作的先后。 We redoubled our efforts, each man working like two. 我们加倍努力，一个人干两个人的活。 3. 名词/代词+过去分词 The job finished, we went home. More time given, we should have done the job much better. *当表人体部位的词做逻辑主语时，不及物动词用现在分词，及物动词用过去分词。 He lay there, his teeth set, his hands clenched, his eyes looking straight up. 他躺在那儿，牙关紧闭，双拳紧握，两眼直视上方。 4. 名词/代词+不定式 We shall assemble at ten forty-five, the procession to start moving at precisely eleven. We divided the work, he to clean the windows and I to sweep the floor. 精神分裂症的发病原因是什么 精神分裂症是一种精神病，对于我们的影响是很大的，如果不幸患上就要及时做好治疗，不然后果会很严重，无法进行正常的工作和生活，是一件很尴尬的事情。因此为了避免患上这样的疾病，我们就要做好预防，今天我们就请广州协佳的专家张可斌来介绍一下精神分裂症的发病原因。 精神分裂症是严重影响人们身体健康的一种疾病，这种疾病会让我们整体看起来不正常，会出现胡言乱语的情况，甚至还会出现幻想幻听，可见精神分裂症这种病的危害程度。 (1)精神刺激：人的心理与社会因素密切相关，个人与社会环境不相适应，就产生了精神刺激，精神刺激导致大脑功能紊乱，出现精神障碍。不管是令人愉快的良性刺激，还是使人痛苦的恶性刺激，超过一定的限度都会对人的心理造成影响。 (2)遗传因素：精神病中如精神分裂症、情感性精神障碍，家族中精神病的患病率明显高于一般普通人群，而且血缘关系愈近，发病机会愈高。此外，精神发育迟滞、癫痫性精神障碍的遗传性在发病因素中也占相当的比重。这也是精神病的病因之一。 (3)自身：在同样的环境中，承受同样的精神刺激，那些心理素质差、对精神刺激耐受力低的人易发病。通常情况下，性格内向、心胸狭窄、过分自尊的人，不与人交往、孤僻懒散的人受挫折后容易出现精神异常。 (4)躯体因素：感染、中毒、颅脑外伤、肿瘤、内分泌、代谢及营养障碍等均可导致精神障碍，。但应注意，精神障碍伴有的躯体因素，并不完全与精神症状直接相关，有些是由躯体因素直接引起的，有些则是以躯体因素只作为一种诱因而存在。 孕期感染。如果在怀孕期间，孕妇感染了某种病毒，病毒也传染给了胎儿的话，那么，胎儿出生长大后患上精神分裂症的可能性是极其的大。所以怀孕中的女性朋友要注意卫生，尽量不要接触病毒源。 上述就是关于精神分裂症的发病原因，想必大家都已经知道了吧。患上精神分裂症之后，大家也不必过于伤心，现在我国的医疗水平是足以让大家快速恢复过来的，所以说一定要保持良好的情绪。 with+复合宾语的用法 一、with的复合结构的构成 二、所谓"with的复合结构”即是"with+复合宾语”也即"with +宾语+宾语补足语” 的结构。其中的宾语一般由名词充当（有时也可由代词充当）；而宾语补足语则是根据 具体的需要由形容词，副词、介词短语，分词短语（包括现在分词和过去分词）及不定式短语充当。下面结合例句就这一结构加以具体的说明。 三、1、with +宾语+形容词作宾补 四、①He slept well with all the windows open.（82 年高考题） 上面句子中形容词open作with的宾词all the windows的补足语， ②It' s impolite to talk with your mouth full of food. 形容词短语full of food 作宾补。Don't sleep with the window ope n in win ter 2、with+宾语+副词作宾补 with Joh n away, we have got more room. He was lying in bed with all his clothes on. ③Her baby is used to sleeping with the light on.句中的on 是副词，作宾语the light 的补足语。 ④The boy can t play with his father in.句中的副词in 作宾补。 3、with+宾语+介词短语。 we sat on the grass with our backs to the wall. his wife came dow n the stairs,with her baby in her arms. They stood with their arms round each other. With tears of joy in her eyes ,she saw her daughter married. ⑤She saw a brook with red flowers and green grass on both sides. 句中介词短语on both sides 作宾语red flowersandgreen grass 的宾补， ⑥There were rows of white houses with trees in front of them.