Independence of P vs. NP in regards to oracle relativizations
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Independence of P vs.NP in regards to oracle relativizations.JERRALD MEEK 1.INTRODUCTION.Previously in “P is a proper subset of NP ”[Meek Article 1,2008]and “Analysis of the Deterministic Polynomial Time Solvability of the 0-1-Knapsack Problem”[Meek Article 2,2008]the present author has demonstrated that some NP-complete problems are likely to have no deterministic polynomial time solution.In this article the concepts of these previous works will be applied to the relativizations of the P vs.NP problem.By constructing examples of ?ve out of the six relativizing oracles from Baker,Gill,and Solovay [1975],it will be demonstrated that inconsistent relativizations do not preclude a proof of P =NP or P =NP .It will then become clear that the P vs.NP question is independent of oracle relativizations.2.PRELIMINARIES.Baker,Gill,and Solovay [1975]have previously shown that there are six types of oracles that can be constructed to examine the P vs.NP problem.These include (1)Oracle A such that P A =NP A .(2)Oracle B such that P B =NP B .
(3)Oracle C such that NP C is not closed under complementation.
(4)Oracle D such that P D =NP D but NP D is closed under complementation.
(5)Oracle E such that P E =NP E and P E =NP E ∩co-NP E .
(6)Oracle F such that P F ?[NP F ∩co-NP F ]?NP .
Theorem 4.4from P is a proper subset of NP .[Meek Article 1,2008]2.1.P =NP Optimization Theorem.
2·Jerrald Meek
The only deterministic optimization of a NP-complete problem that could prove P=NP would be one that can always solve a NP-complete problem by examining no more than a polynomial number of input sets for that problem.
De?nition2.1.Oracle Machine
An oracle machine can be deterministic or non deterministic.The di?erence between an oracle machine and a regular Turing Machine is that an oracle Machine has a yes state,a no state,and a query state.When the oracle machine is placed in the query state after passing a string to the oracle,if that string is an element of the oracle set,then the oracle will place the Turing Machine in the yes state,other wise it will place the Turing Machine in the no state.The processing time required by the oracle can be considered instantaneous.
2.1Encoding Methods.
In this article,two encoding methods will be used.One encoding method will encode the input sets for a problem and is identical to that used by Baker,Gill, and Solovay[1975].This will be identi?ed as“input encoding.”
The second method of encoding will be used to identify a partition of the set of all possible input sets for a problem.This method of encoding will be identi?ed as “partition encoding.”
2.1.1Partition Encoding.The concept of partition encoding is not to encode the input sets for a problem,but an entire partition of the set of all possible input sets.However,for this method to work the partitioning must be identi?ed with a speci?c problem.
The method of partition encoding used in this work will divide partitions by the number of literals in an input set which have a true value.If a problem has n literals,then there will be n+1partitions(one partition has no true literals). Notice that the problem[a∨b∨c]has di?erent input sets that result in true evaluations from the problem[?a∨b∨c]even though the cardinality of these input sets are equivalent.
It will then be necessary that the problem which a partition applies to will need to be encoded.This could be accomplished by representing a literal with a unique unary number terminated with a zero,then literal a is10,b is110,c is1110. The operators can be represented with a unary number that always starts with zero and ends with zero010represents open grouping operators{[,(,{,ect.}, 0110represents close grouping operators{],),},ect.},01110is the negation operator,011110is the inclusive or operator,0111110is the exclusive or operator, 01111110is the and operator,011111110is the equivalence operator,0111111110 is the inequality operator,ect.This encoding could then be extended in a similar manner to allow the encoding of all logical operators.
The encoding of the problem will be pre?xed by a unary number with a termi-nating0representing the number of true literals in the partition.
By this encoding method the partition of the problem[a∨b∨c]with1true literal could be represented by 10,010,10,011110,110,011110,1110,0110 ,while [?a∨b∨c]with2true literals could be represented by
110,010,01110,10,011110,110,011110,1110,0110
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Independence of P vs.NP in regards to oracle relativizations.·3 3.THE RELATIVIZING ORACLES.
Baker,Gill,and Solovay[1975]have shown that the classes of P and NP do not relativize in a consistent manner.The inconsistency of the relativizations of P and NP has often been used to justify arguments that the P vs.NP problem is unusually di?cult at best,or maybe even impossible.
By slightly altering the machine model,we can obtain di?ering answers
to the relativized question.This suggests that resolving the original
question requires careful analysis of the computational power of ma-
chines.It seems unlikely that ordinary diagonalization methods are
adequate for producing an example of a language in NP but not in P;
such diagonalizations,we would expect,would apply equally well to the
relativized classes,implying a negative answer to all relativized P=?
NP questions,a contradiction.[Baker et al.,1975]
Although it is generally accepted that diagonalization is not an e?ective method for solving the P vs.NP problem,it is the purpose of this article to demonstrate that oracle relativizations in no way prevent a solution to the P vs.NP problem. It is also worthwhile to mention that the six oracles of Baker,Gill,and Solovay [1975]do not represent an exclusive set of all oracle relativizations.There has been extensive work to?nd other unusual relativizations of complexity classes in regards to the P vs.NP question.For example,Bennett and Gill[1981]have found that random oracles can produce the relativization P A=NP A=co-NP A.However,the scope of this work will be limited to the six oracles from Baker,Gill,and Solovay [1975]with the expectation that what is learned from these six oracles may be applied to any other relativizing oracle.
3.1A P=NP Oracle.
