xac thuc va chu ky so

8/9/2019 Xac Thuc Va Chu Ky So

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Chn III

Cc H Mt M Cng Khai


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III.1 Nguyn l h mt m cng khai

c im:

Mt m cng khai da trn csca cc hm ton hcch khn hi d a trn h tha th v i ch nhtrong phng php m ho i xng.

M mt cn khai l bt i xn . Tron cch m mtkho cng khai s dng hai kho: kho mt v khocng khai. Vic s dng hai kho khng i xng a

n n ng qu su s c trong n vc an ton t ngtin: tnh ton vn, tnh xc thc, phn phi kho.


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Xu t x:

H m mt kho cng khai c pht trin nhm giiu t hai vn hc t n sinh t hn h m

ho i xng:

Vn th nht: bi ton phn phi kho; Vn th hai: ch k in t.


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V n phn ph i kha:

Trong s m ho truyn thng, qu trnh phn phikho a ra u cu hai ha tham ia vo trao i thntin:

Phi chia s trc kho, kho ny phi c phnph i b ng mt cch no h o h.

Phi s dng trung tm phn phi kho KDC.

V n chk in t: Ch k in t phi c s dng trong cc thng ip

in t v phi c hiu lc tng ng vi ch k trn

giy.


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H mt kha cng khai:

Mi h thng u cui to mt cp kho m ho vii m cc thn i .

Mi h thng u cui cng b mt kho trong cp kho

,kho cn li c gi mt. N u A mu n gi thng ip cho B, A s m ho vn bn

bng kho cng khai ca B.

,mt. Khng mt bn th ba c th gii m c thng.


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S m ha cng khai:

S m mt kho cng khai s dng mt kho mho v m t kho khc c lin uan ii m. Cc thu tton m ho v gii m c mt s c im quan tr ng

sau: Khng th xc nh c kho gii m n u ch bi t

thut ton m ho v kho m ho.

cung cp kh nng s dng bt k mt kho trong cp,

lm kho gii m.


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S m ha:


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S chng thc:


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Cc bc ti n hnh:

1. Mi ngi s dng to mt cp kho c s dng m ha vgii m thng ip.

2. M i ngi dng s giao mt trong hai kha cho ngi ng kkha cng cng hoc mt file c kh nng truy cp. Kha lkha cn khai. Ci cn l i l s c ib m t. h tron hnhtrn gi thit, mi ngi dng s duy tr mt tp cc kha cngkhai thu c t cc ngi dng khc

. ,thng ip bng cch s dng kha cng khai ca Alice.4. Khi Alice nhn c thng ip, c gii m n bng kha ring

ca mnh. Khng c ngi nhn khc c th gii m thng ip v

ch Alice bit kha ring ca Alice.


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M hnh m bo b mt:

K ph m, quanst Y v c quyntruy cp vo KUbnhng khng c

u n tru c vo KRb hoc X, xem c

m phi phc hiX v / hoc KRb.


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M hnh chng thc:Trong trng hp ny, A son mttin nh

n

n B v m h a b

ng kha

ring ca A trc khi truyn n. Bc th ii m thn i bn cchs dng kha cng khai ca A. Biv thng ip c m ha bngkha rin ca A ch A c th sonthng bo ny. V vy, thng ipc m ha ton b phc v nhm t ch k s. N oi ra khn ththay i thng ip m khng cntruy cp vo cc kha ring ca A,

,v ngun v v tnh ton vn dliu.


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M hnh kt hp:

Trong trng hp ny, chng tabt u nh trc bng cch m,

ring ca ngi gi. Cng vic

ny cung cp cc ch k s.p eo, c ng a m a mln na, bng cch s dng khacng khai ca ngi nhn. Cc

bnm c ui cng ch c th cgii m ch bi ngi nhn

,kha. V vy, tnh bo mt ccung cp.


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Cc iu kin cn thit:

Bn B d dng to ra c cp (KUb, KRb) Bn A d dng to ra c C = EKUb(M) Bn B ng g m M = DKRb(C)

i th

khng th

xc

nh

c KR

bkhi bi

t KU

b b Mt trong hai kha c th dng m ha trong khi kha

kia c th dn ii mM = DKRb(EKUb(M)) = DKUb(EKRb(M))


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M ha i xng v m ha cng khai:


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III.2 Thut Ton RSA

H m ha RSA:

xut bi Ron Rivest, Adi Shamir v Len Adleman(MIT) vo nm 1977 H m ha kha cng khai ph dng nh t

M ha khi vi mi khi l mt s nguyn < n n g k bn quyn nm 1983, ht hn nm 2000 An ton v chi h hn tch tha s ca mt s n u n

ln l r t ln


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III.2 Thut Ton RSA

T o kha RSA:

Mi bn t to ra mt cp kha cng khai - kha ringtheo cc bc sau : Chn ng u nhin 2 s nguyn t ln p q Tnh n = pq Tnh n = -1 -1 Chn ngu nhin kha m ha e sao cho 1 < e < (n) v gcd(e,

(n)) = 1 .

