chu ky dien tu rsa va bam sha

7/30/2019 Chu Ky Dien Tu Rsa Va Bam Sha

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Kha lun tt nghip 2011 Tm hiu v ch k in t v ci t chng trnh minh ha.

Mc lc kha lun

Phn 1: Tm hiu lch s v an ton thng tin, chng thc thng tin v ch k in t.............7Phn 2: Ni dung.......................................................................................................................10I. Tng quan v chng thc v an ton thng tin......................................................................10

II. M ha d liu v gii m.................................................................................................12a. Tng quan v m ha d liu v gii m:......................................................................12b. M ha bt i xng (asymmetric)...............................................................................13c. M ha i xng (symmetric)........................................................................................16d. Hm bm (Hashing).......................................................................................................17

- Tnh cht ca hm bm................................................................................................18- Mt s hm bm ni ting...........................................................................................20

+ MD5 (Message Digest)..........................................................................................20+ SHA (Secure Hash Algorithm)...............................................................................26

III. Ch k in t..................................................................................................................281. Tng quan......................................................................................................................28

2. Quy trnh s dng ch k in t..................................................................................303. Mt s s CKT ph bin........................................................................................33a. Rivest Shamir Adleman (RSA)..................................................................................33

- S lc v cc khi nim ton hc dng trong RSA...............................................33- Cch to kha:.........................................................................................................34- Quy trnh thc hin k v xc nhn vn bn...........................................................35

Phn ny c ct b, hy lin h ch ti nhn c bn chi tit hn. ..........36- Tnh bo mt............................................................................................................37- Cc dng tn cng....................................................................................................38

b. H ch k ElGammal.................................................................................................40c. Chun ch k s (DSS) ..............................................................................................44

4. Hm bm v kt hp hm bm vo ch k in t.......................................................47IV. Ci t minh ha s k s RSA kt hp bm SHA....................................................47+ Cc bc thc hin ca chng trnh.............................................................................47

a. Pht sinh kha: ..........................................................................................................47b. K ch k in t: ....................................................................................................48........................................................................................................................................48Phn ny c ct b, hy lin h ch ti nhn c bn chi tit hn. ..........48........................................................................................................................................48- Mt s hm s dng trong chng trnh.....................................................................48- Giao din ca chng trnh:........................................................................................49

Phn 3. Kt lun: ......................................................................................................................50

+ Nhng phn lm c....................................................................................................50+ Nhng phn cha lm c................................................................................................51+ Hng pht trin ca ti.................................................................................................51+ Ti liu tham kho..............................................................................................................51+ Ph lc................................................................................................................................51Demo chng trnh: http://www.mediafire.com/view/?fwtd5cdp500u5xj............................51- Cch lin h ly bi hon chnh: ....................................................................................51

Lin h. Mail: [email protected] or t: 0982.070.520

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ly bi hon chnh. C th ly thm phn code (nguyn code + phn ci t), xin hylin h mail or s t trn lin h ly bi............................................................................51Ph: bi kha lun 50.k, Code: 100.k.....................................................................................51Lin h: mail: [email protected] or 0982.070.520 (c th sms).................................51

Danh mc t vit tt:

- RSA: Rivest Shamir Adleman

- SHA: Secure Hash Algorithm

- MD5: Message Digest

- CKT: Ch k in t

- CA: Certificate Authority - y quyn chng ch

- UCLN: c chung ln nht

Lin h mail: [email protected] or t: 0982.070.520

ly bi + Code y v chi tit hn!


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mailto:[email protected]:[email protected]


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A. PHN M U

1. L do chn ti

Ngy nay, cc ng dng ca cng ngh thng tin ngy cng khng th thiu c

i vi cc thnh phn nh x hi, kinh t, chnh tr, qun s... Mt lnh vc quan

trng m cng ngh thng tin c ng dng rt mnh m v khng th thiu l

lnh vc truyn thng. Rt nhiu thng tin lin quan n nhng cng vic hng ngy

u do my vi tnh qun l v truyn gi i trn h thng mng, ko theo l vn

v xc thc ngun thng tin nhn c. V vn c t ra l lm th no

xc thc c mt cch chnh xc ngun thng tin nhn l ca mt ngi, mt my

ch hay ca mt thc th no gi tin trn h thng mng?


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Hnh 1.1 M hnh trao i thng tin qua mng INTERNET.

V d: khi A gi mt thng tin in t cho D, v gi s B gi mo A cng gi

mt thng tin cho D, hoc mt ngi C trn mi trng truyn bt c v sa i

thng tin A gi sau cc thng tin ny c gi li cho D. Vy thng tin D nhn

c khng chnh xc v ngi nhn D cng khng th xc thc c thng tin

l do ngi A gi, hay B gi, hay mt ngi no khc gi.

Vn t ra l ngi D sau khi nhn c bn tin phi xc thc c rng thng

tin l ca chnh mt i tng c th gi v thng tin khng b tit l cng nh b

khng thay i trn mi trng truyn thng.

ti TM HIU V CH K IN T V CI T CHNG TRNH

MINH HA s tm hiu vn nu trn v ci t chng trnh k s minh ha.

2. Mc ch nghin cu

Tm hiu c s l lun v chng thc thng tin, ch k in t, ci t chng

trnh k s kt hp RSA v hm Bm.

3. Nhim v nghin cu- Nghin cu c s l lun v chng thc thng tin.

- Tm hiu v ch k in t.

- Tm hiu v cc phng thc m ha d liu c bn.


A

INTERNE

TB

D

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- Tm hiu phng thc m ha bt i xng s dng cho ch k in t.

- Tm hiu v hm bm.

- Ci t chng trnh minh ha.

4. i tng nghin cu

- C s l lun v chng thc thng tin.

- H m ha cng khai RSA.

- Hm bm kt hp cho ch k in t.

5. Phm vi nghin cu

Nghin cu l thuyt v chng thc thng tin, h m ha cng khai RSA, hm

bm SHA v ci t chng trnh minh ha.

6. Phng php nghin cu

- Hot ng nghin cu c nhn

- Hot ng nghin cu ti liu

- Trao i vi ging vin hng dn

7. Cu trc kha lun

Gm 5 phn chnh:

1. Phn m u

2. Phn ni dung

3. Phn kt lun

4. Phn ti liu tham kho

5. Phn ph lc

A. Phn m u:


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L do chn ti

Mc ch nghin cu

Nhim v nghin cu

i tng nghin cu Phm vi nghin cu

Phng php nghin cu

Cu trc kha lun

B. Phn ni dung: Gm 2 Phn:

Phn 1: Tm hiu lch s v an ton thng tin, chng thc thng tin, ch k

in t.Phn 2: Ni dung.

I. Tng quan v chng thc v an ton thng tin.

II. M ha d liu v gii m

a. Tng quan v m ha d liu.

b. M ha bt i xng (asymmetric).

c. M ha i xng (symmetric).

d. Hm bm (Hashing)

III. Ch k in t

1. Tng quan

a. Tng quan v ch k in t

b. Quy trnh s dng ch k in t

2. Mt s s CKT ph bin

a. Rivest Shamir Adleman (RSA).


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b. S k s El Gamal.

c. Chun ch k s (DSS)

3. Hm bm v kt hp hm bm vo ch k in t.

IV. Ci t minh ha s k s RSA kt hp bm SHA.

+ Cc bc thc hin ca chng trnh.

+ Ci t chng trnh minh ha bng ngn ng C#.

C. Phn kt lun:

+ Nhng phn lm c

+ Nhng phn cha lm c

+ Hng pht trin ca ti.

- Ti liu tham kho

- Ph lc

Phn 1: Tm hiu lch s v an ton thng tin, chng thc thng tin v ch k

in t.

Nhu cu v bo m an ton thng tin xut hin t rt sm, khi con ngi bit

trao i v truyn a thng tin cho nhau, c bit khi cc thng tin c th

hin di hnh thc ngn ng, th t. Lch s cho ta bit, cc hnh thc mt m

c tm thy t khong bn nghn nm trc trong nn vn minh Ai Cp c i.

Tri qua hng nghn nm lch s, mt m c s dng rng ri trn khp th

gii t ng sang Ty gi b mt cho vic giao lu thng tin trong nhiu lnh

vc hot ng gia con ngi v cc quc gia, c bit trong cc lnh vc qun s,

chnh tr, ngoi giao. Mt m trc ht l mt loi hot ng thc tin, ni dung

chnh ca n l gi b mt thng tin (chng hn di dng mt vn bn).