，介词短语in front of them 作宾补。 4、with+宾词+分词（短语 这一结构中作宾补用的分词有两种，一是现在分词，二是过去分词，一般来说，当分词所表 示的动作跟其前面的宾语之间存在主动关系则用现在分词，若是被动关系，则用过去分词。 ⑦In parts of Asia you must not sit with your feet pointing at another person.（高一第十课），句中用现在分词pointing at…作宾语your feet的补足语，是因它们之间存在主动关系，或者说point 这一动作是your feet发出的。 All the after noon he worked with the door locked. She sat with her head bent. She did not an swer, with her eyes still fixed on the wall. The day was bright,with a fresh breeze（微风）blowing. I won't be able to go on holiday with my mother being ill. With win ter coming on ,it is time to buy warm clothes. He soon fell asleep with the light still bur ning. ⑧From space the earth looks like ahuge water covered globe,with a few patches of land stuk ing out above the water而在下面句子中因with的宾语跟其宾补之间存在被动关系，故用过去分词作宾补： with用法小结 一、with表拥有某物 Mary married a man with a lot of money . 马莉嫁给了一个有着很多钱的男人。 I often dream of a big house with a nice garden . 我经常梦想有一个带花园的大房子。 The old man lived with a little dog on the lonely island . 这个老人和一条小狗住在荒岛上。 二、with表用某种工具或手段 I cut the apple with a sharp knife . 我用一把锋利的刀削平果。 Tom drew the picture with a pencil . 汤母用铅笔画画。 三、with表人与人之间的协同关系 make friends with sb talk with sb quarrel with sb struggle with sb fight with sb play with sb work with sb cooperate with sb I have been friends with Tom for ten years since we worked with each other, and I have never quarreled with him . 自从我们一起工作以来，我和汤姆已经是十年的朋友了，我们从没有吵过架。 四、with 表原因或理由 John was in bed with high fever . 约翰因发烧卧床。 He jumped up with joy . 他因高兴跳起来。 Father is often excited with wine . 父亲常因白酒变的兴奋。 五、with 表“带来”，或“带有,具有”，在…身上，在…身边之意 精神分裂症的病因是什么 精神分裂症是一种精神方面的疾病，青壮年发生的概率高，一般 在16～40岁间，没有正常器官的疾病出现，为一种功能性精神病。 精神分裂症大部分的患者是由于在日常的生活和工作当中受到的压力 过大，而患者没有一个良好的疏导的方式所导致。患者在出现该情况 不仅影响本人的正常社会生活，且对家庭和社会也造成很严重的影响。 精神分裂症常见的致病因素： 1、环境因素：工作环境比如经济水平低低收入人群、无职业的人群中，精神分裂症的患病率明显高于经济水平高的职业人群的患病率。还有实际的生活环境生活中的不如意不开心也会诱发该病。 2、心理因素：生活工作中的不开心不满意，导致情绪上的失控，心里长期受到压抑没有办法和没有正确的途径去发泄，如恋爱失败， 婚姻破裂，学习、工作中不愉快都会成为本病的原因。 3、遗传因素：家族中长辈或者亲属中曾经有过这样的病人，后代会出现精神分裂症的机会比正常人要高。 4、精神影响：人的心里与社会要各个方面都有着不可缺少的联系，对社会环境不适应，自己无法融入到社会中去，自己与社会环境不相 适应，精神和心情就会受到一定的影响，大脑控制着人的精神世界， 有可能促发精神分裂症。 5、身体方面：细菌感染、出现中毒情况、大脑外伤、肿瘤、身体的代谢及营养不良等均可能导致使精神分裂症，身体受到外界环境的 影响受到一定程度的伤害，心里受到打击，无法承受伤害造成的痛苦，可能会出现精神的问题。 对于精神分裂症一定要配合治疗，接受全面正确的治疗，最好的 疗法就是中医疗法加心理疗法。早发现并及时治疗并且科学合理的治疗，不要相信迷信，要去正规的医院接受合理的治疗，接受正确的治 疗按照医生的要求对症下药，配合医生和家人，给病人创造一个良好 的治疗环境，对于该病的康复和痊愈会起到意想不到的效果。 