1 Mc lc 1. C S TON HC.............................................................................................. 3 1.1. L thuyt s......................................................................................................3 1.1.1. Khi nim ng d Modulo......................................................................3 1.1.2. nh l v ng d thc............................................................................3 1.1.3. Khi nim phn t nghch o ..................................................................4 1.1.4. Thut ton Euclide ....................................................................................4 1.1.5. Phn t nguyn thy v logarith ri rc....................................................4 1.1.6. Thng d bc hai v k hiu Legendre .....................................................5 1.1.7. Mt s thut ton kim tra tnh nguyn t ................................................6 1.2. L thuyt v phc tp tnh ton...................................................................6 1.2.1. phc tp tnh ton................................................................................6 1.2.2. Cc lp phc tp .......................................................................................7 1.3. Hm mt pha v hm ca sp mt pha..........................................................8 2. GII THIU V M HA................................................................................. 9 2.1. Cc thut ng ...................................................................................................9 2.2. nh ngha h mt m. .....................................................................................9 2.3. Nhng yu cu i vi h mt m .................................................................10 2.4. Cc phng php m ho...............................................................................10 2.4.1. M ho i xng kho b mt .................................................................10 2.4.1.1. Ni ng dng ...................................................................................11 2.4.1.2. Cc vn i vi phng php m ho i xng .........................11 2.4.2. M ho phi i xng kho cng khai......................................................12 2.4.2.1. Ni ng dng ...................................................................................12 2.4.2.2. iu kin h m ha kha cng khai ...............................................12 2.5. Cc h m ha n gin.................................................................................13 2.5.1. M dch vng...........................................................................................15 2.5.2. M thay th..............................................................................................17 2.5.3. M Apphin ..............................................................................................18 2.5.4. M Vigenre............................................................................................19 2.5.5. M HILL .................................................................................................21 2.5.6. M hon v...............................................................................................23 3. H M HA DES .............................................................................................. 24 3.1. M t DES......................................................................................................24 3.1.1. Thut ton DES.......................................................................................25 2 3.1.2. M t mt vng ca DES........................................................................25 3.1.3. M t hm f .............................................................................................25 3.1.4. M t chi tit cc hm trong DES...........................................................27 3.1.5. Tnh ton bng kha t kha K...............................................................30 3.2. V d...............................................................................................................34 3.3. Tranh lun v DES.........................................................................................37 3.4. DES trong thc t...........................................................................................38 3.5. ng dng ca DES.........................................................................................42 4. M HA KHA CNG KHAI ........................................................................ 43 4.1. Bi ton Logarit ri rc (DL) .........................................................................43 4.2. Cc thut ton cho bi ton Logarit ri rc....................................................43 4.3. H mt RSA...................................................................................................45 4.3.1. nh ngha h mt RSA..........................................................................46 4.3.2. an ton ca h RSA...........................................................................48 4.3.3. Mt s tnh cht ca h RSA ..................................................................48 4.3.4. ng dng ca RSA.................................................................................50 4.4. H mt Elgamal..............................................................................................50 4.5. CC PHNG PHP KIM TRA S NGUYN T LN ......................51 4.5.1. Kim tra Miller-Rabin.............................................................................51 Kim tra Miller-Rabin lp.................................................................................53 4.5.2. Kim tra Fermat ......................................................................................54 4.5.3. Kim tra Solovay-Strassen......................................................................55 5. CH K IN T............................................................................................. 57 5.1. nh