Adaptive Huffman

5/26/2018 Adaptive Huffman

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ADAPTIVE HUFFMAN

David A. Huffman

(9/8/19257/10/1999)


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Topic 9: Adaptive Huffman

Nhm thc hin:L Cao Tin 0612444

Nguyn Khc Tip 0612449

L Phc Trung 0612461

Lu c Tr 0612484


3/79

Adaptive Huffman

Gii thiu: Hn ch ca thut ton Huffman tnh.

tng.

Lch s hnh thnh.

u im.

Thut ton tng qut.

Cy Huffman (ng).

Tnh cht anh em (Sibling property)

Hnh thnh v cp nht cy.

Cc vi phm v cch gii quyt

Thut ton nn (Encoding)

Thut ton gii nn (Decoding)

Demo minh ha.


4/79

Adaptive Huffman - Gii thiu:

Gii thiu: Hn ch ca thut ton Huffman tnh. tng. Lch s hnh thnh. u im.

Thut ton tng qut. Cy Huffman (ng). Tnh cht anh em (Sibling property) Hnh thnh v cp nht cy. Cc vi phm v cch gii quyt

Thut ton nn (Encoding) Thut ton gii nn (Decoding) Demo minh ha.


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Adaptive Huffman - Gii thiu (tt):

Hn ch ca thut ton Huffman tnh. Trong qu trnh nn cn n 2 ln duyt File

Chi ph nn cao.

Cn phi lu tr thng tin gii nn

Lm tng kch thc d liu nn.

D liu nn cn phi c sn

Khng nn c d liu pht sinh theo thigian thc.


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Adaptive Huffman - Gii thiu (tt):

tng: Thut ton ny vn da trn tng ca Huffman l

s dng mt vi bit (bit code) biu din mt k t.

di m bit cho cc k t khng ging nhau:o K t xut hin nhiu lnbiu din bng m ngn.

o K t xut hin t biu din bng m di.

To sn mt cy ti thiu ban u, d liu nn sc cp nht dn vo cy.


7/79

Gii thiu (tt):

Lch s hnh thnh: c xut (c lp) bi Faller (1973) v

Gallager (1978)

Nm 1985 Knuth a ra mt s ci tin v honchnh thut ton. V vy thut ton ny cn cgi l thut ton FGK.

Nm 1987 Vitter trnh by cc ci tin lin quan tivic ti u cy Huffman.


8/79

Gii thiu (tt):

u im: Khng cn tnh trc s ln xut hin ca cc k t.

Qu trnh nn ch cn mt ln duyt file.

Khng cn lu thng tin phc v cho vic gii nn.

Nn online trn d liu pht sinh theo thi gian thc.


9/79

Adaptive Huffman

Gii thiu: Hn ch ca thut ton Huffman tnh. tng. Lch s hnh thnh.

u im. Thut ton tng qut. Cy Huffman (ng). Tnh cht anh em (Sibling property)

Hnh thnh v cp nht cy. Cc vi phm v cch gii quyt

Thut ton nn (Encoding) Thut ton gii nn (Decoding)


10/79

Adaptive Huffman - Thut ton tng qut:

Static Huffman:

o Cy Huffman c to thnh t bng thng k sln xut hin ca cc k t.

Adaptive Huffman:

o Nn online khng c trc bn thng k.

o Phng php: khi to cy ti thiu ban u,

cy s c cp nht dn dn da trn d liu phtsinh trong qu trnh nn hoc gii nn.


11/79


D liupht sinh

Cy

Huffman

D liunn/gii

nn

Khi to cyti thiu

Cp nht cy

Nn/ gii nn

S phi hp gia vic dng cy (cho thut ton nn/gii nn) v cp nht cy


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Cp nht

k t c vo cy

Cy

Huffman

D liunn

c k t c

M ha

(nn k t c)

c != EOF

Khi to cy tithiu

Kt thc

Yes

No

Thut ton nn


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Cy

Huffman

D liugii nn

c

c d liu nnb

Cp nht

k t c vo cy

Gii nn bthnh c

b != EOF

Khi to cyti thiu

Kt thc

Yes

No

Thut ton gii nn


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Adaptive Huffman

Gii thiu: Hn ch ca thut ton Huffman tnh. tng. Lch s hnh thnh. u im.