with复合结构 一. with复合结构的常见形式 1.“with+名词/代词+介词短语”。 The man was walking on the street, with a book under his arm. 那人在街上走着，腋下夹着一本书。 2. “with+名词/代词+形容词”。 With the weather so close and stuffy, ten to one it’ll rain presently. 天气这么闷热，十之八九要下雨。 3. “with+名词/代词+副词”。 The square looks more beautiful than even with all the light on. 所有的灯亮起来，广场看起来更美。 4. “with+名词/代词+名词”。 He left home, with his wife a hopeless soul. 他走了，妻子十分伤心。 5. “with+名词/代词+done”。此结构过去分词和宾语是被动关系，表示动作已经完成。 With this problem solved, neomycin 1 is now in regular production. 随着这个问题的解决，新霉素一号现在已经正式产生。 6. “with+名词/代词+-ing分词”。此结构强调名词是-ing分词的动作的发出者或某动作、状态正在进行。 He felt more uneasy with the whole class staring at him. 全班同学看着他，他感到更不自然了。 7. “with+宾语+to do”。此结构中，不定式和宾语是被动关系，表示尚未发生的动作。 So in the afternoon, with nothing to do, I went on a round of the bookshops. 由于下午无事可做，我就去书店转了转。 二. with复合结构的句法功能 1. with 复合结构，在句中表状态或说明背景情况，常做伴随、方式、原因、条件等状语。With machinery to do all the work, they will soon have got in the crops. 由于所有的工作都是由机器进行，他们将很快收完庄稼。（原因状语） The boy always sleeps with his head on the arm. 这个孩子总是头枕着胳膊睡觉。（伴随状语）The soldier had him stand with his back to his father. 士兵要他背对着他父亲站着。（方式状语）With spring coming on, trees turn green. 春天到了，树变绿了。（时间状语） 2. with 复合结构可以作定语 Anyone with its eyes in his head can see it’s exactly like a rope. 任何一个头上长着眼睛的人都能看出它完全像一条绳子。 【高考链接】 1. ___two exams to worry about, I have to work really hard this weekend.（04北京） A. With B. Besides C. As for D. Because of 【解析】A。“with+宾语+不定式”作状语，表示原因。 2. It was a pity that the great writer died, ______his works unfinished. (04福建) A. for B. with C. from D.of 【解析】B。“with+宾语+过去分词”在句中作状语，表示状态。 3._____production up by 60%, the company has had another excellent year. (NMET) A. As B.For C. With D.Through 【解析】C。“with+宾语+副词”在句中作状语，表示程度。 With复合结构的用法小结 with结构是许多英语复合结构中最常用的一种。学好它对学好复合宾语结构、不定式复合结构、动名词复合结构和独立主格结构均能起很重要的作用。本文就此的构成、特点及用法等作一较全面阐述，以帮助同学们掌握这一重要的语法知识。 一、with结构的构成 它是由介词with或without+复合结构构成，复合结构作介词with或without的复合宾语，复合宾语中第一部分宾语由名词或代词充当，第二 部分补足语由形容词、副词、介词短语、动词不定式或分词充当，分词可以是现在分词，也可以是过去分词。With结构构成方式如下： 1. with或without-名词/代词+形容词； 2. with或without-名词/代词+副词； 3. with或without-名词/代词+介词短语； 4. with或without-名词/代词+动词不定式； 5. with或without-名词/代词+分词。 下面分别举例： 1、She came into the room，with her nose red because of cold.（with+名词+形容词，作伴随状语） 2、With the meal over ，we all went home.（with+名词+副词，作时间状语） 3、The master was walking up and down with the ruler under his arm。