ngha......................................................................................................57 5.2. Hm bm........................................................................................................58 5.3. Phn loi cc s ch k in t.................................................................60 5.3.1. S ch k km thng ip .................................................................61 5.3.2. S ch k khi phc thng ip ........................................................62 5.4. S ch k RSA..........................................................................................63 5.5. S ch k ELGAMAL...............................................................................64 5.5.1. nh ngha...............................................................................................65 5.5.2. an ton ca ch k Elgamal ..............................................................66 5.6. Chun ch k s DSS (Digital Signature Standard) ......................................69 5.6.1. Gii thiu.................................................................................................69 5.6.2. Cc gii thut c bn ca DSS................................................................70 5.6.3. Tnh cht ca ch k ca DSS................................................................72 3 5.6.4. La chn s k kh thi .......................................................................73 5.7. Tn cng ch k in t.................................................................................74 5.8. Kt lun ..........................................................................................................75 6. NG DNG......................................................................................................... 76 7. BI TP, CH THO LUN.................................................................... 77 8. TI LIU THAM KHO.................................................................................. 78 9. PH LC: M NGUN.................................................................................... 80 9.1. M ha dch chuyn .......................................................................................80 9.2. M ha thay th..............................................................................................83 9.3. M ha RSA...................................................................................................88 9.4. Ch k s Elgamal .........................................................................................93 L thuyt mt m v an ton d liu Trang 1 AN TON D LIU TRN MY TNH Ngynay,visphttrinmnhmcacngnghthngtinvicngdngcc cng ngh mng my tnh tr nn v cng ph cp v cn thit. Cng ngh mng my tnh mang li nhng li ch to ln. S xut hin mng Internet cho php mi ngi c th truy cp, chia s v khai thc thng tin mt cch d dng v hiu qu. Cc cng ngh E-mail cho php mi ngi c th gi th cho ngi khc cng nh nhn th ngay trn my tnh ca mnh. Gn y c cng ngh E-business cho php thc hin cc hot ng thng mi trn mng my tnh. Vic ng dng ccmng cc b trong cc t chc, cngty haytrongmt quc gialrtphongph.Cchthngchuyntincaccngnhnghngngycth chuyn hng t la qua h thng ca mnh. Cc thng tin v kinh t, chnh tr, khoa hc x hi c trao i rng ri.Tuynhinlinysinhvnvantonthngtin.cnglmtqutrnhtin trin hp logic: khi nhng vui thch ban u v mt siu xa l thng tin, bn nht nh nhn thy rng khng ch cho php bn truy nhp vo nhiu ni trn th gii, Internet cn cho php nhiu ngi khng mi m t gh thm my tnh ca bn. Thc vy, Internet c nhng k thut tuyt vi cho php mi ngi truy nhp, khai thc, chia s thng tin. Nhng n cng l nguy c chnh dn n thng tin ca bn b h hng hoc ph hu hon ton. Cnhngthngtinvcngquantrngmvicbmthayblmsailchcth nh hng n cc t chc, cc cng ty hay c mt quc gia. Cc thng tin v an ninh qucgia,bmtkinhdoanhhayccthngtintichnhlmctiucacctchc tnh bo nc ngoi v chnh tr hay cng nghip hoc k cpnichung. Bn chng c th lm mi vic c th c c nhng thng tin qu gi ny. Th tng tng nu c k xm nhp c vo h thng chuyn tin ca cc ngn hng th ngn hng s chu nhng thit hi to ln nh mt tin c th dn ti b ph sn. Cha k nu h thng thng tin an ninh quc gia b e do th hu qu khng th lng trc c.Theo s liuca CERT(Computer EmegencyResponseTeam - i cp cumy tnh),slngccvtncngtrnInternetcthngbochotchcnylt hn 200 vo nm 1989, khong 400 vo nm 1991, 1400 vo nm 1993, v 2241 vo nm 1994. Nhng v tn cng ny nhm vo tt c cc my tnh c mt trn Internet,cc my tnh ca tt c cc cng ty ln nh AT&T,IBM, cc trng i hc, cc c quan nh nc, cc t chc qun s, nh bng... Mt s v tn cng c quy m khngl (cti 100.000my tnhb tn cng). Hn na, nhng con s ny ch l phn nica tng bng. Mt phn rt ln cc v tn cng khng c thng bo, v nhiu l do,L thuyt mt m v an ton d liu Trang 2 trongcthknnilobmtuytn,hocnginnhngngiquntrh thng khng h hay bit nhng cuc tn cng nhm vo h thng ca h. Khng ch s lng cc cuc tn cng tng ln nhanh chng, m cc phng php tn cng cng lin tc c hon thin. iu mt phn do cc nhn vin qun tr hthng ckt ni vi Internetngy cng cao cnh gic. Cng theo CERT, nhng cuc tn cng thik 1988-1989 ch yu on tn ngi s dng-mt khu (UserID-password)hocsdngmtslicaccchngtrnhvhiuhnh(security hole) lm v hiuh thng bo v, tuy nhin cc cuc tn cngvo thi gian gn y baogm c cc thao tc nh gi mo a ch IP, theo di thng tintruyn qua mng, chim cc phin lm vic t xa (telnet hoc rlogin). vabomtnhbomtcathngtinlikhnglmgimsphttrinca victraoithngtinqungbtrntoncuthmtgiiphpttnhtlmho thng tin. C th hiu s lc m ho thng tin l che i thng tin ca mnh lm cho k tn cng nu chn c thng bo trn ng truyn th cng khng th c c v phi c mt giao thc gia ngi gi v ngi nhn c th trao i thng tin, l cc c ch m v gii m thng tin.Ngy nay th vicmho tr nn ph cp. Cc cng ty phnmmln trn th giiucnghincuvxydngcccngc,thuttonmhopdngcho thct.Miqucgiahaytchcucnhngcchmhoringbovh thng thng tin ca mnh. Mt s vn an ton i vi nhiu mng hin nay: Mtngidngchuynmtthngbointchomtngisdngkhc. Mt bn th ba trn cng mng LAN ny s dng mt thit b nghe trm gi ly thng bo v c cc thng tin trong . Cng trong tnh hung trn bn th ba chn thng bo, thay i cc thnh phn ca n v sau li gi cho ngi nhn. Ngi nhn khng h nghi ng