Thut ton tng qut. Cy Huffman (ng). Tnh cht anh em (Sibling property) Hnh thnh v cp nht cy. Cc vi phm v cch gii quyt

Thut ton nn (Encoding) Thut ton gii nn (Decoding) Demo minh ha.


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Adaptive Huffman - Cy Huffman (ng):

Mt cy nh phn c n node l c gi l cyHuffman nu tha: Cc node l c trng s wi0, i [1,n] Cc node nhnh c trng s bng tng trng s

ca cc node con ca n. Tnh cht Anh/ em (Sibling Property):o Mi node, ngoi tr node gc u tn ti mt

node Anh/ em (c cng node cha).

o Khi sp xp cc node trong cy theo th ttng dn ca trng s th mi node lun k vinode anh em ca n.


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V d:

W=3

#5

D

W=2

#4

C

W=2

#3

B

W=2

#2

A

W=1

#1

W=4

#6

W=7

#7

E

W=10

#8

RootW=17

#9

Th t #1 #2 #3 #4 #5 #6 #7 #8 #9

Wi 1 2 2 2 3 6 7 10 17

Gi tr A B C D E Root


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Cch thc to cy:b1: Khi to cy ti thiu, ch c node Escape (node 0)

b2: Cp nht tng k t vo trong cy theo qui tc:

Nu k t cha c trong cythm mi node l

Nu k t c trong cytng trng s node ln 1

Cp nht trng s ca cc node lin quan trong cy

Escap

eW = 0


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Thut ton cp nht trng s:

Tng trng s ca node l ln 1.

i t node l n node gc tng trng s cacc node ln 1. Kim tra tnh cht anh em vhiu chnh li cy nu c vi phm.


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W=11

#9

B

W=11

#7

W=25

#10

W=14

#8

L

W=5

#4

W=9

#6

D

W=4

#3

W=5

#5

T

W=3

#2

C

W=2

#1

Tng trng s (1)

Tng trng s (cui)

Tng trng s (2)

Tng trng s (3)

C

W=3

#1

W=15

#8

W=6

#5

W=26

#10


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Khi thm mt node mi hoc tng trng s:

Vi phm tnh cht anh em.

Trn s.


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Vi phm tnh cht anh em:

B

W=17#8

W=29

#9

W=12

#7

L

W=3

#4

W=6

#6

D

W=3

#3

W=6

#5

T

W=3

#2

C

W=3

#1

Tng trng s (1) DW=4

#1


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Hiu chnh cy thatnh cht anh em:

B

W=17#8

W=29

#10

W=12

#8

L

W=3

#4

W=6

#6

D

W=3

#3

W=6

#5

T

W=3

#2

C

W=4

#1

W=6

#5

W=7

#6

W=13

#7

W=30

#9

#1 #4


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Thut ton xc nh node vi phm: Gi x l node hin hnh. So snh x vi cc node tip theo sau (theo th t t

tri sang phi, t di ln trn).

Nu tn ti y sao cho y.Weight < x.Weight th x lnode vi phm. Thut ton hiu chnh cy tha tnh cht anh em:

Gi x l node vi phm. Tm node y xa nht, c trng s cao nht tha

y.Weight < x.Weight

Hon i x v node y trn cy. Cp nht li cc node cha tng ng. Lp li b1 cho n khi khng cn node vi phm.


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Vn trn s:

Qu trnh cp nht cy lm tng trng s cc node.

Trng s node gc tng nhanh c th vt qu khnng lu tr ca kiu d liu.

oKiu int gi tr max = 215- 1

oKiu unsigned int gi tr max = 2161

oKiu long gi tr max = 231 - 1


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V d v trn s:

Node gc ang c gi tr

trng s ti a m kiu dliu c th biu din.

