Building-block supply in genetic programming

Building-block supply in genetic programming
Building-block supply in genetic programming

Building-Block Supply in

Genetic Programming

Kumara Sastry

Una-May O’Reilly

David E.Goldberg

David Hill

IlliGAL Report No.2003012

April,2003

Illinois Genetic Algorithms Laboratory(IlliGAL) Department of General Engineering University of Illinois at Urbana-Champaign

117Transportation Building

104S.Mathews Avenue,Urbana,IL61801

Building-Block Supply in Genetic Programming

Kumara Sastry

Illinois Genetic Algorithms Laboratory,and

Department of Material Science&Engineering

University of Illinois at Urbana-Champaign

ksastry@https://www.360docs.net/doc/7012284132.html,

Una-May O’Reilly

Arti?cial Intelligence Lab Massachusetts Institute of Technology

unamay@https://www.360docs.net/doc/7012284132.html,

David E.Goldberg

Illinois Genetic Algorithms Laboratory

Department of General Engineering University of Illinois at Urbana-Champaign

deg@https://www.360docs.net/doc/7012284132.html,

David Hill

Department of Civil and Environmental Engineering

University of Illinois at Urbana-Champaign

djhill1@https://www.360docs.net/doc/7012284132.html,

May5,2003

Abstract

This paper analyzes building block supply in the initial population for genetic programming.

Facetwise models for the supply of a single schema as well as for the supply of all schemas in a

partition are developed.An estimate for the population size,given the size(or size distribution)

of trees,that ensures the presence of all raw building blocks with a given error is derived using

these facetwise models.The facetwise models and the population sizing estimate are veri?ed

with empirical results.

1Introduction

Genetic programming(GP)practitioners are often frustrated by the lack of theory available to guide them in selecting parameters for applied problems.They also lack a foundation of knowledge to explain many of their empirical?ndings regarding e?ective population sizes.To date,fundamental reasons for why varying population size while maintaining a constant number of?tness evaluations for a run results in di?erent evolutionary trajectories and endpoints,have not been revealed.

The purpose of this paper is to start addressing this lack of theory by providing an estimate of the population size necessary to solve a given GP problem.It is hoped that like in genetic algorithm(GA)theory(Goldberg,2002),the availability of a population sizing equation will be a valuable tool to aid GP practitioners in their e?orts to understand how GP processes information. In addition,it may indicate how to adapt GP to be more competent.

The?rst step towards understanding population sizing is to tackle the issue of building-block (BB)supply.We forgo a temporal approach which would assume that the recombination,mutation and other diversity-generating operators will create and maintain su?cient BB diversity on an

appropriate time scale.Instead,using a spatial approach,we estimate the population size required to ensure diversity and the number of BBs present in the initial population.

The objective of this study is to develop facetwise models for supply of BBs and to estimate the population size required to guarantee the presence of all raw BBs for a given tree size(or size distribution)in the initial GP population.Though ensuring BB growth supersedes BB supply in the subsequent population,BB growth will be extremely di?cult if BB supply is not ensured. Also,while decision making usually governs population sizing,it is sometimes governed by BB supply.In such cases a facetwise model of BB supply is necessary for ensuring a successful GP design.Furthermore,understanding initial supply of BBs is essential for developing a practical population-sizing model.

This paper is structured as follows.We start with a brief literature review of BB supply. Section3provides background and states key assumptions made in this study.Details of an expression mechanism and test problems are provided in section4,followed by facetwise models for BB supply in section5.Section6outlines some thoughts on handling BB supply for real GP expressions.Finally,summary and conclusions are presented.

2Brief Literature Review

The GP community is interested in identifying strategies to size populations,in order to estimate the computational e?ort required to solve particular problems with GP;however,few studies have addressed this topic,thus far.One approach has been suggested by Langdon and Poli(2002),but it has not been fully developed.This approach employs the methodology used by Poli(2000)for the sizing of populations in GA.In this method,Poli used Stephens and Waelbroeck’s(1999)concept of transmission probability to develop a recursive conditional schema theory that allows for the prediction of the probability of reaching a solution to a problem in a?xed number of generations. An expression for the transmission probability for standard GP was developed by Langdon and Poli(2002).However,the expression is very di?cult to evaluate.

The methodological and analytic foundation for our approach to deciphering selectorecombina-tive GA(Goldberg,2002)and GP(Goldberg&O’Reilly,1998;O’Reilly&Goldberg,1998)has been stated before.Put succinctly,our approach is to analyze and understand GP’s simple mech-anisms before its complex ones.We predict that lessons learned from experimentation and theory on a simple case will lead to insight,and possibly,carry over to more complex cases.Therefore,we start by analyzing building-block supply in GP’s initial population,before the activity of crossover and selection.

While building-block supply has been largely ignored in GP literature,many researchers have studied the BB supply in GAs.Holland(1975)estimated the number of BBs that receive at least a speci?ed number of trials using Poisson distribution.A later study(Goldberg,1989)calculated the same quantity more exactly using binomial distribution and studied their e?ects on population sizing in serial and parallel computation.Reeves(1993)proposed a population sizing model for supply of alphabets with?xed cardinality.Recently,Goldberg,Sastry,and Latoza(2001)developed facetwise models for ensuring BB supply in the initial population for genetic algorithms.They considered a population of?xed-length strings consisting alphabets of arbitrary cardinalityχ.They predicted that the population size required to ensure the presence of all competing building blocks with a tolerance of =1/m is given by n=χk(k logχ+log m),whereχis the alphabet cardinality, k is BB size,and m is the number of BBs.

This paper follows a similar methodology along the lines of Goldberg,Sastry,and Latoza(2001) and develops facetwise models for predicting the probability of the presence of a single schema as

F

(a)(b)

(c)

(d)

(e)

(f)

(g)

Figure1:The smallest tree fragments in GP.Fragments(c)and(d)have mirrors where the child is2nd parameter of the function.Likewise,fragment(f)has mirror where1st and2nd parameters of the function are reversed.

well as all schemas in a given partition.Before developing the models,we present some background and state assumptions used in the modeling procedure.