g tr khi nhnra thng bo l v l, v c th thc hin vi hnhng da trn cc thnh phn sai ny em li li ch cho bn th ba. Ngi dng log vo mt server m khng s dng mt khu c m ho. Mt ngi khc ang nghe trm trn ng truyn v bt c mt khu logon ca ngi dng, sau c th truy nhp thng tin trn server nh ngi s dng. Mt ngi qun tr h thng khng hiu v kha cnh an ton v yu cu ca h thngvvtnhchophpngidngkhctruynhpvothmcchacc thng tin h thng. Ngi dng pht hin ra h c th c c cc thng tin h thng v c th dng n phc v cho li ch ca mnh.L thuyt mt m v an ton d liu Trang 3 1.C S TON HC Trong phn ny s trnh by v mt s c s ton hc ca m ha, iu ny s gip ta nm c mt cch chi tit hn v cc phng php m ha. 1.1.L thuyt s 1.1.1.Khi nim ng d Modulo nh ngha 1: Gi s a v b l cc s nguyn v m l mt s nguyn dng. Khi ta vita b(mod m) nu b-achiaht cho m. Mnh a b(modm) c gi l a ng d vi b theo moun m. Gischiaavbchomvtathucthngnguynvphnd,ccphnd nm gia 0 v m-1, ngha l a = q1*m + r1 v b = q2*m + r2 trong 0 r1 m-1 v 0 r2 m-1. Khi c th d dng thy rng a b(mod m) khi v ch khi r1 = r2 .Ta s dng k hiu a mod m xc nh phn d khi a c chia cho m (chnh l gi tr r1 trn). Nh vy: ab(mod m) khi v ch khi (a mod m) = (b mod m). Php rt gn, thay a bng a mod m th ta ni rng a c rt gn theo modulo m. Nhnxt:Nhiungnnglptrnhcamytnhxcnhamodmlphnd trong di-m+1,,m-1 c cng du vi a. V d -18 mod 7 s l 4, gi tr ny khc vi gi tr 3 l gi tr c xc nh theo cng thc trn. Tuy nhin, thun tin ta s xc nh a mod m lun l mt s khng m. By gi ta c th nh ngha s hc modulo m: Zm c coi l tp hp {0,1,,m-1} ctrangbhaiphptoncngvnhn.ViccngvnhntrongZm cthchin gingnhcngvnhnccsthcngoitrmtimlccktqucrtgn theo moun m. 1.1.2.nh l v ng d thc nh l 1: ng d thc ax b (mod m) ch c mt nghim duy nht x Zm vi mi b Zm khi v ch khi UCLN(a,m) = 1. Ta gi s rng, UCLN(a,m) = d >1. Khi , vi b = 0 th ng d thc ax 0 (mod m) s c t nht hai nghim phn bit trong Zm l x = 0 v x = m/d. L thuyt mt m v an ton d liu Trang 4 1.1.3.Khi nim phn t nghch o nh ngha 2: Gi s a Zm. Phn t nghch o (theo php nhn) ca a l phn t a-1 Zm sao cho aa-1 a-1a 1 (mod m). V d Z10, a=5, suy ra a-1=9. Bngcclluntngtnhtrn,cthchngtrng.acnghchotheo moun m khi v ch khi UCLN(a,m) = 1, v nu nghch o ny tn ti th n phi l duynht.Tacngthyrng,nub=a-1tha=b-1.Numlsnguyntthmi phn t khc khng ca Zm u c nghch o.1.1.4.Thut ton EuclideCho hai s t nhin a, n. K hiu (a,n) l c s chung ln nht ca a,n; (n) l s cc s nguyn dng< n v nguyn t vin, khngmt tnh tng qut gi s n > a. ThuttonEuclidetmUCLN(a,n)cthchinbngmtdyccphpchialin tip sau y: t r0 = n, r1 = a, r0 = q1r1 + r2 , 0 < r2 < r1 r1 = q2r2 + r3 , 0 < r3 < r2 rm-2 = qm-1rm-1 + rm , 0 < rm < rm-1 rm-1 = qmrm Thut ton phi kt thc mt bc th m no . Ta c: (n,a) = (r0,r1) = (r1,r2) = = (rm-1,rm) = rm Vytatmcrm =(n,a).MrngthuttonEuclidebngcchxcnhthm dy s t0, t1,,tm : t0 = 0, t1 = 1, tj = tj-2 qj-1tj-1 mod r0 , nu j 2 , ta d chng minh bng qui np rng:rj tjr1 (mod r0) Do , nu (n,a) = 1, th tm = a-1 mod n 1.1.5.Phn t nguyn thy v logarith ri rc Cho s n nguyn dng. Ta bit rng tp cc thng d thu gn theo moun n (tc l tp cc s nguyn dng < n v nguyn t vi n) lp thnh mt nhm vi php nhn L thuyt mt m v an ton d liu Trang 5 mod n, ta k hiu l Zn* . Nhm c cp (s phn t) l (n). Mt phn t g Zn* c cp m, nu m l s nguyn dng b nht sao cho gm = 1 trong Zn*. Theo mt nh l i s, ta c m |(n) (k hiu m l c s ca (n)) v vy vi mi b Zn* ta lun c:b(n) 1 (mod n) Nu p l s nguyn t, th do (p) = p-1, nn ta c vi mi b nguyn t vi p bp-1 1 (mod p) (1) Nu b c cp p-1, th p-1 l s m b nht sao cho c cng thc (1), do cc phn t b, b2,, bp-1 u khc nhau, v lp thnh Zp*. Ni cch khc, b l mt phn t sinh, hay nh thng gi l phn t nguyn thy ca Zp* ; v khi Zp* l mt nhm cyclic. Trong l thuyt s, ngi ta chng minh c cc nh l sau y: Vi mi s nguyn t p, Zp* l nhm cyclic, v s cc phn t nguyn thy ca Zp* bng (p-1) Nu g l phn t nguyn thy theo moun p, th = gi, vi mi i m (i,p-1) = 1, cng l phn t nguyn thy theo moun p 1.1.6.Thng d bc hai v k hiu Legendre Cho p l mt s nguyn t l, v x l mt s nguyn dng p-1. x c gi l mt thng d bc hai theo moun p, nu phng trnh: y2 x (mod p) c nghim. Ta c tiu chun Euler sau y: x l thng d bc hai theo moun p, nu v ch nu x(p-1)/2 1 (mod p) Tiu chun c chng minh nh sau: Gi s c x y2 (mod p). Khi c: x(p-1)/2 (y2)(p-1)/2 yp-1 1 (mod p) ; Ngc li, gi s rng x(p-1)/2 1 (mod p). Ly b l mt phn t nguyn thy (mod p), ta c x bi (mod p) vi s i no .Ta c: x(p-1)/2 (bi)(p-1)/2 (mod p) bi(p-1)/2 (mod p) V b c cp p-1, do p-1 phi l c s ca i(p-1)/2, suy ra i phi l s chn, v cn bc hai ca x l bi/2.Gisplsnguyntl.Vimia0,tanhnghakhiu |||

\|paLegendrenh sau: asad|||

\|pa = 0 nu a 0 (mod p) 1 nu a l thng d bc hai theo mod p-1 nu a khng l thng d bc hai theo mod pL thuyt mt m v an ton d liu Trang 6 Ta c tnh cht quan trng sau y: nu p l s nguyn t l th vi mi s nguyn a 0, ta c: a(p-1)/2 (mod p). 1.1.7.Mt s thut ton kim tra tnh nguyn t Taphtbiumtstnhchtsauy,chnglcschovicphttrinmts thut ton xc sut th tnh nguyn t ca cc s nguyn.Solovay_Strassen :Nu n l s nguyn t, th vi mi 1 a n-1: ||

\|na a(n-1)/2 (mod n). Nu n l hp s th: |{a: 1 a n-1,||

\|na a(n-1)/2 (mod n)}| (n-1)/2 Solovay_Strassen (ci tin bi Lehmann): Nu n l s nguyn t, th vi mi 1 a n-1: a(n-1)/2 1 (mod n); Nu n l hp s th: |{a: 1 a n-1, a(n-1)/2 1(mod n)}| (n-1)/2 1.2.L thuyt v phc tp tnh ton 1.2.1. phc tp tnh ton L thuyt thut ton v cc hm tnh ra i t nhng nm 30 t nn mng cho cc nghin cu v cc vn tnh c, gii c, v thu c nhiu kt qu rt quan trng. Nhng t ci tnh c mt cch tru tng, tim nng n vic tnh c trong thc t ca khoa hc tnh ton bng my tnh in t l mt khong cch rt ln. L thuyt v phc tp tnh ton c nghin cu bt u t nhng nm 60 b p cho khongtrng , cho ta nhiutri thc c bn, ng thi c nhiu ng dng thc t rt phong ph.