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Sut my nghn nm lch s, cc thng bo, th t c truyn a v trao i

vi nhau thng l cc vn bn, tc l c dng cc dy k t trong mt ngn ng

no . V vy, cc thut ton lp mt m thng cng n gin l thut ton xo

trn, thay i cc k t c xc nh bi cc php chuyn dch, thay th hay honv cc k t trong bng k t ca ngn ng tng ng, kha mt m l thng tin

dng thc hin php lp mt m v gii mt m c th, th d nh s v tr i

vi php chuyn dch, bng xc nh cc cp k t tng ng i vi php thay th

hay hon v,... Mt m cha phi l mt khoa hc, do cha c nhiu kin thc

sch v li, tuy nhin hot ng bo mt v thm m trong lch s cc cuc u

tranh chnh tr, ngoi giao v qun s th ht sc phong ph, v mt m c nhiu

tc ng rt quan trng a n nhng kt qu lm khi c ngha quyt nh trong

cc cuc u tranh . Do trong mt thi gian di, bn thn hot ng mt m cng

c xem l mt b mt, nn cc ti liu k thut v mt m c ph bin n nay

thng ch ghi li cc kin thc kinh nghim, thnh thong mi c mtvi "pht

minh"nhcc h mt m Vigenre vo th k 16 hoc h mt m Hillra i nm

1929 l cc h m thc hin php chuyn dch (i vi m Vigenre) hay php thay

th (m Hill) ng thi trn mt nhm k t ch khng phi trn tng k t ringr. [2 - tr12,13]

Bc sang th k 20, vo nhng thp nin u ca th k. S pht trin ca cc k

thut biu din, truyn v x l tn hiu c tc ng gip cho cc hot ng lp

v gii mt m t th cng chuyn sang c gii ha ri in t ha. Cc vn bn,

cc bn mt m trc y c vit bng ngn ng thng thng nay c chuyn

bng k thut s thnh cc dy tn hiu nh phn, tc cc dy bit, v cc php bin

i trn cc dy k t c chuyn thnh cc php bin i trn cc dy bit, hay cc

dy s, vic thc hin cc php lp m, gii m tr thnh vic thc hin cc hm s

s hc. Ton hc v k thut tnh ton bt u tr thnh cng c cho vic pht trin

khoa hc v mt m. Khi nim trung tm ca khoa hc mt m l khi nim b



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mt. l mt khi nim ph bin trong i sng, khi nim b mt thot u c

gn vi khi nim ngu nhin, ri v sau trong nhng thp nin gn y, vi khi

nim phc tp, c th hn l khi nim phc tp tnh ton. [2 tr13]

Nm 1978,Rivest, Shamirv Adleman tm ra mt h mt m kha cng khai v

mt s ch k in t hon ton c th ng dng trong thc tin, tnh bo mt v

an ton ca chng c bo m bng phc tp ca mt bi ton s hc ni

ting l bi ton phn tch s nguyn thnh cc tha s nguyn t. Sau pht minh ra

h mt m (m nay ta thng gi l h RSA), vic nghin cu pht minh ra

cc h mt m kha cng khai khc, v ng dng cc h mt m kha cng khai vo

cc bi ton khc nhau ca an ton thng tin c tin hnh rng ri, l thuyt

mt m v an ton thngtin tr thnh mt lnh vc khoa hc c pht trin nhanh

trong vi ba thp nin cui ca th k 20, li cun theo s pht trin ca mt s b

mn ca ton hc v tin hc [2 tr96].

Con ngi s dng cc hp ng di dng in t t hn 100 nm nay vi

vic s dng m Morse v in tn. Vo nm 1889, ta n ti cao bang New

Hampshire (Hoa k) ph chun tnh hiu lc ca ch k in t. Tuy nhin, ch

vi nhng pht trin ca khoa hc k thut gn y th ch k in t mi i vo

cuc sng mt cch rng ri.

Vo thp k 1980, cc cng ty v mt s c nhn bt u s dng my fax

truyn i cc ti liu quan trng. Mc d ch k trn cc ti liu ny vn th hin

trn giy nhng qu trnh truyn v nhn chng hon ton da trn tn hiu in t.

Hin nay, ch k in t c th bao hm cc cam kt gi bng email, nhp cc s

nh dng c nhn (PIN) vo cc my ATM, k bng bt in t vi thit b mn

hnh cm ng ti cc quy tnh tin, chp nhn cc iu khon ngi dng (EULA)

khi ci t phn mm my tnh, k cc hp ng in t online...[7]


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Phn 2: Ni dung.

I. Tng quan v chng thc v an ton thng tin.

Chng ta ang sng trong mt thi i bng n thng tin. Nhu cu trao i thng

tin v cc phng tin truyn a thng tin pht trin mt cch nhanh chng. V

cng vi s pht trin , i hi bo v tnh b mt v chng thc ngun thng tin

cng cng ngy cng to ln v c tnh ph bin. C nhiu bi ton khc nhau v yucu an ton thng tin ty theo nhng tnh hung khc nhau.

V d trong thc t, mt s bi ton chung nht m ta thng gp l nhng bi

ton sau y:


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- Bo mt: Gi thng tin c b mt i vi tt c mi ngi, tr mt t ngi c

thm quyn c c, bit thng tin .

- Ton vn thng tin: Bo m thng tin khng b thay i hay xuyn tc bi

nhng k khng c thm quyn hoc bng nhng phng tin khng c php.

- Nhn thc mt thc th: Xc nhn danh tnh ca mt thc th, chng hn mt

ngi, mt my tnh cui trong mng, mt th tn dng,...

- Nhn thc mt thng bo: Xc nhn ngun gc ca mt thng bo c gi

n.

- Ch k: Mt cch gn kt mt thng tin vi mt thc th, thng dng trong

bi ton nhn thc mt thng bo cng nh trong nhiu bi ton nhn thc khc.

- y quyn: Chuyn cho mt thc th khc quyn c i din hoc c lm

mt vic g .

- Cp chng ch: Cp mt s xc nhn thng tin bi mt thc th c tn nhim.

- Bo nhn: Xc nhn mt thng bo c nhn hay mt dch v c thc

hin.

- Lm chng: Kim th vic tn ti mt thng tin mt thc th khc vi ngi

ch s hu thng tin .

- Khng chi b c: Ngn nga vic chi b trch nhim i vi mt cam kt

c (th d k vo mt vn bn).

- n danh: che giu danh tnh ca mt thc th tham gia trong mt tin trnh no

(thng dng trong giao dch tin in t).

- Thu hi: Rt li mt giy chng ch hay y quyn cp.

C s ca cc gii php cho cc bi ton k trn l cc phng php mt m, c

bit l mt m kha cng khai.


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Trong th gii s, c 3 cch xc thc mt ngi hoc mc tin cy ca mt

thng tin trn my tnh. Mt l Th thng hnh (Pass Card) m nc ta hin nay

cha ph bin. Hai l Password, cch ny s dng tn truy nhp (User Name) v

mt khu (Password) cung cp cho cc giao din ng nhp xc thc thng tin.Th ba, dng ch k in t (Digital Signature). [2 tr19]

II. M ha d liu v gii m

a. Tng quan v m ha d liu v gii m:

S pht trin chng mt ca Internet tc ng n c cng vic kinh doanh v

ngi tiu dng vi s ha hn v vic thay i cch m con ngi sng v lm

vic. Nhng mi lo ngi ln nht c cp n l vic bo mt trn Internet, c

bit khi cc thng tin mang tnh nhy cm v ring t c gi i trn mng.

M ha l ngnh nghin cu cc thut ton v phng thc m bo tnh b mt

v (thng l di dng cc vn bn lu tr trn my tnh). Cc sn phm ca lnh

vc ny l cc h m mt, cc hm bm, cc h ch k in t, cc c ch phn

phi, qun l kha v cc giao thc mt m.

C rt nhiu thng tin m chng ta khng mun ngi khc bit khi gi i nh:

thng tin v Credit-Card, thng tin v kinh doanh ca cng ty, thng tin v ti

khon c nhn, thng tin v c nhn nh s chng minh th, s th...

Qu trnh m ho trong my tnh da vo khoa hc v mt m (Cryptography)

c con ngi s dng t lu i. Trc thi i s ho, ngi s dng mt m

nhiu nht vn l chnh ph, ch yu trong mc ch qun s. Hu ht cc phng

php m ho c dng hin nay da vo cc my tnh, n gin l do cc m docon ngi sinh ra rt d b ph bi cng c my tnh. Cc h thng m ho trong

my tnh ph bin nht thuc mt trong hai loi sau:

* M ho vi kho i xng (Symmetric-key Encryption)


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* M ho vi kho cng khai (Public-key Encryption)

Gn y nht l cc s kin lin quan ti cc hm bm MD5 (mt hm bm thuc

h MD do Ron Rivest pht trin) vSHA. Mt nhm cc nh khoa hc ngi Trung

Quc (Xiaoyun Wang, Yiqun Lisa Yin, Hongbo Yu) pht trin cc phng php

cho php pht hin ra cc ng ca cc hm bm c s dng rng ri nht

trong s cc hm bm ny. y l mt s kin ln i vi ngnh mt m hc do s

ng dng rng ri v c th xem l cn quan trng hn bn thn cc h m mt ca

cc hm bm. Do s kin ny cc hng vit phn mm ln (nh Microsoft) v cc

nh mt m hc khuyn co cc l p trnh vin s dng cc hm bm mnh hn

(nh SHA-256, SHA-512) trong cc ng dng.

Ngy nay kh c th tm thy cc ng dng trn my tnh li khng s dng ti

cc thut ton v cc giao thc mt m hc. Ti cc ng dng cho my tnh c nhn

(Desktop Applications) cho ti cc chng trnh h thng nh h iu hnh

(Operating Systems) hoc cc ng dng mng nh Yahoo Messenger hoc h c s

d liu u c s dng cc thut ton m ha mt khu ngi dng bng mt h m

hoc mt hm bm no . c bit vi s pht trin mnh m ca thng mi in

t cc m hnh ch k in t ngy cng ng vai tr tch cc cho mt mi trng

an ton cho ngi dng [1 tr17].

b. M ha bt i xng (asymmetric).

Mt m ha kha cng khai l mt dng mt m ha cho php ngi s dng trao

i cc thng tin mt m khng cn phi trao i cc kha chung b mt trc .

iu ny c thc hin bng cch s dng mt cp kha c quan h ton hc vi

nhau l kha cng khai v kha c nhn (hay kha b mt).