The P=NP Optimization Theorem requires that no more than a polynomial number of input sets can be examined by a deterministic polynomial time search algorithm.By the same logic it is obvious that a deterministic oracle machine can not request more than a polynomial number of queries in polynomial time. Baker,Gill,and Solovay[1975]de?ne the oracle A such that A={ i,x,0n : NP A i accepts x in This de?nition produces an oracle such that P A=NP A.However,this de?nition does not really say anything about how the oracle functions.In this work,the oracle will be created by partition encoding strings,a string will be included in oracle A if there exists at least one element of the partition represented by the string which results in the problem associated to the string evaluating true. —Let L(X)={x:there exists y∈X such that|y|=|x|}for any oracle X.—Let x be the set of all problems in L(X). —Let S be the set such that S i is the set of all possible input sets for x i when 1≤i≤|x|. The oracle set is created by the following algorithm. (1)Let i=1. (2)Let e=1. ArXiv,Vol.V,No.N,Month20YY. 4·Jerrald Meek (3)If x i evaluates true with input S i e ,then add the partition encoding o f S i e →x i to set A,and let e=|S i|. (4)If e<|S i|,then increment e and continue at step3. (5)If e=|S i|and i<|x|,then increment i and continue at step2. (6)If e=|S i|and i=|x|,then the set A has been found. The goal was to produce an example of an oracle set such that P A=NP A.Here, the oracle A is actually such that P A=NP.Such an oracle would also create the condition that P A=NP A.Another method could be used to create an oracle such that P X=PSPACE,which would also satisfy P X=NP X,except that NP= NP X.There are multiple such oracles,however according to Mehlhorn[1973]the set of computable oracles such that P A=NP A is meager. Notice that in this example the set of all possible input sets will be exponential in size,and therefore some partitions for any problem will be exponential in size. It is then easy to see that the problem of creating the oracle set(even when the oracle is created to only work for one problem)is a member of the FNP complexity class(when P A=NP). The process that a Deterministic Oracle Machine with oracle set A uses to solve problem x i which is in NP and has n literals is the algorithm. (1)Let e=0. (2)Encode the partition for problem x i for input sets having e true literals. (3)Pass the partition encoding from step2to the oracle and enter the query state. (4)If the machine is in the yes state then halt in an accepting state. (5)If the machine is in the no state and e step2. (6)If the machine is in the no state and e=n then halt in a non-accepting state. The algorithm of the oracle machine solves any NP problem in polynomial time on a Deterministic Oracle Machine.A Non Deterministic Oracle Machine could also use this algorithm to solve the same set of problems in polynomial time.Therefore, P A=NP A. This method is dependant on an oracle set creation algorithm requiring deter-ministic exponential time for each problem that the oracle is capable of solving.If so desired,the time required to compute the oracle set could be treated as separate to the process of solving the problem. Another option is that the Deterministic Oracle Machine could be networked to both an oracle and a Non Deterministic Turing Machine.In this model the non deterministic machine generates the oracle set which it sends to the oracle.This method will allow the entire process to require polynomial time. A third method is that we simply assume that the oracle set somehow creates itself magically. Regardless of the method used to generate the oracle set,all three methods involve somehow concealing the majority of the work.This does not in any way reduce the complexity of the problem;it only removes the di?culty of the problem by placing the majority of the computational burden upon the process of oracle set creation. ArXiv,Vol.V,No.N,Month20YY. Independence of P vs.NP in regards to oracle relativizations.·5 In this example the oracle A was created such that P A=NP.It would have been valid to create oracle set A such that P A=PSPACE.However,it can not be expected that such an oracle would be any easier to create. 3.2A P=NP Oracle. The method of producing oracle set B as described by Baker,Gill,and Solovay [1975]is similar to the algorithm: —Let L(X)={x:there exists y∈X such that|y|=|x|}for any oracle X.—Let x be the set of all problems in L(X). —Let S be the set such that S i is the set of all possible input sets for x i when 1≤i≤|x|. —Let B(i)represent the elements of B placed in B prior to stage i. —Let p i(n)<|S i|be a polynomial function of n that represents the number of computations preformed by P X i and NP X i for all oracles X.If p i(n)computations have been preformed,and no accepting element of S i has been found,then the algorithm halts in the non-accepting state without examining all elements of S i. Create the oracle with the algorithm. (1)Let i=1. (2)If P B i(i)rejects x i then the next unexamined element of S i is an element of B. (3)If i<|x|then increment i and continue at step2. (4)If i=|x|then the oracle set B has been found. The process used by an oracle machine with oracle set B for?nding a solution to a problem x i in NP with n i literals is preformed by the following algorithm. (1)Search for an element of S i which results in x i evaluating true.Terminate the search if the time limit p i(n)has been reached. (2)If a solution to x i was found then,halt in an accepting state. (3)If no solution to x i was found and all elements of S i were examined in step1, then halt in a non accepting state. (4)If no solution to x i has been found and not all elements of S i have been exam- ined,then query the oracle B with an unexamined element of S i. (5)If the machine is in the yes state then halt in an accepting state. (6)If the machine is in the no state then halt in a non accepting state. This algorithm will allow for a polynomial time solution on a Non Deterministic Oracle Machine NP B because such a machine will always?nd a solution in step one for any problem in NP if one exists,and halt in step2or3. However,a Deterministic Oracle Machine may not?nd the solution in step1. When a deterministic machine queries the oracle the oracle only contains one set of the proper cardinality which may or may not be an accepting input set(because the result of the input set is not a requirement for membership in B).The oracle will then place the machine in the yes state if the input set passed to the oracle is equivalent to the one input set of proper cardinality that is an element of B.It is then the case that the Deterministic Oracle Machine may terminate in an accepting ArXiv,Vol.V,No.N,Month20YY. 6·Jerrald Meek state when there is no accepting input to the problem.On the other hand if the deterministic oracle machine fails to?nd a solution in step1,and the input set passed to the oracle is not an element of B,then the machine may terminate in a non-accepting state when an accepting input set may exist.It is then clear that oracle B does not produce a reliable polynomial time solution