\|pa L thuyt mt m v an ton d liu Trang 7 phc tp (v khng gian hay thi gian) ca mt qu trnh tnh ton l s nh hay s cc php ton c thc hin trong qu trnh tnh ton . phctptnhtoncamtthuttonchiulmthmsf,saochovi mi n, f(n) l l s nh hay s cc php ton ti a m thut ton thc hin qu trnh tnh ton ca mnh trn cc d liu vo c ln n. phc tp tnh ton ca mt bi ton (ca mt hm) c nh ngha l phc tp ca mt thut ton tt nht c th tm c gii bi ton (hay tnh hm) . Mt bi ton c cho bi: Mt tp cc d liu vo YMt cu hi dng R(I)? vi I Y, li gii bi ton l ng hay khng V d: Bi ton ng d bc hai oD liu: Cc s nguyn dng a,b,c oCu hi: C hay khng s x < c sao cho x2 a mod b ? Bi ton hp s oD liu: S nguyn dng N oCu hi: C hay khng hai s m,n > 1 sao cho N = mn ? 1.2.2.Cc lp phc tp Ta nh ngha P l lp cc bi ton c phc tp thi gian l a thc tc lp cc bi ton m i vi chng c thut ton gii bi ton trong thi gian a thc. MtlpquantrngccbitoncnghincunhiullpNP,tcccbi tonmivichngcthuttonkhngnnhgiitrongthigianathc. Thut ton khng n nh l mt m hnh tnh ton tru tng, c gi nh l sau mi bc c th c mt s hu hn bc c la chn ng thi tip sau. Nhiu bi ton c chng t l thuc lp NP, nhng cha ai chng minh c l chng thuc lp P hay khng. V mt vn cho n nay vn cn m, cha c li gii l: NP = P ? Mt cch trc gic, lp NP bao gm cc bi ton kh hn phc tp hn cc bi ton thuc lp P, nhng iu c v hin nhin trc gic vn cha c chng minh hay bc b. L thuyt mt m v an ton d liu Trang 8 Gi s NP P, th trong NP c mt lp con cc bi ton c gi l NP_y , l nhng bi ton m bn thn thuc lp NP, v mi bi ton bt k thuc lp NP u c th qui dn v bi ton bng mt hm tnh c trong thi gian a thc. Cho n nay, ngi ta chng minh c hng trm bi ton thuc nhiu lnh vc khc nhau l NP_y . Bi ton ng d bc hai k trn l NP_y , bi ton hp s khng l NP_y , nhng cha tm c mt thut ton lm vic trong thi gian a thc gii n. 1.3.Hm mt pha v hm ca sp mt pha Hm f(x) c gi l hm mt pha, nu tnh y = f(x) l d, nhng vic tnh ngcx=f-1(y) l rt kh. C th hiu d l tnh c trong thi gian a thc (vi a thc bc thp), v kh l khng tnh c trong thi gian a thc. Vd:Hmf(x)=gx(modp)(plsnguynt,glphntnguynthy theomounp)lhmmtpha.Vbitxtnhf(x)lkhngin,nhngbit f(x) tnh x th vi cc thut ton bit hin nay i hi mt khi lng tnh ton c O(exp(lnp lnlnp)112) php tnh (nu p l s nguyn t c 200 ch s thp phn, th khi lng tnh ton trn i hi mt my tnh 1 t php tnh/giy lm vic khng ngh trong khong 3000 nm) Hm f(x) c gi l hm ca sp mt pha, nu tnh y = f(x) l d, tnh x = f-1(y) l rt kh, nhng c ca sp z tnh x = fz-1(y) l d Vd:Chon=pqltchcahaisnguyntln,alsnguyn,hm f(x)=xa(mod n) l hm ca sp mt pha, nu ch bit n v a th tnh x = f-1(y) l rtkh,nhngnubitcasp,chnghnhaithascan,thstnhcf-1(y) kh d. Trn y l hai th d in hnh, v cng l hai trng hp c s dng rng ri v hm mt pha v hm ca sp mt pha. V y l nhng im then cht ca l thuyt mt m kha cng khai, nn vic tm kim cc loi hm mt pha v ca sp mt pha c nghin cu rt khn trng, vn naytuy c t cmt s kt qu, nhng vic tm kim vn tip tc, y hng th nhng cng y kh khn. L thuyt mt m v an ton d liu Trang 9 2.GII THIU V M HA 2.1.Cc thut ng 1.Hmt mltphpccthuttonvccthtckthpcheduthngtin cng nh lm r n. 2.Mt m hc nghin cu mt m bi cc nh mt m hc, ngi vit mt m v cc nh phn tch m. 3.Mholqutrnhchuynthngtincthcgilbnrthnhthngtin khng th c gi l bn m. 4.Gii m l qu trnh chuyn ngc li thng tin c m ho thnh bn r. 5.Thut ton m ho l cc th tc tnh ton s dng che du v lm r thng tin. Thut ton cng phc tp th bn m cng an ton. 6.Mtkholmtgitrlmchothuttonmhochytheocchringbitv sinh ra bn r ring bit tu theo kho. Kho cng ln th bn m kt qu cng an ton. Kch thc ca kho c o bng bit. Phm vi cc gi tr c th c ca kho c gi l khng gian kho. 7.Phn tch m l qu trnh hay ngh thut phn tch h mt m hoc kim tra tnh ton vn ca n hoc ph n v nhng l do b mt. 8.Mt k tn cng l mt ngi (hay h thng) thc hin phn tch m lm hi h thng.Nhngktncnglnhngkthcmivochuynngikhc,cctay hacker,nhngknghetrmhaynhngcctnngngkhc,vhlmnhng vic thng gi l cracking 2.2.nh ngha h mt m. H mt m: l mt h bao gm 5 thnh phn (P, C, K, E, D) tho mn cc tnh cht sau P ( Plaintext ) l tp hp hu hn cc bn r c th. C ( Ciphertext ) l tp hp hu hn cc bn m c th. K ( Key ) l tp hp cc bn kho c th. E ( Encrytion ) l tp hp cc qui tc m ho c th. D ( Decrytion ) l tp hp cc qui tc gii m c th. Chng ta bit mt thng bo thng c t chc di dng bn r. Ngi gi s lmnhim v m ho bn r, kt qu thu c gi l bn m. Bn m ny c gi i trn mt ng truyn ti ngi nhn sau khi nhn c bn m ngi nhn gii m L thuyt mt m v an ton d liu Trang 10 n tm hiu ni dung. D dng thy c cng vic trn khi s dng nh ngha h mt m : EK( P) = C v DK( C ) = P 2.3.Nhng yu cu i vi h mt m Cungcpmtmccaovtincy,tnhtonvn,skhngtchivsxc thc. tincy:cungcpsbmtchoccthngbovdliuclubngvic che du thng tin s dng cc k thut m ha. Tnhtonvn:cungcpsbomvittcccbnrngthngbocnli khng thay i t khi to ra cho n khi ngi nhn m n. Tnh khng t chi: c th cung cp mt cch xc nhn rng ti liu n t ai ngay c khi h c gng t chi n. Tnhxcthc:cungcphaidchv:utinlnhndngngungccamt thngbovcungcpmtvisbomrngnlngsthc.Thhail kimtractnhcangianglogonmththngvsautiptckimtra ctnhcahtrongtrnghpaicgngtnhinktnivgidngl ngi s dng 2.4.Cc phng php m ho2.4.1.M ho i xng kho b mt Thut ton i xng hay cn gi thut ton m ho c in l thut ton m ti khomhocthtnhtonractkhogiim.Trongrtnhiutrnghp, kho m ho v kho gii m l ging nhau. Thut ton ny cn c nhiu tn gi khc nh thut ton kho b mt, thut ton kho n gin, thut ton mt kho. Thut ton ny yu cu ngi gi v ngi nhn phi tho thunmt khotrc khi thng bo c gi i, v kho ny phi c ct gi b mt. an ton ca thut ton ny vn ph thuc vo kho, nu l ra kho ny ngha l bt k ngi no cng c th m ho v gii m thng bo trong h thng m ho. S m ho v gii m ca thut ton i xng biu th bi : EK( P ) = C v DK( C ) = P L thuyt mt m v an ton d liu Trang 11 Trong hnh v trn th : K1c th trng K2, hocK1 c th tnh ton t K2, hoc K2 c th tnh ton t K1. 2.4.1.1.Ni ng dng Sdngtrongmitrngmkhonddngcchuynnhltrongcng mt vn phng. Cng dng m ho thng tin lu tr trn a. 2.4.1.2.Cc vn i vi phng php m ho i xng Ccphngmhocinihingimhovngigiimphicng chung mt kho. Khi kho phi c gi b mt tuyt i, do vy ta d dng xc nh mt kho nu bit kho kia. H m ho i xng khng bo v c s an ton nu c xc sut cao kho ngi gi b l. Trong h kho phi c gi i trn knh an ton nu kch tn cng trn knh ny c th pht hin ra kho. Vn qun l v phn phi kho l kh khn v phc tp khi s dng h m ho c in. Ngi gi v ngi nhn lun lun thng nht vi nhau v vn kho. Vic thay i kho l rt kh v d b l. Khuynh hng cung cp kho di m n phi c thay i thng xuyn cho mi ngi trong khi vn duy tr c tnh an ton ln hiu qu chi ph s cn tr rt nhiu ti vic pht trin h mt m c in.

