Ma Khoi Tuyen Tinh

7/31/2019 Ma Khoi Tuyen Tinh

1/19

CHNG 3

M KHI TUYN TNH

3.1 Gii thiu

M khi tuyn tnh l mt lp m c dng rt ph bin trong vic chng nhiu. Loim ny c xy dng da trn cc kt qu ca i s tuyn tnh. y chng ta cngch nghin cu v m nh phn.

nh ngha

Mt m khi c chiu di n gm 2k t m c gi l m tuyn tnh C(n, k) nu v chnu 2k t m hnh thnh mt khng gian vect con k chiu ca khng gian vect n chiu

gm tt c cc vect n thnh phn trn trng GF(2).Trng GF(2) (Galois Field (2)) l trng nh phn ng thi php cng l php cngmodul 2 (k hiu l ), cn php nhn l php v (AND). C th

M tuyn tnh C(n, k) c mc ch m ho nhng khi tin (hay thng bo) k bit thnhnhng t m n bit. Hay ni cch khc trong n bit ca t m c cha k bit thng tin. Cc

phn tip theo sau s trnh by cch biu din m, cch m ho cc thng bo thnh tm, cch gii m t t m thnh thng bo, cch pht hin sai v sa sai.

Qui c

n gin sau ny chng ta s vit du + thay cho du v du + s c hiu theong cnh.

3.2 Cc khi nim v nguyn l hot ng

Cch biu din m Ma trn sinh

M tuyn tnh C(n, k) l mt khng gian con k chiu ca mt khng gian vect n thnhphn. Do vy c th tm c k t m c lp tuyn tnh trong C chng hn (g0, g1, ..., gk-1) sao cho mi t m trong C l mt t hp tuyn tnh ca k t m ny:

v = a0g0 + a1g1 + ... + ak-1gk-1

vi ai{0, 1} vi mi i = 0, 1, ..., k1.


2/19

t k t m c lp tuyn tnh ny thnh nhng hngchng ta c c mt ma trn cpk n nh sau .

Vi gi = (gi0, gi1, , gi(n-1)), vi i = 0, 1, , k1.

Cch m ho

Nu u = (a0, a1, , ak-1) l thng tin cn c m ho th t m v tng ng vi u c tabng cch ly u nhn vi G.

hay

v = a0g0 + a1g1+ + ak-1gk-1

V cc t m tng ng vi cc thng bo c sinh ra bi G theo cch nh trn nn Gc gi l ma trn sinh (generating matrix) ca b m.

V d 3.1

Cho ma trn sinh ca mt m tuyn tnh (7, 4) sau


3/19

Nu u = (1101) l thng tin cn m ho th t m tng ng s l .

v = 1.g0 + 1.g1 + 0.g2 + 1.g3

= (1100101)

Ch

1.Bt k k t m c lp tuyn tnh no cng c th c dng lm ma trn sinh chob m. Hay ni cch khc cc ma trn sinh khc nhau c th biu din cng mt b mtuyn tnh (hay cn gi l khng gian m) nh nhau, hay ngc li mt b m tuyn tnhc th c nhiu ma trn sinh khc nhau biu din.

2.Tng ng vi mi ma trn sinh chng ta c mt php m ho. C ngha l ng vi haima trn sinh khc nhau chng ta c hai php m ho khc nhau. V vy vi cng mt bm tuyn tnh vic chn ma trn sinh no l rt quan trng v n quyt nh vic nh xthng bo no thnh t m no.

Cch gii m

y chng ta s trnh by cch gii m t t m v thng tin ban u. Ly ma trn sinhnh trong V d 1.1. Chng ta gi thng bo l u = (a0, a1, a2, a3) v t m tng ngl v = (b0, b1, b2, b3, b4, b5, b6). Chng ta c h phng trnh sau lin h gia u v v.

v = u G

Suy ra

b0 = a0 + a1 + a3 (1)

b1 = a0 + a2 (2)

b2 = a1 + a3 (3)

b3 = a0 + a1 (4)

b4 = a1 (5)

b5 = a2 (6)

b6 = a2 + a3 (7)


4/19

gii cc ai theo cc bj chng ta ch cn chn bn phng trnh n gin nht gii.Chng hn chng ta chn cc phng trnh (4), (5), (6), (7) chng ta s gii c

a0 = b3 + b4

a1 = b4

a2 = b5

a3 = b5 + b6

H phng trnh trn c gi l h phng trnh gii m. D nhin ng vi cc ma trnsinh khc nhau cho d biu din cng mt b m chng ta s c cc h phng trnh giim khc nhau.