to the problem x i. It is then true that P B=NP B.However this is only because the oracle B is dysfunctional.Oracle B does not actually help a Non Deterministic Oracle Machine to solve a NP problem,because the non deterministic machine has the ability to solve the problem without entering the query state.The oracle is then never queried,and the dysfunctionality of the oracle does not interfere with the operation of the Non Deterministic Oracle Machine.However,a Deterministic Oracle Machine requires the assistance of the oracle to overcome the P=NP Optimization Theorem.It is then the case that a dysfunctional oracle will prevent a Deterministic Oracle Machine from solving a problem in polynomial time while not a?ecting the performance of a Non Deterministic Oracle Machine. It then becomes easy to see why P A=NP A oracle sets are meager[Mehlhorn 1973].The oracle set must be carefully fabricated in order to create such a condition. If great care has not been taken to ensure that the oracle will be of assistance to a deterministic machine,then the oracle will be dysfunctional. 3.3An Oracle such that NP is not closed under complementation. A complexity class is closed under complementation if the complements of all prob-lems in that class are members of the same class.Baker,Gill,and Solovay[1975] have shown that there exists an oracle such that one or more problems in NP C have a complement that is not in NP C. —Let L(X)={x:there exists y∈X such that|y|=|x|}for any oracle X.—Let x be the set of all problems in L(X). —Let S be the set such that S i is the set of all possible input sets for x i when 1≤i≤|x|. —Let C(i)represent the elements of C placed in C prior to stage i. —Let p i(n)<|S i|be a polynomial function of n that represents the maximum number of computations preformed by P X i and NP X i for all oracles X.If p i(n) computations have been preformed,and no accepting element of S i has been found,then the algorithm halts in the non-accepting state without examining all elements of S i. Create the oracle with the algorithm. (1)Let i=1. (2)If NP C(i) i accepts x i then any one element of S i that is accepted by x i is added to the set C. (3)If i<|x|then increment i and continue at step2. (4)If i=|x|then the oracle set C has been found. The set C is a set containing exactly one accepting input set for each problem described by L(X)that has an accepting input set. ArXiv,Vol.V,No.N,Month20YY. Independence of P vs.NP in regards to oracle relativizations.·7 The process of?nding a solution to a problem x i which is in NP with oracle set C is preformed with the following algorithm. (1)Pass each element of S i to the oracle and enter the query state. (2)If the machine is in the yes state for any query,then halt in an accepting state. (3)If the machine is in the no state for all queries,then halt in a non accepting state. For a Non Deterministic Oracle Machine,all elements of S i can be evaluated simultaneously.However,a Deterministic Oracle Machine must iterate threw them one at a time.It is then easy to see that P C=NP C. To create the oracle setˉC the compliment of C,use the algorithm. (1)Let i=1. (2)If NPˉC(i) rejects x i then add all elements of S i to the setˉC. i (3)If i<|x|then increment i and continue at step2. (4)If i=|x|then the oracle setˉC has been found. With the oracle setˉC a Deterministic Oracle Machine can solve the compliment of any NP problem by performing one query.This is becauseˉC contains all input sets for any problem in co-NP that has at least one input set that results in the problem evaluating true.If the oracle places the machine in the yes state for any input set then an accepting input set exists for the co-NP problem,although the accepting input set may not be represented by the string queried.If the oracle places the machine in the no state then no accepting input set exists for the co-NP problem.It is then the case that PˉC?co-NP. However,this situation requires that the oracle sets C andˉC are created non deterministically.It is then the case that the work of solving these problems has been done by the creation of the oracle sets,and this allows the Deterministic Oracle Machine to avoid the limitations of the P=NP Optimization Theorem when solving co-NP problems. 3.4A P=NP oracle such that NP is closed under complementation. —Let L(X)={x:there exists y∈X such that|y|=|x|}for any oracle X.—Let x be the set of all problems in L(X). —Let S be the set such that S i is the set of all possible input sets for x i when 1≤i≤|x|. —Let D(i)represent the elements of D placed in D prior to stage i. —LetˉD(i)represent the elements ofˉD placed inˉD prior to stage i. —Let p i(n)<|S i|be a polynomial function of n that represents the maximum number of computations preformed by P X i and NP X i for all oracles X.If p i(n) computations have been preformed,and no accepting element of S i has been found,then the algorithm halts in the non-accepting state without examining all elements of S i. Construct the oracle sets D andˉD with the algorithm. (1)Let i=1. ArXiv,Vol.V,No.N,Month20YY. 8·Jerrald Meek (2)Let n=1. (3)If n is even,then let e=1. (4)If n is even and S n e /∈ˉD(i)then?nd the pre?x u o f S n e with length|S n e |/2. (5)If n is even and S n e /∈ˉD(i)and u is an input set for x|u|and NP D(n) i rejects x|u|,then S n e ∈D. (6)If n is even and S n e /∈ˉD(i)and e<|S n|then increment e and continue at step 4. (7)If n is even and S n e /∈ˉD(i)and e=|S n|then break to step10. (8)If n is odd and all elements ofˉD(n)have cardinality less than n,and p i(n)< 2(n?1)/2,then add toˉD all strings queried by P D(n) i for problem x n.If P D(n) i rejects x n,then add to D the next element of S n not queried. (9)If n is odd,then increment i. (10)If n<|x|,then increment n and continue at step3. (11)If n=|x|,then the sets D andˉD have been found. In steps4and5,elements are added to D if their pre?x is rejected.Notice that if f(x)has no input set resulting in a true evaluation,then f(x)∧g(x)has no input set resulting in a true evaluation.It is then the case that in step4and5,elements are only added to D if they represent a problem that has no accepting input set. In step8,an input set is added to D if P D(n) i rejects the problem which that input set belongs to.However,on the?rst iteration,D(n)will be an empty set. An empty oracle set is equivalent to having no oracle,and so the polynomial time algorithm will not check all possible input sets for the?rst problem examined. For the third problem examined,D(n)may contain an input set if it belongs to a problem that always evaluates false.It is then easy to see that step8adds elements to D that may or may not cause a problem to evaluate true. Because the elements of D do not have any consistent representation,it then follows that the set D will result in a dysfunctional oracle.A deterministic machine requires a functional oracle set to be able to solve all NP problems,but a non deterministic machine does not rely upon the oracle set when solving NP problems. It then follows that P D=NP D. In step8,the requirement that all elements ofˉD(n)must have cardinality less than n will always evaluate true.The statement p i(n)<2(n?1)/2will evaluate true for the?rst values of n up to some limit.For problems less than this limit,a polyno-mial number of elements will be added toˉD.The elements will be added regardless of weather or not they result in a true evaluation of the co-NP problem associated to them.It is then clear to see that the setˉD will also result in a dysfunctional oracle,and PˉD will not produce reliable results for any co-NP problem. 