Bn rM hoGii mBn r Bn m Kho Hnh 2.1. M ho vi kho m v kho gii ging nhau K1 K2 L thuyt mt m v an ton d liu Trang 12 2.4.2.M ho phi i xng kho cng khai Vo nhng nm 1970 Diffie v Hellman pht minh ra mt h m ho mi c gi l h m ho cng khai hay h m ho phi i xng. Thut ton m ho cng khai khc bit so vi thut ton i xng. Chng c thit k sao cho kho s dng vo vic m ho l khc so vi kho gii m. Hn na kho gii m khng th tnh ton c t kho m ho. Chng c gi vi tn h thng m hocngkhaibivkhomhocthcngkhai,mtngibtkcths dng kho cng khai m ho thng bo, nhng ch mt vi ngi c ng kho gii m th mi c kh nng gii m. Trong nhiu h thng, kho m ho gi l kho cng khai (public key), kho gii m thng c gi l kho ring (private key). Trong hnh v trn th : K1 khng th trng K2, hocK2 khng th tnh ton t K1. c trng ni bt ca h m ho cng khai l c kho cng khai (public key) v bn tin m ho (ciphertext) u c th gi i trn mt knh thng tin khng an ton. 2.4.2.1.Ni ng dngS dng chyu trn cc mng cng khai nh Internet khi m kho chuyn tng i kh khn.2.4.2.2.iu kin h m ha kha cng khai Diffie v Hellman xc inh r cc iu kin camt hm ho cng khai nh sau: Bn rM hoGii mBn r Bn m Kho gii m k2 Hnh 2.2. M ho vi kho m v kho gii khc nhau Kho m ha k1 L thuyt mt m v an ton d liu Trang 13 1.Vic tnh ton ra cp kho cng khai KA v b mt KB da trn c s cc iu kinbanuphicthchinmtcchddng,nghalthchintrong thi gian a thc. 2.Ngi gi A c c kho cng khai ca ngi nhn B v c bn tin P cn gi i th c th d dng to ra c bn m C. C = EKA (P) Cng vic ny cng trong thi gian a thc. 3.NginhnBkhinhncbntinmhaCvikhobmtkBthcth gii m bn tin trong thi gian a thc. P = DkB (C) = DKB[EKA(P)] 4.NukchbitkhocngkhaiKAcgngtnhtonkhobmtthkhi chngphinguvitrnghpnangii,trnghpnyihinhiu yu cu khng kh thi v thi gian. 5.Nu k ch bit c cp (KA,C) v c gng tnh ton ra bn r P th gii quyt bi ton kh vi s php th l v cng ln, do khng kh thi.2.5.Cc h m ha n gin i tng c bn ca mt m l to ra kh nng lin lc trn mt knh khng mt chohaingisdng(tmgilAlicevBob)saochoiphng(Oscar)khng th hiu c thng tin c truyn i. Knh ny c th l mt ng dy in thoi hoc mt mng my tnh. Thng tin m Alice mun gi cho Bob (bn r) c th l mt vn bn ting Anh, cc d liu bng s hoc bt c ti liu no c cu trc tu . Alice smhobnrbngmtkhacxcnhtrcvgibnmktqutrn knh. Oscarc bn m thu trm c trn knh song khng th xc nh ni dung ca bn r, nhng Bob (ngi bit kho m) c th gii m v thu c bn r. Ta s m t hnh thc ho ni dung bng cch dung khi nim ton hc nh sau: nh ngha: Mt h mt l mt b 5 (P,C,K,E,D) tho mn cc iu kin sau: P l mt tp hu hn cc bn r c th. C l mt tp hu hn cc bn m c th. K (khng gian kho) l tp hu hn cc kho c th. i vi mi k K c mt quy tc m ek: P C v mt quy tc gii m tng ng dk D. Mi ek: P Cv dk: C P l nhng hm m: dk(ek (x)) = x vi mi bn r x P. L thuyt mt m v an ton d liu Trang 14 Trongtnhcht4ltnhchtchyunht.Nidungcanlnumtbnrx c m ho bng ek v bn m nhn c sau c gii m bng dk th ta phi thu c bn r ban u x. Alice v Bob s p dng th tc sau dng h mt kho ring. TrctinhchnmtkhongunhinkK.iunycthchinkhih cngmt ch v khng b Oscar theo di hoc khi h cmt knh mttrong trng hp h xa nhau. Sau gi s Alice mun gi mt thng bo cho Bob trn mt knh khng mt v ta xem thng bo ny l mt chui: x = x1,x2 ,. . .,xn

vi s nguyn n 1 no . y mi k hiu ca mi bn r xi P , 1 i n. Mi xi s c m ho bng quy tc m ek vi kho k xc nh trc . Bi vy Alice s tnh yi = ek(xi), 1 i n v chui bn m nhn c: y = y1, y2 ,. . ., yn

s c gi trn knh. Khi Bob nhn c y1,y2 ,. . .,yn anh ta s gii m bng hm gii m dk v thu c bn r gc x1,x2 ,. . .,xn. Hnh di l mt v d v mt knh lin lc Hnh 3.3. Knh lin lc R rng l trong trng hp ny hm m ho phi l hm n nh ( tc l nh x 1-1), nu khng vic gii m s khng thc hin c mt cch tng minh. V dy = ek(x1) = ek(x2) trong x1 x2 , th Bob s khng c cch no bit liu s phi gii m c x1 hay x2 . Ch rng nu P = C th mi hm m ho l mt php hon v, tc l nu tp cc bnmvtpccbnrlngnhtthmimthmmslmtsspxpli (hay hon v ) cc phn t ca tp ny. Oscar B gii m Bm ho Bob Alice Knh an ton Ngun kho L thuyt mt m v an ton d liu Trang 15 Do cc v d ca chng ta xt trn tp d liu l bng ch ci nn chng ta coi bng ch ci ting Anh l tp hp gm 26 gi tr nh sau. ABCDEFGHIJKLM 0123456789101112 NOPQRSTUVWXYZ 13141516171819202122232425 2.5.1.M dch vng M dch vng c xc nh trn Z26 (do c 26 ch ci trn bng ch ci ting Anh) mc d c th xc nh n trn Zm vi modulus m tu . D dng thy rng, MDV s to nn mt h mt nh xc nh trn, tc l dk (ek(x)) = x vi mi x Z26 . nh ngha: Mt h mt l mt b 5 (P,C,K,E,D)Gi s P = C = K = Z26 vi 0 k 25 , nh ngha: ek(x) = x +k mod 26 v dk(y) = y + (-k) mod 26(x,y Z26) -k l phn t i vi k trong Z26, v d phn t i ca 3 l 