Nghiên cứu lý thuyết mã nén văn bản dựa theo mô hình markov

Dynanic Markov Coding... 1 Để cho đơn giản ta gọi tắt nửa đoạn dạng [x,y là khoảng.. 2 Nh- vậy điều quan trọng là làm thế nào để luôn có thể xác định đ-ợc xác suất xuất hiện của chữ tiếp

tin

c

-

-

, c

2/1

3/1

a

b

5/3

7/1

7/6

5/2

a

b

5/2

5/2

3/2

5/3

c

5/1

5/2

3/1

t

A

) ( ) (

( ) - x

ha

n

m 1

1 log

1

1 log p k

19 101

26 log 101

26 101

26 log 101

26 101

30 log 101

30

Markov A Markov (1856-1922) đưa ra

ij 1

m

1 j ij p

) p

(

} i

| j { P

p(ijk) k 0

k.

j ) k ( i )

l k (

n n

Markov ergodic

1.1.

i 1

5 4 3 2 1

5 4 3 2 1

0 0 75 0 0

25

.

0

25 0 0 0 75

.

0

0

0 25 0 0 0

75

.

0

75 0 0 25 0 0

0

0 75 0 0 25

4 1

1/4

1/4

1/4 3/4

H×nh 1.1

1 log w E

m 1 i

i E =

ij 2 m

1 i m

1 j ij i

w

1 log w

A

ij 2 m

1 i m

1 j ij i

w

1 log w

c

2 i

c

3 i

c

4 i

c

H×nh 1.2

146)

n

) ( p log

19 101

26 log 101

26 101

26 log 101

26 101

30 log 101

30

, b, c, d” Tuy nhiên

12 7 30

; p12 =

75

26 26 ) 12 7 30 (

21 11

p p

p p

=

49

75 126

75 0

2

1 2

1

0 75

26

1 75

H×nh 1.3

Nguån 2 Entropy=1.55

1

E1 =

30

26 12 7 30 log 26 12 7 30

19

7

7 19 log 7 19

n

var a, b, Ea, Eb, Pa, Pb: extended;

H×nh 1.4a

(Dynanic Markov Coding.)

1.5

1 = 2/3, p2 = 1/3

= 2/3 log2(3/2)+ 1/3 log2(3) 0.918 11%

da:=da+1;

end else

begin write(f,b);

db:=db+1;

end end;

close(f);

clrscr;

E:=(da/640000*ln(640000/da)+ db/640000*ln(640000/db))/ln(2); writeln(' ty le nen con = ',round(E/8*100), '%');

2/13/1

b

a b

a

p

pp

p2/13/2

2/13/1

n pa pb 1

7/3

3/2

2/1

2/1

3/1

H×nh 1.7

var f:file of byte;

'a':begin if random<1/3 then

begin write(f,a);M:='a'; end

else

begin write(f,b);M:='b'; end;

end;

'b':begin if random<1/2 then

begin write(f,a);M:='a'; end

else

begin write(f,b);M:='b'; end;

end;

end;

a

b

5/3

7/1

7/6

5/2

H×nh 1.8

a b

a

p

pp

p7/15/3

7/65/2

pa pb 1

51/30

'a':begin if random<2/5 then

begin write(f,a);M:='a';Na:=Na+1; end

else

begin write(f,b);M:='b';Nb:=Nb+1; end;

end;

'b':begin if random<6/7 then

begin write(f,a);M:='a';Na:=Na+1; end

3/15/25/2

3/25/15/3

c b a

c b a

ppp

ppp

05/20

3/15/25/2

3/25/15/3

n pa pb pc 1

77/35

pa , pb 30/77, pc 12/77var

5/2

3/2

5/3

c

5/1

5/2

3/1

H×nh 1.9

else if R<3/5 then begin write(f,b);M:='b';Nb:=Nb+1;end

else begin write(f,c);M:='c';Nc:=Nc+1;end

i i

1

) (

) (

) ( L ) ( p n

1

) (

3 2 2 1

k k

i

0, i 2 3 k

1 2

1 2

3 4 2

S:=S*2;

if S>1 then begin S:=S-1;

code[i]:=code[i]+'1' end

else code[i]:=code[i]+'0' end;

end;

i i

i

i 2 i

i

2

p

1 log 1 r p

1 log

m

1 i

i

i r p r

m

1

2 i m

1 log

tư (e,0 3) (a,0.2) (o,0.2) (i

{a,o}

end;

} y:=a[m-1].prob;inc(a[m-1].prob,a[m].prob);

} k:=m-1;

while (k>1)and (a[k].prob>a[k-1].prob) do begin

x:=a[k-1]; a[k-1]:=a[k]; a[k]:=x; k:=k-1;

1 log

m

i i i

p

1 log

End

sau

2.3

:

: Sx:=a[k];

for i:=1 to n do a[i].code:='';

if a[i].prob>a[i-1].prob then begin x:=a[i];a[i]:=a[i-1];a[i-1]:=x;end;

end;

end

aaaaaabcdefghaahhaaaaagabghabaecdcaaadaecccccccccghaacbgbchaecbdhabdehahcghghaebcd

101110011000111001000110100011000100011010

c

1 Để cho đơn giản ta gọi tắt nửa đoạn dạng [x,y) là khoảng

2 Nh- vậy điều quan trọng là làm thế nào để luôn có thể xác định đ-ợc xác suất xuất hiện của chữ tiếp theo?

1 i i

p ) a b (

j

) , [ j 1 j

m 2

1 0

[0, 1)