3.5A P=NP oracle such that P=[NP∩co-NP]. Baker,Gill,and Solovay[1975]construct the oracle set E by adding elements to the oracle set A which was previously constructed such that P A=NP A.The additional elements in E will result in P E=NP E.The trick is to arrange the additional elements such that if a problem is in NP,and the compliment of the problem is also in NP,then both of these problems are solvable in polynomial time by the oracle P E. ArXiv,Vol.V,No.N,Month20YY. Independence of P vs.NP in regards to oracle relativizations.·9 In the description of creating oracle E ,it is allowed to start with “any oracle A such that P A =NP A .”Here,the oracle A that was previously constructed in this article will be used.However,remember that oracle A did not contain strings representing input sets,but instead partition encoding was used.It will then be the case that the same encoding speci?cation will need to be used for the additional elements of E . —Let L (X )={x :there exists y ∈X such that |y |=|x |}for any oracle X .—Let x be the set of all problems in L (X ). —Let S be the set such that S i is the set of all possible input sets for x i when 1≤i ≤|x |. —Let e (n )be a function such that e (0)=0and e (x )=22e (x ?1)←x >1. —Let E (0)=A and let E (n )be the set of elements placed in E prior to stage n .—Let r be the set of all possible values for j,k ←j ∈N,k ∈N,j =k . —Let r j (i )denote the j element of set r i ,and r k (i )denote the k element of r i .—Let p i (n )<|S i |be a polynomial function of n that represents the maximum number of computations preformed by P X i and NP X i for all oracles X .If p i (n )computations have been preformed,and no accepting element of S i has been found,then the algorithm halts in the non-accepting state without examining all elements of S i . Create the oracle set E with the algorithm. (1) Let n =1.(2) Let i =1.(3) Let l ∈S n .(4)If e (n ?1)<(log 2|l |)≤e (n )≤max p r j (i )(l ),p r k (i )(l ) neither NP E (n )r j (i )or NP E (n ) r k (i ).accepts x n ,then j,k is unsatis?ed. (5)If p i (e (n ))≥2e (n ),then i,i is unsatis?ed. (6)If j,k was not determined unsatis?ed in step 4,and i,i was not determined unsatis?ed in step 5,then increment i . (7)If j,k was not determined unsatis?ed in step 4,and i,i was determined unsatis?ed in step 5,and P E (n )i rejects x n ,then add to E the encoding of x n with the next unevaluated element of S n . (8)If j,k was not determined unsatis?ed in step 4,and i,i was determined unsatis?ed in step 5,then increment i . (9)If i <|N |then increment n and continue at step 3. (10)If i =|N |then the set E has been found. Notice that when i,i is unsatis?ed,then P E (n )i will be run on x n .The set E (n )will only contain elements relevant to x n if it was inherited from set A .It is then the case that P E (n )i will only reject x n when x n has no accepting input set.In step 7,if i,i is unsatis?ed,and P E (n )i rejects x n ,then the input set encoding for x n will always be an encoding that represents a partition containing no accepting input set.It will then be the case that P E will accept x n when a Non Deterministic ArXiv,Vol.V,No.N,Month 20YY. 10·Jerrald Meek Oracle Machine would halt in a non accepting state without querying the oracle. Therefore,P E=NP E because the oracle E is dysfunctional when handling some problems. Letκbe the set of all problems in NP which have complements in NP.When x n∈κ,then j,k will be satis?ed.When j,k is satis?ed then no strings applicable to x n will be added to E.It is then the case that the subset of E that applies to all elements ofκwill be identical to the subset of A that applies to all elements ofκ. So P E=NP E when the problem being evaluated is an element ofκ. 3.6An oracle set such that P F?[NP F∩co-NP F]?NP. Baker,Gill and Solovay[1975]do not lay out the speci?c method for creating oracle F.However they do provide an outline which includes creating two languages L1(F)and L2(F)such that L(F)=[L1(F)∪L2(F)].Two oracle sets must also be created by two di?erent methods,and F is the union of these two oracle sets. However,notice that if P F?[NP F∩co-NP F],and[NP F∩co-NP F]?NP, then P F?of NP. It has already been demonstrated that an oracle set can be created such that P A=NP.In other words the set of problems solvable by a Deterministic Oracle Machine with oracle set A in polynomial time is equivalent to the set of problems solvable by a Non Deterministic Turing Machine in polynomial time. It is then the case that the only new condition of oracle set F is that[NP F∩co-NP F]contains all problems in NP.This is the same thing as saying that NP is strictly contained within co-NP. Earlier,it was shown that an oracle set C could be created such that NP C is not closed under complementation.The reason that oracle set C allowed this condition was because it resulted in all problems in co-NP being solvable in polynomial time by PˉC,while problems in NP where not solvable in polynomial time by P C. It is then easy to see that an oracle F could be created such that all problems in NP are solvable in polynomial time by P F,and all problems in co-NP are solvable in polynomial time by PˉF.Under such an oracle the complexity classes NP,and co-NP are both contained by P. If co-NP?PˉF,and NP?P F,then P F?[NP∩co-NP].All problems in both NP and co-NP have polynomial time solutions on a Deterministic Oracle Machine with oracle set F.It then follows that all problems in NP and co-NP have polynomial time solutions on a Non Deterministic Oracle Machine with oracle set F. 4.ORACLE RELATIVIZATIONS AND P VS.NP SOLUTION METHODS. Here,the relation of oracle relativizations to the P vs.NP question will be exam-ined by creating an analog of the P vs.NP question.Two complexity classes will be created,PΛand NPΛ;the question will be asked,is PΛ=NPΛ? To create the analog complexity classes,one problem that is obviously in P will be excluded from PΛ.Any problem in P will due,here the problem used will be the set-sum problem.The set-sum problem will be de?ned as: —Let S be a set of real numbers. —Let r be the number of elements in S. ArXiv,Vol.V,No.N,Month20YY. Independence of P vs.NP in regards to oracle relativizations.