Hintngtrn s

sxuthinkhi ta

tngtrngsca

btk node l no

W=120

#7

L

W=35

#4

W=65

#6

D

W=30

#3

W=55

#5

T

W=30

#2

C

W=25

#1

B

W=135

#8

W=255

#9


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Thut ton x l trn s:

Khi cp nht trng s, kim ta trng s node gc.

Nu trng s node gc > MAX_VALUE:o Gim trng s cc node l trong cy (chia cho 2).

o Cp nht lai trng s cac node nhnh.

o Kim tra tnh cht anh em v hiu chnh li cy.

(Do php chia cho 2 lm mt phn d ca s nguyn)


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X l trn s: (tt)

W=120

#7

L

W=35

#4

W=65

#6

D

W=30

#3

W=55

#5

T

W=30

#2

C

W=25

#1

B

W=135#8

W=255

#9

W=12 W=15 W=15 W=17

W=27W=32

W=67

W=116

W=59


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Adaptive Huffman


tng.

Lch s hnh thnh.

u im. Thut ton tng qut.

Cy Huffman (ng).






Demo minh ha.


29/79

Adaptive Huffman

Thut ton nn:

initialize_model();

while(c != EOF)

{

c = getchar(inputfile);

encode(c, outputfile);

update_model(c);}


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Adaptive Huffman

Thutton m ha k tc:

encode (c, outputfile)

Nuc chac trong cy

oDuytcy T tm m bit caEscape ghi lnoutputfile.

oGhi tip8 bit m ASCII cac ln outputfile.

oThm node c vo cy v cpnhtlicy.

Nuc c trong cy:

o Duytcy T tm m bit cac v ghi ln outputfile

o Tngtrngscanode c ln 1 v cpnhtlicy


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Adaptive Huffman


tng.

Lch s hnh thnh.

u im. Thut ton tng qut.

Cy Huffman (ng).






Demo minh ha.


32/79

Adaptive Huffman

Thut ton gii nn:

inputfile: d liu dng nn

outputfile: d liu gii nn

initialize_model();

while((c = decode(inputfile)) != EOF)

{putchar(c, outputfile);

update_model(c);

}


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Adaptive Huffman

Thut ton gii m k t c:decode(inputfile)

b = getchar(inputfile);

Bt u t v tr hin ti trn inputfile.

Ly tng bit ca b, duyt trn cy (b==0: tri, b==1:phi).

o Nu i n node l x, return x.char

o Nu i n node Escape: c = ly 8 bit tip theo t inputfile

Thm c vo cy, cp nht li cy.

return c


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DEMO MINH HA


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Adaptive HuffmanV d

Gisc dliunhsau:

f = abacbdc


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Adaptive HuffmanV d (tt)

B1: Khi to cy v a k t a vo cy:a

a

W=1E#1

#3

#2

W = 1

input = 01100001


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B2: Thm k t b vo trong cyb cha c trong cy

Ta thm b vo thnh node l

trn cy

input = 01100001

a

W=1E

W=1 #3

#2#1

0

0 01100010


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b

a

W=1E

W=1 #3

#2#1

b

W=1

#3

#1 #2

W = 1


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a

W=1

W=1

b

W=1

#3

#1 #2

W = 1

E

#4

#5


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Cp nht trng s ca cc node c

lin quan

a

W=1

W=1

b

W=1

#3

#1 #2

W = 1

E

#4

#5W = 2


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Adaptive HuffmanV d (tt)a

B3: thm k t a vo cy.

a tn ti trong cy

Tng trng s ca node a thm 1

Cp nht li trng s cacc node lin quan

a

W=1

W=2

b

W=1

#3

#1 #2

W = 1

E

#4

#5

W=2

W = 3

input = 01100001001100010

1

1


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B4: thm k t c vo cy.

c cha c trong cy

Ta thm c vo thnh node l

trn cy aW=2

W=3

b

W=1

#3

#1 #2

W = 1

E

#4

#5

input = 011000010011000101

0

0

00 01100010


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c

aW=2

W=3

#4

#5

W=1

bW=1

E#1 #2

#3

c

W=1E#1 #2

#3 W = 1


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a

W=2

W=3

#4

#5

W=1

b

W=1#2

#3W=1

c

W=1E#1 #2

#3

#4

#6#5

#7


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lin quan aW=2

W=3

#6

#7

W=1

b

W=1#4

#5

W=1

c

W=1E#1 #2

#3

W = 2

W = 4


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Adaptive HuffmanV d (tt)b

B5: thm k t b vo cy.