Thut ng mt m ha kha bt i xng thng c dng ng ngha vi mt

m ha kha cngkhai mc d hai khi nim khng hon ton tng ng. C

nhng thut ton mt m kha bt i xng khng c tnh cht kha cng khai v b



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mt nh cp trn m c hai kha (cho m ha v gii m) u cn phi gi b

mt. [1 tr77]

Trong mt m ha kha cng khai, kha c nhn phi c gi b mt trong khi

kha cng khai c ph bin cng khai. Trong 2 kha, mt dng m ha v

kha cn li dng gii m. iu quan trng i vi h thng l khng th tm ra

kha b mt nu ch bit kha cng khai.

H thng mt m ha kha cng khai c th s dng vi cc mc ch:

- M ha: gi b mt thng tin v ch c ngi c kha b mt mi gii m c.

- To ch k s: cho php kim tra mt vn bn c phi c to vi mt kha

b mt no hay khng.

- Tha thun kha: cho php thit lp kha dng trao i thng tin mt gia 2

bn.

Thng thng, cc k thut mt m ha kha cng khai i hi khi lng tnh

ton nhiu hn cc k thut m ha kha i xng nhng nhng li im m chng

mang li khin cho chng c p dng trong nhiu ng dng.

S ra i ca khi nim h mt m kho cng khai l mt tin b c tnh cht

bc ngot trong lch s mt m ni chung, gn lin vi s pht trin ca khoa hc

tnh ton hin i. Ngi ta c th xem thi im khi u ca bc ngot l s

xut hin tng ca W. DiffievM.E. Hellman c trnh by vo thng su nm

1976 ti Hi ngh quc gia hng nm caAFIPS (Hoa k) trongbi Multiuser

cryptographic techniques. [1 tr78]

Mt nm sau, nm 1977, R.L. Rivest, A. Shamir v L.M. Adleman xut mt h

c th v mt m kho cng khai m an ton ca h da vo bi ton kh phn

tch s nguyn thnh tha s nguyn t, h ny v sau tr thnh mt h ni ting

v mang tn l h RSA, c s dng rng ri trong thc tin bo mt v an ton



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thng tin.Cng vo thi gian , M.O. Rabin cng xut mt h mt m kho

cng khai da vo cng bi ton s hc kh ni trn. Lin tip sau , nhiu h mt

m kha cng khai c xut, m kh ni ting v c quan tm nhiu l cc

h: h McEliece c a ra nm 1978 da trn NP(kh) ca bi ton gii mi vi cc h m cyclic tuyn tnh, h Merkle- Hellman da trn tnh NP- y

ca bi ton xp ba l (knapsack problem), h mt m ni ting ElGamalda trn

kh ca bi ton lgaritri rc, h ny v sau c m rng pht trin nhiu

h tng t da trn kh ca cc bi ton tng t lgaritri rc trn cc cu

trc nhm cyclic hu hn, nhm cc im nguyn trn ng cong eliptic, v.v...

tng bo mt, h mt mElGamalcn dng vi t cch u vo cho thut ton

lp mt m ca mnh, ngoi kho cng khai v bn r, mt yu t ngu nhin c

chn tu , iu lm cho h mt m tr thnh mt h mt m xc sut kho cng

khai. Mt s h mt m xc sut kho cng khai cng c pht trin sau bi

Goldwasser-Micali v Blum-Goldwasser. [1 tr79]

Khng phi tt c cc thut ton mt m ha kha bt i xng u hot ng

ging nhau nhng phn ln u gm 2 kha c quan h ton hc vi nhau: mt cho

m ha v mt gii m. thut ton m bo an ton th khng th tm ckha gii m nu ch bit kha dng m ha. iu ny cn c gi l m ha

cng khai v kha dng m ha c th cng b cng khai m khng nh hng

n b mt ca vn bn m ha.

Cc thng tin m kha th ch c ngi s hu mi bit. Tn ti kh nng mt

ngi no c th tm ra c kha b mt. Khng ging vi h thng mt m s

dng mt ln (one-time pad) hoc tng ng, cha c thut ton m ha kha bti xng no c chng minh l an ton trc cc tn cng da trn bn cht ton

hc ca thut ton. Kh nng mt mi quan h no gia 2 kha hay im yu

ca thut ton dn ti cho php gii m khng cn ti kha hay ch cn kha m

ha vn cha c loi tr. An ton ca cc thut ton ny u da trn cc c



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lng v khi lng tnh ton gii cc bi ton gn vi chng. Cc c lng

ny li lun thay i ty thuc kh nng ca my tnh v cc pht hin ton hc

mi.[2- tr18]

Mc d vy, an ton ca cc thut ton mt m ha kha cng khai cng tng

i m bo. Nu thi gian ph mt m (bng phng php duyt ton b) c

c lng l 1000 nm th thut ton ny hon ton c th dng m ha cc

thng tin v th tn dng - R rng l thi gian ph m ln hn nhiu ln thi gian

tn ti ca th (vi nm). [2 tr21]

c. M ha i xng (symmetric).

Trong mt m hc, cc thut ton kha i xng (ting Anh: symmetric-key

algorithms) l mt lp cc thut ton mt m ha trong cc kha dng cho vic

mt m ha v gii m c quan h r rng vi nhau (c th d dng tm c mt

kha nu bit kha kia). [8]

Kha dng m ha c lin h mt cch r rng vi kha dng gii m c

ngha chng c th hon ton ging nhau, hoc ch khc nhau nh mt bin i n

gin gia hai kha. Trn thc t, cc kha ny i din cho mt b mt c phnhng bi hai bn hoc nhiu hn v c s dng gi gn s b mt trong knh

truyn thng tin.

Thut ton i xng c th c chia ra lm hai th loi, mt m lung (stream

ciphers) v mt m khi (block ciphers). Mt m lung m ha tng bit ca thng

ip trong khi mt m khi gp mt s bit li v mt m ha chng nh mt n v.

C khi c dng thng l cc khi 64 bit. Thut ton tiu chun m ha tn tin(Advanced Encryption Standard), cNIST cng nhn thng 12 nm 2001, s

dng cc khi gm 128 bit. [8]

Cc thut ton i xng thng khng c s dng c lp. Trong thit k ca

cc h thng mt m hin i, c hai thut ton bt i xng v thut ton i xng



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c s dng phi hp tn dng cc u im ca c hai. Nhng h thng s

dng c hai thut ton bao gm SSL (Secure Sockets Layer), PGP (Pretty Good

Privacy) v GPG (GNU Privacy Guard)... Cc thut ton cha kha bt i xng

c s dng phn phi cha kha mt cho thut ton i xng c tc caohn.

Mt s v d cc thut ton i xng ni ting bao gm Twofish, Serpent, AES

(cn c gi l Rijndael), Blowfish, CAST5, RC4, Tam phn DES (Triple DES),

v IDEA (International Data Encryption Algorithm - Thut ton mt m ha d

liu quc t). [8]

Cc thut ton i xng ni chung i hi cng sut tnh ton t hn cc thut tonkha bt i xng. Trn thc t, mt thut ton kha bt i xng c khi lng

tnh ton nhiu hn gp hng trm, hng ngn ln mt thut ton kha i xng c

cht lng tng ng.

Hn ch ca cc thut ton kha i xng bt ngun t yu cu vs phn hng

cha kha b mt, mi bn phi c mt bn sao ca cha. Do kh nng cc cha kha

c th b pht hin bi i th mt m, chng thng phi c bo an trong khi

phn phi v trong khi dng. Hu qu ca yu cu v vic la chn, phn phi v

lu tr cc cha kha mt cch khng c li, khng b mt mt l mt vic lm kh

khn, kh c th t c mt cch ng tin cy.

m bo giao thng lin lc an ton cho tt c mi ngi trong mt nhm gm

n ngi, tng s lng cha kha cn phi c l n(n-1)/2.

Cc thut ton kha i xng khng th dng cho mc ch xc thc hay mc chchng thoi thc.

d. Hm bm (Hashing)

- Tng quan v hm bm


17


18/51


Trong ngnh mt m hc, mt hm bm mt m hc (ting Anh: Cryptographic

hash function) l mt hm bm vi mt s tnh cht bo mt nht nh ph hp

vic s dng trong nhiu ng dng bo mt thng tin a dng, chng hn nh chng

thc v kim tra tnh nguyn vn ca thng ip. Mt hm bm nhn u vo lmt xu k t di (hay thng ip) c di ty v to ra kt qu l mt xu k t

c di c nh, i khi c gi l tm tt thng ip (message digest) hoc ch

k s (digital fingerprint) [1 tr109].

Hm bm l cc thut ton khng s dng kha m ha ( y ta dng thut

ng bm thay cho m ha), n c nhim v lc (bm) thng ip c a

vo theo mt thut ton h mt chiu no , ri a ra mt bn bm gi l vn bn

i din c kch thc c nh.Do ngi nhn khng bit c ni dung hay

di ban u ca thng ip c bm bng hm bm.

Gi tr ca hm bm l duy nht, v khng th suy ngc li c ni dung thng

ip t gi tr bm ny. [1 tr109]

- Tnh cht ca hm bm

Tnh ng : Theo nguyn l Diricle: nu c (n+1) con th c b vo n cichung th phi tn ti t nht mt ci chung m trong c t nht l hai con th

chung. R rng vi khng gian gi tr Bm nh hn rt nhiu so vi khng gian tin

v mt kch thc th chc chn s tn ti ng , ngha l c hai tinx # xm gi

tr Bm ca chng l ging nhau, tc h(x) = h(x) [1 - 109].

Sau y chng ta s xt cc dng tn cng c th c, t rt ra cc tnh cht ca

hm Bm:Tnh cht 1: Hm bm khng va chm yu.