·11—Let M be a real number. r S i=M i=1 Given the value of S and M,determine if the equality evaluates true. Notice that the set-sum problem is similar to the0-1-Knapsack problem,and could be solved by?nding the sum of all subsets of S,but only evaluate true if the one subset of S that contains all elements of S sums to M.In the analog complexity model,it will be assumed that this is the only known method of solving set-sum. An analog complexity model will be created which ignores the existence of some NP problems(such as NP-complete problems).The analog complexity classes will be de?ned as: —Let NPΛcontain all problems known to be in P assuming P=NP. —Let PΛcontain all problems in NPΛexcept the set-sum problem. In the analog computational model NPΛcontains all problems known to be solv-able in non deterministic polynomial time,while PΛcontains all problems known to be solvable in deterministic polynomial time.In the analog question PΛ=NPΛassume that no method has yet been found to prove that set-sum is a member of PΛ,yet set-sum has not been conclusively excluded from PΛ. Now the following questions may be asked. —Does there exist an oracle such that P AΛ=NP AΛ? —Does there exist an oracle such that P BΛ=NP BΛ? —Does there exist an oracle such that NP CΛis not closed under complementation?—Does there exist an oracle such that P DΛ=NP DΛbut NP DΛis closed under com-plementation? —Does there exist an oracle such that P FΛ?[NP FΛ∩co-NP FΛ]?NPΛ? It should be easy to see that the answer to all of these questions is yes.If inconsistent relativizations are an indication of the di?culty of the PΛvs.NPΛproblem,then the problem of proving the set-sum problem solvable in deterministic polynomial time must be at least as di?cult as proving that NP-complete problems are solvable in deterministic polynomial time. Is the set of computable oracles such that P XΛ=NP XΛmeager?If so,then does that indicate a greater likelihood that PΛ=NPΛ? With this analog it happens to be the case that PΛ=NPΛ.If the P vs.NP ques-tion is dependant on oracle relativizations,then the analog PΛvs.NPΛquestion should indicate that P=NP.If P vs.NP is dependant on oracle relativizations, then it should be expected that P=NP could be proven by?nding the relation be-tween the polynomial time solvability of set-sum and the relativizations of PΛand NPΛ.If such a relation exists,then it should provide an indication of how to prove P=NP.However,if no such relation exists,then there must exist no dependancy between oracle relativizations and the solution to the P vs.NP question. Obviously,the proof that PΛ=NPΛcan be accomplished by eliminating the evaluation of all subsets of S for the set-sum problem and only evaluating the one ArXiv,Vol.V,No.N,Month20YY. 12·Jerrald Meek relevant subset.It is known that the only relevant subset is the subset equivalent to S.This is known because the original de?nition of the problem says it is so.Is there some way that the oracle relativizations also identify the one relevant subset of S?If so then the relativizations should identify a limited partition of relevant subsets for the Knapsack problem. It is then easy to see that oracle relativizations only tell us that the complexity of any given problem can be hidden by oracle set creation.Essentially,using an oracle to solve a NP-complete problem in deterministic polynomial time is the reverse operation of taking a problem from the P complexity class and pretending that it has no deterministic polynomial time solution.When applied to a NP-complete problem,which may actually have no deterministic polynomial time solution,an oracle such that P A=NP A allows us to pretend that NP-complete problems have deterministic polynomial time solutions. 5.CONCLUSION. In this article all six oracles form Baker,Gill and Solovay[1975]have been examined. As a result,the following things can be said about the oracles. —Oracle set A such that P A=NP A is a functional oracle which allows a Deter-ministic Oracle Machine to solve any NP problem with a polynomial number of queries. —Oracle set B such that P B=NP B is a dysfunctional oracle which fails to allow a Deterministic Oracle Machine to solve all NP B problems with a polynomial number of queries. —Oracle set C such that NP C is not closed under complementation is a dysfunc-tional oracle such that a Deterministic Oracle Machine can not solve all NP problems with a polynomial number of queries,although a Deterministic Oracle Machine could solve any co-NP problem with one query. —Oracle set D such that P D=NP D but NP D is closed under complementation is a dysfunctional oracle with a dysfunctional complement.It is then the case that a Deterministic Oracle Machine can not solve any NP problem or any co-NP problem in polynomial time. —Oracle set E such that P E=NP E and P E=[NP E∩co-NP E]is a dysfunctional oracle that allows a Deterministic Oracle Machine to solve a NP problem in polynomial time when the compliment of the problem is also in NP. —Oracle set F such that P F?[NP F∩co-NP F]?NP is a functional oracle which allows a Deterministic Oracle Machine to solve all problems in NP and all problems in co-NP in polynomial time. If P=NP then all problems in NP will be solvable in deterministic polynomial time with or without an oracle set.It would then be the case that a Deterministic Oracle Machine would only be dependant on an oracle set to solve a problem in NP or co-NP if the machine is not using an optimal algorithm.It is possible to create an algorithm for any NP problem such that a solution would require exponential time on a Non Deterministic Turing Machine.This however does not disqualify the problem’s membership in NP.Likewise,just because a NP problem has a determin-ArXiv,Vol.V,No.N,Month20YY. Independence of P vs.NP in regards to oracle relativizations.·13 istic exponential time solution,this does not disqualify that problem’s membership in P.Therefore,the oracle relativizations are no indicator of P vs.NP. If P=NP then any problem not ordinarily in P will be solvable in polynomial time by a Deterministic Oracle Machine only when the oracle set is functional for that problem.In this case if P=NP then the Baker,Gill,and Solovay model works exactly as would be expected. The meagerness of computable P X=NP X oracle sets also seems not to indicate anything about the P vs.NP problem.That is,we should expect a functional oracle for any purpose to be meager.All of these observations lead to the conclusion that the P vs.NP question is independent of oracle relativizations. 