M tuyn tnh h thng, ma trn sinh h thng

Mt m tuyn tnh C(n, k) c gi l m tuyn tnh h thng nu mi t m c mttrong hai dng sauDng 1: T m bao gm phn thng tin k bit i trc v phn cn li (gm n k bit) isau (phn ny cn c gi l phn d tha hay phn kim tra).

k bit thng tin nk bit kim tra

Dng 2: Ngc ca dng 1, t m bao gm phn kim tra i trc v phn thng tin isau.

nk bit kim tra k bit thng tin

T iu kin v dng t m ca m tuyn tnh h thng, chng ta cn xc nh dng cama trn sinh tng ng. i vim tuyn tnh h thng dng 1, p ng iu kin can ma trn sinh phi c dng nh sau:


5/19

trong k ct u ca ma trn to thnh mt ma trn n v cn n k ct sau tu viPij= 0 hoc 1. Vi dng ny chng ta thy khi thng bo u c m hothnh t m v

bng cng

thc v = u G th k bit thng bo ca u s tr thnh k bit u ca t m v p ng yu

cu ca m tuyn tnh h thng. Ma trn c dng trn ca m tuyn tnh h thng cgi l ma trn sinh h thng v c th biu din n ginnh sau:

Gkn = [Ikk | Pk(n-k) ]trong Ikk l ma trn n v kch thc k k.

Tng t i vi m tuyn tnh h thng c dng 2 th ma trn sinh h thng phi cdngGkn = [Pk(n-k) | Ikk ]

Qui c

Nu khng c pht biu g khc th khi dng m tuyn tnh h thng chng ta s dng mtuyn tnh h thng dng 1.

V d 3.2

Ma trn sinh h thng cho m tuyn tnh h thng tng ng vi m tuyn tnh trong

V d 3.1 l

Nu u = (1101) l thng tin cn m ho th t m tng ng s l v = u Ght =(1101000). Tng t nu u = (0110) th v = 0110100.

M tuyn tnh h thng c li im l gip cho vic gii m t t m thnh thng bo

nhanh chng bng cch ly k bit u hay k bit cui ca t m tu theo m thuc dng 1hay 2 m khng phi gii h phng trnh nh i vi ma trn sinh bnh thng. Chnghn i vi m trong V d 3.2 nu chng ta nhn c t m v = (0101110) th d dngxc nh c thng bo tng ng l u = 0101.

Ch , t mt ma trn sinh kch thc k n chng ta c th dng cc php bin i scp trn hng (nhn mt hng vi mt h s khc 0, thay mt hng bng cch cng hng


6/19

vi mt hng khc) chng ta c th bin i thnh mt ma trn c k ct to thnh mtma trn n v.

V d 3.3

Cho ma trn sinh sau

Bng cch thc hin cc php bin i hng nh bn di

chng ta c trong ma trn mi G c cc ct 1, ct 4, ct 6 v ct 3 to thnh mt ma trnn v (cc ct c nh s t tri sang phi v bt u bng 1).

3.3 Vn pht hin sai v sa sai

Nguyn l pht hin sai rt n gin nh sau: Kim tra xem t hp nhn c phi l t mhay khng, nu khng th t hp nhn l sai.

Vic kim tra ny c th c thc hin bng cch so trng t hp nhn c vi cc tm. Nh vy vic kim tra ny s tn mt s bc bng vi s lng cc t m.

Tng t i vi vic sa sai chng ta c nguyn l sau: Kim tra xem t hp nhn ckhong cch Hamming gn vi t m no nht, nu gn vi t m no nht th t m chnh l t m ng c pht i. Nguyn l ny c gi l nguyn l khong cch

Hamming ti thiu. Vic kim tra ny tn mt s bc bng vi s lng cc t m.Tuy nhin i vi m tuyn tnh, da vo cc tnh cht ca m chng ta s c cch phthin sai v sa sai hiu qu hn. Cc phn tip theo s trnh by ln lt v vn ny,nhng trc ht chng ta s trnh by mt s kin thc ton hc cn thit cho vic chngminh mt s kt qu trong loi m ny.


7/19

Khng gian b trc giao

Cho S l mt khng gian con k chiu ca khng gian n chiu V, Sdl tp tt c cc vecttrong V sao cho u S, v Sd, th u v = 0 (php nhn y l php nhn v hngca hai vect) th Sdl mt khng gian con ca V v c s chiu l n k. Sdc gi l

khng gian b trc giao ca S v ngc li.