18


19/51


Hm bm h l khng va chm yu nu khi cho trc mt bc in x, khng th

tin hnh v mt tnh ton tm ra mt bc in x x m h(x) = h(x). [1 -

tr110]

V d: Ngi A gi cho B (x,y) viy = SigA(h(x)). Nhng trn ng truyn, tin

b ly trm. Tn trm, bng cch no tm c mt bn thng ipx c h(x) =

h(x) m x x. Sau , tn trm a x thay th x ri truyn tip cho ngi B.

Ngi B nhn c v vn xc thc c thng tin ng n.

trnh tn cng trn, hm bm phi khng va chm yu.

Tnh cht 2: Hm bm khng va chm mnh

Hm bm h l khng va chm mnh nu khng c kh nng tnh ton tm ra hai

bc thng ip x v x m x x v h(x) = h(x). [1 tr110]

V d: u tin, tn gi mo tm ra c hai bc thng ip x v x (x x) m c

h(x) = h(x) (ta coi bc thng ip x l hp l, cn x l gi mo). Tip theo, tn

trm a cho ng A v thuyt phc ng ny k vo bn tm lc h(x) nhn c

y. Khi (x, y) l bc in gi mo nhng hp l.

trnh kiu tn cng ny, hm h phi tha mn tnh khng va chm mnh

Tnh cht 3: Hm bm mt chiu.

Hm bm h l mt chiu nu khi cho trc mt bn tm lc thng bo z, khng

th thc hin v mt tnh ton tm bc in x sao cho h(x) = z. [1 tr110]

Vic gi mo cc ch k trn bn tm lc thng boz ngu nhin thng xy ra

vi s ch k. Gi s tn gi mo tnh ch k trn bn tm lc thng bo z ngu

nhin nh vy. Sau anh ta tmx sao choz = h(x). Nu lm c nh vy th (x,y)

l bc in gi mo hp l. trnh c tn cng ny, h cn tho mn tnh cht

mt chiu:



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- Mt s hm bm ni ting

+ MD5 (Message Digest)

Ronald Rivest l ngi pht minh ra cc hm Bm MD2, MD4 (1990) v MD5

(1991). Do tnh cht tng t ca cc hm Bm ny, sau y chng ta s xem xt

hm Bm MD5, y l mt ci tin ca MD4 v l hm Bm c s dung rng ri

nht, nguyn tc thit k ca hm bm ny cng l nguyn tc chung cho rt nhiu

cc hm bm khc [1 tr111].

a. Miu t MD5:

u vo l nhng khi 512 bit, c chia cho 16 khi con 32 bit. u ra ca thut

ton l mt thit l p ca 4 khi 32 bit to thnh mt hm Bm 128 bitduy nht.

u tin, ta chia bc in thnh cc khi 512 bit, vi khi cui cng (t l x v x

< 512bit) ca bc in, chng ta cng thm mt bit 1 vo cui ca x, theo sau l

cc bit 0 c di cn thit (512 bit). Kt qu l bc in vo l mt chui M

c di chia ht cho 512, v vy ta c th chia M ra thnh cc N khi con 32 bit

(N khi ny s chia ht cho 16).

By gi, ta bt u tm ct ca bc in vi 4 khi 32 bitA, B, C v D (c xem

nh thanh ghi) :

A = 0x01234567

B = 0x89abcdef

C = 0xfedcba98

D = 0x76543210.

Ngi ta thng gi A, B, C, D l cc chui bin s (chaining variables).

Bc in c chia ra thnh nhiu khi 512 bit, mi khi 512 bitli c chia ra

16 khi 32 biti vo bn vng l p ca MD5. Gi s ta t a, b, c v d thay cho A,



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B, C v D i vi khi 512 bitu tin ca bc in. Bn vng l p trong MD5 u

c cu trc ging nhau. Mi vng thc hin 16 ln bin i: thc hin vi mt hm

phi tuyn ca 3 trong 4 gi tr a, b, c v d; sau n cng kt qu n gi tr th 4,

tip cng vi mt khi con 32 bitv mt hng s. Sau , n dch tri mt lngbit thay i v cng kt qu vo mt trong 4 gi tr a, b, c hay d. Kt qu cui cng

l mt gi tr mi c thay th mt trong 4 gi tr a, b, c hay d.

Hnh 1.2 s vng lp chnh ca MD5

C bn hm phi tuyn, mi hm ny c s dng cho mi vng:

F(X,Y,Z ) = (X and Y) or ((not X) and Z)

G(X,Y,Z ) = ((X and Z) or (Y and (not Z)))

H(X,Y,Z ) = X xor Y xor Z

I(X,Y,Z ) = Y xor (X or (not Z)).

Nhng hm ny c thit k sao cho cc bit tng ng ca X, Y v Z l c l p

v khng u tin, v mi bit ca kt qu cng c l p v ngang bng nhau.

Nu Mj l mt biu din ca khi con th j (j = 16) v


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FF(a,b,c,d,Mj,s,ti) c biu din a = b + ((a + F(b,c,d) + Mj + ti)


23/51


GG (a, b, c, d, M1, 5, 0x61e2562)

GG (d, a, b, c, M6, 9, 0xc040b340)

GG (c, d, a, b, M11, 14, 0x265e5a51)

GG (b, c, d, a, M0, 20, 0xe9b6c7aa)

GG (a, b, c, d, M5, 5, 0xd62f105d)

GG (d, a, b, c, M10, 9, 0x02441453)

GG (c, d, a, b, M15, 14, 0xd8a1e681)

GG (b, c, d, a, M4, 20, 0xe7d3fbc8)

GG (a, b, c, d, M9, 5, 0x21e1cde6)

GG (d, a, b, c, M14, 9, 0xc33707d6)

GG (c, d, a, b, M3, 14, 0xf4d50d87)

GG (b, c, d, a, M8, 20, 0x455a14ed)

GG (a, b, c, d, M13, 5, 0xa9e3e905)

GG (d, a, b, c, M2, 9, 0xfcefa3f8)

GG (c, d, a, b, M7, 14, 0x676f02d9)

GG (b, c, d, a, M12, 20, 0x8d2a4c8a).

Vng 3:

HH (a, b, c, d, M5, 4, 0xfffa3942)

HH (d, a, b, c, M8, 11, 0x8771f681)

HH (c, d, a, b, M11, 16, 0x6d9d6122)

HH (b, c, d, a, M14, 23, 0xfde5380c)

HH (a, b, c, d, M1, 4, 0xa4beea44)

HH (d, a, b, c, M4, 11, 0x4bdecfa9)



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HH (c, d, a, b, M7, 16, 0xf6bb4b60)

HH (b, c, d, a, M10, 23, 0xbebfbc70)

HH (a, b, c, d, M13, 4, 0x289b7ec6)

HH (d, a, b, c, M0, 11, 0xeaa127fa)

HH (c, d, a, b, M3, 16, 0xd4ef3085)

HH (b, c, d, a, M6, 23, 0x04881d05)

HH (a, b, c, d, M9, 4, 0xd9d4d039)

HH (d, a, b, c, M12, 11, 0xe6db99e5)

HH (c, d, a, b, M15, 16, 0x1fa27cf8)

HH (b, c, d, a, M2, 23, 0xc4ac5665).

Vng 4:

II (a, b, c, d, M0, 6, 0xf4292244)

II (d, a, b, c, M7, 10, 0x432aff97)

II (c, d, a, b, M14, 15, 0xab9423a7)

II (b, c, d, a, M5, 21, 0xfc93a039)

II (a, b, c, d, M12, 6, 0x655b59c3)

II (d, a, b, c, M3, 10, 0x8f0ccc92)

II (c, d, a, b, M10, 15, 0xffeff47d)

II (b, c, d, a, M1, 21, 0x85845dd1)

II (a, b, c, d, M8, 6, 0x6fa87e4f)

II (d, a, b, c, M15, 10, 0xfe2ce6e0)

II (c, d, a, b, M6, 15, 0xa3013414)

II (b, c, d, a, M13, 21, 0x4e0811a1)



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II (a, b, c, d, M4, 6, 0xf7537e82)

II (d, a, b, c, M11, 10, 0xbd3af235)

II (c, d, a, b, M2, 15, 0x2ad7d2bb)

II (b, c, d, a, M9, 21, 0xeb86d391).

Nhng hng s ti c chn theo quy lut sau: bc th i gi tr t i l phn

nguyn ca 232*abs(sin(i)), trong i = [0..63] c tnh theo radian.

Sau tt c nhng bc ny a, b, c v d ln lt c cng vi A, B, C v D cho

kt qu u ra, v thut ton tip tc vi khi d liu 512 bittip theo cho n ht

bc in. u ra cui cng l mt khi 128 bitca A, B, C v D, y chnh l hm

Bm nhn c [1 tr111 > tr115].

b. Tnh bo mt trong MD5:

Ron Rivest phc ho nhng ci tin ca MD5 so vi MD4 nh sau:

- Vng th 4 c thm vo (cnMD4 ch c 3 vng).

- Mi bc c cng thm mt hng s duy nht.

- Hm G vng 2 thay i t ((X and Y) or (X and Z) or (Y and Z)) thnh ((X

and Z) or (Y and (not Z))) nhm gim tnh i xng ca G (gim tnh tuyn

tnh).

- Mi bc c cng kt qu ca bc trc n, lm cc qu trnh c tnh

lin kt, ph thuc ln nhau.

- Vic cc khi con b thay i khi vo vng 2 v vng 3 lm cho khun dng

cu trc vng l p thay i theo.

- S lng lng bit dch tri ca mi vng c ti u v cc bc dch

mi vng l khc nhau.