23, phn t i ca 15 l 11 xt trong Z26. Nhnxt:Trongtrnghpk=3,hmtthngcgilmCaesartng c Julius Caesar s dng. TassdngMDV(vimodulo26)mhomtvnbntingAnhthng thng bng cch thit lp s tng nggia cc k t v cc thng d theo modulo 26 nh sau: A 0,B 1, . . ., Z 25. V d 1: Gi s kho cho MDV l K = 11 v bn r l: wewillmeetatmidnight Trc tin bin i bn r thnh dy cc s nguyn nh dng php tng ng trn. Ta c: 2242281111124419 01912831386719 sau cng 11 vo mi gi tr ri rt gn tng theo modulo 26 71571922222315154 114231914241917184 Cui cng bin i dy s nguyn ny thnh cc k t thu c bn m sau: HPHTWWXPPELEXTOYTRSE gi m bn m ny, trc tin, Bob s bin i bn m thnh dy cc s nguyn ritrigitrcho11(rtgntheomodulo26)vcuicngbinilidyny thnh cc k t. L thuyt mt m v an ton d liu Trang 16 Nhn xt: Trong v d trn , ta dng cc ch in hoa cho bn m, cc ch thng cho bn r tin phn bit. Quy tc ny cn tip tc s dng sau ny. Nu mt h mt c th s dng c trong thc t th n pho tho mn mt s tnh cht nht nh. Ngay sau y s nu ra hai trong s : 1. Mi hm m ho eK v mi hm gii m dK phi c kh nng tnh ton c mt cch hiu qu. 2. i phng da trn xu bn m phi khng c kh nng xc nh kho K dng hoc khng c kh nng xc nh c xu bn r x. Tnh cht th hai xc nh (theo cch kh mp m) tng tng "bo mt". Qu trnh th tnh kho K (khi bit bn m y) c gi l m thm (sau ny khi nim ny s c lm chnh xc hn). Cn ch rng, nu Oscar c th xc nh c K th anh ta c thgiimcy nh Bobbngcch dng dK. Bivy, vic xc nh K ch t cng kh nh vic xc nh bn r x. Nhnxt:MDV(theomodulo26)lkhngantonvncthbthmtheo phng php vt cn. Do ch c 26 kho nn d dng th mi kho dK c th cho ti khi nhn c bn r c ngha. iu ny c minh ho theo v d sau: V du 2: Cho bn m JBCRCLQRWCRVNBJENBWRWN ta s th lin tip cc kho gii m d0 ,d1 .. . v y thu c: j b c r c l q r w c r v n b j e n b w r w n i a b q b k p q v b q u m a i d m a v q v m h z a p a j o p u a p t l z h c l z u p u l g y z o z i n o t z o s k y g b k y t o t k j x y n y h m n s y n r j e x f a j x s n s j e w x m x g l m r x m q i w e z i w r m r i d v w l w f k l q w l p h v o d y h v q l q h c u v k v e j k p v k o g u c x g u p k p g b t u j u d i j o u j n f t b w f o j o f a s t i t c h i n t i m e s a v e s n i n e Ti y ta xc nh c bn r v dng li. Kho tng ng K = 9. Trung bnh c th tnh c bn r sau khi th 26/2 = 13 quy tc gii m. Nh ch ra trong v d trn , iu kin mt h mt an ton l php tm kho vt cn phi khngththchinc;tckhnggiankhophirtln.Tuynhin,mtkhng gian kho ln vn cha m bo mt. L thuyt mt m v an ton d liu Trang 17 2.5.2.M thay th Mthmtnitingkhclhmthayth.Hmtnycsdnghng trmnm.Trchich"cryptogram"trongccbibolnhngvdvMTT. Trn thc t MTT c th ly c Pv Cu l b ch ci ting anh, gm 26 ch ci. Ta dng Z26 trong MDV v cc php m v gii m u l cc php ton i s. Tuy nhin, trong MTT, thch hp hn l xem php m v gii m nh cc hon v ca cc k t. nh ngha: Mt h mt l mt b 5 (P,C,K,E,D)Cho P =C = Z26 . Kcha mi hon v c th ca 26 k hiu 0,1, . . . ,25 Vi mi php hon v K , ta nh ngha: e(x) = (x) vd(y) = -1(y) trong -1 l hon v ngc ca . Sau y l mt v d v php hon v ngu nhin to nn mt hm m ho (cng nh trc, cc k hiu ca bn r c vit bng ch thng cn cc k hiu ca bn m l ch in hoa). abcdeFghijklm XNYAHPOGZQWBT nopqrStuvwxyz SFLRCVMUEKJDI Nh vy, e (a) = X, e (b) = N,. . . . Hm gii m l php hon v ngc. iu ny c thc hin bng cch vit hng th hai ln trc ri sp xp theo th t ch ci. Ta nhn c: ABCDEFGHIJKLM dlryvOhezxwpt NOPQRSTUVWXYZ b gfjqNmuskaci Bi vy d (A) = d, d(B) = 1, . . . Bi tp: gii m bn m sau bng cch dng hm gii m n gin: M G Z V Y Z L G H C M H J M Y X S S F M N H A H Y C D L M H A. M kho ca MTT l mt php hon v ca 26 k t. S cc hon v ny l 26!, ln hn410 26lmtsrtln.Bivy,phptmkhovtcnkhngththchin L thuyt mt m v an ton d liu Trang 18 c, thm ch bng my tnh. Tuy nhin, sau ny s thy rng MTT c th d dng b thm bng cc phng php khc. 2.5.3.M Apphin MDV l mt trng hp c bit ca MTT ch gm 26 trong s 26! cc hon v c th ca 26 phn t. Mt trng hp c bit khc ca MTT l m Affine c m t di y. trong m Affine, ta gii hn ch xt cc hm m c dng: e(x) = ax + b mod 26, a,bZ26.CchmnycgilcchmAffine(chrngkhia=1,tac MDV). vic gii m c th thc hin c, yu cu cn thit l hm Affine phi l n nh. Ni cch khc, vi bt k y Z26, ta mun c ng nht thc sau: ax + b y (mod 26) phi c nghim x duy nht. ng d thc ny tng ng vi: ax y+(-b) (mod 26) Lu : -b l phn t i ca b trong Z26. V y thay i trnZ26 nn y+(-b) cng thay i trn Z26 . Bi vy, ta ch cn nghin cu phng trnh ng d: ax y (mod 26) (y Z26 ). Ta bit rng, phng trnh ny c mt nghim duy nht i vi mi y khi v ch khi UCLN(a,26)=1(yhmUCLNlcchunglnnhtcaccbincan). Trc tin ta gi s rng, UCLN(a,26) = d >1. Khi , ng d thc ax 0 (mod 26) sctnhthainghimphnbittrongZ26lx=0vx=26/d.Trongtrnghp ny, e(x) = ax + b mod 26 khng phi l mt hm n nh v bi vy n khng th l hm m ho hp l. Gii thch theo mt cch khc nh sau:Php lp m c cho bi mt hm apphin dng:e(x) = ax + b mod 26 c c php gii m tng ng, tc l cho phng trnh sau c nghim:ax + b = c mod 26 c li gii i vi x (vi bt k c cho trc), theo mt nh l s hc, iu kin cn v l a nguyn t vi 26, tc l UCLN(a,26) = 1. Khi UCLN(a,26)=1 th c:a-1Z26 sao cho a.a-1=a-1.a=1 mod 26.v do nu y=ax+b mod 26 th x=a-1(y-b) mod 26 v ngc li L thuyt mt m v an ton d liu Trang 19 nh ngha: Mt h mt l mt b 5 (P,C,K,E,D) Cho P = C = Z26 v K = { (a,b) Z26 Z26 : UCLN(a,26) =1 } Vi k = (a,b) K, ta nh ngha: ek(x) = ax +b mod 26 vdk(y) = a-1(y-b) mod 26, x,y Z26 V d: Gi s k = (7,3). Nh nu trn, 7-1 mod 26 = 15. Hm m ho l eK(x) = (7x+3) mod 26 V hm gii m tng ng l: dK(x) = 15(y-3) mod 26 = (15y -19) mod 26=(15y +7) mod 26 7 l phn t i ca 19. y, tt c cc php ton u thc hin trn Z26. Ta s kim tra liu dK(eK(x)) = x vi mi xZ26khng? Dng cc tnh ton trn Z26 , ta c :dK(eK(x)) =dK(7x+3) =15(7x+3)-19 = x +45 - 19 = x. minh ho, ta hy m ho bn r "hot". Trc tin bin i cc ch h, o, t thnh cc thng du theo modulo 26. Ta c cc s tng ng l 7, 14 v 19. By gi s m ho: 7x7 +3 mod 26= 52 mod 26 = 0 7 x14 + 3 mod 26 = 101 mod 26 =23 7 x19 +3 mod 26 = 136 mod 26 = 6 Bi vy 3 k hiu ca bn m l 0, 23 v 6 tng ng vi xu k t AXG. Vic gii m s do bn thc hin nh mt bi tp. 2.5.4.M VigenreTrong c hai h MDV v MTT (mt khi kho c chn) mi k t s c nh x vo mt k t duy nht. V l do , cc h mt cn c gi h thay th n biu. By gi ta s trnh by mt h mt khng phi l b ch n, l h m Vigenre ni ting. Mt m ny ly tn ca Blaise de Vigenre sng vo th k XVI. S dng php tng ng A 0, B 1, . . . , Z 25 m t trn, ta c th gn cho mi kha K vi mt chui k t c di m c gi l t kho. Mt m Vigenre s m ho ng thi m k t: Mi phn t ca bn r tng ng vi m k t.nh ngha: Mt h mt l mt b 5 (P,C,K,E,D) Chomlmtsnguyndngcnhno.nhnghaP=C=K= (Z26)m . Vi kho K = (k1, k2, . . . ,km) ta xc nh : eK(x1, x2, . . . ,xm) = (x1+k1, x2+k2, . . . , xm+km) L thuyt mt m v an ton d liu Trang 20 v dK(y1, y2, . . . ,ym) = (y1-k1, y2-k2, . . . , ym-km) trong tt c cc php ton c thc hin trong Z26 V d: Gi s m =6 v t kho l CIPHER. T kho ny tng ng vi dy s K = (2,8,15,7,4,17). Gi s bn r l xu:thiscryptosystemisnotsecure Tasbiniccphntcabnrthnhccthngdtheomodulo26,vit chng thnh cc nhm 6 ri cng vi t kho theo modulo 26 nh sau: Bi vy, dy k t tng ng ca xu bn m s l: V P X Z G I A X I V W P U B T T M J P W I Z I T W Z T giimtacthdngcngtkhonhngthaychocng,tatrchontheo modulo 26. Ta thy rng cc t kho c th vi s di m trong mt m Vigenre l 26m, bi vy, thm ch vi cc gi tr m kh nh, phng php tm kim vt cn cng yu cu thi gian kh ln. V d, nu m = 5 th khng gian kho cng c kch thc ln hn 1,1107.Lngkhonylnngaenngavictmkhobngtay(ch khng phi dng my tnh). Trong h mt Vigenre c t kho di m,mi k t c th c nh x vo trong m k t c th c (gi s rng t kho cha m k t phn bit). Mt h mt nh vy c gi l h mt thay th a biu (polyalphabetic). Ni chung, vic thm m h thay th a biu s kh khn hn so vic thm m h n biu. 19 7818217241519141824 2815741728157417 21 152325680238212215 18 194128181314191842 2815741728157417 20 119191291522815819 20 174 2815 22 2519 L thuyt mt m v an ton d liu Trang 21 2.5.5.M HILL Trong phn ny s m t mt h mt thay th a biu khc c gi l mt m Hill. Mt m ny do Lester S.Hill a ra nm 1929. Gi s m l mt s nguyn dng, t P = C =(Z26)m . tng y l ly m t hp tuyn tnh ca m k t trong mt phn t ca bn r to ra m k t mt phn t ca bn m. V d nu m = 2 ta c th vit mt phn t ca bn r l x = (x1,x2) v mt phn t ca bn m l y = (y1,y2). y, y1cng nh y2 u l mt t hp tuyn tnh ca x1v x2. Chng hn, c th lyy1 = 11x1+ 3x2

y2 = 8x1+ 7x2

Tt nhin c th vit gn hn theo k hiu ma trn nh sau 1 2 1 211 8(y ) ( )3 7y x x| |= |\ Ni chung, c th ly mt ma trn K kch thc m m lm kho. Nu mt phn t hng i v ct j ca K l ki,,j th c th vitK = (ki,,j), vi x = (x1, x2, . . . ,xm) P v K K , ta tnh y = eK(x) = (y1, y2, . . . ,ym) nh sau: 1,1 1,11 1,1 ,...( , ..., y ) (,...,) ... ... ......m mm m mk ky x xk k| | |= | |\ Ni mt cch khc y = xK. Chng ta ni rng bn m nhn c t bn r nh php bin i tuyn tnh. Ta s xt xem phi thc hin gii m nh th no, tc l lm th no tnh x t y. Bn lm quen vi i s tuyn tnh sthy rngphi dngma trn nghch o K-1 gi m. Bn m c gii m bng cng thc y K-1 . nh ngha: Mt h mt l mt b 5 (P,C,K,E,D) Cho m l mt s nguyn dng c nh. Cho P = C = (Z26 )mv cho K= { cc ma trn kh nghch cp m m trn Z26} Vi mt kho K Kta xc nheK(x) = xK v dK(y) = yK -1