Cng b kha m ha cng khai KU = {e, n} Gi b mt kha ii m rin KR = {d, n}

Cc gi tr b mt p v q b hy b


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III.2 Thut Ton RSA

Th c hi n RSA:

m ha 1 thng bo M, bn gi thc hin Ly kha cng khai ca bn nhn KU = {e, n}n = mo n

gii m bn m ha C, bn nhn thc hin S dn kha rin KR = d, n Tnh M = Cd mod n

Lu l thng bo M phi nh hn n


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III.2 Thut Ton RSA

Tnh kh thi ca RSA:

Theo nh l Euler a, n : gcd(a, n) = 1 a(n) mod n = 1 n s c c s nguy n ng n n n v nguy n c ng

nhau vi n i vi RSA c

n = pq vi p v q l cc s nguyn t (n) = (p - 1)(q - 1) = M < n

C th suy ra Cd mod n = Med mod n = Mk(n) + 1 mod n = M mod n = M


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III.2 Thut Ton RSA

V d t o kha RSA:

Chn 2 s nguyn t p = 17 v q = 11 Tnh n = pq = 17 11 = 187 Tnh (n) = (p - 1)(q - 1) = 16 10 = 160 Ch

n e : gcd(e, 160) = 1 v 1 < e < 160; l

y e = 7

Gi tr d = 23 v 23 7 = 161 = 1 160 + 1 Cn b kha cn khai KU = 7 187 Gi b mt kha ring KR = {23, 187}

Hy b cc gi tr b mt p = 17 v q = 11


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III.2 Thut Ton RSA

V d to kha RSA:


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III.2 Thut Ton RSA

Ch n tham s RSA:

Cn chn p v q ln Thng chn e n h Thng c th chn cng gi tr ca e cho t t c ngi

dng,

c coi l qu nh Thn ch n e = 216 - 1 = 65535 Gi tr ca d s ln v kh on


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III.2 Thut Ton RSA

An ton ca RSA:

Kha 128 bit l mt s gia 1 v mt s rt ln340.282.366.920.938.000.000.000.000.000.000.000.000 ao n u s nguy n g a v s n y

n / ln(n) = 2128 / ln(2128)

3.835.341.275.459.350.000.000.000.000.000.000.000 Cn bao nhiu thi gian nu mi giy c th tnh c

1012 s

n , , , , , nm o ng tr u n tu c av tr) An ton nhn cn hn nhn im u


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III.2 Thut Ton RSA

Ph m RSA:

Phng php vt cn Th tt c cc kha ring c th

Phng php phn tch ton hc Phn n thnh tch 2 s nguyn t p v q Xc nh trc tip (n) khng thng qua p v q Xc nh trc tip d khng thng qua (n)

Da trn vic o thi gian gii m C th ngn nga bng cch lm nhiu


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III.3 H Trao i Kha Diffie-Hellman

Gii thu t m t m kha cn khai u tin

xut bi Whitfield Diffie v Martin Hellman vo nm1976 Malcolm Williamson (GCHQ - Anh) pht hin trc m y nm

nhng n nm 1997 mi cng b

cc kch thng tin khng an ton Kha b mt c tnh ton bi c hai bn An ton ph thuc vo phc tp ca vic tnh log ri

rc


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Thi t l Diffie-Hellman:

Cc bn thng nht vi nhau cc tham s chung q l mt s nguyn t ln m nguy n c n c a q mod q, 2 mod q,..., p-1 mod q l cc s nguyn giao hon ca cc s t

1 n q - 1 n

Chn ngu nhin lm kha ring XA < q Tnh kha chun Y = XA mod

Bn B Chn ngu nhin lm kha ring XB < q

Xn a c ung B = mo q


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Trao i kha Diffie-Hellman:

Tnh ton kha b mt Bn A bit kha ring XA v kha cng khai YBX= B mo q

Bn B bit kha ring XB v kha cng khai YA

K = YAX

B mod q Chng minh

YAXB mod q = (XA mod q)XB mod q

= XAXB= XBXA mod q= (XB mod q)XA mod q= B mo q


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V d Diffie-Hellman:

Alice v Bob mun trao i kha b mt Cng chn q = 353 v = 3 Chn ng u nhin cc kha ring

Alice chn XA = 97, Bob chn XB = 233

YA = 397 mod 353 = 40 (Alice) YB = 3233 mod 353 = 248 (Bob)

Tn ton a mt c ung K = YBXA mod 353 = 24897 mod 353 = 160 (Alice)

= XB = 97 =


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H n ch ca kha cn khai:

Tc x l Cc gii thut kha cng khai ch yu dng cc php nhnc m n n u so v c c g u x ng

Khng thch hp cho m ha thng thng

Thng dng trao i kha b mt u phin truyn tin Tnh xc thc ca kha cng khai

Bt c ai cng c th to ra mt kha cng b l ca mt

Chng no vic gi mo cha b pht hin c thc c nidung cc thng bo gi cho ngi kia

n m o n ng ng ng a ng t n


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Chn IV

Xc Thc v ChK S


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IV.1 Cc Hnh ng Tn Cng

Gii phng ni dung thngip: Pht hnh ni dung thng ip ti mtngi hoc mt ti n trnh no m khng shu kha mt thch hp.

Phn tch ti: xc nh tn sut, thi gian k t ni, s lng v chiudi ca thn i ia 2 bn.

Gi mo: chn tin nhn vo mng t mt ngun gian ln, chng hn nhto ra cc thng

ip c m

c

ch x

u b

i ng

i c th

m quy

n.

ay n ung : ay c c n ung c a t n n n c , ao g mchn, xa, hon v, v sa i

Thay i tht: l bt ky mt s sa i no cho mt chui cc thng

ip gia cc bn, bao gm chn, xa, v sp xp li. Thay i thi gian: lm tr hoc pht li cc tin nhn.

nhn cc thng ip ch.


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IV.1 Cc Vn Xc Thc

Cc tiu chun cn xc minh

Thng bo c ngun gc r rng chnh xc Ni dung thng bo ton vn, khng b thay i Thng bo c gi ng trnh t v thi i m

Mc ch chng li tn cng ch ng (xuyn tc d

Cc phng php xc thc thng bo M ho thn bo 1 S dng m xc thc thng bo(2) S dng hm bm(3)


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IV.1 Cc Vn Xc Thc

Xc thc bng cch m ha: S dng m ha i xng

Thng bo gi tng ngun v ch c ngi gi mi bit

Ni dung khng th b thay i v vn bn th c cu trc nht

nhc g n c n s v m a n n ng ayi trnh t v thi im nhn c

S dng m ha kha cng khai

Khng ch xc thc thng bo m cn to ch k s Phc tp v mt thi gian hn m ha i xng


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IV.1 Cc Vn Xc Thc

Xc thc dng m Checksum:

Dng m xc thc thng bo(MAC Message AuthenticationCode) L khi c kch thc nh c nh gn vo thng bo to ra

t thng bo vkhabmt chung n n n c n c ng g u r n ng o v o

so xem MAC c chnh xc khng

gii ngc.


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IV.1 Cc Vn Xc Thc

Xc th c dn m Checksum:

C th c nhiu thng bo c cng chung MAC Nhn nu bit 1 thn bo v MAC r t kh tm ra

mt thng bo khc cng MAC

Cc thng bo c cng xc sut to ra MAC p ng 3 tiu chun xc thc


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IV.1 Cc Vn Xc Thc

3 tiu chun xc thc:

M: Thng ip

K: Kha bo mt

c chia s MAC: M xc

thc thng ip


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IV.1 Cc Vn Xc Thc

Ti sao dng MAC:

Nhiu trng hp ch cn xc thc, khng cn m ha tnthi gian v ti nguyn Thng bo h th ng Chng trnh my tnh

vic t chc linh hot hn

Chng hn mi chc nng thc hin mt tng ring

C n m bo tnh ton vn ca thng bo trong su t thigian tn ti khng ch khi lu chuyn


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IV.1 Cc Vn Xc Thc

Xc thc dng hm bm:

M

t hm bm nh

n

u vo l m

t thng bo c

di

ty v to ra kt qu l mt xu k t c di cnh,c g t m t t t ng o o c c s .