26/51


Nm 1993, den BoervBosselaers tm ra ng trong vic s dng hm nn

(vng 2 v 3) ca MD5. iu ny ph v quy lut thit k MD5 l chng li s

ng , nhng MD5 vn l hm Bm c s dng rng ri hin nay [1 tr115].

+ SHA (Secure Hash Algorithm)

Nm 1995, t chc NIST cng NSA thit k ra thut ton hm Bm an ton

(SHA) s dng cho chun ch k in t DSS. SHA c thit k da trn nhng

nguyn tc ca MD4/MD5, to ra 160 bitgi tr Bm [1 tr116].

a. Miu t SHA:

Cng ging vi MD5, bc in c cng thm mt bit 1 v cc bit 0 cui bc

in bc in c th chia ht cho 512. SHA s dng 5 thanh ghi dch:

A = 0x67452301

B = 0xefcdab89

C = 0x98badcfe

D = 0x10325476

E = 0xc3d2e1f0

Bc in c chia ra thnh nhiu khi 512 bit. Ta cng t l a, b, c, d v e thay

cho A, B, C, D v E i vi khi 512 bitu tin ca bc in. SHA c bn vng

l p chnh vi mi vng thc hin 20 ln bin i: bao gm thc hin vi mt hm

phi tuyn ca 3 trong 5 gi tr a, b, c, d v e; sau c ng c cng v dch nh

trong MD5.

SHA xc l p bn hm phi tuyn nh sau:

ft(X,Y,Z) = (X and Y) or ((not X) and Z) vi 0 t 19

ft(X,Y,Z) = X xor Y xor Z vi 20 t 39

ft(X,Y,Z) = (X and Y) or (X and Z) or (Y and Z) vi 40 t 59



27/51


ft(X,Y,Z) = X xor Y xor Z vi 60 t 79.

Bn hng s s dng trong thut ton l:

Kt = 21/2 /4 = 0x5a827999 vi 0 t 19

Kt = 31/2 /4 = 0x6ed9eba1 vi 20 t 39

Kt = 51/2 /4 = 0x8f1bbcdc vi 40 t 59

Kt = 101/2 /4 = 0xca62c1d6 vi 60 t 79.

Cc khi bc in c m rng t 16 khi con 32 bit(M0 n M15) thnh 80

khi con 32 bit(W0 n W79) bng vic s dng thut ton m rng:

Wt = Mt vi 0 t 15

Wt = (Wt-3 xorWt-8 xorWt-14 xorWt-16) vi 16 t 79.

Nu gi Wt l biu din ca khi con th t ca bc in c m rng, v


28/51


b. Tnh bo mt trong SHA:

hiu r hn v tnh bo mt ca SHA, ta hy so snh SHA vi MD5 c th

tm ra nhng im khc nhau ca hai hm Bm ny:

- MD5 v SHA u cng thm cc bit gi to thnh nhng khi chia ht cho

512 bit, nhng SHA s dng cng mt hm phi tuynfcho c bn vng.

- MD5 s dng mi hng s duy nht cho mi bc bin i, SHA s dng mi

hng s cho mi vng bin i, hng s dch ny l mt s nguyn t i vi ln

ca t (ging viMD4).

- Trong hm phi tuyn th 2 ca MD5 c s ci tin so vi MD4, SHA th s

dng li hm phi tuyn ca MD4, tc (X and Y) or (X and Z) or (Y and Z).

- Trong MD5 vi mi bc c cng kt qu ca bc trc . S khc bit i

vi SHA l ct th 5 c cng (khng phi b, c hay d nh trongMD5), iu ny

lm cho phng php tn cng ca Boer-Bosselaers i vi SHA b tht bi (den

Boer va Bosselaers l hai ngi ph thnh cng 2 vng cui trongMD4).

Cho n nay, cha c mt cng b no c a ra trong vic tn cng SHA, bi

v di ca hm Bm SHA l 160 bit, n c th chng li phng php tn cng

bng vt cn (k c birthday attack) tt hn so vi hm Bm MD5 128 bit [1

tr117].

III. Ch k in t

1. Tng quan

Trong cuc sng hng ngy, ta cn dng ch k xc nhn cc vn bn ti liuno v c th dng con du vi gi tr php l cao hn i km vi ch k.

Cng vi s pht trin nhanh chng ca cng ngh thng tin, cc vn bn ti liu

c lu di dng s, d dng c sao chp, sa i. Nu ta s dng hnh thc



29/51


ch k truyn thng nh trn s rt d dng b gi mo ch k. Vy lm sao c

th k vo cc vn bn, ti liu s nh vy?

Cu tr li l s dng ch k in t! Ch k in t i km vi cc thng tin

ch s hu v mt s thng tin cn thit khc s tr thnh Chng ch in t.

Ch k in t (ting Anh: electronic signature) l thng tin i km theo d liu

(vn bn, hnh nh, video...) nhm mc ch xc nh ngi ch ca d liu .

Mt s ch k in t l b 5 (P, A, K, S, V) tho mn cc iu kin di y:

1) P l tp hu hn cc bc in (thng i p, bn r) c th.

2) A l tp hu hn cc ch k c th.

3) K l tp khng gian kho (tp hu hn cc kho c th).

4) Vi mi kho K k tn ti mt thut ton k SigK Sv mt thut ton xc

minh VerK V. Mi Sigk: P A v verK: P x A {TRUE, FALSE} l nhng hm

sao cho mi bc in x P v mi ch k y A tho mn phng trnh di y:

True nu y = sig(x)

Ver (x, y) =

False nuy sig(x).

Vi miK k, hm SigK v VerK l cc hm a thc thi gian. Hm VerK s l hm

cng khai cn hm SigK l b mt. Khng th d dng tnh ton gi mo ch k

ca B trn bc inx, ngha l vix cho trc ch c B mi c th tnh c y

Ver(x, y) = TRUE. Mt s ch k khng th an ton v iu kin v mt ngi C

no c th kim tra tt c ch s y trn bc in x nh dng thut ton Ver()cng khai cho ti khi anh ta tm thy ch k ng. V th, nu c thi gian, C

lun c th gi mo ch k ca B. Nh vy mc ch ca chng ta l tm cc s

ch k in t an ton v mt tnh ton [1 tr116].


29


30/51


Ch k in t c s dng trong cc giao dch in t. Xut pht t thc t, ch

k in t cng cn m bo cc chc nng: Xc nh c ngi ch ca mt d

liu no v d vn bn, nh, video, ... d liu c b thay i hay khng.

Hai khi nim ch k s (digital signature) v ch k in t thng c dng

thay th cho nhau mc d chng khng hon ton c cng ngha. Ch k s ch l

mt tp con ca ch k in t (ch k in t bao hm ch k s).

Mt ch k in t s l mt ch k s nu n s dng mt phng php m ha

no m bo tnh ton vn (thng tin) v tnh xc thc. V d nh mt bn d

tho hp ng son bi bn bn hng gi bng email ti ngi mua sau khi c k

(in t). [1- tr117]

Mt vn bn c k c th c m ha khi gi nhng iu ny khng bt buc.

Vic m bo tnh b mt v tnh ton vn ca d liu c th c tin hnh c lp.

2. Quy trnh s dng ch k in t

Ch k in t hot ng da trn h thng m ha kha cng khai. H thng m

ha ny gm hai kha, kha b mt v kha cng khai. Mi ch th c mt cp kha

nh vy, ch th s gi kha b mt, cn kha cng khai ca ch th s c a

ra cng cng bt k ai cng c th bit. Nguyn tc ca h thng m ha kha

cng khai l, nu m ha bng kha b mt th ch kha cng khai mi gii m

ng thng tin c, v ngc li, nu m ha bng kha cng khai, th ch c kha

b mt mi gii m ng c.

Ngoi ra, ch k cn m bo pht gic c bt k s thay i no trn d liu

c k. k ln mt vn bn, phn mm k s nghin ( crunch down) dliu gi gn bng mt vi dng, c gi l thng bo tm tt, bng mt tin

trnh c gi l k thut bm, ri to thnh ch k in t. Cui cng, phn

mm k tn s gn ch k in t ny vo vn bn.



31/51


V d: Gi s bn A c ti liu P cn k. Bn A s thc hin bm vn bn thnh

mt bn tm lc X, sau dng kha b mt ca mnh k ln bn tm lc X

c vn bn ch k in t Y, sau gi ti liuPkm theo ch k Ycho A.

Gi s B mun xc nhn ti liu P l ca A, vi ch k l bn m Y. Bn B s

dng kha cng khai ca A xc nhn ch k Yca A k trn vn bn P gi c

ng hay khng, nu xc nhn ng th ch k Ychnh l A k trn vn bn P,

ngc li th khng phi hoc bn k c thay i.

Mt s trng hp xy ra vi ch k in t, cng ging nh cc trng hp xy

ra vi ch k truyn thng. V d: Khi ti liu TL ca A b thay i ( d ch mt k

t, mt du chm, hay mt k hiu bt k), khi B xc nhn, anh ta s thy bn giim khc vi ti liu TL ca anh A. B s kt lun rng ti liu b thay i,

khng phi l ti liu anh A k.

Trng hp khc, nu A l kha b mt, ngha l vn bn ti liu ca A c th

k bi ngi khc c kha b mt ca A. Khi mt ai xc nhn ti liu c cho

l ca A k, ch k vn l hp l, mc d khng phi chnh A k. Nh vy, ch k

ca A s khng cn gi tr php l na. Do , vic gi kha b mt l tuyt i

quan trng trong h thng ch k in t.