3Preliminaries

In this section,we present de?nitions and concepts that underpin our analysis of BB supply in GP.

3.1GP Tree Composition

Most GP implementations reported in the literature use parse trees to represent candidate programs in the population(Langdon&Poli,2002).We have assumed this representation in our analysis. To simplify the analysis further,we consider the following:

1.A primitive set of the GP tree is F∪T,where F denotes the set of functions(interior nodes

to a GP parse tree),and T denotes the set of terminals(leaf nodes in a GP parse tree).

2.The cardinality of F=χf and the cardinality of T=χt.

3.The arity of all functions in the primitive set is two:All functions are binary and thus parse

trees generated from the primitive set are binary.

We believe that our analysis could be extended to primitive sets containing functions with arity greater than two(non-binary trees).We also note that our assumption closely matches a common GP benchmark,symbolic regression,which frequently has arithmetic functions of arity two.

3.2Translating GA Schemas to GP Schemas Isn’t Straightforward

Schemas are similarity templates that describe sets of solutions that share a common feature.The GP literature contains several alternative de?nitions of schemas(Koza,1992;Altenberg,1994; O’Reilly&Oppacher,1995;Whigham,1995;Rosca,1997;Langdon&Poli,2002).Per O’Reilly and Oppacher(1995),a GP schema is a multiset of subtrees and tree fragments with nodes denoted as functions,terminals or don’t care symbols.Tree fragments are trees with at least one leaf that is a“don’t care”symbol which can be matched by any subtree(including subtrees with only one node).

3.2.1Tree Fragments

While in general tree fragments refer to a multiset of tree patterns or tree templates,we restrict ourselves to a single tree pattern.A tree fragment pattern has each of its nodes labeled with the

function symbol,F,or terminal symbol,T.However,it does not have an absolute position or positional anchor.Figure1shows the fragments our analysis focuses on.Along the edge between a function and its child node,a numeral denotes what parameter of the function the child node is (i.e.the?rst or second argument in the case of a binary function).A tree fragment has a length or size;that is,its number of nodes,k=N t+N f,where N t and N f are the number of terminal and functional nodes in the tree fragment,respectively.Furthermore,the total number of possible instances of a tree fragment is given by

κ=χN f f?χN t t(1) For example,for the tree fragment P b(fragment with only terminal),N f=0,and N t=1,and therefore,the total number of instances of P b isχt.

Since a tree fragment is not anchored to a position of a tree,there can be none or more than one instance of a fragment in a single tree.Yet,the smallest fragments P a and P b appear at least once or twice in a tree respectively.Assuming a single tree of size s1and the tree properties listed in section3.1,Table3.2.1provides estimates(derived by probability of frequency)of the average quantity of tree-fragment instances,φ.In other words,φcounts the expected number of tree-fragment instances,given the tree size(or size distribution),in the population.

P Descriptionφκ

P a function1

2(s?1)χf

P b terminal1

2(s+1)χt

P c one terminal that is the?rst parameter of a binary function 1

4

(s+1)χf·χt

P d a function at the root and a function as its ?rst parameter 1

4

(s?3)χ2f

P e a function at the root and2terminals as its parameters 1

8

(s+1)2

(s?1)

χf·χ2t

P f a function at the root and1terminal as the ?rst parameter and one function as it second

parameter.1

8

(s+1)(s+3)

(s?1)

χ3f

P g a binary function at the root and2functions as its parameters 1

8

(s?3)2

(s?1)

χ2f·χt

Table1:Designations,P i,and descriptions of tree fragments considered in the BB supply models, the quantity of fragments,φ,and the number of competing schemas in the fragments,for a binary tree of size s.See also Figure1.

3.2.2The Tree Fragments are not Enough:How are They Expressed?

While tree fragments are the parts of a physical tree,and counting number of instances of tree frag-ments can itself be important,what is more important are those tree fragments that get expressed. The expression mechanism dictates what the building blocks of a problem are and therefore a?ects the BB supply.Speci?cally,we are interested in expression of small tree fragments into partially correct subfunctions.Let us consider,for example,symbolic regression of1+x+x2+x3.Early on 1It should be noted here that the average tree size of a population can be calculated for popular initialization schemes(Koza,1992),or initialization schemes such as PCT1or PCT2(Luke,2000)can be used to generate a population which conform to an expected tree size.

in the GP run,it is important to get the constant and the linear part of the symbolic equation right.

Therefore,all the tree fragments that contribute to the correct constant and linear subfunctions

are important and their supply is critical in the initial population.

We illustrate the methodology to incorporate expression mechanism in BB supply models by

using a simple expression mechanism,called ORDER,which is explained in the next section.We

choose ORDER because while it models some of the GP behavior(Goldberg&O’Reilly,1998;O’Reilly

&Goldberg,1998),the expression mechanism can be analyzed in a straightforward manner.

4ORDER Expression Mechanism

ORDER is a simple,yet intuitive expression mechanism which makes it amenable to analysis and modeling(Goldberg&O’Reilly,1998;O’Reilly&Goldberg,1998).The primitive set of ORDER

consists of the primitive JOIN of arity two and complimentary primitive pairs X i,ˉX i ,i=0,1,···, of arity one.A candidate solution of the ORDER problem is a binary tree with JOIN primitive at the

internal nodes and either X i’s orˉX i’s at its leaves.The candidate solution’s expression is determined

by parsing the program tree inorder(from left to right).The program expresses the value X i if,

during the inorder parse,a X i leaf is encountered before its complementˉX i.Furthermore,only

unique primitives are expressed in ORDER during the inorder parse.