6.VERSION HISTORY. The author wishes to encourage further feedback which may improve,strengthen, or perhaps disprove the content of this article.For that reason the author does not publish the names of any speci?c people who may have suggested,commented,or criticized the article in such a way that resulted in a revision,unless premission has been granted to do so. arXiv Current Version 20May08Submitted to arXiv. —Clari?cations were made for some statements. —The Oracle relativizations and P vs.NP solution methods section was added.—Refrance to[Bennett and Gill,1981]and[Mehlhorn1973]were added. arXiv Version2 16May08The article was withdrawn due to misleading statements and incomplete research. arXiv Version1 14May08Submitted to arXiv. REFERENCES Baker,T.,Gill,J.,and Solovay,R.1975.Relativizations of the P=?NP question.SIAM J. Comput.4,(Dec.),431-442.MR395311.Zbl0323.68033. Bennett,C.,and Gill,J.1981.Relative to a random oracle A,P A=NP A=co-NP A with probability1.SIAM https://www.wendangku.net/doc/e616010054.html,put.10,(Feb.),96-113.MR0605606.Zbl0454.68030. Meek,J.2008.P is a proper subset of NP.arXiv:0804.1079Article1in series of4. Meek,J.2008.Analysis of the deterministic polynomial time solvability of the0-1-Knapsack problem.arXiv:0805.0517Article2in series of4. Mehlhorn,K.1973.On the size of sets of computable functions.14th Annual Symp.on Automata and Switching Theory(Univ.Iowa,Iowa City,Iowa),190-196.MR429513. Received xx/2008;revised xx/2008;accepted xx/2008 ArXiv,Vol.V,No.N,Month20YY. 比较级与最高级 1.as...as 与(not) as(so)...as as...as...句型中，as的词性 第一个as是副词，用在形容词和副词的原级前，常译为“同样地”。第二个as是连词，连接与前面句子结构相同的一个句子(相同部分常省略)，可译为“同..... He is as tall as his brother is (tall) . (后面的as 为连词) 只有在否定句中，第一个as才可换为so 改错: He is so tall as his brother.（X） 2.在比较状语从句中，主句和从句的句式结构一般是相同的 与as...as 句式中第二个as一样，than 也是连词。as和than这两个连词后面的从句的结构与前面的句子大部分情况下结构是相同的，相同部分可以省略。 He picked more apples than she did. 完整的表达为: He picked more apples than she picked apples. 后而的picked apples和前面相同，用did 替代。 He walked as slowly as she did.完整表达为: He walked as slowly as she walked slowly. she后面walked slowly与前面相同，用did替代。 3.谓语的替代 在as和than 引导的比较状语从句中，由于句式同前面 主句相同，为避免重复，常把主句中出现而从句中又出现的动词用do的适当形式来代替。 John speaks German as fluently as Mary does. 4.前后的比较对象应一致 不管后面连词是than 还是as,前后的比较对象应一致。The weather of Beijing is colder than Guangzhou. x than前面比较对象是“天气”，than 后面比较对象是“广州”，不能相比较。应改为: The weather of Bejing is colder than that of Guangzhou. 再如: His handwriting is as good as me. 应改为: His handwriting is as good as mine. 5.可以修饰比较级的词 常用来修饰比较级的词或短语有: Much,even,far,a little,a lot,a bit,by far,rather,any,still,a great deal等。 by far的用法: 用于强调，意为“...得多”“最最...”“显然”等，可修饰形容词或副词的比较级和最高级，通常置于其后，但是若比较级或最高级前有冠词，则可置于其前或其后。 The way 的用法 Ⅰ常见用法： 1)the way+ that 2)the way + in which（最为正式的用法） 3)the way + 省略（最为自然的用法） 举例：I like the way in which he talks. I like the way that he talks. I like the way he talks. Ⅱ习惯用法： 在当代美国英语中，the way用作为副词的对格，“the way+ 从句”实际上相当于一个状语从句来修饰整个句子。 1)The way =as I am talking to you just the way I’d talk to my own child. He did not do it the way his friends did. Most fruits are naturally sweet and we can eat them just the way they are—all we have to do is to clean and peel them. 2）The way= according to the way/ judging from the way The way you answer the question, you are an excellent student. The way most people look at you, you’d think trash man is a monster. 3）The way =how/ how much No one can imagine the way he missed her. 4）The way =because 人教版（新目标）初中英语形容词与副词的比较级与最高级 （一）规则变化： 1．绝大多数的单音节和少数双音节词，加词尾-er ，-est tall—taller—tallest 2．以不发音的e结尾的单音节词和少数以-le结尾的双音节词只加-r，-st nice—nicer—nicest , able—abler—ablest 3．以一个辅音字母结尾的重读闭音节词或少数双音节词，双写结尾的辅音字母，再加-er，-est big—bigger—biggest 4．以辅音字母加y结尾的双音节词，改y为i再加-er，-est easy—easier—easiest 5．少数以-er，-ow结尾的双音节词末尾加-er，-est clever—cleverer—cleverest, narrow—narrower—narrowest 6．其他双音节词和多音节词，在前面加more，most来构成比较级和最高级 easily—more easily—most easily （二）不规则变化 常见的有： good / well—better—best ; bad (ly)/ ill—worse—worst ; old—older/elder—oldest/eldest many / much—more—most ; little—less—least ; far—farther/further—farthest/furthest 用法： 1．原级比较：as + adj./adv. +as（否定为not so/as + adj./adv. +as）当as… as中间有名字时，采用as + adj. + a + n.或as + many / much + n. This is as good an example as the other is . I can carry as much paper as you can. 表示倍数的词或其他程度副词做修饰语时放在as的前面 This room is twice as big as that one. 倍数+as+adj.+as = 倍数+the +n.+of Your room is twice as larger as mine. = Your room is twice the size of mine. 2．比较级+ than 比较级前可加程度状语much, still, even, far, a lot, a little, three years. five times,20%等 He is three years older than I (am). 表示“（两个中）较……的那个”时，比较级前常加the（后面有名字时前面才能加冠词） He is the taller of the two brothers. / He is taller than his two brothers. Which is larger, Canada or Australia? / Which is the larger country, Canada or Australia? 可用比较级形式表示最高级概念，关键是要用或或否定词等把一事物（或人）与其他同类事物（或人）相分离 He is taller than any other boy / anybody else. 大多数形容词有三种形式,原级,比较级和最高级, 以表示形容词说明的性质在程度上的不同。 形容词的原级: 形容词的原级形式就是词典中出现的形容词的原形。例如: poor tall great glad bad 形容词的比较级和最高级: 形容词的比较级和最高级形式是在形容词的原级形式的基础上变化的。分为规则变化和不规则变化。 规则变化如下: 1) 单音节形容词的比较级和最高级形式是在词尾加 -er 和 -est 构成。 great (原级) (比较级) (最高级) 2) 以 -e 结尾的单音节形容词的比较级和最高级是在词尾加 -r 和 -st 构成。wide (原级) (比较级) (最高级) 3)少数以-y, -er, -ow, -ble结尾的双音节形容词的比较级和最高级是在词尾加 -er 和 -est 构成。 clever(原级) (比较级) (最高级) 4) 以 -y 结尾,但 -y 前是辅音字母的形容词的比较级和最高级是把 -y 去掉,加上 -ier 和-est 构成. happy (原形) (比较级) (最高级) 5) 以一个辅音字母结尾其前面的元音字母发短元音的形容词的比较级和最高级是双写该辅音字母然后再加 -er和-est。 big (原级) (比较级) (最高级) 6) 双音节和多音节形容词的比较级和最高级需用more 和 most 加在形容词前面来构成。 beautiful (原级) (比较级) (比较级) difficult (原级) (最高级) (最高级) 常用的不规则变化的形容词的比较级和最高级: 原级------比较级------最高级 good------better------best many------more------most much------more------most bad------worse------worst far------farther, further------farthest, furthest 形容词前如加 less 和 least 则表示"较不"和"最不 形容词比较级的用法: 形容词的比较级用于两个人或事物的比较,其结构形式如下: 主语+谓语(系动词)+ 形容词比较级+than+ 对比成分。也就是, 含有形容词比较级的主句+than+从句。注意从句常常省去意义上和主句相同的部分, 而只剩下对比的成分。 The way的用法及其含义（二） 二、the way在句中的语法作用 the way在句中可以作主语、宾语或表语： 1.作主语 The way you are doing it is completely crazy.你这个干法简直发疯。 The way she puts on that accent really irritates me. 她故意操那种口音的样子实在令我恼火。The way she behaved towards him was utterly ruthless. 她对待他真是无情至极。 Words are important, but the way a person stands, folds his or her arms or moves his or her hands can also give us information about his or her feelings. 言语固然重要,但人的站姿,抱臂的方式和手势也回告诉我们他(她)的情感。 2.作宾语 I hate the way she stared at me.我讨厌她盯我看的样子。 We like the way that her hair hangs down.我们喜欢她的头发笔直地垂下来。 You could tell she was foreign by the way she was dressed. 从她的穿著就可以看出她是外国人。 She could not hide her amusement at the way he was dancing. 她见他跳舞的姿势，忍俊不禁。 3.作表语 This is the way the accident happened.这就是事故如何发生的。 Believe it or not, that's the way it is. 信不信由你, 反正事情就是这样。 