Trong trng hp v d trn, A c mt cp kha c th k trn vn bn, ti liu

s. Tng t nh vy, B hay bt c ai s dng ch k in t, u c mt cp kha

nh vy. Kha b mt c gi ring, cn kha cng khai c a ra cng cng.

Vy vn t ra l lm th no bit mt kha cng khai thuc v A, B hay mt

ngi no ?

Hn na, gi s trong mi trng giao dch trn Internet, cn s tin cy cao, A

mun giao dch vi mt nhn vt X. X v A cn trao i thng tin c nhn cho

nhau, cc thng tin gm h tn, a ch, s in thoi, email Vy lm sao A

c th chc chn rng mnh ang giao dch vi nhn vt X ch khng phi l ai



32/51


khc gi mo X? Chng ch s c to ra gii quyt vn ny! Chng ch s

c c ch xc nhn thng tin chnh xc v cc i tng s dng chng ch s.

Thng tin gia A v X s c xc nhn bng mt bn trung gian m A v X tin

tng.

Bn trung gian l nh cung cp chng ch s CA ( Certificate Authority). CA c

mt chng ch s ca ring mnh, CA s cp chng ch s cho A v X cng nh

nhng i tng khc.

Tr li vn trn, A v X s c cch kim tra thng tin ca nhau da trn chng

ch s nh sau: khi A giao dch vi X, h s chuyn chng ch s cho nhau, ng

thi h cng c chng ch s ca CA, phn mm ti my tnh ca A c c ch kim tra chng ch s ca X c hp l khng, phn mm s kt hp chng ch s

ca nh cung cp CA v chng ch ca X thng bo cho A v tnh xc thc ca

i tng X.

Nu phn mm kim tra v thy chng ch ca X l ph hp vi chng ch CA, th

A c th tin tng vo X.

C ch ch k in t v chng ch s s dng cc thut ton m ha m bokhng th gi mo CA cp chng ch khng hp php, mi chng ch gi mo

u c th d dng b pht hin.

Tr li vi vic k vn bn, ti liu, kha b mt s dng k cc vn bn, ti

liu ca ch s hu. Nh cp trong v d trn, gi s A mun gi mt vn

bn km vi ch k ca mnh trn vn bn , A s dng kha b mt m ha thu

c bn m vn bn, bn m chnh l ch k in t ca A trn vn bn.Khi A gi vn bn v ch k, ngi khc c th xc nhn vn bn ca mnh vi

thng tin y v ch s hu, A s gi c chng ch ca mnh i km vi vn bn.

Gi s X nhn c vn bn A gi km vi chng ch, khi X c th d dng

xc nhn tnh hp php ca vn bn .


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3. Mt s s CKT ph bin

a. Rivest Shamir Adleman (RSA)

- S lc v cc khi nim ton hc dng trong RSA.

* S nguyn t (prime)

S nguyn t l nhng s nguyn ch chia chn c cho 1 v cho chnh n.

V d : 2, 3, 5, 7, 11, 13, 17, 23...

* Khi nim nguyn t cng nhau (relatively prime or coprime).

Vi hai s nguyn dng a v b. Ta k hiu UCLN(a,b): c chung ln nht ca a

v b. n gin ta k hiu UCLN(a,b) = (a,b)

V d :

(4,6)=2

(5,6)=1

Hai s a v b gi l nguyn t cng nhau khi (a,b)=1

V d : 9 v 10 nguyn t cng nhau v (9,10)=1* Khi nim modulo

Vi m l mt s nguyn dng. Ta ni hai s nguyn a v b l ng dvi nhau

+ modulo m, nu m chia ht hiu (a-b) (vit l m|(a-b) )

K hiu a b (mod m) [5]

Nh vy a b (mod m ) khi v ch khi tn ti s nguyn k sao cho: a = b + k*m

V d: 13 3 (mod10) v 13= 3 + 1*10

* Phi Hm EULER

nh ngha: Phi HmEuler (n) c gi tr ti n bng s cc s khng vt qu

n v nguyn t cng nhau vi n. [5]


33


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V d : (5) = 4 , (6) = 2 ,(10) = 4

* Mt s nh l c bn

nh l Euler: Nu m l s nguyn dng v P nguyn t cng nhau vi m th

P(m) 1 (mod m) [5]

Vy nu m v p nguyn t cng nhau . Ta ts = (m) th Ps 1 (mod m)

Suy ra vi: a= 1 + k*s

Ta c : Pa P*(Ps)k P*1k (mod m) P (mod m) Vi e l s nguyn dng

nguyn t cng nhau vi s ,tc l (e,s)=1. Khi tn ti mt nghch o dca e

modulos tc l e*d 1 (mod s) ; e*d = 1 + k*s t E(P) C Pe (mod m)

tD(C) Cd(mod m) Ta thy D(C) Cd(Pe (mod m))d (mod m) Pe*d (mod m) P(1+k*s) (mod m) P.(Ps)k(mod m)P.(1)k(mod m) P (mod m)

V d : m = 10 , P = 9 ta c (10,9)=1, s = (10) = 4, e = 7, ta c (7,4) = 1.

Nghch o ca (7 modulo 4) l: d = 3, v 7*3 =1 + 5*4 Lc ta c: E(P) C

Pe 97 4.782.969 9 (mod10) => C=9 D(C) Cd 93 729 9 (mod 10) Vy D

chnh l hm ngc ca E. y l c s cho vic xy dng thut tonRSA.

Tnh (m) khi bit m. Chng ta c nh l sau y: Gi s m = p1a1

*p2a2

* *pkak

.Khi . (m) =( p1a1 p1(a1-1))* *(pkak pk(ak-1))

V d: m= 10 Ta phn tch 10 =2*5=> (10) =( 21 20) *(51 50) = 1*4 = 4.

- Cch to kha:

Chng ta cn to ra mt cp kha K v Xc nhn theo cc bc sau:

Bc1. Chn 2 s nguyn t ln p v q vi (p # q), la chn ngu nhin v c

lp.

Bc2. Tnh s hm modulo ca h thng: n= p*q.

Bc3. Tnh: Gi tr hm s le: (n)= (p-1)(q-1).


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Bc4. Chn mt s t nhin kha m esao cho (1


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Phn ny c ct b, hy lin h ch ti nhn c bn

chi tit hn.

Hm trn c th tnh d dng s dng phng php tnh hm m ( theo mun) bng

(thut ton bnh phng v nhn). Cui cng ta c bn k c hay bn ch k in t v

gi cho i tc.

* Xc nhn (Gii m).

Sau khi nhn c bn ch k in t, ngi nhn cn phi xc nhn ch k trn

vn bn l ng ngi k bng cch xc nhn bn k vi kha cng khai ca ngi

k vi cng thc sau.

VerK(m,c) = TRUEm ce (mod n)vi x, y Zn.

Qu trnh gii m hot ng v ta c

Ce (md)e mde (mod n);

Do: ed 1 (modp-1) v ed 1 (modq-1), (theo nh l Fermat nh) nn:

Mde m (mod p);

vmde m (mod q);

Dop v q l hai s nguyn t cng nhau, p dng nh l s d Trung Quc, ta c:

Mde m (mod pq);

hay:

Ce m (mod n);

Thng thng, ch k c kt hp vi hm m ho cng khai. Gi s A mun

gi mt bc in c m ho v c k n cho B. Vi bn rx cho trc,

A s tnh ton ch k ca mnhy = SigA(m) v sau m ho c x v y s dng

kho cng khai eB ca B, kt qu nhn c l z = eB(m, c). Bn mz s c gi

ti B, khi B nhn cz, u tin anh ta gii m vi hm gii m dB ca mnh



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nhn c (m, c). Sau anh ta dng hm xc minh cng khai ca A kim tra

xem VerA(m,c) = TRUEhay khng [1].

Song nu u tin A m ho m, ri sau mi k ln bn m nhn c th sao?

Khi , A s tnh:

c = SigA(eB(m))

A s truyn c p (z, c) ti B, B s gii m z v nhn c m, sau xc minh ch

k c trn m nh dng VerA. Mt vn ny sinh nu A truyn (m, c) kiu ny th

mt ngi th ba C c th thay ch kc ca A bng ch k ca chnh mnh.

c = SigC(eB(m))

Ch rng, C c th k ln bn m eB(m) ngay c khi anh ta khng bit bn r m.

Khi nu C truyn (z, c) n B, ch k ca C c B xc minh bng VerC v do

, B cho rng bn r x xut pht t C. Do kh khn ny, hu ht ngi s dng

c khuyn ngh k trc khi m [1 tr103].

- Tnh bo mt.

Bi ton bo mt ca h ch k RSA l trnh trng hp ngi ngoi c th tnhra gi tr d b mt (gi tr k hay m ha) khi bit c gi tr xc nhn e (cng

khai).

an ton ca h thng k RSA da trn 2 vn ca ton hc: Bi ton phn

tch ra tha s nguyn t cc s nguyn ln v bi ton RSA. Nu 2 bi ton trn l

kh (khng tm c thut ton hiu qu gii chng) th khng th thc hin

c vic ph m ton b i vi RSA.

Bi ton RSA l bi ton tnh cn bc e mun n (vi n l hp s): Tm s m sao

cho me=c mod n, trong (e, n) chnh l kha cng khai v c l bn m. Hin nay

phng php trin vng nht gii bi ton ny l phn tch n ra tha s nguyn t.

Khi thc hin c iu ny, k tn cng s tm ra s m b mt d t kha cng



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khai v c th gii m theo ng quy trnh ca thut ton. Nu k tn cng tm c

2 s nguyn tp v q sao cho: n =pq th c th d dng tm c gi tr (p-1)(q-1)

v qua xc nh dt e. Cha c mt phng php no c tm ra trn my tnh

gii bi ton ny trong thi gian a thc (polynomial-time). Tuy nhin ngi tacng cha chng minh c iu ngc li (s khng tn ti ca thut ton).