Building blocks in ORDER are the sets of primitives that are part of the subfunctions that reduce

error(alternatively improve?tness).In this study,we consider two test problems that use ORDER

expression mechanism:1.UNITATION:where each primitive X i is a BB,and2.DECEPTION:where

k primitives form a BB.The following sections describe these two test problems.

4.1UNITATION

In UNITATION,for each X i(orˉX i)that is expressed,an equal unit of?tness value is accredited.

That is,

f1(x i)= 1if x i∈{X1,X2,···,X }

0otherwise

.(2) The?tness function for ORDER is then de?ned as

F(x)=

i=1f1(x i),(3)

where x is the set of primitives expressed by the tree.The output for optimal solution of a -primitive UNITATION problem is{X1,X2,···,X },and its?tness value is .

4.2DECEPTION

In DECEPTION,the primitives are divided into m subgroups,each subgroup consisting of k primitives. The?tness of each subgroup is computed using the following trap function(Goldberg,1987;Deb &Goldberg,1993):

f k(u(x1,x2,···,x k))= 1.0u=k

(1.0?δ) 1?u k?1 u

u(x1,x2,···,x k)=

k

i=1f1(x i),(5)

where x i is the i th primitive,δis the di?erence in the functional value between the correct BB and its deceptive attractor.The ?tness function of a candidate solution (tree)is then given by

F (x )=f k (u (x 1,x 2,···,x k ))+f k (u (x k +1,···,x 2k ))+

···+f k

u (x (m ?1)k +1,···,x mk )

,

(6)

where,F is the ?tness function,x is the expressed primitives,m is the number of BBs,and =mk .

5Facetwise Models of Building-Block Supply

In this section,we develop facetwise models for building-block supply for ORDER expression.First we start with addressing the supply of a single BB in a given partition.Then we extend the model to ensure the supply of all schemas in a partition.We then use the facetwise models to derive a population-sizing model dictated by BB supply.The models developed in this section are veri?ed with empirical results for UNITATION and DECEPTION along the way.

5.1Supply of a Single Building Block

Assuming trees of size,s ,and that the expression mechanism used is ORDER ,the probability that a primitive expressed by a tree is given by

p X exp

i

=p

#ofX i ≥1,#of ˉX i =0

+p

X i appears before ˉX i

,=n l

j =1

n l n l ?j

2j ?1

?2

n l ?j 1

j

,

=1

2 n l [ n l ?( ?2)n l ],=

12 1? 1?2

n l

(7)

where n l =(s +1)/2,is the number of leaf nodes in the tree,and =χt .

Assuming that primitives are expressed independent of each other,the probability that a order k BB (without loss of generality,we will consider X 1X 2···X k )is expressed by a tree is given by

p X exp

1···k

=p (X i is expressed)k ,=

12

1? 1?2

n l k

.(8)

The probability that the BB is not expressed by a tree is then given by

p X not

exp 1···k

=1?p X exp 1···k

,=1? 12

1? 1?

2

n l k

(9)

The probability that a BB is not expressed by any of the n individuals in the population is

given by

p X exp=0

1···k

= p X not

exp

i ···k

n

,

=

1?

12

1?

1?2 n l k n

(10)

P r o b . t h a t X i i s e x p r e s s e d , p X i

e x p >= 1

Figure 2:Veri?cation of the facetwise model for a single BB supply (Equation 11)with empirical results for the UNITATION problem for di?erent tree heights,h ,as a function of population size,n .The empirical results depict the proportion of runs having at least one copy of a particular schema out of 1000trials.

Therefore the probability that a order-k BB is expressed by at least one individual in the population is given by

p X exp ≥1

1···k

=1?p X exp=01···k

,=1?

1?

12

1?

1?2

n l k n

(11)

The model for single BB success given by Equation 11is compared to empirical results for the

UNITATION problem (k =1)in Figure 2,and for the DECEPTION problem (k =4)in Figure 3.The empirical results are for full trees,therefore,n l =2h .The results show that the empirical results agree with the models.

Using the approximation,(1?r/s )s ≈e ?r ,and recognizing that this approximation is su?-

P r o b . t h a t X i

i s e x p r e s s e d , p

X

i

e x p >= 1

Figure 3:Veri?cation of the facetwise model for a single BB supply (Equation 11)with empirical results for the DECEPTION problem for di?erent tree heights,h ,as a function of population size,n .The empirical results depict the proportion of runs having at least one copy of a particular schema out of 1000trials.

ciently accurate even for modest values of s ,we can simplify Equation 11as follows:

p X exp ≥11···k

≈1?exp

?n 2

?k

exp ?k exp ?

2n l

.(12)

When n l ,p X exp ≥1i

≈1?exp

?n 2?k

.In other words,the probability of a BB being expressed

by at least one individual,given a population size,increases with the tree size and saturates as 2h > ,as shown in Figure 4for UNITATION problem.

5.2Supply for Partition Success

When solving real-world problems,one does not have prior knowledge about a particular schema being superior to others in a partition.Hence it is necessary to ensure that all competing schemas in a partition are present.The decision process would then be able to consider all the relevant

P r o b . t h a t X i i s e x p r e s s e d , p X i

e x p >= 1

Figure 4:Veri?cation of the facetwise model for a single BB supply (Equation 11)with empirical results for the UNITATION problem for di?erent population size,n ,as a function of tree height,h .The empirical results depict the proportion of runs having at least one copy of a particular schema out of 1000trials.

alternative schemas.Therefore,in this section we extend the model developed in the previous section to ensure the presence of at least one copy of all the competing schemas (both the primitive and its complement)in a partition.

For ORDER ,we are interested in the probability that all the 2k possible schemas are present in the population.Assuming that individual schema success values are independent,the probability for partition success is given by

p s =

p X exp ≥1

1···k

2k

,

=

1?

1?

1

2

1+

1?2

n l k n 2k

.(13)

P r o b . f o r p a r t i t i o n s u c c e s s , p s

Figure 5:Veri?cation of the models for BB partition success (Equations 16,and 14)with empirical results for UNITATION problem for di?erent tree heights,h ,and problem sizes, ,as a function of population size,n .The empirical results depict the proportion of runs having at least one copy of a primitive and its complement in the population out of 1000trials.