（with+名词+介词短语，作伴随状语。）The teacher entered the classroom with a book in his hand. 4、He lay in the dark empty house，with not a man ，woman or child to say he was kind to me.（with+名词+不定式，作伴随状语）He could not finish it without me to help him.（without+代词+不定式，作条件状语） 5、She fell asleep with the light burning.（with+名词+现在分词，作伴随状语）Without anything left in the cupboard，shewent out to get something to eat.（without+代词+过去分词，作为原因状语） 二、with结构的用法 在句子中with结构多数充当状语，表示行为方式，伴随情况、时间、原因或条件（详见上述例句）。 With结构在句中也可以作定语。例如： 1.I like eating the mooncakes with eggs. 2.From space the earth looks like a huge water-covered globe with a few patches of land sticking out above the water. 3.A little boy with two of his front teeth missing ran into the house. 三、with结构的特点 1. with结构由介词with或without+复合结构构成。复合结构中第一部分与第二部分语法上是宾语和宾语补足语关系，而在逻辑上，却具有主谓关系，也就是说，可以用第一部分作主语，第二部分作谓语，构成一个句子。例如：With him taken care of，we felt quite relieved.（欣慰）→（He was taken good care of.）She fell asleep with the light burning. →（The light was burning.）With her hair gone，there could be no use for them. →（Her hair was gone.） 2. 在with结构中，第一部分为人称代词时，则该用宾格代词。例如：He could not finish it without me to help him. 四、几点说明： 1. with结构在句子中的位置：with 结构在句中作状语，表示时间、条件、原因时一般放在 with[wIT] prep.1.与…(在)一起，带着：Come with me. 跟我一起来吧。/ I went on holiday with my friend. 我跟我朋友一起去度假。/ Do you want to walk home with me? 你愿意和我一道走回家吗 2.(表带有或拥有)有…的，持有，随身带着：I have no money with me. 我没有带钱。/ He is a man with a hot temper. 他是一个脾气暴躁的人。/ We bought a house with a garden. 我们买了一座带花园的房子。/ China is a very large country with a long history. 中国是一个具有历史悠久的大国。3.(表方式、手段或工具)以，用：He caught the ball with his left hand. 他用左手接球。/ She wrote the letter with a pencil. 她用铅笔写那封信。4.(表材料或内容)以，用：Fill the glass with wine. 把杯子装满酒。/ The road is paved with stones. 这条路用石头铺砌。5.(表状态)在…的情况下，…地：He can read French with ease. 他能轻易地读法文。/ I finished my homework though with difficulty. 虽然有困难，我还是做完了功课。6.(表让步)尽管，虽然：With all his money, he is unhappy. 尽管他有钱，他并不快乐。/ With all his efforts, he lost the match. 虽然尽了全力，他还是输了那场比赛。7.(表条件)若是，如果：With your permission, I’ll go. 如蒙你同意我就去。8.(表原因或理由)因为，由于：He is tired with work. 他工作做累了。/ At the news we all jumped with joy. 听到这消息我们都高兴得跳了起来。9.(表时间)当…的时候，在…之后：With that remark, he left. 他说了那话就离开了。/ With daylight I hurried there to see what had happened. 天一亮我就去那儿看发生了什么事。10. (表同时或随同)与…一起，随着：The girl seemed to be growing prettier with each day. 那女孩好像长得一天比一天漂亮。11.