That's the way I look at it, too. 我也是这么想。 That was the way minority nationalities were treated in old China. 那就是少数民族在旧中 一、规范化管理 规范化管理就是从企业生产经营系统的整体出发，对各环节输入的各项生产要素、转换过程、产出等制订制度、规程、指标等标准（规范），并严格地实施这些规范，以使企业协调统一地运转。 实行规范化管理在理论和实践中都证明是极为重要的。首先，这是现代化大生产的客观要求。现代企业是具有高度分工与协作的社会化大生产，只有进行规范化管理，才能把成百上千人的意志统一起来，形成合力为实现企业的目标而努力工作。其次，实行规范化管理是变人治为法治的必然选择。每个员工都有干好本职工作的愿望，但在没有“干好”的标准的情况下，往往凭领导者的主观印象进行考核和奖惩，难免出现在管理中时紧时松、时宽时严的现象，并很容易挫伤员工的积极性。 按照统一的规范进行严格管理，人和人之间可以公正比较、平等竞争。最后，实行规范化管理是提高员工总体素质的客观要求。规范使员工明确企业对自己的要求，有了努力的标准，必然能逐步提高自己的素质；员工还可以对照规范进行自我管理。因为规范是在系统原则下设计出来的，管理人员依据规范进行管理，也能提高立足本职、纵观全局的管理水平。 二、标准化管理 所谓标准，是指依据科学技术和实践经验的综合成果，在协商的基础上，对经济、技术和管理等活动中，具有多样性的、相关性征的重复事物，以特定的程序和形式颁发的统一规定。标准可分为技术标准和管理标准两大类。 技术标准是对技术活动中，需要统一协调的事物制定的技术准则。它是根据不同时期的科学技术水平和实践经验，针对具有普遍性和重复出现的技术问题，提出的最佳解决方案。 管理标准是企业为了保证与提高产品质量，实现总的质量目标而规定的各方面经营管理活动、管理业务的具体标准。若按发生作用的范围分，标准又可分为国际标准、国家标准、部颁标准和企业标准。以生产过程的地位分，又有原材料标准、零部件标准、工艺和工艺装备标准、产品标准等。在标准化工作中，又通常把标准归纳为：基础标准、产品标准、方法标准和卫生安全标准。 英语比较级和最高级的用法归纳 在学习英语过程中，会遇到很多的语法问题，比如比较级和最高级的用法，对于 这些语法你能够掌握吗？下面是小编整理的英语比较级和最高级的用法，欢迎阅读！ 英语比较级和最高级的用法 一、形容词、副词的比较级和最高级的构成规则 1.一般单音节词和少数以-er，-ow结尾的双音节词，比较级在后面加-er，最高级 在后面加-est; (1)单音节词 如：small→smaller→smallest short→shorter→shortest tall→taller→tallest great→greater→greatest (2)双音节词 如：clever→cleverer→cleverest narrow→narrower→narrowest 2.以不发音e结尾的单音节词，比较在原级后加-r，最高级在原级后加-st; 如：large→larger→largest nice→nicer→nicest able→abler→ablest 3.在重读闭音节(即：辅音+元音+辅音)中，先双写末尾的辅音字母，比较级加-er，最高级加-est; 如：big→bigger→biggest hot→hotter→hottest fat→fatter→fattest 4.以“辅音字母+y”结尾的双音节词，把y改为i，比较级加-er，最高级加-est; 如：easy→easier→easiest heavy→heavier→heaviest busy→busier→busiest happy→happier→happiest 5.其他双音节词和多音节词，比较级在前面加more，最高级在前面加most; 如：bea utiful→more beautiful→most beautiful different→more different→most different easily→more easily→most easily 注意：(1)形容词最高级前通常必须用定冠词 the，副词最高级前可不用。 例句： The Sahara is the biggest desert in the world. (2) 形容词most前面没有the，不表示最高级的含义，只表示"非常"。 It is a most important problem. =It is a very important problem. 定冠词the的用法： 定冠词the与指示代词this ,that同源,有“那(这)个”的意思,但较弱,可以和一个名词连用,来表示某个或某些特定的人或东西. (1)特指双方都明白的人或物 Take the medicine.把药吃了. (2)上文提到过的人或事 He bought a house.他买了幢房子. I've been to the house.我去过那幢房子. (3)指世界上独一无二的事物 the sun ,the sky ,the moon, the earth (4)单数名词连用表示一类事物 the dollar 美元 the fox 狐狸 或与形容词或分词连用,表示一类人 the rich 富人 the living 生者 (5)用在序数词和形容词最高级,及形容词等前面 Where do you live?你住在哪? I live on the second floor.我住在二楼. That's the very thing I've been looking for.那正是我要找的东西. (6)与复数名词连用,指整个群体 They are the teachers of this school.(指全体教师) They are teachers of this school.(指部分教师) (7)表示所有,相当于物主代词,用在表示身体部位的名词前 She caught me by the arm.她抓住了我的手臂. (8)用在某些有普通名词构成的国家名称,机关团体,阶级等专有名词前 the People's Republic of China 中华人民共和国 the United States 美国 (9)用在表示乐器的名词前 She plays the piano.她会弹钢琴. (10)用在姓氏的复数名词之前,表示一家人 the Greens 格林一家人(或格林夫妇) (11)用在惯用语中 in the day, in the morning... the day before yesterday, the next morning... in the sky... in the dark... in the end... on the whole, by the way... 学校后勤管理精细化体现“细”“实”“精” 大气，即大度，海纳百川的气度。做人大气方能容天下之事，运筹帷幄，成就大事。精细，精心细致。事无大小，惟有精细，才能事半功倍。成功源于大气、精细。 在学校的后勤管理工作中,没有精细化管理就谈不上有效的后勤管理。学校后勤管理工作，就是需要大家齐心协力，精细工作，用心工作，做好“三服务”，优质、安全为学校提供保障。 何谓精细化管理？精细化管理就是落实管理责任，变一人操心为大家操心，将管理责任具体化，明确化。它要求每一个人都要第一次就把工作做到位，对工作负责，对岗位负责，人人都管理，处处有管理，事事见管理。精细化管理要求层层完善、系统健康，权力层层有，任务个个担，责任人人负。在日常管理中，它要求每一个步骤都要精心，每一个环节都要精细，每一项工作都是精品，树立精心是态度，精细是过程，精品是成绩的服务态度。 精细化管理必须有制度，有过程，有监管，有反馈，有整改，最核心的是落实。它要求后勤工作人人有事做，事事有人做，要用心去做好每件细枝末节的事情。要将平凡事情做好，将简单事情做精，将小事做实。 精细化管理就是人要在“勤”字上多下功夫，下硬功夫，下实功夫。确定目标，分解任务，责任到人。不管是领导还是从业人员应该在校园特别是容易被忽略的地方多走走，在自己每天工作记录本上多记记，在大会小会上多讲讲，对全校教职工特别是后勤人员多交流，好的经验要多宣传，反面教训要作案例多分析，让大家引以为戒。 对如何实施后勤精细化管理，关键要在“细化”“务实”“精致”三个方面下功夫,精细化管理体现在“细”字上，落实在“实”字上，成效在“精”字上。 一、精细管理体现在“细”字上. 为教育教学、科研及师生的生活服务，是学校后勤的根本任务。这是由学校后勤管理工作的服务性特点所决定的。如何服务呢?为保证学校各项工作的正常运转，首先必须为学校教育、教学、科研提供基础设施、物质条件和相应的管理，同时还必须为全体师生员工提供吃、住、行的条件，以保证员工的正常生活。就其内容来说，学校后勤工作的服务包括资产管理服务、校舍管理服务、生活后勤服务、基本建设服务以及财务管理服务等诸多方面。这就要求后勤管理者对人、财、物进行计划、组织、监督、指挥、评估，使之最大限度地满足教育、教学、科研和师生员工的生活需要，为培养人才创造条件，为教育教学、科研及师生员工等提供优质的服务。要达到这样要求，后勤服务就必须做到主动、热情、周到、细致。要善于抓小事、抓细节。重视细节、善于抓住带有倾向性的小事和细节，这实质是提倡一种认真的态度和科学的精神。在学校的后勤工作中没有一件事情小到不值得去做，也没有一个细节应该被忽略或忽视，后勤管理工作尤为突出的：环境布置、卫生打扫、水电维修、校舍维护，桌椅门窗维修、伙食供应、食品卫生、校园安全、车辆安全、物品采购及供应、绿化养护等等。件件都是不起眼的琐碎小事，而这些繁杂小事如不认真对待或不及时处理，小到影响教学环境或某一方面、某些人或某个人，大到影响正常的教学秩序，甚至还会造成大的事故或隐患。所以必须认真细致地对待每个细节，在细字上下功夫。通过对在精细化管理方面的讨论，本组统一了认识，明确了责任。达成了以下共识。 “细”必须健全管理制度。做任何事都必须行之有据，按章办事，有规可依，后勤服务作为一项基础性和保障性的工作是学校工作中不可缺少的重要组成部分。要做好、做细这方面的工作必须健全操作性强的规章制度，制度本身就是管理的一个重要部分，俗话说“不怕做不到就怕想不到。”只有想得细、想得全，才能做得到位。 学校后勤部门修订制订了各方面的规章制度，有各工种岗位的岗位职责，内容涵盖了学 More than的用法 A. “More than+名词”表示“不仅仅是” 1)Modern science is more than a large amount of information. 2)Jason is more than a lecturer; he is a writer, too. 3) We need more than material wealth to build our country.建设我们国家，不仅仅需要物质财富. B. “More than+数词”含“以上”或“不止”之意，如： 4)I have known David for more than 20 years. 5)Let's carry out the test with more than the sample copy. 6) More than one person has made this suggestion. 不止一人提过这个建议. C. “More than+形容词”等于“很”或“非常”的意思，如： 7)In doing scientific experiments, one must be more than careful with the instruments. 8)I assure you I am more than glad to help you. D. more than + (that)从句，其基本意义是“超过（=over）”，但可译成“简直不”“远非”.难以，完全不能（其后通常连用情态动词can） 9) That is more than I can understand . 那非我所能懂的. 10) That is more than I can tell. 那事我实在不明白。 11) The heat there was more than he could stand. 那儿的炎热程度是他所不能忍受的 此外，“more than”也在一些惯用语中出现，如： more...than 的用法 1. 比……多，比……更 He has more books than me. 他的书比我多。 He is more careful than the others. 他比其他人更仔细。 2. 与其……不如 He is more lucky than clever. 与其说他聪明，不如说他幸运。 He is more （a）scholar than （a）teacher. 与其说他是位教师，不如说他是位学者。 注：该句型主要用于同一个人或物在两个不同性质或特征等方面的比较，其中的比较级必须用加more 的形式，不能用加词尾-er 的形式。 No more than/not more than 1. no more than 的意思是“仅仅”“只有”“最多不超过”，强调少。如： --This test takes no more than thirty minutes. 这个测验只要30分钟。 --The pub was no more than half full. 该酒吧的上座率最多不超过五成。-For thirty years，he had done no more than he （had）needed to. 30年来，他只干了他需要干的工作。 2. not more than 为more than （多于）的否定式，其意为“不多于”“不超过”。如：Not more than 10 guests came to her birthday party. 来参加她的生日宴会的客人不超过十人。 比较： She has no more than three hats. 她只有3顶帽子。（太少了） She has not more than three hats. 她至多有3顶帽子。（也许不到3顶帽子） I have no more than five yuan in my pocket. 我口袋里的钱最多不过5元。（言其少） I have not more than five yuan in my pocket. 我口袋里的钱不多于5元。（也许不到5元） more than, less than 的用法 1. （指数量）不到，不足 It’s less than half an hour’s drive from here. 开车到那里不到半个钟头。 In less than an hour he finished the work. 没要上一个小时，他就完成了工作。 2. 比……（小）少 She eats less than she should. 她吃得比她应该吃的少。 Half the group felt they spent less than average. 半数人觉得他们的花费低于平均水平。 more…than,/no more than/not more than (1)Mr．Li is ________ a professor; he is also a famous scientist． (2)As I had ________ five dollars with me, I couldn’t afford the new jacket then． (3)He had to work at the age of ________ twelve． (4)There were ________ ten chairs in the room．