Using the approximation (1?r/s )s ≈e ?r ,the above equation can be further approximated as

p s ≈exp

?2k exp

?n 2?k exp

?k exp

?

2h +1

.(14)

It should be noted that the independence assumption of individual schema success is an ap-proximation and a more exact model can be derived which is illustrated for BB of unit size (k =1).

p s =

n i =2

2i

?2

n

n ?i

p X exp

i

i

1?2p X exp

i

n ?i

,(15)

The above equation can be rearranged as follows:

p s =

n ?2 j =0

n

j

2p

X exp i

n ?j

1?2p

X exp i

j

P r o b . f o r p a r t i t i o n s u c c e s s , p s

Figure 6:Veri?cation of the BB partition success model (Equation 14)with empirical results for DECEPTION problem for di?erent tree heights,h ,and problem sizes, ,as a function of population size,n .The empirical results depict the proportion of runs having at least one copy of a primitive and its complement in the population out of 1000trials.

?2

n ?2 j =0

n j

p

X exp i

n ?j

1?2p

X exp i

j

,

=1+

1?2p X exp

i

n

?2

1?p X exp

i

n

.

(16)

Equations (16),and (13)are compared with empirical results in Figure 5.The ?gures show

that the approximate model (Equation 13)agrees with Equation 16for higher population sizes and larger tree sizes.The partition success model (Equation 13)is compared with the empirical results for DECEPTION with k =4,in Figure 6.Both Figures 5,and 6clearly validate the BB supply model.

5.3Population Sizing for Building-Block Supply

The facetwise model derived in the previous section will be rearranged in this section to estimate the population size required to ensure the presence of all BBs of a partition for ORDER ,given the

problem size is ,and the tree height is h.Assuming that we can tolerate a probability of not having all BBs in a given partition,and setting p s to1? ,we can rewrite Equation14,

1? =exp ?2k exp ?n2?k exp ?k exp ?2n l .(17)

Taking logarithms on both sides of the above equation and using the approximation,ln(1? )≈? , we get

=2k exp ?n2?k exp ?k exp ?2h+1 .(18)

After taking logarithms on both sides of the above equation and rearranging the resulting equation, we can write

n=2k(k ln2?ln )exp ?k exp ?2n l .(19) If we assume tree size to be big enough(n l ),then the above equation can be simpli?ed as n≈2k(k ln2?ln ).Furthermore,if we assume that the supply error is inversely proportional to the number of BBs,m,i.e., =1/m,

n≈2k(k ln2+ln m).(20) It is interesting to note that the above population-sizing equation for BB supply in DECEPTION is identical to that developed by Goldberg,Sastry,and Latoza(2001)for selectorecombinative GAs. 6Some Thoughts On Modeling Realistic GP Expressions

The last section developed BB supply models for ORDER expression mechanism and veri?ed it for two test problems for di?erent parameter values.This section provides a brief outline on how to develop BB supply models for realistic GP expressions.First we start by addressing the supply of raw tree fragments,or in other words,we consider that every tree fragment in the tree is expressed.

6.1Tree Fragment Supply

6.1.1Single BB Success

The probability that a tree does not contain a partition,P i,is given by

p(#ofP i=0)= 1?1κ φ(21)

Recall that the values forκ,andφfor di?erent partitions are given in Table3.2.1.From the above equation,we can write the probability that the population contains at least one copy of the partition,P i,as

p k=1? 1?1κ φ n.(22)

Using the approximation,(1?r/s)s≈e?r,and recognizing that this approximation is su?ciently accurate even for modest values of s,we can write

p k≈1?exp ?nφκ .(23)

Furthermore,from table3.2.1,we can see thatφ≈2?k s,where k=N t+N f.Substituting this approximation forφin the above equation,we get

p k≈1?exp ?n·sκ·2 .(24)

It should be noted that that the approximation forφis an underestimation for the tree fragments, P b,P c,P e and P f,and an overestimation for the tree fragments,P a,P d,and P g.

6.1.2Partition Success

Similar to the previous section,we assume that the schema partition success values are independent. Then the probability of at least one success of each of theκschemas,p s is given by p s=pκk:

p s= 1?exp ?n·sκ·2 κ,(25)

≈exp ?κexp ?n·sκ·2 .(26) 6.1.3Population Sizing for Partition Success

We now proceed to model the population size required to ensure the presence of all order-k tree fragments.Assuming that we can tolerate a probability of not having all BBs in a given partition, and setting p s to1? ,we can rewrite equation26,

1? =exp ?κexp ?n·sκ·2 (27)

Taking logarithm on both sides and using the approximation ln(1? )≈? ,for small values of , gives

=κexp ?n·sκ·2k (28) Solving the above equation for n yields

n=1

2

2kκ(logκ?log ).(29)

Recall thatκ=χN f fχN t t,and k=N f+N t.Then we can rewrite the above equation as

n=1

s

(2χf)N f 2χN t t [N f lnχf+N t lnχt?ln ](30)

This relation can be further simpli?ed if we assume that the supply error is inversely proportional to the number of BBs,m,i.e., =1/m.Then the equation may be rewritten as

n=1

s

(2χf)N f 2χN t t [N f lnχf+N t lnχt+ln m](31)

6.2Incorporating Expression

While counting the tree fragments may be useful enroute with proper expression model as in section5,on its own it is not realistic.Therefore,we have to compute the combined probability that a tree fragment is present in the population and that it expresses a correct subfunction:

p(BB is present)=p(fragment is present)p(expression)(32) In the above equation we assume that the events that a tree being present in the population and it being expressed are independent.It should be noted that this assumption becomes more accurate as the population size increases.The probability of a tree fragment being present in the population, p(fragment is present)=p k,and is given by equation24,and the expression model is incorporated by the term p(expression).For example,in the symbolic regression example of1+x+x2+x3, the probability of expression incorporates the probability of di?erent tree fragments expressing the linear and constant subfunctions.