Da trn kt qu ny chng ta suy ra rng vi ma trn G bt k kch thc k n vi khng c lp tuyn tnh lun tn ti ma trn H kch thc (n k) n vi (n k) hng clp tuyn tnh sao cho G HT = 0, trong HTl ma trn chuyn v ca ma trn H. Hayni cch khc cc vect hng ca H u trc giao vi cc vect hng ca G.

Cch pht hin sai

ng dng kt qu trn vo vn pht hin sai, chng ta thy rng

Nu v l mt t m c sinh ra t ma trn sinh G c ma trn trc giao tng

ng l H th do v l mt t hp tuyn tnh ca cc vect hng ca G nnv H

T= 0

V ngc li nu v HT= 0 th v phi l mt t hp tuyn tnh ca cc vect hng ca Gdo v l mt t m.

Syndromevect sa sai (corrector)v H

Tthng c gi l syndrome hay vect sa sai ca v v k hiu l s(v). Vy

chng ta c

v l t m khi v ch khi s(v) = 0

Vi tnh cht ny chng ta thy H c th c s dng kim tra mt t hp c phi lt m khng hay ni cch khc H c th c dng pht hin sai. V l do ny m matrn H cn c gi l ma trn kim tra.

Ma trn kim tra

Ma trn kim tra ca mt b m c ma trn sinh Gkn l ma trn H c kch thc (n k) n sao cho

G HT

= 0


8/19

V d 3.4

Tm ma trn kim tra tng ng vi ma trn sinh trong V d 3.1.

Chng ta thy ma trn kim tra cn tm phi c kch thc 3 7. Gi h = (a0, a1, a2, a3,a4, a5, a6) l mt hng bt k ca H. V h trc giao vi mi vect hng ca G nn chngta c h bn phng trnh sau tng ng vi bn hng ca G:

a0 + a1 + a3 = 0

a0 + a2 + a3 + a4 = 0

a1 + a5 + a6 = 0

a0 + a2 + a6 = 0

Vn l by gi chng ta lm sao tm c 3 vect hng c lp tuyn tnh l nghimca h phng trnh trn. Ch , h phng trnh trn c th cho php chng ta gii bn

bin theo ba bin cn li. Chng hn chng ta gii a3, a4, a5, a6 theo a0, a1, a2 nh sau:

a3 = a0 + a1

a4 = a1 + a2a5 = a0 + a1 + a2 a6 = a0 + a2

Ch cc php cng, +, y chnh l php cng, , trong GF(2) nh qui c vtrong GF(2) php tr hon ton ging php +.

By gi chng ta cho (a0, a1, a2) ln lt cc gi tr (1, 0, 0), (0, 1, 0), (0, 0, 1) th chngta s xc nh c (a3, a4, a5, a6) ln lt nh sau (1, 0, 1, 1), (1, 1, 1, 0), (0, 1, 1, 1).Vy chng ta c ma trn H nh sau:


9/19

Ch

Cc ma trn kim tra khc nhau ca cng mt b m u c kh nng kim tra nh nhautc l u c th gip chng ta pht hin mt t hp c phi l t m hay khng.

i vi ma trn sinh h thng th vic xc nh ma trn kim tra d hn nhiu, da trnb sau:

B Nu ma trn sinh h thng ca mt m tuyn tnh h thng c dng

Gkn = [Ikk | Pk(n-k)]

th

H(n-k)n

= [Pk(n-k)

T

| I(n-k)(n-k)

]

l mt ma trn kim tra ca m.

Tng t nu ma trn sinh c dng

Gkn = [Pk(n-k) | Ikk]

th ma trn kim tra c dng

H(n-k)n = [I(n-k)(n-k) | Pk(n-k)T]

trong I(n-k)(n-k)l ma trn n v kch thc (nk) (nk), cn Pk(n-k)Tl ma trn

chuyn v ca ma trn Pk(n-k)

Chng minh

Xt ma trn sinh h thng c dng 1


10/19

Xt ma trn H sau

Ta chng minh

G HT

= 0

chng minh iu ny ta chng minh gi hj = 0 i = 0, , k1, j = 0, , nk1 trong gi = (gi0, , gi(n-1)) l hng i ca G cn hj = (hj0, , hj(n-1)) l hng j ca ma trn H.

Tht vy ta c

Chng minh tng t cho dng cn li ca G.