However, the number of the children is twelve． (5)If you tel l your father what you’ve done, he’ll be ________ angry． (6)－What did you think of this novel? －I was disappointed to find it ________ interesting ________ that one． 倍数表达法 1. “倍数+形容词（或副词）的比较级+than+从句”表示“A比B大（长、高、宽等）多少倍” This rope is twice longer than that one.这根绳是那根绳的三倍（比那根绳长两倍）。The car runs twice faster than that truck.这辆小车的速度比那辆卡车快两倍（是那辆卡车的三倍）。 2. “倍数+as+形容词或副词的原级+as+从句”表示“A正好是B的多少倍”。 “tｈｅwａｙ+从句”结构的意义及用法 首先让我们来看下面这个句子: Ｒead the foｌloｗｉnｇpassagｅａnd talkａbout ｉt wi ｔh your classmaｔes．Try tｏｔeｌl whaｔyｏu thiｎk ｏf Tom aｎd ｏｆｔhe ｗay the chiｌｄrenｔrｅated ｈim. 在这个句子中，ｔhe waｙ是先行词,后面是省略了关系副词that或in whｉch的定语从句。 下面我们将叙述“the ｗaｙ+从句”结构的用法。 1.tｈe wａy之后，引导定语从句的关系词是tｈat而不是hoｗ,因此，<<现代英语惯用法词典＞＞中所给出的下面两个句子是错误的：Ｔhｉs is ｔｈeｗayｈowｉtｈａppened. This is the way how he ａlwａｙs treats me. 2．在正式语体中,thaｔ可被in which所代替;在非正式语体中，ｔhat则往往省略。由此我们得到theｗay后接定语从句时的三种模式：1) tｈe ｗaｙ＋ｔhａt-从句２）the way +ｉn whiｃh－从句3) the ｗay +从句 例如：Ｔｈe way(in whiｃh ，thａt) ｔhｅｓｅｃomｒade ｓloｏｋａｔｐrobｌems is wrong．这些同志看问题的方法 不对。 Ｔｈｅｗａy(that ,ｉn ｗhiｃh)yoｕ’ｒe doinｇit is comple ｔeｌy crａzy.你这么个干法,简直发疯。 Wｅａdmiｒｅd him for thｅｗay ｉｎｗhｉch he fａｃeｓdiｆficultieｓ． Waｌlａｃe ａｎd Ｄarwiｎｇreed ｏn the way ｉｎwhi ｃh ｄｉfｆｅrｅnt forms ｏf life had ｂegｕn.华莱士和达尔文对不同类型的生物是如何起源的持相同的观点。 Thｉs is ｔｈe ｗaｙ（tｈａｔ) hｅdｉd it. I lｉkeｄｔhe waｙ(ｔhat) ｓｈeｏｒganized ｔhe ｍeetｉng． 3.theｗay(tｈat)有时可以与hｏw(作“如何”解）通用。例如： That’s tｈe ｗaｙ(tｈaｔ) shｅspoke. = Tｈat’s how shｅspoke. 电厂精细化管理内容 精细化管理从以下八个方面做起： 一、从问题抓起，有问题后分析问题存在的原因，找出解决办法，并最终的解决问题。经常我们听到的汇报是：没问题，事实真是这样吗？经过现场的考察，我们不是没问题，而是问题多的是，大家都认为领导干部希望听到的是没问题，所以有了问题也不说，只要能撑就撑，能拖就拖，等到实在撑不住了，就两手一摊没办法，问题越来越多。造成了困难多、问题多，解决办法少；说得多，做得少。到最后造成事故发生的频率越来越高。管理上不要怕问题暴露，有问题就如实地反映，尽早尽快地想办法去处理解决。只有不断把这些问题解决，机组的故障率才会不断下降。 二、从考核奖惩抓起，对于先进的埋头苦干的员工，一定给予奖励，对于扯皮，消极怠工的员工给予惩罚；奖罚分明，公平公正。目前我们虽然有考核，但力度不大，制度还不够完善，要让我们的绩效真正地发挥作用，要真正做到人尽其才，那就必须不断完善我们的考核机制。 三、从机制抓起，由于一些客观的原因，造成我们有的人忙得很，有的人成天没事做。地龙整个片区必须形成一个整体，形成一个战斗团队，要求不分专业、不分运行检修，有困难大家一起解决，管理上一定要杜绝人浮于事的现象，杜绝混日子、不作为等现象，杜绝专业之间的推诿扯皮现象，发现一起处理一起；一项工作从筹备、实施、验收到完工，始终保持同样的温度，直至画上完美的句号，严格按照制度进行落实好。 四、从现场抓起，管理干部必须经常深入生产现场，掌握第一手资料，绝不允许只听汇报，只有现场调查之后才有发言权，现场的文明生产、设备的健康状况必须亲临现场检查，到自己所辖的每一个车间、每一个角落、甚至每个墙角、定时的去看看。随时把握你责任区的一切动态，跟踪落实好，有困难时有无及时提出，寻找对策，及时解决；这样才能将每项工作落到实处。 五、从环节抓起，要想把计划的工作做好，就必须从每个细小的环节抓起，计划做好了，落实到什么程度，备品备件的跟踪、人员的安排、工作任务的布置、时间的安排等等，必须做全面考虑，完成工作的过程中，必须做到抓住每一个环节，同时要一环扣一环，环环相扣，中间不允许脱扣，这样才能提高整个工作效率。一件工作的进行注意每个环节，将最重要的安全、质量、时间、卫生等因素融入到每个工作的环节中，赢在细节。 六、从计划抓起，开始工作前要制定好周密的计划，将一切的意外因素全盘考虑，组织措施要严密周到，执行计划要合理安排，如果做事没有计划，整天忙忙碌碌，到最后也不知道忙的什么，因此我要求每位管理人员做事要有计划，只有计划做好之后，工作才会井井有条，到什么时间该做什么工作，这样才是一个比较正规的管理。 七、从规范抓起，任何工作的开展都有规范，不拍脑门开展工作，按照事物的客观规律开展工作，不违背电力工作的行规做事, 管理要规范，电力生产“两票三制”的执行必须规范，因此从基础工作做起，平时的工作必须按照部门的要求严格执行，减少随意性，这样才能减少故障的发生。 八、从要求抓起，做任何工作必须要有严标 初中英语比较级和最高级讲解与练习 形容词比较级和最高级 一．绝大多数形容词有三种形式,原级,比较级和最高级, 以表示形容词说明的性质在程度上的不同。 1. 形容词的原级: 形容词的原级形式就是词典中出现的形容词的原形。例如: poor tall great glad bad 2. 形容词的比较级和最高级: 形容词的比较级和最高级形式是在形容词的原级形式的基 础上变化的。分为规则变化和不规则变化。 二．形容词比较级和最高级规则变化如下: 1) 单音节形容词的比较级和最高级形式是在词尾加-er 和-est 构成。 great (原级) greater(比较级) greatest(最高级) 2) 以-e 结尾的单音节形容词的比较级和最高级是在词尾加-r 和-st 构成。 wide (原级) wider (比较级) widest (最高级) 3) 少数以-y, -er, -ow, -ble结尾的双音节形容词的比较级和最高级是在词尾加 -er 和-est构成。 clever(原级) cleverer(比较级) cleverest(最高级)， slow(原级) slower(比较级) slowest (最高级) 4) 以-y 结尾,但-y 前是辅音字母的形容词的比较级和最高级是把-y 去掉,加上-ier 和-est 构成. happy (原形) happier (比较级) happiest (最高级) 5) 以一个辅音字母结尾其前面的元音字母发短元音的形容词的比较级和最高级是双写该 辅音字母然后再加-er和-est。 原形比较级最高级原形比较级最高级 big bigger biggest hot hotter hottest red redder reddest thin thinner thinnest 6) 双音节和多音节形容词的比较级和最高级需用more 和most 加在形容词前面来构 成。 原形比较级最高级 careful careful more careful most careful difficult more difficult most difficult delicious more delicious most delicious 7)常用的不规则变化的形容词的比较级和最高级: 原级比较级最高级 good better best 好的 well better best 身体好的 bad worse worst 坏的 ill worse worst 病的 many more most 许多 much more most 许多 few less least 少数几个 little less least 少数一点儿 （little littler littlest 小的） far further furthest 远（指更进一步，深度。亦可指更远） far farther farthest 远（指更远，路程） 表示“方式”、“方法”，注意以下用法： 1.表示用某种方法或按某种方式，通常用介词in(此介词有时可省略)。如： Do it (in) your own way. 按你自己的方法做吧。 Please do not talk (in) that way. 请不要那样说。 2.表示做某事的方式或方法，其后可接不定式或of doing sth。 如： It’s the best way of studying [to study] English. 这是学习英语的最好方法。 There are different ways to do [of doing] it. 做这事有不同的办法。 3.其后通常可直接跟一个定语从句(不用任何引导词)，也可跟由that 或in which 引导的定语从句，但是其后的从句不能由how 来引导。如： 我不喜欢他说话的态度。 正：I don’t like the way he spoke. 正：I don’t like the way that he spoke. 正：I don’t like the way in which he spoke. 误：I don’t like the way how he spoke. 4.注意以下各句the way 的用法： That’s the way (=how) he spoke. 那就是他说话的方式。 Nobody else loves you the way(=as) I do. 没有人像我这样爱你。 The way (=According as) you are studying now, you won’tmake much progress. 根据你现在学习情况来看，你不会有多大的进步。 2007年陕西省高考英语中有这样一道单项填空题： ——I think he is taking an active part insocial work. ——I agree with you_____. A、in a way B、on the way C、by the way D、in the way 此题答案选A。要想弄清为什么选A，而不选其他几项，则要弄清选项中含way的四个短语的不同意义和用法，下面我们就对此作一归纳和小结。 一、in a way的用法 表示：在一定程度上，从某方面说。如： In a way he was right.在某种程度上他是对的。注：in a way也可说成in one way。 二、on the way的用法 1、表示：即将来(去)，就要来(去)。如： Spring is on the way.春天快到了。 I'd better be on my way soon.我最好还是快点儿走。 Radio forecasts said a sixth-grade wind was on the way.无线电预报说将有六级大风。 2、表示：在路上，在行进中。如： He stopped for breakfast on the way.他中途停下吃早点。 We had some good laughs on the way.我们在路上好好笑了一阵子。 3、表示：(婴儿)尚未出生。如： She has two children with another one on the way.她有两个孩子，现在还怀着一个。 She's got five children，and another one is on the way.她已经有5个孩子了，另一个又快生了。 三、by the way的用法 什么是精细化管理 精细化管理就是落实管理责任〃将管理责任具体化、明确化〃它要求每一个管理者都要到位、尽职。第一次就把工作做到位〃工作要日清日结〃每天都要对当天的情况进行检查〃发现问题及时纠正〃及时处理等等。 精细化管理是社会分工精细化、以及服务质量精细化对现代管理的必然要求。现代管理学认为〃科学化管理有三个层次：第一个层次是规范化〃第二层次是精细化〃第三个层次是个性化。 精细化管理是一项比较复杂的系统工程〃管理要素多〃对管理层、执行层和相应的管理制度层有着极其严格的要求。所以〃要实施精细化管理〃必须着力解决好观念、载体和制度创新三个问题。 观念问题〃包括管理层和执行层的观念。其一是管理层的观念。管理层在研究、规划和部署管理工作中〃要确立精益求精的思想〃充分考虑服务对象的需求和执行层的能力〃真正把上级要求和自身实际充分结合起来〃形成本单位具体而不复杂的贯彻落实措施；其二是执行层的观念。执行层能不能准确地领会、精确地贯彻落实上级的要求〃能不能在工作中不打折扣〃能不能真正负起责任〃做到一丝不苟、精益求精〃都是对其观念和态度的极大挑战。 载体问题〃管理层在研究、设计载体时〃要克服不顾实际贪多求全的思想。当前〃部分基层单位总是抱怨载体多、不好落实〃其实问题出在自身的思路不够清晰、没有在深刻领会与充分结合上下足工夫等方面。实施精细化管理必须解决这些问题。 制度创新问题〃考核历来是企业管理工作的难题〃如果具备了精细化管理的观念和思路〃制度创新必然能够走出一条新路〃考核的问题也将迎刃而解。精细化管理对制度的要求几乎达到苛刻的地步〃每一项工作、每一个细节、每一个流程〃随时随地都要有相应的制度来制约和考核。制度到位与否〃直接影响精细化管理的程度。所以说〃制度创新是实施精细化管理最为核心、最为细致也是最大的难点所在〃是必须解决的关键性问题。 要真想贯彻落实公司的“公司化运作、集约化发展、精细化管理”的工作思路〃适应公司发展的要求〃不断提高公司整体管理水平〃必须先要做份《精细化管理工作实施方案》。 实施精细化管理〃能使不断提升公司整体管理水平的重要举措〃也是针对公司管理现状〃切实解决目前存在问题的重要工作方法。其指导思想是：在认真学习精细化管理方法论的基础上〃通过深入分析公司各类管理工作中存在的问题〃以全面提升执行力、提高效率和效益为出发点〃围绕公司中心工作〃运用科学适当的管理工具和方法〃把“精、准、细、严”落实到管理工作的每个环节〃不断提高公司整体管理水平〃内强素质〃外树形象〃培育精细文化〃为公司的可持续发展而努力。《精细化管理工作实施方案》还明确提出了总体目标、具体英语中的比较级与最高级 详解
The way常见用法
人教版(新目标)初中英语形容词与副词的比较级与最高级
英语中的比较级和最高级
The way的用法及其含义(二)
何为规范化、标准化、精细化管理
英语比较级和最高级的用法归纳
(完整版)the的用法
学校后勤管理精细化体现“细”“实”“精”
英语比较级和最高级的用法
“the way+从句”结构的意义及用法
精细化管理内容
初中英语比较级和最高级讲解与练习
way 用法
什么是精细化管理