7Conclusions

In this paper,a detailed analysis of building-block supply in the initial population of GP using ORDER expression has been presented.Two facetwise models are derived,one for ensuring the supply of a single schema in a partition,and the other for ensuring the supply of all competing schemas in a partition for problems which employ ORDER expression mechanism.The latter model has been employed to estimate the population size required to ensure the presence of at least one copy of all raw BBs of a partition in the initial population.The population sizing model indicates that there is a minimum tree size dependent on the problem size.Furthermore,the models suggest that when the tree size is greater than the problem size,the population size required on BB supply grounds is2k(k lnχ+ln m).This study also shows that the population size required to ensure the presence of all instances of tree fragments(assuming that all of them are expressed)is 1

(2χf)N f(2χt)N t[N f lnχf+N t lnχf+ln m].

s

Acknowledgments

We thank Martin Martin,Sean Luke,Terry Soule,and Bill Langdon.We also thank Gerulf Peder-sen,Ying-Ping Chen,and Tian-Li Yu for their insightful comments and suggestions.

This work was sponsored by the Air Force O?ce of Scienti?c Research,Air Force Materiel Command,USAF,under grant F49620-00-0163and F49620-03-1-0129,and the National Science Foundation under grant DMI-9908252,and CSE fellowship,UIUC.The https://www.360docs.net/doc/7012284132.html,ernment is autho-rized to reproduce and distribute reprints for government purposes notwithstanding any copyright notation thereon.The views and conclusions contained herein are those of the authors and should not be interpreted as necessarily representing the o?cial policies or endorsements,either expressed or implied,of the AFOSR,the NSF,or the https://www.360docs.net/doc/7012284132.html,ernment.

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史上-最全-wordExcel使用技巧大全(超全)

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word中80个常见的操作技巧(很强大哦)

答:格式从“页眉”改为“清除格式”,就在“格式”快捷工具栏最左边;选 中页眉文字和箭头,格式-边框和底纹-设置选无 8. 问:页眉一般是---------,上面写上题目或者其它,想做的是把这根线变 为双线,WORD 中修改页眉的那根线怎么改成双线的? 答:按以下步骤操作去做: ●选中页眉的文字,包括最后面的箭头●格式-边框和底纹●选线性为双线的 ●在预览里,点击左下小方块,预览的图形会出现双线●确定▲上面和下面自 己可以设置,点击在预览周围的四个小方块,页眉线就可以在不同的位置 9. 问:Word 中的脚注如何删除?把正文相应的符号删除,内容可以删除,但 最后那个格式还在,应该怎么办? 答:步骤如下:1、切换到普通视图,菜单中“视图”――“脚注”,这时最 下方出现了尾注的编辑栏。2、在尾注的下拉菜单中选择“尾注分隔符”,这 时那条短横线出现了,选中它,删除。3、再在下拉菜单中选择“尾注延续分 隔符”,这是那条长横线出现了,选中它,删除。4、切换回到页面视图。尾 注和脚注应该都是一样的 10. 问:Word 里面有没有自动断词得功能?常常有得单词太长了,如果能设置 下自动断词就好了 答:在工具―语言―断字―自动断字,勾上,word 还是很强大的 11. 问:如何将word 文档里的繁体字改为简化字? 答:工具―语言―中文简繁转换 12. 问:怎样微调WORD 表格线?WORD 表格上下竖线不能对齐,用鼠标拖动其 中一条线,可是一拖就跑老远,想微调表格竖线让上下对齐,请问该怎么办 答:选定上下两个单元格,然后指定其宽度就可以对齐了,再怎么拉都行press"Alt",打开绘图,其中有个调整坐标线,单击,将其中水平间距与垂直

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标准件中英文对照知识

常用钢材理论重量表 [2004-12-22 8:51:47] 计算公式: t(重量kg) = A(断面积mm²)×L(长度m) ×p(密度g/cm³) ×1/1000 注:(1) 型材制造中有允许偏差值,故上式只做估算之用。 (2) 关于p值,钢材通常取7.85 . 螺丝产品名称中英文对照(台湾版)螺丝螺 [2004-12-22 8:51:24] 六角螺絲(栓) HEX HEAD CAP SCREWS(HEX BOLTS) 六角機械螺絲HEX HEAD MACHINE BOLTS 六角木牙螺絲HEX LAG BOLTS 四角螺絲SQUARE HEAD BOLTS T頭螺絲T HEAD BOLTS 馬車螺絲CARRIAGE BOLTS 環首螺絲EYE BOLTS 內六角孔螺絲HEX SOCKET CAP SCREWS 固定螺絲SET SCREWS 螺椿栓STUD BOLTS 螺旋椿SCREW STUDS 輪殼螺栓WHEEL BOLTS 翼形螺絲WING SCREWS 自攻螺絲SELF TAPPING SCREWS 自削螺絲THREAD CUTTING SCREWS 鑽尾螺絲SELF DRILLING SCREWS 旋入螺絲DRIVE SCREWS 機械螺絲MACHINE SCREWS 木螺絲WOOD SCREWS 家具螺絲FURNITURE SCREWS 塑板螺絲CHIPBOARD SCREWS 牆用螺絲DRYWALL SCREWS 基礎螺栓FOUNDATION BOLTS U型螺栓U BOLTS 勾頭螺栓HOOK BOLTS 套掛螺絲TOGGLE BOLTS 突緣螺絲FLANGE BOLTS 軌道螺栓(魚尾螺絲) TRACK BOLTS 耐候鋼螺絲CORTEN STEEL HEA VY HEX BOLTS 扭矩控制螺栓T.C.BOLTS