Kh nng chng nhiu tng ng

Chng ta bit rng kh nng pht hin sai v sa sai ca mt m tuyn tnh ph thucvo khong cch Hamming ca b m. V vy chng ta nh ngha rng

Hai m tuyn tnh C(n, k) c gi l c kh nng chng nhiu tng ng nu chngc cng khong cch Hamming.

T y chng ta dn n b sau

B

Nu hon v hai ct ca mt ma trn sinh s to ra mt b m mi c kh nng chngnhiu tng ng vi b m c. Ni cch khc vic hon v hai ct ca ma trn sinhkhng lm thay i kh nng chng nhiu.p dng iu ny nn ngi ta thng dng php hon v ny cng vi cc php bin i


11/19

s cp trn hng to ra nhng ma trn sinh hiu qu hn trong vic m ho v gii mnhng kh nng chng nhiu vn khng thay i.

Cch tnh khong cch Hamming ca b m

Chng ta bit khong cch Hamming cahai t m bng trng s ca tng hai t m. M do i vi m tuyn tnh tng hai t m l mt t m nn t y chng ta suy rakhong cch Hamming ca hai t m bng trng s ca mt t m no . Khi qut lnchng ta c b sau:

B

Khong cch Hamming ca mt m tuyn tnh bng trng s nh nht khc 0 ca b m.

V vy tnh khong cch Hamming ca mt m tuyn tnh chng ta s tm t m no

khc khng m c trng s nh nht.Ngoi ra tnh khong cch Hamming ca mt m tuyn tnh chng ta cn c mt cchc pht biu thng qua b sau:

B

Gi H l ma trn kim tra ca mt m tuyn tnh, nu mt t m c trng s d th tn tid ct ca H c tng bng 0.

H qu

Nu trong ma trn kim tra H ca mt m tuyn tnh s ct ph thuc tuyn tnh nh nhtl d th khong cch Hamming ca b m bng d.

V d 3.5

Xt ma trn H trong V d 3.4

Chng ta thy s ct ph thuc tuyn tnh ca H l 3, c th l cc ct s 3, 4 v 6 (ch sc tnh bt u t 1). V vy b m tng ng c khong cch Hamming d = 3, do m c th pht hin sai 2 bit v sa sai c 1 bit.


12/19

Cch sa sai

Trc ht chng ta s nh ngha khi nim vect li (error pattern vector) v khi nimtp gii m (decoding set) hay i lc cn gi l tp coset l nhng khi nim lm nntng cho vic sa sai.

Vect li

L vect biu din cc v tr li gia t m truyn v t hp nhn, mi v tr li c biudin bng bit 1, cn cc v tr cn li s c gi tr 0.

Nu t m c truyn i l w, vect li l e v vect nhn l v th chng ta c:

v = w + e

e = v + w

V d nu t m w = 1011011, vect li l e = 0010100 c ngha l sai v tr s 3 v 5(tnh t tri, bt u bng 1) th vect nhn s l v = w + e = 1001111. Ngc li nu tm w = 0110010 cn vec t nhn l v = 0010011 th vect li l e = w + v = 0100001 cngha l c sai xy ra cc v tr s 2 v s 7.

Chng ta thy trng s ca vect li biu din khong cch Hamming gia t m pht vt hp nhn. Khi nim trn gi cho chng ta mt iu nh sau, nu vect nhn l v thchng ta c th tnh c vect li tng ng vi mi t m bng cch cng v vi lnlt cc t m v ri da vo nguyn l khong cch Hamming tithiu chng ta thy

rng vect li no c trng s nh nht th t m tng ng chnh l t m c phti. Chng ta s hnh thc ho iu ny bng khi nim sau:

Tp gii m Coset

Cho S l mt khng gian con cc t m ca khng gian V, coset ca mt phn t z Vi vi S c k hiu l z + S v c nh ngha nh sau

z + S = {z + w: w S}

B

Tp coset z + S c cc tnh cht sau.

(1)z z + S. L do ny v t m 0...0 S nn z = z + 0...0 S.

(2)Nu z S th z + S = S. L do ny l v tng ca hai t m l mt t m. Hn na z +wi z + wjvi wi wjnn khng c hai phn t no ca z + S ging nhau.


13/19

(3)Nu v z + S th v + S = z + S. Tht vy v v = z + wi vi mt wi no S. V vyv + S = z + wi + S. M wi + S = S nn v + S = z + S.

(4) Nu v z + S th v + S v z + S ri nhau. Tht vy nu v + S v z + S khng ri nhau.Suy ra tn ti u v + S v u z + S. M u v + S suy ra u + S = v + S. Tng t u z +

S suy ra u + S = z + S. V vy v + S = z + S m v v + S suy ra v z + S. iu ny muthun. V vy v + S v z + S phi ri nhau.

