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第五章  学科中的数学方法 PowerPoint PPT Presentation


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第五章  学科中的数学方法. 在计算学科中,采用的数学方法主要是离散数学方法。本章首先简单介绍数学的基本特征及数学方法的作用。然后,介绍计算学科中常用的数学概念和术语,包括集合,函数和关系,代数系统(含群、环、格、布尔代数,布尔代数与数字逻辑电路等),字母表、字符串和语言,定义、定理和证明,必要条件和充分条件、证明方法、递归和迭代、公理化方法等内容。最后,介绍计算学科中的形式化方法,包括形式系统的组成、基本特点和局限性,形式化方法的定义,以及形式规格和形式验证等内容。. 5.1 引 言.

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第五章  学科中的数学方法

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5.1


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What is itHow to do it


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5.2

3

1.

2

3


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5.3

3

1


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2

3


5 4 5 4 1

5.4 5.4.1

1

2

3

1

15

A=12345

2

{ 0112358132134 }

3

{Fn|Fn+2=Fn+1+FnF0=0F1=1n0}


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3

1

ABABCABCABxxAxB

5.1 A={abcd}B={bde}AB

ABabcde

2

ABABSABS=AB=xxAxB

5.2 A={abcd}B={bde}AB

AB={ac}

3

ABCABCABxxAxB


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5.3 A={x | x >5}B={x | x < 1}B

ABx | x >5x | x < 1x5 < x < 1

4

IAIIAA

=IA=xxIxA

5.4 I=x5x5A=x0x1

=IA=x5x01x5

5

A1A2AnRene Descartes

A1A2An={a1a2anaiAii12n}

A1A2Ann

5.5 A={123}B={ab}AB

AB={1a1b2a2b3a3b}


5 4 2

5.4.2

1

fa b

2

k5.1

5.1

5.1


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3

3

1aaRa

2abaRbbRa

3abcaRbbRcaRc

ARA

5.6 N3R

:

R=(ab)abNa b3N

1

aNa a = 03N

2

abnNa b = 3n3N

b a = 3( n)3N


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3

abcnmNa b = 3nb c = 3m(a b)+(b c)= 3n + 3ma c = 3(n+m)3N

5.7 e1e2e3

coe1 co e1e2 co e2e3 co e3e1 co e2e2 co e1e1 co e3e3 co e1e2 co e1e1 co e3e2 co e3


5 4 3

5.4.3

AAAAAAnAAn

Af1f2fn<Af1f2fn>

4

1.

1 ABCfABCABC

1*AxyAx*yA*A

2*AxyA x*y=y*x*

3*A xyzA x*y*c=x*y*z*


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4*Axyzx*yz=x*yx*zxy*z=x*zy*z*

5*Ael Aa Ael*a=aelA*er Aa*er=aerA*eAeaAe*a=a*e=a

6*Al AaAl*a=ll A*rAaAa*r =rrA*AaA*a=a*=

7*AeaAal-1Aal-1*a=eal-1*al-1 aar-1Aa*ar-1 =ear-1*ar-1 a


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2.

19E.Galois


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1 <S*>S*S*<S*>

2 <S*>S*S

1*

2*

<S*>


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3 <G*>G*G

1*

2*

3e

4xGx-1

<G,*>

5.1

5.1


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3.

1ax=b1/aax+b=c-b1/aR<R*>2<R*->

4.

4 <A,>A<A,>

5.

013


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5 <A,*,,>*AA01AxyzA

1x*y=y*xxy=yx ()

2x*(yz)=(x*y) (x*z) x (y*z)=(xy)*(xz) ()

3x*0=x x1=x ()

4x*x=1 xx=0 ()

5x* (y*z)=(x*y)*z x (yz)=(xy) z ()

<A, * ,,>*01

G.Boole1847

1938C.E.ShannonASymbolicAnalysisofRelayandSwitchingCircuits

1033+-35.1


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5.1

33333


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ABCFF=f(ABC)

35.2

5.2


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6.*

n

X YFn


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5.3


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5.23

5.2

Cn-1Cn5.4

5.4


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Cn

5.3


5 4 4

5.4.4

09AZaz+*/

abc123


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1aa=a = a

2aa

312


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={ab}{abaabbababbba}anbnn1

0123


5 4 5

5.4.5

5.8

1.