word操作小技巧,很实用的

问:WORD里边怎样设置每页不同的页眉?如何使不同的章节显示的页眉不同? 答:分节,每节可以设置不同的页眉。文件——页面设置——版式——页眉和页脚——首页不同 问:请问word中怎样让每一章用不同的页眉?怎么我现在只能用一个页眉,一改就全部改了? 答:在插入分隔符里,选插入分节符,可以选连续的那个,然后下一页改页眉前,按一下“同前”钮,再做的改动就不影响前面的了。简言之,分节符使得它们独立了。这个工具栏上的“同前”按钮就显示在工具栏上,不过是图标的形式,把光标移到上面就显示出”同前“两个字来了 问:如何合并两个WORD文档,不同的页眉需要先写两个文件,然后合并,如何做? 答:页眉设置中,选择奇偶页不同/与前不同等选项 问:WORD编辑页眉设置,如何实现奇偶页不同?比如:单页浙江大学学位论文,这一个容易设;双页:(每章标题),这一个有什么技巧啊? 答:插入节分隔符,与前节设置相同去掉,再设置奇偶页不同 问:怎样使WORD文档只有第一页没有页眉,页脚? 答:页面设置-页眉和页脚,选首页不同,然后选中首页页眉中的小箭头,格式-边框和底纹,选择无,这个只要在“视图”——“页眉页脚”,其中的页面设置里,不要整个文档,就可以看到一个“同前”的标志,不选,前后的设置情况就不同了。 问:如何从第三页起设置页眉? 答:在第二页末插入分节符,在第三页的页眉格式中去掉同前节,如果第一、二页还有页眉,把它设置成正文就可以了 ●在新建文档中,菜单—视图—页脚—插入页码—页码格式—起始页码为0,确定; ●菜单—文件—页面设置—版式—首页不同,确定; ●将光标放到第一页末,菜单—文件—页面设置—版式—首页不同—应用于插入点之后,确定。 第2步与第三步差别在于第2步应用于整篇文档,第3步应用于插入点之后。这样,做两次首页不同以后,页码从第三页开始从1编号,完成。 问:WORD页眉自动出现一根直线,请问怎么处理?

haso标准件中英文对照

HASCO Z标准件系列 Z00 导柱Guide pillar Z01 导柱Guide pillar Z011 导柱Guide pillar Z012 导柱Guide pillar Z013 导柱Guide pillar Z014 导柱Guide pillar Z0141 导柱连接端End piece Z0142 导柱连接端End piece Z015 导柱Guide pillar Z0151 导柱连接端(Z015/…专用)Pillar adapter Z0152 导柱连接端Pillar adapter Z02 顶杆Ejector rod Z022 导管Guide sleeve Z03 导柱Guide pillar Z05 圆锥形管精定位Locating unit, round Z055 圆形垫片Spacer Z056 圆形垫件Spacer Z06 平锥形导块Locating unit Z07 长方形导块Square guide bar Z08 直身方形管位导块Pre-centering unit Z081 方形垫片Spacer disc Z091 方形定位锁块(公)Square interlock Z092 方形定位锁块(母)Square interlock Z093 方形定位锁块(母)Square interlock Z10 导套Guide bush Z10W 自润滑导套Guide bush Z11 导套Guide bush Z11W 自润滑导套Guide bush Z12 滚珠导套Ball guide bush Z13W 自润滑导套Self lubric.Guide bush Z15W 自润滑垫片Self lubric. Flat stock Z16W 自润滑垫片Self lubric. Guide rail Z17 长方形导块(Z07/...专用)Guide retainer Z20 定板导套Centering sleeve Z25 定位销Dowel pin Z26 定位销Dowel pin Z28 六角螺母Hexagon nut Z281 六角螺母Hexagon nut Z282 六角螺母Hexagon nut Z285 螺母Nut for T-slots Z286 螺母Nut for T-slots Z30 胚头内六角螺丝Socket head cap screw

Word文档使用技巧方法大全(超全)

Word文档及快捷键 使用技巧方法大全 以下分别介绍word 2003、2000、2007、2010. 一、打造Word右键菜单 使用Word2000/2002编辑文档时,如果你经常要用到“首字下沉”命令,可以按下述方法将该命令加到“文字”右键菜单中: 1.在“工具”菜单中,单击“自定义”。 2.单击“工具栏”选项,单击“快捷菜单”,则“快捷菜单”工具栏出现在屏幕上。 3.单击“命令”选项,在“类别”框中单击“格式”,在“命令”框中,找到并单击“首字下沉”,然后用左键拖动“首字下沉”选项到“快捷菜单”工具栏的“文字”按钮上,此时将打开其下拉列表,继续拖动鼠标至下拉列表的“文字”处,再拖动鼠标到“文字”子菜单的最顶端,并松开鼠标左键。 4.单击“关闭”按钮。 现在,当你在文档中的任意文本处,单击鼠标右键,则弹出快捷菜单的顶端将是“首字下沉”命令,方便多了吧! 二、如果希望Word每次启动时都能自动打开某个文档,可以通过简单的宏命令来实现这个功能,方法是 2.在“录制宏”对话框中,在“宏名”输入框中输入要建立的宏的名称,如“auto”,点击“确定 3.从菜单中单击“文件”,点击最近打开文件列表中显示的某一个文件名 在Word2002中想保存多篇打开的文档,可以先按住Shift键,然后单击“文件”菜单,这时原来的“保存”命令就变成了“全部保存”命令,单击它就可以保存所有打开的Word文档。也可以先按住Shift键,然后用鼠标单击常用工具栏上的“保存”按钮,这时“保存”按钮的图标就变成“全部保存”的图标,松开鼠标,即保存完毕。 你可以一次性关闭所有打开的文档,方法是:按住Shift键,单击“文件”菜单,在“文件”菜单中将出现“全部关闭”选项,单击该命令即可一次性关闭所有打开的文档,且在关闭文档前Word将提示你保存所作的改动。 对于打开的多个Word文档,如果只想关闭其中的几个文档(不是全部关闭),可以按下Ctrl键再单击任务栏上要关闭的文档的图标,图标都呈凹下状态,在图标上单击鼠标右键,选择“关闭组”命令,可将所选中的文档全部关闭。 用这种方法还可以将所选文档最小化、还原或最大化。并且此法也适用于其它的0ffice组件,如Excel、PowerPoint、Access等。