Ch mi coset c s phn t ng bng s phn t ca tp S v tt c cc phn t z + w(w S) l khc nhau. V vy, vi b trn chng ta suy ra s cc coset ca V bng s

phn t ca V chia cho s phn t ca S. C th vi V l mt khng gian 2nvect, cn Sc 2

kvect th s tp coset s l 2n-k.

Vi cc khi nim trn chng ta a ra s gii m theo nguyn l khong cchHamming ti thiu nh sau.

(1) Vi mi vect nhn v chng ta s c mt tp coset tng ng l v + S.

(2) Trong tp ny chn phn t c trng s nh nht, chng hn l z. Phn t nythng c gi l coset leader.

(3) Thng bo t m c truyn chnh l w = v + z.

R rng coset leader chnh l vect li m b gii m theo nguyn l khong cch tithiu gn cho v. Ch rng tt c cc thnh vin ca tp coset v + S c cng tp cosetnh v

(theo tnh cht 3). Nh vy, tt c cc thnh vin ca mt tp coset c cng vect lichnh l coset leader ca tp. V vy nu chng ta nhn bit c mt vect nhn thuctp coset no th cng n vi coset leader ca tp s c tm ng c pht itng ng. l l do ti sao tp coset cn c gi l tp gii m.

Vn by gi l lm sao xc nh mt cch nhanh nht mt vect nhn tng ngvi vect li no hay chnh l vi coset leader no. May mn thay c mt cchrt d thc hin iu ny, ci m c pht biu thng qua b sau

B

Cc phn t ca mt tp coset c cng mt syndrome nh nhau. Cc tp coset khc nhauc cc syndrome khc nhau.


14/19

Chng minh

Tht vy chng ta c vi wi S ths(v + wi) = (v + wi) H

T= (v H

T) + (wi H

T) = (v H

T) + 0 = (v H

T) = s(v)

V vy cc phn t ca tp coset v + S c cng syndrome nh nhau v d nhin c cngsyndrome ca coset leader ca tp.

Mt khc nu u v + S. Gi s s(u) = s(v), suy ra

u HT

= v HT

T y suy ra

(u + v) H

T

= 0

Suy ra u + v = wivi wil mt t m no . iu ny suy ra u = v + wic ngha lu v + S (mu thun). iu ny hon tt chng minh ca chng ta.

Vy mi coset c t trng bng mt syndrome hay mt vect sa sai duy nht. Ccvect sa sai ny c kch thc l n k. Vy c mt s tng ng mt mt gia 2nktp coset vi 2nk vect c chiu di l n k.

ng dng iu ny chng ta thy bn nhn s ch cn gi mt bng bao gm 2nk dy cchiu di n k, mi dy tng ng vi mt vect li chnh l coset leader ca cc tp

coset. Vi mi vect nhn v, b gii m s tnh s(v) = v HTri tm trong bng xcnh vect li e tng ng vi s(v). Cui cng b gii m s thng bo t m ng c

pht i l w = v + e.

t e = (a1, a2, ..., an), v gi cc ct ca H ln lt bng h1, h2, ..., hn th chng ta c

C ngha l s(e) bng tng nhng ct nhng v tr tng ng vi nhng v tr bng 1ca e. Chng hn nu v tr li sai trong khi truyn l 3 th syndrome ca vect nhntng ng s bng ct s 3 ca ma trn kim tra H. Chi tit hn chng ta xem v d sauy.

V d 3.6

Xt mt m tuyn tnh C(7, 4) c ma trn sinh h thng nh sau


15/19

T y chng ta c mt ma trn kim tra ca b m trn nh sau:

B m trn c khong cch Hamming d = 3. V vy c th pht hin sai 2 bit v sa sai

c 1 bit.

Gi s ng truyn sai ti a 1 bit. V vect nhn l v = 1010110, hy tm t m ng c pht i. gii bi ny chng ta tnh

trong hi l ct th i ca H (tnh t tri v bt u bng 1).

Chng ta thy s(v) h1nn suy ra li sai v tr s 1, v vy t m ng c pht il

w = v + e = 1010110 + 1000000 = 0010110

T tng ny gi cho chng ta mt loi m cho php pht hin sai 1 bit nhanh nht.M ny c tn gi l m tuyn tnh Hamming.