1

2

3

2.

1

2

3

4


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5.9

5.10

5.11

5.12


5 4 6

5.4.6

necessary conditionsufficient condition

pqpqpqqppqpq

pqpqqp

A={x|p}B={x|q}ABx AxAxBABpq5.2(a)

pq, pqA=B, 5.2(b)ABBAaAB(b)


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5.13

5.14

5.2 AB


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5.15 x0x100

5.16 x=2x2=4


5 5 5 5 1

5.5 5.5.1

1

pqpq

5.17 pp2

pp=2kkp2=22k2p22

2

pqqpqppq

5.18 p2p

pkp=2k+1p2=4k2+4k+1=22k2+2k+1p221


5 5 2

5.5.2

5.19

500Pythagoras

Hippasus 15L.Da.Vinci


5 5 3

5.5.3

1

2

3

4

1

2

nP(n)P(n)

P(1)

i1P(i)P(i+1)


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3

P(1)()(P(n)P(n+1))P(n)

4

5.20 P(n)1nn

1+3+5++(2n1)=n2

n=11=12

k1P(k)

1+3+5++(2k1)= k2

1+3+5++(2k1)+(2(k+1)1) =

k2+2k+1=(k+1)2

P(k)P(k+1)


5 5 4

5.5.4

1

xP(x)xP(x)xP(x)

2

P(a)a

BArmstrong


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5.6

an1an


5 6 1

5.6.1

1

2030POST

1

2

3

4

5

6


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2

5.21 56

66+6=1212+6=1818+6=2424+6=30

564636261616+6=1212+6=1818+6=2424+6=30

564663616630

an=Can-1+g(n)234

C{g(n)}an-1ana1


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3

1

5.22 251123an=2an1+1

a1=2

an=2an1+1n=234

2

5.23 F(n)=n!

F(n)=n!

F(0)=1

F(n)=nF(n1)n=123


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3

5.24 G

S0A1 SG

A01

A0A1

L(G)=0n1n|n101020


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4

Ackermann1928


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:

Begin

if m=0 then n+1

else if n=0 then A(m-1,1)

else A(m-1,A(m,n-1))

End

5.25 A(1,2)

: A(1,2)= A(0, A(1,1)) =

A(0, A(0, A(1,0)) =

A(0, A(0, A(0,1)) =

A(0, A(0,2)) =

A(0,3) =

4


5 6 2

5.6.2

X0=aXn+1=f(n)


5 6 21

5.6.2

X0=aXn+1=f(n)


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5.7

5.7.1

1

T=<CPS>

1T

2C

3P

4S


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2


5 7 2

5.7.2

1

23

1


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2

3


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3

BArmstrong

5.6.3

5.26

1

n111

5.27

55119465

1

2

3

4

5


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1

2

3

4

5

5.28

10080,00023,00060,000 167,000167,000167,000


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8167,000238,000405,000


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5.8


5 8 1

5.8.1

1

1

2

3

4

4


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2

1

2


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3

1

1931

2

CALCSALC


5 8 2

5.8.2

1.

2060E. W.DijkstraC.HoareD.S.Scott

2.

What to doHow to do


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3.

4.

1


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2

IBMCICS9T800Inmos12

3

2090

Defense Standard0055Def Stan00-55

0056Def Stan 00-56


5 8 3

5.8.3

1.

2070ZavePAISleyPetri

2.


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1

ADTADTADTADTADTADTADTZLOTOSVDMLarch

2

HornalwayssometimeshencefortheventuallyCTLRTILTCTLTPTL

3.

ESM/RTTLTRIO+TROL


5 8 4

5.8.4

1.

MfM f

M fCTLLTL

OBDDC


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2.

ff

F

ACL2EvesLPNqthmReveRRLCoqHOLLEGOLCFNuprlAnalyticaMathematicaPVSStep

D


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3.

5.2


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5.2

HOL-VossVossProverPVSSTePCadence SMVCVC


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5.9

1,2,3,


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20048IEEE/CSACMSE2004


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