To ra mt gi tr bm c kch thc cnh t thng

h = H(M) Hm bm khn cn i b mt Gi tr bm gn km vi thng bo dng kim tra tnh

ton vn ca thng bo B t k s thay i M no d nh cng to ra mt gi tr h

khc


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IV.1 Cc Vn Xc Thc

Xc thc dng hm bm:


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IV.1 Cc Vn Xc Thc

Xc thc dng hm bm:


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IV.1 Cc Vn Xc Thc

Yu c u i vi hm bm:

C th p dng vi thng bo M c di bt k To ra gi tr bm h c di cnh H(M) d dng tnh c vi bt k M no

T h rt kh tm c M sao cho H(M) = h n m c u T M1 rt kh tm c M2 sao cho H(M2) = H(M1)

Tnh chn xun t u Rt kh tm c (M1, M2) sao cho H(M1) = H(M2)

Tnh chng xung t mnh


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IV.1 Cc Vn Xc Thc

Yu c u i vi hm bm:

C th p dng vi thng bo M c di bt k To ra gi tr bm h c di cnh H(M) d dng tnh c vi bt k M no T

h r

t kh tm

c M sao cho H(M) = h

n m c u T M1 rt kh tm c M2 sao cho H(M2) = H(M1)

Tnh chn xun t u Rt kh tm c (M1, M2) sao cho H(M1) = H(M2)

Tnh chng xung t mnh


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IV.2 Vn An Ton Ca Hm Bm & MAC

Kiu t n cng ngy sinh:

Nghch l ngy sinh Trong 23 ngi, xc sut tm ra 1 ngi khc c cng ngys n v

Xc sut 2 trong 23 ngi c cng ngy sinh l 50% Cch thc tn cn m bmmbi t

To r a 2m/2bin th ng ngha ca thng bo hp l To r a 2m/2bin th ca thng bo gi mo

(xc sut > 0,5 theo nghch l ngy sinh) ngi gi k bin th hp l, ri dng ch k gn vo bin

th gi mo


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IV.2 Vn An Ton Ca Hm Bm & MAC

Vi h m bm, n lc ph thuc d im ca m bm phc tp ca tnh mt chiu v tnh chng xung t yu l 2m; ca

tnh ch ng xung t mn h l2m 128 bit c thph c, thng dng 160 bit

Vi MAC, n lc ph thuc vo di k ca kha v di nca MAC phc tp l min(2k, 2n) t nht phi l 128 bit

Ki u t n cng dng k thut Hm bm thng gm nhiu vng nh m ha khi nn c th


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IV.3 Chk s

nhn mungyhi cho nhau Bn nhn gi mo thng bo ca b n gi Bn gi ch i l gi thng bo n bn nhn

Ch k s khng nhng gip xc thc thng bo m cn

Chc nng ch k s Xc minh tc gi v thi im k thng bo

Xc thc ni dung thng bo L cn c gii quyt tranh chp


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IV.3 Chk s

Yu c u i vi chk s :

Ph thuc vo thng bo c k C s dng thng tin ring ca ngi gi trnh gi mo v ch i b

Tng i d to ra

Rt kh gi mo Bng cch to thng bo khc c cng ch k s

Bng cch to ch k s theo mun cho thng bo Thun tin trong vic lu tr


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IV.3 Chk s

Chk s tr c ti : Ch

lin quan

n b n g

i v bn nh

n

Vi mt m kha cng khai Dng kha ring k ton b thng bo hoc gi tr bm C th m ha s dng kha cng khai ca bn nhn

uan trn l k tr c m ha sau Ch c tc dng khi kha ring ca bn gi c m

bo an ton

Bn g c t g vm t a r ng Cn b xung thng tin thi gian v bo mt kha kp thi Kha ring c thb mt tht

K c p c th gi thng bo vi thng tin thi gian sai lch


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IV.3 Chk s

C s tham gia ca mt bn tr ng ti Nhn thng bo c ch k s t bn gi, kim tra tnh hp lca n

B xung thng tin thi gian v gi n bn nhn

Cn c b n gi v bn nhn t in tng C th ci t vi m ha i xng hoc m ha kha

cng khai Bn trng ti c th c php nhn thy hoc khng ni


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IV.3 Chk s

Cc k thut chk s gin tip:(a) M ha i xng, trng ti th y thng bo

(1) X A : M EKXA[IDX H(M)]2 AY : E ID M E ID H M T

AY XA(b) M ha i xng, trng ti khng thy thng bo

(1) X A : IDX EKXY

[M] EKXA

[IDX H(EKXY

[M])] KAY X KXY KXA X KXY

T(c) M ha kha cng khai, trng ti khng thy thng bo

(1) X

A : IDX EKRX[IDX EKUY[EKRX[M])](2) AY : EKRA[IDX EKUY[EKRX[M]] T]K hiu : X = Bn i M = Thn bo

Y = Bn nhn T = Nhn thi gian

A = Trng ti

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