Ti thi im nm 2005, s ln nht c th c phn tch ra tha s nguyn t c

di 663bitvi phng php phn tn trong khi kha ca RSA c di t 1024

ti 2048 bit. Mt s chuyn gia cho rng kha 1024 bitc th sm b ph v (cng

c nhiu ngi phn i vic ny). Vi kha 4096 bit th hu nh khng c kh

nng b ph v trong tng lai gn. Do , ngi ta thng cho rng RSA m bo

an ton vi iu kin n c chn ln. Nu n c di 256 bithoc ngn hn,

n c th b phn tch trong vi gi vi my tnh c nhn dng cc phn mm c

sn. Nu n c di 512 bit, n c th b phn tch bi vi trm my tnh ti thi

im nm 1999. Mt thit b l thuyt c tn l TWIRL do Shamirv Tromerm t

nm 2003 t ra cu hi v an ton ca kha 1024 bit. V vy hin nay ngi

ta khuyn co s dng kha c di ti thiu 2048 bit.

Nm 1993, Peter Shorcng b thut ton Shorch ra rng: My tnh lng t(trn l thuyt) c th gii bi ton phn tch ra tha s trong thi gian a thc. Tuy

nhin, my tnh lng t vn cha th pht trin c ti mc ny trong nhiu

nm na [2 tr117].

- Cc dng tn cng

* Phn phi kha

Cng ging nh cc thut ton m ha khc, cch thc phn phi kha cng khai

l mt trong nhng yu t quyt nh i vi an ton ca RSA. Qu trnh phn

phi kha cn chng li c tn cng ng gia (man-in-the-middle attack). Gi

s k xu (C) c th gi cho Ngi gi thng tin(A) mt kha bt k v khin (A)

tin rng l kha (cng khai) ca i tc(B). ng thi (C) c kh nng c c



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thng tin trao i gia (A) v (B). Khi , (C) s gi cho (A) kha cng khai ca

chnh mnh (m (A) ngh rng l kha ca (B)). Sau , (C) c tt c vn bn

m ha do (A) gi, gii m vi kha b mt ca mnh, gi 1 bn copy ng thi m

ha bng kha cng khai ca (B) v gi cho (B). V nguyn tc, c (A) v (B) ukhng pht hin ra s can thip ca ngi th ba. Cc phng php chng li dng

tn cng ny thng da trn cc chng thc kha cng khai (digital certificate)

hoc cc thnh phn ca h tng kha cng khai (public key infrastructure - PKI).

[6]

* Tn cng da trn thi gian

Vo nm 1995,Paul Kocherm t mt dng tn cng mi ln RSA: Nu k tn

cng nm thng tin v phn cng thc hin m ha v xc nh c thi gian

gii m i vi mt s bn m la chn th c th nhanh chng tm ra kha d. Dng

tn cng ny c th p dng i vi h thng ch k in t s dng RSA. Nm

2003, Dan Boneh v David Brumley chng minh mt dng tn cng thc t hn:

Phn tch tha s RSA dng mng my tnh (My ch web dngSSL). Tn cng

khai thc thng tin r r ca vic ti u ha nh l s d Trung quc m nhiu ng

dng thc hin. chng li tn cng da trn thi gian l m bo qu trnh gii m lun din

ra trong thi gian khng i bt k vn bn m. Tuy nhin, cch ny c th lm

gim hiu sut tnh ton. Thay vo , hu ht cc ng dng RSA s dng mt k

thut gi l che mt. K thut ny da trn tnh nhn ca RSA: thay v tnh cd mod

n, u tin chn mt s ngu nhin rv tnh (rec)d mod n. Kt qu ca php tnh ny

l rm mod n v tc ng ca r s c loi b bng cch nhn kt qu vi nghch

o ca r. i vi mi vn bn m, ngi ta chn mt gi tr ca r. V vy, thi

gian gii m s khng cn ph thuc vo gi tr ca vn bn m. [6]


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* Tn cng bng phng php la chn thch nghi bn m.

Nm 1981, Daniel Bleichenbacherm t dng tn cng la chn thch nghi bn

m (adaptive chosen ciphertext attack) u tin c th thc hin trn thc t i vi

mt vn bn m ha bng RSA. Vn bn ny c m ha da trn tiu chunPKCS #1 v1, mt tiu chun chuyn i bn r c kh nng kim tra tnh hp l

ca vn bn sau khi gii m.

Do nhng khim khuyt ca PKCS #1,Bleichenbacherc th thc hin mt tn

cng ln bn RSA dng cho giao thc SSL (tm c kha phin). Do pht hin

ny, cc m hnh chuyn i an ton hn nh chuyn i m ha bt i xng ti u

(Optimal Asymmetric Encryption Padding) c khuyn co s dng. ng thi

phng nghin cu ca RSA cng a ra phin bn mi ca PKCS #1 c kh nng

chng li dng tn cng ni trn. [6]

b. H ch k ElGammal

H ch k ElGammal c a ra vo 1985. Mt phin bn sa i h ny c

Hc vin Quc gia tiu chun v k thut (NIST) a ra nh mt chun ca ch k

in t [1 tr103].

H ch k ElGammal c thit k ring bit cho mc ch ch k, tri ngc

vi RSA thng c s dng cho c mc ch m ho cng khai v ch k. H

ch k ElGammal l khng xc nh, ngha l c rt nhiu gi tr ch k cho cng

mt bc in cho trc. Thut ton xc minh phi c kh nng nhn bt k gi tr

ch k no nh l vic xc thc. S ch kElGammal c miu t nh sau:

Cho p l mt s nguyn t nh l bi ton logarit ri rc trong Zp, Zp* l mt

phn t nguyn t vP = Zp*, A = (Zp*)*Zp-1, v nh ngha:

K = {(p, , a, ) : a (mod p)}

trong gi tr p, v l cng khai, cn a l b mt [1 tr103].



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ViK = (p, , a, ) v chn mt s ngu nhin k Zp-1*, nh ngha:

SigK(x, k) = (, )

trong : = k mod p

= (x - a* )k-1 mod (p 1).

Vix, Zp* v Zp-1, nh ngha:

Ver(x, , ) = TRUE x (mod p).

Nu ch k l ng th vic xc nhn thnh cng khi:

ak (mod p)

x (mod p).

trong : a + k x (mod p -1).

B s tnh ton ch k bng vic s dng c gi tr b mt a (mt phn ca kho) v

s b mt ngu nhin k (gi tr k bc in). Vic xc minh c th thc hin c

ch vi cc thng tin c cng khai:

V d:

Chng ta chn p = 467, = 2, a = 127. Ta tnh: = a mod p = 2127 mod 467 = 132.

By gi B mun k ln bc in x = 100 v anh ta chn mt gi tr ngu nhin k =

213 (ch l UCLN(213, 466) = 1 v 213-1 mod 466 = 431). Sau tnh:

= 2213 mod 467 = 29

= (100 127*29)431 mod 466 = 51.

Bt c ai cng c th kim tra ch k ny bng cch tnh:

132292951 189 (mod 467)

2100 189 (mod 467).



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Gi s k th ba C mun gi mo ch k ca B trn bc in x m khng bit s

b mt a. Nu C chn mt gi tr v c gng tm , anh ta phi tnh mt hm

logaritri rc logx-. Mt khc, nu u tin anh ta chn c gng tm th

anh ta phi tnh x

= x

(mod p). C hai vic ny u khng th thc hin c [1].

Tuy nhin c mt l thuyt m C c th k ln mt bc in ngu nhin bng cch

chn ng thi , v x. Cho i, j l s nguyn vi 0 i, j p - 2, v UCLN(j, p - 1)

= 1. Sau tnh:

= ij mod p

= - j-1 (mod p-1)

x = - ij-1 (mod p-1).

Nh vy, ta xem (, ) l gi tr ch k cho bc in x. Vic xc minh s thc hin

nh sau:

V d:

Nh v d trn, ta chn p = 467, = 2, = 132. K th ba C s chn i = 99 v j =

179. Anh ta s tnh:

= 299132179 mod 467 = 117

= -117*151 mod 466 = 41

x= 99*44 mod 466 = 331



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C p gi tr (117, 41) l gi tr ch k cho bc in 331. Vic xc minh c thc

hin nh sau:

13211711741 303 (mod 467)

2331 303 (mod 467).

Mt phng php th hai c th gi mo ch k l s dng li ch k ca bc

in trc , ngha l vi c p (, ) l gi tr ch k ca bc in x, n s c C

k cho nhiu bc in khc. Cho h, i v j l cc s nguyn, trong 0 i, j, h p-2

v UCLN(h-j, p-1) = 1.

= hij mod p

= (h - j)-1 mod (p-1)

x = (hx + i)(h - j)-1 mod (p-1).

Ta c th kim tra: = x mod p. V do , (, ) l c p gi tr ch k ca bc

inx.

iu th ba l vn sai lm ca ngi k khi s dng cng mt gi tr k trong

vic k hai bc in khc nhau. Cho (, 1) l ch k trn bc in x1 v (, 2) l

ch k trn bc in x2. Vic kim tra s thc hin:

1 x1 (mod p)

2 x2 (mod p).

Do : x1-x2 y1-y2 mod p.

t = k

, khi : x1 - x2 = k(1 - 2) (mod p-1).