(表伴随或附带情况)同时：I slept with the window open. 我开着窗户睡觉。/ Don’t speak with your mouth full. 不要满嘴巴食物说话。12.赞成，同意：I am with you there. 在那点上我同你意见一致。13.由…照看，交…管理，把…放在某处：I left a message for you with your secretary. 我给你留了个信儿交给你的秘书了。/ The keys are with reception. 钥匙放在接待处。14 (表连同或包含)连用，包含：The meal with wine came to ￡8 each. 那顿饭连酒每人8英镑。/ With preparation and marking a teacher works 12 hours a day. 一位老师连备课带批改作业每天工作12小时。15. (表对象或关系)对，关于，就…而言，对…来说：He is pleased with his new house. 他对他的新房子很满意。/ The teacher was very angry with him. 老师对他很生气。/ It’s the same with us students. 我们学生也是这样。16.(表对立或敌对)跟，以…为对手：The dog was fighting with the cat. 狗在同猫打架。/ He’s always arguing with his brother. 他老是跟他弟弟争论。17.(在祈使句中与副词连用)：Away with him! 带他走！/ Off with your clothes! 脱掉衣服！/ Down with your money! 交出钱来！ 【用法】1.表示方式、手段或工具等时(=以，用)，注意不要受汉语意思的影响而用错搭配，如“用英语”习惯上用in English，而不是with English。2.与某些抽象名词连用时，其作用相当于一个副词：with care=carefully 认真地/ with kindness=kindly 亲切地/ with joy=joyfully 高兴地/ with anger=angrily 生气地/ with sorrow=sorrowfully 悲伤地/ with ease=easily 容易地/ with delight=delightedly 高兴地/ with great fluency =very fluently 很流利地3.表示条件时，根据情况可与虚拟语气连用：With more money I would be able to buy it. 要是钱多一点，我就买得起了。/ With better equipment, we could have finished the job even sooner. 要是设备好些，我们完成这项工作还要快些。4.比较with 和as：两者均可表示“随着”，但前者是介词，后者是连词：He will improve as he grows older. 随着年龄的增长，他会进步的。/ People’s ideas change with the change of the times. 时代变了，人们的观念也会变化。5.介词with和to 均可表示“对”，但各自的搭配不同，注意不要受汉语意思的影响而用错，如在kind, polite, rude, good, married等形容词后通常不接介词with而接to。6.复合结构“with+宾语+宾语补足语”是一个很有用的结构，它在句中主要用作状语，表示伴随、原因、时间、条件、方式等；其中的宾语补足语可以是名词、形容词、副词、现在分词、过去分词、不定式、介词短语等：I went out with the windows open. 我外出时没有关窗户。/ He stood before his teacher with his head down. 他低着头站在老师面前。/ He was lying on the bed with all his clothes on. 他和衣躺在床上。/ He died with his daughter yet a schoolgirl. 他去世时，女儿还是个小学生。/ The old man sat there with a basket beside her. 老人坐在那儿，身边放着一个篮子。/ He fell asleep with the lamp burning. 他没熄灯就睡着了。/ He sat there with his eyes closed. 他闭目坐在那儿。/ I can’t go out with all these clothes to wash. 要洗这些衣服，我无法出去了。这类结构也常用于名词后作定语：The boy with nothing on is her son. 没穿衣服的这个男孩子是她儿子。 （摘自《英语常用词多用途词典》金盾出版社） - 1 - 精神分裂症应该怎么治疗 1、坚持服药治疗 服药治疗是最有效的预防复发措施临床大量统计资料表明，大多数精神分裂症的复发与自行停药有关。坚持维持量服药的病人复发率为40%。而没坚持维持量服药者复发率高达80%。因此，病人和家属要高度重视维持治疗。 2、及时发现复发的先兆，及时处理 精神分裂症的复发是有先兆的，只要及时发现，及时调整药物和剂量，一般都能防止复发，常见的复发先兆为：病人无原因出现睡眠不好、懒散、不愿起床、发呆发愣、情绪不稳、无故发脾气、烦躁易怒、胡思乱想、说话离谱，或病中的想法又露头等。