供应商选择和管理程序-中英文对照

供应商选择和管理程序-中英文对照

————————————————————————————————作者:————————————————————————————————日期:

XXX Supplier Selection and Management Procedure 供应商选择和管理程序 Checked by/校准:Approved by/批准:Doc. No./文件编号:Ver. No./版本号:

XXX Checked by/校准: Approved by/批准: Doc. No./文件编号: Ver. No./版本号: Audit and Approval 审核及批准 Audit and Approval 审核及批准 Signature 签 字 Date 日 期 Remark 备 注 General Manager 总经理 √ Production Dept 制造部 √ R&D Dept. 产品开发部 √ Quality Dept. 质量部 √ Purchasing Dept. 采购部 √ Finance Dept. 财务部 √ HR Dept. 人力资源部 Sales Dept. 销售部

XXX 1 Purpose 目的: To define the methodology to select, assess, approve new suppliers and manage supplier to ensure uniformity across the group and that all purchased product conforms to the purchase requirements of NAILI. 定义如何选择、评估、批准和管理供应商。 2 Scope 范围: All parts and services contribute to products that sold to Customers including Raw material, Subcontracting parts, Services and Other Products. 所有作用于销售给客户产品的原材料、零部件和服务,包括外包合同、零件、服 务和其他的产品。 3 Definition定义: None 无 4. Responsibility职责: Purchasing department is responsible for sourcing, preliminarily assessing, recommending, auditing, manage new suppliers and steer price determination. 采购部负责寻找,初评,推荐,审核,管理供应商及跟进价格确认流程。 R&D department is responsible for audit new suppliers to confirm their technical capability. 产品开发部负责审核新供应商以确认其技术能力. Quality department is responsible for preliminarily assessing, audit new suppliers to confirm their quality assurance capability and incoming control。 质量部负责初评,审核新供应商,确认其品质保证能力及来料控制. 5. Procedure内容: 5.1 New supplier development 5.1.1 Sourcing for new supplier寻找新供应商 5.1.1.1Purchasing department gets purchasing request and needs for new supplier. All departments can request a need. 采购部得到采购需求,需要发展新供应商。公司任何部门都可以提出采购需求。Checked by/校准:Approved by/批准:Doc. No./文件编号:Ver. No./版本号:

WORD使用技巧(骨灰级合集)

目录 1.问:WORD里边怎样设置每页不同的页眉?如何使不同的章节显示的页眉不同? (4) 2.问:请问word 中怎样让每一章用不同的页眉?怎么我现在只能用一个页眉,一改就全部改了? (4) 3. 问:如何合并两个WORD 文档,不同的页眉需要先写两个文件,然后合并,如何做? (4) 4. 问:WORD 编辑页眉设置,如何实现奇偶页不同? 比如:单页浙江大学学位论文,这一个容易设;双页:(每章标题),这一个有什么技巧啊? (4) 5. 问:怎样使WORD 文档只有第一页没有页眉,页脚? (5) 6. 问:如何从第三页起设置页眉? (5) 7. 问:WORD 页眉自动出现一根直线,请问怎么处理? (5) 8. 问:页眉一般是---------,上面写上题目或者其它,想做的是把这根线变为双线,WORD 中修改页眉的那根线怎么改成双线的? (6) 9. 问:Word 中的脚注如何删除?把正文相应的符号删除,内容可以删除,但最后那个格式还在,应该怎么办? (6) 10. 问:Word 里面有没有自动断词得功能?常常有得单词太长了,如果能设置下自动断词就好了 (6) 11. 问:如何将word 文档里的繁体字改为简化字? (7) 12. 问:怎样微调WORD 表格线?WORD 表格上下竖线不能对齐,用鼠标拖动其中一条线,可是一拖就跑老远,想微调表格竖线让上下对齐,请问该怎么办?7 13. 问:怎样微调word 表格线?我的word 表格上下竖线不能对齐,用鼠标拖动其中一条线,可是一拖就跑老远,我想微调表格竖线让上下对齐,请问该怎么办? (7) 14. 问:怎么把word 文档里已经有的分页符去掉? (7) 15. 问:Word 中下标的大小可以改的吗? (8) 16. 问:Word 里怎么自动生成目录啊 (8) 17. 问:Word 的文档结构图能否整个复制? 论文要写目录了,不想再照着文档结构图输入一遍,有办法复制粘贴过来吗? (8) 18. 问:做目录的时候有什么办法时右边的页码对齐?比如:1.1 标题..........11.2 标题...............2. (8) 19. 问:怎样在word 中将所有大写字母转为小写?比如一句全大写的转为全小写的 (8) 20. 问:在存盘的时候,出现了问题,症状如下:磁盘已满或打开文件过多,不能保存,另开新窗口重存也不管用。如何解决? (8) 21. 问:WORD 中的表格一复制粘贴到PPT 中就散掉了,怎么把WORD 里面的表格原样粘贴到PPT 中? (9) 22. 问:有没有办法将PPT 的文字拷入WORD 里面? (9) 23. 问:word 中图片的分栏如何处理?假如有:1 2 图3 4 这样的结构,我想实现:1 3 图(要横跨两栏)2 4 但是,试了半天总是:1 2 图3 4 怎么办呀?help! (9) 24. 问:用word 写东西时字距老是变动,有时候自动隔得很开,有时候进入下

最新史上最全最经典的WORD使用技巧(文件处理的超级技巧!!真的好经典!!!)