By gi t d = UCLN(1 - 2, p - 1). V d | (1 - 2) v d | (p - 1) nn n cng

chia ht cho (x1 - x2). Ta t tip:

x = (x1-x2) /d.


43


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= (1- 2)/d.

p = (p-1)/d.

Cui cng, ta c:x k (mod p). V UCLN(, p) = 1 nn ta c:

= ()-1 mod p

Nh vy, gi tr k s c xc nh nh sau:

k = x (mod p) = x + ip (mod p)

Vi 0 i d-1, ta c th tm c gi tr k duy nht bng hm kim tra:

k mod p. [1 tr104 > tr106]

c. Chun ch k s (DSS)

Chun ch k in t (DSS) c sa i t h ch kElGammal. N c cng

b ti hi ngh Tiu chun x l thng tin Lin Bang (FIPS) vo 19/05/1994 v tr

thnh chun vo 01/12/1994. DSS s dng mt kho cng khai kim tra tnh

ton vn ca d liu nhn c v ng nht vi d liu ca ngi gi. DSS cng

c th s dng bi ngi th ba xc nh tnh xc thc ca ch k v d liu

trong n. u tin chng ta hy tm hiu ng c ca s thay i ny, sau s tm

hiu thut ton ca DSS [1 tr106].

Trong rt nhiu trng hp, mt bc in c th c m ho v gii m mt ln,

v vy n p ng cho vic s dng ca bt k h thng bo mt no c bit l an

ton lc bc in c m ho. Ni cch khc, mt bc in c k m nhim

chc nng nh mt vn bn hp php, chng hn nh cc bn hp ng, v vy n

cng ging nh vic cn thit xc minh ch k sau rt nhiu nm bc in c

k. iu ny rt quan trng cho vic phng nga v an ton ca ch k c

a ra bi mt h thng bo mt. V h ch k ElGammal khng m nhn c

iu ny, vic thc hin ny cn mt gi tr ln modulo p. Tt nhin p nn c t nht



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512 bit, v nhiu ngi cho rng di ca p nn l 1024 bit nhm chng li vic

gi mo trong tng lai [1 tr107].

Tuy nhin, ngay c mt thut ton modulo 512 bitdng k cng phi thc hin

vic tnh ton n 1024 bit. Cho ng dng tim nng ny, c rt nhiu card thng

minh c a ra, nhm thc hin mt ch k ngn hn nh mong mun. DSS

sa i h ch kElGammal cho ph hp theo cch ny mt cch kho lo, mi

160 bitbc in c k s dng mt ch k 320 bit, nhng vic tnh ton c

thc hin vi 512 bit modulo p. Cch ny c thc hin nh vic chia nh Zp*

thnh cc trng c kch thc 2160. Vic thay i ny s lm thay i gi tr :

= (x + )k-1

mod(p - 1).

iu ny cng lm cho gi tr kim tra cng thay i:

x (mod p). (1.1)

Nu UCLN(x + , p - 1) = 1 th s tn ti -1 mod (p - 1), do (1.1) s bin i

thnh:

x-1-1 (mod p). (1.2)

y chnh l s i mi ca DSS. Chng ta cho q l mt s nguyn t 160 bitsao

cho q | (p-1), v l mt s th q ca 1 mod p, th v cng l s th q ca 1 mod

p. Do , v c th c ti gin trong modulo p m khng nh hng g n

vic xc minh ch k. S thut ton nh sau:

Cho p l mt s nguyn t 512 bit trong trng logarit ri rc Zp; q l mt s

nguyn t 160 bitv q chia ht (p-1). Cho Zp*; P = Zp*, A = Zq*Zq , v nh

ngha:

K = {(p, q, , a, ) : a (mod p)}

trong gi trp, q, v l cng khai, cn a l b mt.



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ViK = (p, , a, ) v chn mt s ngu nhin k (1 k q-1), nh ngha:

sigK(x, k) = (, )

trong : = (k mod p) mod q

= (x + a*)k-1 mod q.

Vix Zp* v , Zq, vic xc minh c thc hin bng cch tnh:

e1 = x -1 mod q

e2 = -1 mod q

Ver(x, , ) = TRUEe1e2 ( mod p) mod q = .

Ch rng, vi DSS th # 0 (mod q) v gi tr: -1 mod q cn cho vic xc minh

ch k (iu ny cng tng t nh vic yu cu UCLN(, p-1) = 1 (1.1)

(1.2)). Khi B tnh mt gi tr 0 (mod q) trong thut ton k, anh ta nn b n i

v chn mt s ngu nhin k mi.

V d:

Chng ta chn q = 101 v p = 78*q + 1 = 7879 v g = 3 l m t nguyn t trongZ7879. V vy , ta c th tnh:

= 378 mod 7879 = 170.

Chn a = 75, do : = a mod 7879 = 4567.

By gi, B mun k mt bc in x = 1234, anh ta chn mt s ngu nhin k = 50.

V vy :

k-1 mod 101 = 99.

Tip : = (17050 mod 7879) mod 101 = 2518 mod 101 = 94

= (1234 + 75*94)99 mod 101 = 97.



47/51


C p ch k (94, 97) cho bc in 1234 c xc thc nh sau:

-1 = 97-1 mod 101 = 25

e1 = 1234*25 mod 101 = 45

e2 = 94*25 mod 101 = 27

(17045456727 mod 7879) mod 101 = 2518 mod 101 = 94.

K t khi DSS c xut vo nm 1991, c nhiu ph bnh a ra. Chng

hn nh kch c ca moduloe p b c nh 512 bit, iu m nhiu ngi khng

mun. V vy, NIST thay i chun ny c th thay i kch thc moduloe

(chia bi 64) thnh mt dy t 512 n 1024 bit[1 tr107 > tr108].

4. Hm bm v kt hp hm bm vo ch k in t.

Kt hp hm bm vo ch k in t: T file cn gi ban u, chng tas s

dng hm bm bm (m ha) thnh chui k t c di c nh, vi hm bm

MD5 cho ta chui c di 128 bit, hm bm SHA cho ta chui c di 160 bit

gi l bn tm lc. Sau dng k s k ln bn tm lc tr thnh bn ch k

in t, tip theo gi cho ngi nhn hai file l file cn gi v bn ch k in t.Khi ngi nhn nhn c s thc hin Xc nhn ch k xc nhn ngi gi

ng thi dng hm bm - bm file gi km sau so snh vi bn tm lc xc

nh thng tin c s thay i hay khng.

IV. Ci t minh ha s k s RSA kt hp bm SHA.

+ Cc bc thc hin ca chng trnh.

a. Pht sinh kha:

T cp s nguyn t bt k ban u, chng trnh s thc hin tnh ton a ra

cp kha cng khai (e, n) v kha b mt (d, n). Sau kha cng khai c tit l

ra cng cng, kha b mt c gi li.


47


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b. K ch k in t:

....................................................................

Phn ny c ct b, hy lin h ch ti nhn c bn

chi tit hn.

.............................................................

- Mt s hm s dng trong chng trnh.

// Kt s nguyn t

public Boolean ktnt(long k)

{

if (k < 2) return false;

for (int i = 2; i = b) max = a;

else max = b;

return max;

}

// Tm c chung ln nht ca 2 s

public long ucln(long a, long b)

{

int ucln, r;

while (b != 0)

{

r = a % b;

a = b;


48


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b = r;

}

ucln = a;

return ucln;

}

//Tnh nghch o ca a trong Zb

public long nd(long a, long b)

{

long kq, i = 1;

while ((((i * b) + 1) % a) != 0)

{

i++;

}

kq = ((i * b) + 1) / a;

return kq;

}

//Tinh x^y mod N (tnh theo dng s d)

public static long tinh1(long x, long y, long n)

{

long kq;

kq = x % n;

for (long f = 1; f < y; f++){

kq = (kq * x ) % n;

}

return kq;

}

- Giao din ca chng trnh:

K:


49


50/51


Xc nhn ch k:

Phn 3. Kt lun:

+ Nhng phn lm c.

- tm hiu, nghin cu c s l lun v an ton, chng thc thng tin.

- tm hiu v ch k in t.

- tm hiu v cc phng thc m ha d liu c bn, tm hiu v hm bm.

- tm hiu v phng thc m ha bt i xng s dng cho ch k in t.


50


51/51


- tm hiu v cc h ch k in t c bn v h ch k RSA ng dng cho

ch k in t.

- ci t thnh cng chng trnh minh ha k s.

+ Nhng phn cha lm c

- Do nng lc v thi gian c hn nn trong qu trnh nghin cu cn nhiu phn

cn cha thc hin c nh chng trnh cha thc hin c vi s nguyn t

ln, hm bm s dng trong chng trnh cha c ci t bng thut ton.

- Chng trnh ci t cha c tnh ng dng vo thc t.

+ Hng pht trin ca ti.

- C th nghin cu su hn cc vn a ra v thc hin hon thin cc chc

nng ca chng trnh ng dng c vo trong i sng, phc v nhu cu v

mc ch ca ngi s dng.

+ Ti liu tham kho

+ Ph lc

Demo chng trnh: http://www.mediafire.com/view/?fwtd5cdp500u5xj

- Cch lin h ly bi hon chnh:

ly bi hon chnh. C th ly thm phn code (nguyn code +

phn ci t), xin hy lin h mail or s t trn lin h ly bi.

Ph: bi kha lun 50.k, Code: 100.k

Lin h: mail: [email protected] or 0982.070.520 (c th
http://www.mediafire.com/view/?fwtd5cdp500u5xjmailto:[email protected]://www.mediafire.com/view/?fwtd5cdp500u5xjmailto:[email protected]

chu ky dien tu rsa va bam sha

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