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# A perfect random number generator

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Texte intégral

(1)

(2)

(3)

(4)

n

(5)

0

0

(6)

0

0

n

r0

n

n

n+p

n

n+p

(7)

1x1

x22

xSS

x1+....+xS=y

x11

2x2

xSS

q

(8)

qd

e

e

2

P e−L(I)pe−D l

Dll2Dpe

2

(9)

0

0

0

0

1

(10)

1

q

(11)

m

m

m

m

p

2

2

s

s

s

1

s

s

s

1

2

p

p

s

n

n

n

n

+

qd

qd

q

q

q2

n

n

n+2

n+1

n

1

2

m

m

1

1

m

m

(12)

2

2

2

G

G

G

G

G

G

G

n

2

p

n

2

p

n

n+j2

2

n+jp

p

n

1

2

1

2

1

2

n

n

n

n

n

n

1

2

1

2

n

D

n

D

n

n

P

n

P

n

np

np

p!(nn!p)!

P

n

P

m P

P

n

P

n

2s

s2

(13)

## Introduction

n

n

n

n

n

n

n

(14)

n

n

n

n

(15)

n

1

2

n

n+1

e

n+1

1

n

e

n+1

1

n

n

n

n

n

n

(16)

n

n

n

n

n

n

n

n

n

n

n

n

n

1

2

p

s

1

1

2

p

1

2

p

n

2

p

n

n+j2

2

n+jp

p

m

m

m

(17)

n

s

n,j(.)

2

p

n

m

n+js

s

0

s

2

n,j(.)

2

p

n,j(.)

2

p

0

s

n

2

p

0

n

n

n

n

0

n

1

2

n+2

n+1

n

0

n0

n0+1

m

m

m

m

(18)

1

1

m

1

qd

q

qd

qd

qd

1

2

q

1

2

qd

q

qd

q

q

n

G

G

G

G

n

q

n

n

100

102340

n

50

20

(19)

n

n

n

n

n

n

n

20

340

n

n

q

n

n

j

0

1

1

1

1

1m

2

1

0

0

1

1

3

1m

2

1

1m

3

3

3

S

i=1

m

q0

(20)

q0

ns

ns

0

1m

1

0

n,j(.)

2

p

n,j(.)

2

p

0

2

p

0

0

1

0

0

q0/2

1

2

0

### N ,

1To consider the models A(j) is equivalent to consider the modelsE3(j) : cf properties of functionsTqin section 1.3

(21)

2

q0

0

0

2

p

0

0

0

e

0

s

s

e

e

0

1

1

0

p

p

1

p

p

0

e

p

p

p2

e

eC

0

1

0

s

s

0

eC

pC

pC

2

eC

0

0

e

p

e

p

e

p

0

(22)

0

0

0

1

1

0

1

0

0

(23)

(24)

## Quality of obtained sequences

n

n

n

n

n

n

n

n

n

n

n

n

n

(25)

s

e

e

4

N4

n=1

I1

n

I2

n+1

Ip

n+p

4

n

e

41/2

1

2

p

n

2

4

t1

t2

tm

s

ts

s

n

t1

t2

tm

2

2

e

2

5

n−t

(26)

n

N1/2(PσeL(I))

e

2

1/2

e

e

1/2

e

n

n

ts

1

2

p

s

q

q

q

q

1/2

e

n

n

1n

2n

3n

4n

n

1n

n

1n

4n

n

s

1

2

3

4n

n

e

4

N4

n=1

Bo1

n

Bo2

n+1

Bop

n+p

i

n

ts

1

p

1/2

e

(27)

n

s

n+j2

n+j3

n+jp

n

e

n

n+j2

n+jp

e

n

n+j2

n+jp

n

n+j2

n+jp

n+js

n

n+j2

n+jp

n+1

1

n

1

p

p

n

1

2

n

n+1

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n+1

1

n

e

n+1

1

n

n+1

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n

(28)

n

n

n

n

n

nT

n

nT

n

nT

n

nT

n

n

n

n

n

n

n

(29)

n

n

n

n

na

n

n

n

0

t(n+1)

t(n)

t(n+1)

t(n)

nT

n

(30)

q

n

q

n

n+j2

2

n+jp

p

0

0

q

n

n+j2

2

n+jp

p

q

n

0

### answer this question without making a study which proves it.

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