b tn ti trong cy

Tng trng s ca node b thm 1

Cp nht li trng s cacc node lin quan

a

W=2

W=4

#6

#7

W=2

b

W=1#4

#5

W=1

c

W=1E#1 #2

#3

W = 2

W = 3

input = 0110000100110001010001100010

0

1

01


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Vi phm tnh cht anh em

Ta phi xoay li cyCp nht li trng s ca cc

Node lin quan.

a

W=2

W=4

#6

#7

#5

#4

W=3

b

W=2W=1

c

W=1E#1 #2

#3

W=5


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B6: thm k t d vo cy.

d cha c trong cy

Ta thm d vo thnh node l

trn cy

W=3

W=5

#6

#7

a

W=2

b

W=2#4

#5

W=1

c

W=1E#1 #2

#3

input = 011000010011000101000110001001

1

0

0

10001100100


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Adaptive HuffmanV d (tt)d

d

W=1#1

#3

#2

W = 1

W=3

W=5

#6

#7

a

W=2

b

W=2#4

#5

c

W=1

W=1

E#1 #2

#3

E


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W=1

d

W=1E#1 #2

#3

W=3

W=5

#6

#7

a

W=2

b

W=2#4

#5

cW=1

W=1

#2

#3

#4

#5 #6

#7#8

#9


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W=1

d

W=1E#1 #2

#3

W=3

W=5

#8

#9

a

W=2

b

W=2#6

#7

cW=1

W=1

#4

#5

Cp nht li trng s ca ccnode c lin quan

W=4

W=6

W=2


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W=1

d

W=1E#1 #2

#3

W=4

W=6

#8

#9

a

W=2

b

W=2#6

#7

W=2

#4

#5

B7: thm k t c vo cy.c tn ti trong cy

Tng trng s ca node c thm 1

c

cW=1W=2

input = 01100001001100010

1000110001001100

01100100

1

1

0

101


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W=1

d

W=1E#1 #2

#3

W=4

W=6

#8

#9

a

W=2

b

W=2#6

#7

W=2

#4

#5

cW=2

Vi phm tnh cht anh emTa phi xoay li cy

Cp nht li trng s ca cc nodelin quan v hiu chnh li cy.

W=3


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W=4

W=6

#8

#9

a

W=2

b

W=2#6

#7

#5

W=1

d

W=1E#1 #2

#3

W=3

#4c

W=2


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W=4

W=6

#8

#9

a

W=2

b

W=2#6

#7

#5

W=1

d

W=1E#1 #2

#3

W=3

#4

c

W=2

Vi phm tnh cht anh emTa hiu chnh li cy


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W=4

W=6

#8

#9

a

W=2

b

W=2#6

#7

#5W=1

dW=1E#1 #2

#3

W=3

#4c

W=2

kim tra trng s ca ccnode lin quan.

input = 01100001001100010100011000100110001100100101


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Adaptive Huffman

Gii nn


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B1: c 8bit u tin t file input a

a

W=1E#1

#3

#2

W = 1

input = 01100001001100010100011000100110001100100101

a

output = a


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B2: c bit tip theo, duyt cyCp nht k t b vo trong cy

Di chuyn u c n v tr tip theo

a

W=1E

W=1 #3

#2#1

input = 0110000100110001010001100010011000110010010

fseek

0

b

output = a b


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b

a

W=1E

W=1 #3

#2#1

b

W=1

#3

#1 #2

W = 1


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a

W=1

W=1

b

W=1

#3

#1 #2

W = 1

E

#4

#5


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lin quan

a

W=1

W=1

b

W=1

#3

#1 #2

W = 1

E

#4

#5W = 2


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B3:c bit tip theo v duyt cy.Tng trng s ca node a thm 1

Cp nht li trng s ca

cc node lin quanDi chuyn u c n

v tr tip theo

a

W=1

W=2

b

W=1

#3

#1 #2

W = 1

E

#4

#5

W=2

W = 3

1

input = 0110000100110001010001100010011000110010010

fseekoutput = ab a


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B4: c bit tip theo v duyt cy.