-

a

0.100 0.101 0.102 0.103 0.104 0.105 0.106 0.107 0.108 0.109 0.110

c

0.1020 0.1021 0.1022 0.1023 0.1024 0.1025 0.1026 0.1027 0.1028 0.1029 0.1030

g

H×nh 3.3

- : 0.10, 0.11, 0.12, , 0.19, 0.20 0.10

a

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

a

0.100 0.101 0.102 0.103 0.104 0.105 0.106 0.107 0.108 0.109 0.110

c

0.1020 0.1021 0.1022 0.1023 0.1024 0.1025 0.1026 0.1027 0.1028 0.1029 0.1030

g

H×nh 3.4

[ i, i+1

ơhơn

0.10260 0.10261 0.10262 0.10263 0.10264 0.10265 0.10266 0.10267 0.10268 0.10269 0.10270

b

0.160 0.168 0.176 0.184 0.192 0.200 0.208 0.216 0.224 0.232 0.240

a

0.1600 0.1021 0.1022 0.1023 0.1024 0.1025 0.1026 0.1027 0.1028 0.1029 0.1920

c

H×nh 3.5

end

L:=0;rank:=1;state:=1;T1:='';

} S:=random;

if S<4/26 then

begin T1:=T1+'d';State:=4;Rank:=Rank*4/26;

end else begin T1:=T1+'a';

begin T1:=T1+'d';

State:=4;L:=L+4/20*Rank;Rank:=Rank*1/20;

end else if S<12/20 then

begin T1:=T1+'b'; State:=2;L:=L+5/20*Rank;Rank:=Rank*7/20; end

else

begin T1:=T1+'a';State:=3;L:=L+12/20*Rank;Rank:=Rank*8/20; end;

Str(L:22:19,V1);Str(L+Rank:22:19,V2);V:='['+V1+V2+')';

Str(L+Rank/2:22:19,VL);Val(VL,L,m);

} State:=1;S:=0;Rank:=1;T2:='';

if State=4 then

if (L-S)/Rank<4/20 then

begin T2:=T2+'c';State:=1;Rank:=Rank*4/20;

end else

if (L-S)/Rank<5/20 then

begin T2:=T2+'d';State:=4;S:=S+4/20*Rank;Rank:=Rank*1/20; end

else

if (L-S)/Rank<12/20 then begin

T2:=T2+'b';State:=2;S:=S+5/20*Rank;Rank:=Rank*7/20; end

else begin T2:=T2+'a';

[0.058907264212235128, 0.058907264212235135)

bacdcbacbacbacddbdcdcbacbacdbdcdcbacbacdcbacbacbacbacbacdcbac

[0.553712632942476848, 0.553712632942476853)

dcbacbacbacdcbacbacbacbacbacbacbacdcdcddbdcdcbacbacbacbacbacbacdcbac [0.048585947986781746, 0.048585947986781755)

bacbacbacdcbacbacbacbacbacbacdcbacbacdcbacbacbacbacbacbacbacbacbacbacbacbacbacbacbacdbdcdcbacbacbac

[0.902281543958376885, 0.902281543958376893)

LZW )

:

Nh©n v©t cña c©u truyÖn

Néi dung c©u chuyÖn

tõ ch÷ c¸i

z z

z1 k k 1

n 3 2

1 s s s s

z z

z1 k k 1

n 1

L m 2 m 1 m L n 2 n 1

n z z s s s z

L n 1 n

n z z

Thay n n+L

n 3 2

1 s s s

s (m1, L1) (m2, L2)(m3, L3) (mr, Lr

z z

n z z z z z

L n 1 n n

1 z z

z (m1,L1) (m2,L2)(m3,L3) (mr,Lr

(Huffman, Fano, )

Khôn k

ro

Output 0+a 1+a 0+b 3+a 4+a 5+a 4+b

4.1 (0,a)(1,a)(0,b)(3,a)(4,a)(5,a)(4,b)

(empty)

T

aabababaaababb aabababaaababb aabababaaababb aabababaaababb aabababaaababb aabababaaababb

0+a 0+b 1+a 3+a 5+a 2+b 4+b 1+b

Var

S :array[255 15000] of word;C :array[255 15000] of byte;

for m:=256 to H do write(g,S[m]);x:=a;write(g,x); close(g);

assign(g,'C.nen'); reset(g); assign(f,'D.txt'); rewrite(f);

read(g,S[256]); write(f,byte(S[256]));

m:=256;

while not eof(g) do

1

3 Donald Samuel Ornstein and Benjamin Weiss (1993), Entropy and Data

Compression Schemes, IEEE Transactions on Information

Theory, Vol.39, No.1, January , pages 78-83

4 Gyula O H Katona, Tibor O H Nemetz (1976), Huffman Codes and

Self-Information, IEEE Transactions on Information Theory,

Vol.22, No.3, May , pages 337-339

5 Ian H Witten, Radford M Neal (1987), and John G Cleary, Arithmetic

coding for data compression, Communicatio ns of the ACM,

June , Volume 30, Number 6, pages 520-540

6 I E Witten, R M Neal, J G Cleary (1990), Text Compression,

Prentice Hall

7 Nelson Mark (1991), The Data Compression Book, M&T Books,

8 Obert J McEliece (1993), The Theory of Information and Coding,

Cambridge University Press