WORD文档使用技巧 1. 问:WORD 里边怎样设置每页不同的页眉?如何使不同的章节显示的页眉不同? 答:分节,每节可以设置不同的页眉。文件――页面设置――版式――页眉和页脚――首页不同 2. 问:请问word 中怎样让每一章用不同的页眉?怎么我现在只能用一个页眉,一改就全部改了? 答:在插入分隔符里,选插入分节符,可以选连续的那个,然后下一页改页眉前,按一下“同前”钮,再做的改动就不影响前面的了。简言之,分节符使得它们独立了。这个工具栏上的“同前”按钮就显示在工具栏上,不过是图标的形式,把光标移到上面就显示出”同前“两个字来 3. 问:如何合并两个WORD 文档,不同的页眉需要先写两个文件,然后合并,如何做?答:页眉设置中,选择奇偶页不同/与前不同等选项 4. 问:WORD 编辑页眉设置,如何实现奇偶页不同? 比如:单页浙江大学学位论文,这一个容易设;双页:(每章标题),这一个有什么技巧啊? 答:插入节分隔符,与前节设置相同去掉,再设置奇偶页不同 5. 问:怎样使WORD 文档只有第一页没有页眉,页脚? 答:页面设置-页眉和页脚,选首页不同,然后选中首页页眉中的小箭头,格式-边框和底纹,选择无,这个只要在“视图”――“页眉页脚”,其中的页面设

置里,不要整个文档,就可以看到一个“同前”的标志,不选,前后的设置情况就不同了 6. 问:如何从第三页起设置页眉? 答:在第二页末插入分节符,在第三页的页眉格式中去掉同前节,如果第一、二页还有页眉,把它设置成正文就可以了 ●在新建文档中,菜单―视图―页脚―插入页码―页码格式―起始页码为0,确定;●菜单―文件―页面设置―版式―首页不同,确定;●将光标放到第一页末,菜单―文件―页面设置―版式―首页不同―应用于插入点之后,确定。第2 步与第三步差别在于第2 步应用于整篇文档,第3 步应用于插入点之后。这样,做两次首页不同以后,页码从第三页开始从1 编号,完成。 7. 问:WORD 页眉自动出现一根直线,请问怎么处理? 答:格式从“页眉”改为“清除格式”,就在“格式”快捷工具栏最左边;选中页眉文字和箭头,格式-边框和底纹-设置选无 8. 问:页眉一般是---------,上面写上题目或者其它,想做的是把这根线变为双线,WORD 中修改页眉的那根线怎么改成双线的? 答:按以下步骤操作去做: ●选中页眉的文字,包括最后面的箭头●格式-边框和底纹●选线性为双线的●在预览里,点击左下小方块,预览的图形会出现双线●确定▲上面和下面自己可以设置,点击在预览周围的四个小方块,页眉线就可以在不同的位置

haso标准件中英文对照

HASCO Z标准件系列 Z00 导柱Guide pillar Z01 导柱Guide pillar Z011 导柱Guide pillar Z012 导柱Guide pillar Z013 导柱Guide pillar Z014 导柱Guide pillar Z0141 导柱连接端End p iece Z0142 导柱连接端End p iece Z015 导柱Guide pillar Z0151 导柱连接端(Z015/…专用)Pillar ada pter Z0152 导柱连接端Pillar ada pter Z02 顶杆Ejector rod Z022 导管Guide sleeve Z03 导柱Guide pillar Z05 圆锥形管精定位Locating unit, round Z055 圆形垫片Spacer Z056 圆形垫件Spacer Z06 平锥形导块Locating unit Z07 长方形导块Square guide bar Z08 直身方形管位导块Pre-centering unit Z081 方形垫片Sp acer disc Z091 方形定位锁块(公)Square interlock Z092 方形定位锁块(母)Square interlock Z093 方形定位锁块(母)Square interlock Z10 导套Guide bush Z10W 自润滑导套Guide bush Z11 导套Guide bush Z11W 自润滑导套Guide bush Z12 滚珠导套Ball guide bush Z13W 自润滑导套Self lubric.Guide bush Z15W 自润滑垫片Self lubric. Flat stock Z16W 自润滑垫片Self lubric. Guide rail Z17 长方形导块(Z07/...专用)Guide retainer Z20 定板导套Centering sleeve Z25 定位销Dowel pin Z26 定位销Dowel pin Z28 六角螺母Hexagon nut Z281 六角螺母Hexagon nut Z282 六角螺母Hexagon nut Z285 螺母Nut for T-slots Z286 螺母Nut for T-slots Z30 胚头内六角螺丝Socket head cap screw

紧固件-中英文对照

No.xx简写中文 1 screw scr螺丝 2 nut nut螺母 3 washer wash垫圈 4 head hd头 5 thread thre螺纹 6 coating coat镀层 7 rivet riv铆钉 8 standoff stdoff内外六角钉 9 driver driv穴 10 hexagon hexa六角 11 tooth tooth齿 12 lock lck锁 13 spring spr弹簧 14 wave wave波浪型 15 pitch pitch牙距 16 partial thread PT半螺纹 17 PT PT一种螺纹形式 18 Full Thread FT全螺纹 19 silver silv银 20 cone cone锥形

21 Eco-syn Eco 30度牙的一种螺纹 22 ThreadForming TF自攻螺纹 23 Self Tapping self tapp自切屑螺纹 24 pin pin销 25 shaft Shaf轴 26 steel St铁 27 stainless steel SS不锈钢 28 Brass(Cu) Bras(Cu)铜 29 Aluminum AL铝 30 Rubber Rub橡胶 31 machine screw mech scr机制螺丝 32 SEM screw SEM Scr组合螺丝 33 Material material材料 34 Grade GD等级 35 Finish Finish表面处理 36 Point Point点 37 Clamp Clam夹子 38 internal Int内 39 external ext外 40 thickness /厚度 41 length /xx

Word的进阶使用技巧

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论文排版超有用的word小技巧 [图片]

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