Ta thm c vo thnh node l

trn cy

Di chuyn u c nv tr tip theo

a

W=2

W=3

b

W=1

#3

#1 #2

W = 1

E

#4

#5

0

0

input = 0110000100110001010001100010011000110010010

fseek

c

output = aba c


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c

aW=2

W=3

#4

#5

W=1

bW=1

E#1 #2

#3

c

W=1E#1 #2

#3 W = 1


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a

W=2

W=3

#4

#5

W=1

b

W=1#2

#3W=1

c

W=1E#1 #2

#3

#4

#6#5

#7


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lin quan aW=2

W=3

#6

#7

W=1

b

W=1#4

#5

W=1

c

W=1E#1 #2

#3

W = 2

W = 4


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B5: c bit tip theo v duyt cy..

Tng trng s ca node b thm 1

Cp nht li trng s ca

cc node lin quanDi chuyn u c n

v tr tip theo

a

W=2

W=4

#6

#7

W=2

b

W=1#4

#5

W=1

c

W=1E#1 #2

#3

W = 2

W = 3

0

1

input = 0110000100110001010001100010011000110010010

fseek

output = abac b


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Vi phm tnh cht anh em

Ta phi xoay li cyCp nht li trng s ca cc

Node lin quan.

a

W=2

W=4

#6

#7

#5

#4

W=3

b

W=2W=1

c

W=1E#1 #2

#3

W=5


70/79


B6: c bit tip theo v duyt cy..

Ta thm d vo thnh node l

trn cy W=3

W=5

#6

#7

a

W=2

b

W=2#4

#5

W=1

c

W=1E#1 #2

#3

1

0

0

input = 0110000100110001010001100010011000110010010

fseek

d

Di chuyn u c nv tr tip theo

output = abacb d


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Adaptive HuffmanV d (tt)d

d

W=1#1

#3

#2

W = 1

W=3

W=5

#6

#7

a

W=2

b

W=2#4

#5

cW=1

W=1

E#1 #2

#3

E


72/79


W=1

d

W=1E#1 #2

#3

W=3

W=5

#6

#7

a

W=2

b

W=2#4

#5

c

W=1

W=1

#2

#3

#4

#5 #6

#7#8

#9


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W=1

d

W=1E#1 #2

#3

W=3

W=5

#8

#9

a

W=2

b

W=2#6

#7

c

W=1

W=1

#4

#5

Cp nht li trng s ca ccnode c lin quan

W=4

W=6

W=2


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W=1

d

W=1E#1 #2

#3

W=4

W=6

#8

#9

a

W=2

b

W=2#6

#7

W=2

#4

#5

B7: c bit tip theo v duyt cy...Tng trng s ca node c thm 1

c

W=1W=2

input = 01100001001100010

1000110001001100

01100100101

1

1

0

fseek

output = abacbd c


75/79


W=1

d

W=1E#1 #2

#3

W=4

W=6

#8

#9

a

W=2

b

W=2#6

#7

W=2

#4

#5

c

W=2

Vi phm tnh cht anh emTa phi xoay li cy

Cp nht li trng s ca cc nodelin quan v hiu chnh li cy.

W=3


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W=4

W=6

#8

#9

a

W=2

b

W=2#6

#7

#5

W=1

d

W=1E#1 #2

#3

W=3

#4c

W=2


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W=4

W=6

#8

#9

a

W=2

b

W=2#6

#7

#5

W=1

d

W=1E#1 #2

#3

W=3

#4c

W=2

Vi phm tnh cht anh emTa hiu chnh li cy


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W=4

W=6

#8

#9

a

W=2

b

W=2#6

#7

#5W=1

dW=1E#1 #2

#3

W=3

#4c

W=2

kim tra trng s ca ccnode lin quan.

output = abacbdc


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Adaptive Huffman

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