M tuyn tnh Hamming

M tuyn tnh Hamming l m c ma trn H c tnh cht gi tr ca ct hi bng i (i = 1, 2,... )


16/19

V d ma trn sau biu din m tuyn tnh Hamming C(7, 4).

B

Cc m tuyn tnh Hamming u c khong cch Hamming d = 3. V vy c th phthin sai 2 bit v sa sai 1 bit.

Chng minh

D thy vi cch nh ngha ca ma trn H, chng ta thy s ct ph thuc tuyn tnh tnht ca H l 3 (chng hn 3 ct u). V vy m Hamming c d = 3.

M tuyn tnh Hamming cho php chng ta sa sai mt bit mt cch n gin nh sau.

(1) Tnh syndrome s(v) ca vect nhn

(2) i chui nh phn tng ng ra gi tr thp phn, kt qu i s l v tr li sai xy ra.

(3) Sa sai v tr li sai tng ng.

Chng hn ly m tuyn tnh Hamming C(7, 4) nh trn nu s(v) = 101 th v tr li sai l5 (v 1012 i ra thp phn bng 5).

Mt bi ton cn li t ra y i vi m tuyn tnh Hamming l, xc nh ma trnsinh G thc hin vic m ho v gii m. lm iu ny ngi ta thng ly k trongn v tr cha cc bit thng tin, cc bit cnli s lm cc bit kim tra. Chi tit hnchng ta xem v d sau y.

V d 3.7

Xt m tuyn tnh Hamming C(7, 4) c cc bit thng tin nm cc v tr 3, 5, 6, 7. Hyxc nh ma trn sinh G ca b m.

Gi w = (a1, a2, a3, a4, a5, a6, a7) l mt t m. Chng ta c h phng trnh sau cdn ra t cng thc w HT = 0.


17/19

a4 + a5 + a6 + a7 = 0

a2 + a3 + a6 + a7 = 0

a1 + a3 + a5 + a7 = 0

T y chng ta suy ra cng thc tnh cc bit kim tra a1, a2, a4 theo cc bit thng boa3, a5, a6, a7 nh sau

a1 = a3 + a5 + a7

a2 = a3 + a6 + a7

a4 = a5 + a6 + a7

Cng thc ny cho php chng ta m ho c thng bo u thnh t m w. Chng hnnu

Bin i cng thc trn thnh dng w = u G cho php chng ta tm G c d dng:

a1 = b1 + b2 + b4a2 = b1 + b3 + b4

a3 = b1

a4 = b2 + b3 + b4

a5 = b2

a6 = b3

a7 = b4

T y chng ta suy ra c ma trn sinh G nh sau:


18/19

3.4 Mt s gii hn

Gii hn trn Hamming v s lng t m

nh l 3.1

Nu b m {w1, , wM} gm cc t m nh phn chiu di n c th sa sai cc li sai t bit, th s lng t m M phi tho bt ng thc sau:

Chng minh

i vi mi t m wi nh ngha mt tp Ai bao gm tt c cc dy vj c khong cchHamming so vi wi nh hn hay bng t. Tng s dy trong Ai c cho bng

c th sa sai c cc li sai t bit th cc tp A1, A2, , AM phi ri nhau. Mtng s phn t ca tt c cc tp Ai phi nh hn hoc bng tng s dy c chiu di n.

V vy

T y suy ra iu phi chng minh.

Gii hn di v s lng bit kim tra

nh l 3.2

Nu mt m tuyn tnh C(n, k) c th sa sai cc li sai t bit, th s bitkim trar = nk phi tho bt ng thc sau:


19/19

Chng minhi vi mi vect li e chng ta c mt syndrome (hay vect sa sai) tng ng. Vy

sa sai tt c cc li sai t bit th chng ta c vect sa

sai. M tng s vect sa sai l bng 2nk = 2r. Tht vy v cc vect thuc cng mt tpcoset th c cng mt vect sa sai. Do s lng vect sa sai bng vi s lng tpcoset tc bng 2nk. V vy chng ta phi c

iu ny hon tt chng minh.

Ch nh l ny ch l iu kin cn ch khng . Chng hn vi n = 10, t = 2 chngta suy ra t nh l trn rng r 6 l cn thit. Tuy nhin, chng ta c th kim tra lirng sa c cc li sai 2 bit th phi c t nht r = 7.

Ch gii hn di v s lng bit kim tra i vi m tuyn tnh ging vi gii hntrn v s lng t m trong nh l A. Tht vy i vi m tuyn tnh C(n, k) theonh l A chng ta c

Documents

Ma Khoi Tuyen Tinh