这时就应该及时就医，调整治疗病情波动时的及时处理可免于疾病的复发。 3、坚持定期门诊复查 一定要坚持定期到门诊复查，使医生连续地、动态地了解病情，使病人经常处于精神科医生的医疗监护之下，及时根据病情变化调整药量。通过复查也可使端正人及时得到咨询和心理治疗解除病人在生活、工作和药物治疗中的各种困惑，这对预防精神分裂症的复发也起着重要作用。 4、减少诱发因素 家属及周围人要充分认识到精神分裂症病人病后精神状态的薄弱性，帮助安排好日常的生活、工作、学习。经常与病人谈心，帮助病人正确对待疾病，正确对待现实生活，帮助病人提高心理承受能力，学会对待应激事件的方法，鼓励病人增强信心，指导病人充实生活，使病人在没有心理压力和精神困扰的环境中生活。 首先是性格上的改变，塬本活泼开朗爱玩的人，突然变得沉默寡言，独自发呆，不与人交往，爱干净的人也变的不注意卫生、生活 懒散、纪律松弛、做事注意力不集中，总是和患病之前的性格完全 相悖。 再者就是语言表达异常，在谈话中说一些无关的谈话内容，使人无法理解。连最简单的话语都无法准确称述，与之谈话完全感觉不 到重心。 第三个就是行为的异常，行为怪异让人无法理解，喜欢独处、不适意的追逐异性，不知廉耻，自语自笑、生活懒散、时常发呆、蒙 头大睡、四处乱跑，夜不归宿等。 还有情感上的变化，失去了以往的热情，开始变的冷淡、对亲人不关心、和友人疏远，对周围事情不感兴趣，一点消失都可大动干戈。 最后就是敏感多疑，对任何事情比较敏感，精神分裂症患者，总认为有人针对自己。甚至有时认为有人要害自己，从而不吃不喝。 但是也有的会出现难以入眠、容易被惊醒或睡眠不深，整晚做恶梦或者长睡不醒的现象。这些都有可能是患上了精神分裂症。 1.加强心理护理 心理护理是家庭护理中的重要方面，由于社会上普遍存在对精神病人的歧视和偏见，给病人造成很大的精神压力，常表现为自卑、 抑郁、绝望等，有的病人会因无法承受压力而自杀。家属应多给予 些爱心和理解，满足其心理需求，尽力消除病人的悲观情绪。病人 生活在家庭中，与亲人朝夕相处，接触密切，家属便于对病人的情感、行为进行细致的观察，病人的思想活动也易于向家属暴露。家 属应掌握适当的心理护理方法，随时对病人进行启发与帮助，启发 病人对病态的认识，帮助他们树立自信，以积极的心态好地回归社会。 2.重视服药的依从性 精神分裂症病人家庭护理的关键就在于要让病人按时按量吃药维持治疗。如果不按时服药，精神病尤其是精神分裂症的复发率很高。精神病人在医院经过一系统的治疗痊愈后，一般需要维持2~3年的 精神分裂症患者在怎样的情况下会自杀 精神分裂症是最常见的一种精神病。早期主要表现为性格改变，如不理采亲人、不讲卫生、对镜子独笑等。病情进一步发展，即表现为思维紊乱，病人的思考过程缺乏逻辑性和连贯性，言语零乱、词不达意。精神分裂症患者随时有可能出现危险行为，这主要是指伤人毁物、自伤自杀和忽然出走。这些危险行为是受特定的精神症状支配的.那么精神分裂症患者在什么情况下会自杀呢？ 被害妄想：这是所有精神病人最常见的症状之一，多数病人采取忍耐、逃避的态度，少数病人也会“先下手为强”，对他的“假想敌”主动攻击。对此，最重要的是弄清病人的妄想对象，即：病人以为是谁要害他。假如病人的妄想对象是某个家里人，则应尽量让这位家属阔别病人，至少不要让他与病人单独在一起。 抑郁情绪：精神分裂症病人在疾病的不同时期，可能出现情绪低落，甚至悲观厌世。特别需要留意的是，有相当一部分自杀成功的病人，是在疾病的恢复期实施自杀行为的。病人在精神病症状消除以后，因自己的病背上了沉重的思想包袱，不能正确对待升学、就业、婚姻等现实问题，感到走投无路，因此选择了轻生。对此，家属一定要防患于未然，要尽早发现病人的心理困扰，及时疏导。 对已经明确表示出自杀观念的病人，家属既不要惊慌失措，也不要躲躲闪闪，要主动与病人讨论自杀的利弊，帮助病人全面、客观地评估现实中碰到的各种困难，找出切实可行的解决办法。 另外，这种病人在自杀之前，是经过周密考虑，并且做了充分预备的，例如写遗书、收拾旧物、向家人离别、选择自杀时间、预备自杀工具等。这类病人的自杀方式也是比较温顺的，多数是服药自杀。因此，他需要一定的时间来积攒足足数目的药物，这时就能看出由家属保管药品的重要性了。只要家属密切观察病人的情绪变化，是不难早期发现病人的自杀企图的。 药源性焦虑：抗精神病药的副作用之一是可能引起病人莫名的焦躁不安、手足无措，并伴有心慌、出汗、恐惧等。这些表现多是发作性的，多数发生在下午到傍晚时分，也有的病人在打长效针以后的2?3天内出现上述表现。这种时间上的规律性，有助于家属判定病人的焦虑情绪是否由于药物所致。病人急于摆脱这种强烈的痛苦，会出现冲动伤人或自伤，这些行为只是为了发泄和解脱，并不以死为终极目的。家属可以在病人发作时，给他服用小剂量的安定类药物，或者在医生的指导下，调整抗精神病药的剂量或品种，这样就可以有效地控制病人的焦虑发作。 极度兴奋：病人的精神症状表现为严重的思维紊乱、言语杂乱无章、行为缺乏目的性，这类病人也可能出现自伤或伤人毁物。由于病人的兴奋躁动是持续性的，家属有充分的思想预备，一般比较轻易防范。家属要保管好家里的刀、剪、火、煤气等危险物品，但最根本的办法，是使用大剂量的、具有强烈镇静作用的药物来控制病人的兴奋。假如在家里护理病人确有困难，则可以强制病人住院治with复合结构专项练习96126
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