Tt c cc php ton c thc hin trong Z26 V d:Gi s c kha11 83 7K| |= |\ L thuyt mt m v an ton d liu Trang 22 T cc tnh ton trn ta c: 17 1823 11K| |= |\ Giscnmhobnr"July".Tachaiphntcabnrmho:(9,20) (ng vi Ju) v (11,24) (ng vi ly). Ta tnh nh sau: 11 8(9, 20) (99 60, 72 140) (3, 4)3 7| | = + + = |\ V11 8(11, 21) (121 72, 88 168) (11, 22)3 7| | = + + = |\ Bi vy bn m ca July l DELW. gii m Bob s tnh: 7 18(3, 4) (9, 20)23 11| | = |\ v 7 18(11, 22) (11, 24)23 11| | = |\ Nh vy Bob nhn c bn ng. Cho ti lc ny ta ch ra rng c th thc hin php gii m nu K c mt nghch o. Trn thc t, php gii m l c th thc hin c, iu kin cn l K phi c nghch o. ( iu ny d dng rt ra t i s tuyn tnh s cp, tuy nhin s khng chng minh y). Bi vy, chng ta ch quan tm ti cc ma trn K kh nghich. Tnh kh nghch ca mt ma trn vung ph thuc vo gi tr nh thc ca n. trnh s tng qut ho khng cn thit, ta ch gii hn trong trng hp 22. nh ngha : nh thc ca ma trn A = (a,i j ) cp 2 2 l gi trdet A = a1,1 a2,2 - a1,2 a2,1 Mtma trnthc K l c nghch o khi vch khi nhthc can khc 0. Tuy nhin, iu quan trng cn nh l ta ang lm vic trn Z26 . Kt qu tng ng l ma trn K c nghch o theo modulo 26 khi v ch khi UCLN(det K,26) = 1. nh l: Gi s A = (ai j) l mt ma trn cp 2 2 trn Z26 sao cho det A = a1,1a2,2 - a1,2 a2,1 c nghch o. Khi 2,2 1,21 12,1 1,1(det )a aA Aa a | |= |\ Tr li v d xt trn . Trc ht ta c: 11 83 7Det| | |\ =(11.7-8.3) mod 26 = 77 - 24 mod 26 = 53 mod 26 =1 V 1-1 mod 26 = 1 nn ma trn nghch o l 111 8 7 8 7 183 7 3 11 23 11 | | | | | |= = |||\ \ \ (do theo modulo 26) y chnh l ma trn c trn. L thuyt mt m v an ton d liu Trang 23 2.5.6.M hon v Tt c cc h mt tho lun trn t nhiu u xoay quanh php thay th: cc k t ca bn r c thay th bng cc k t khc trongbn m. tng ca MHV l gi cc k t ca bn r khng thay i nhng s thay i v tr ca chng bng cch sp xpliccktny.MHV(cncgilmchuynv)cdngthng trm nm nay. Tht ra th s phn bit gia MHV v MTT c Giovani Porta ch ra t 1563. nh ngha hnh thc cho MHV c nu ra bn di. Khng ging nh MTT, y khng c cc php ton i s no cn thc hin khi m ho v gii m nn thch hp hn c l dng cc k t m khng dng cc thng d theo modulo 26. Di y l mt v d minh ho nh ngha: Mt h mt l mt b 5 (P,C,K,E,D) Cho m l mt s nguyn dng xc nh no . Cho P = C = (Z26 )mv cho Kgm tt c cc hon v ca {1, . . ., m}. i vi mt kho ( tc l mt hon v) ta xc nh e(x1, . . . , xm ) = (x(1), . . . , x(m))v d(x1, . . . , xm ) = (y-1(1), . . . , y-1(m)) trong -1 l hon v ngc ca V d: Gi s m = 6 v kho l php hon v ( ) sau: Khi php hon v ngc -1 s tng ng nh trn: By gi gi s c bn r Shesellsseashellsbytheseashore Trc tin ta nhm bn r thnh cc nhm 6 k t: shesel | lsseas | hellsb | ythese | ashore By gi mi nhm 6 ch ci c sp xp li theo php hon v , ta c: EESLSH | SALSES | LSHBLE | HSYEET | HRAEOS Nh vy bn m l EESLSH SALSES LSHBLE HSYEET HRAEOS Nh vy bn m c m theo cch tng t bng php hon v o -1. 123456 351642 123456 361524 Hon v Hon v -1 L thuyt mt m v an ton d liu Trang 24 Thc t m hon v l trng hp c bit ca mt m Hill. Khi cho php hon v ca tp {1, . . . ,m}, ta c th xc nh mt ma trn hon v m m thch hpK = { ki,j} theo cng thc:,1 neu j= (i)0 neu nguoc laii jk = (ma trn hon v l ma trn trong mi hng v mi ct ch c mt s "1", cn tt cccgitrkhculs"0".Tacththucmtmatrnhonvtmatrn n vbng cch hon v cc hng hoc ct). Ddngthyrng,phpmHilldngmatrnKtrnthcttngngvi php m hon v dng hon v . Hn na K-1= K -1 tc ma trn nghch o ca K l ma trn hon vxc nh theo hon v -1. Nh vy, php giim Hill tng ng vi php gii m hon v. i vi hon v c dung trong v d trn, cc ma trn hon v kt hp l: Bn c th kim tra thy rng, tch ca hai ma trn ny l mt ma trn n v. 3.H M HA DES Ngy 15.5.1973. U ban tiu chun quc gia M cng b mt khuyn ngh cho cc h mt trong H s qun l lin bang. iu ny cui cng dn n s pht trin ca Chun m d liu (DES) v n tr thnh mt h mt c s dng rng ri nht trnthgii.DEScIBMphttrinvcxemnhmtcibincuhmt LUCIPHER.LnutinDESccngbtrongHsLinbangvongy 17.3.1975.Saunhiucuctrnhluncngkhai,DEScchpnhnchnlm chun cho cc ng dng khng c coi l mt vo 5.1.1977. K t c 5 nm mt ln, DES li c U ban Tiu chun Quc gia xem xt li. Ln i mi gn y nht ca DES l vo thng 1.1994 v tip ti s l 1998. Ngi ta on rng DES s khng cn l chun sau 1998. 3.1.M t DES K =0 0 1 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 00010 v K-1 = 0 0 1 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 01000 L thuyt mt m v an ton d liu Trang 25 M t y ca DES c nu trong Cng b s 46 v cc chun x l thng tin Lin bang (M) vo 15.1.1977. DES m ho mt xu bt x ca bn r di 64 bng mt kho 54 bt. Bn m nhn c cng l mt xu bt c di 48. Trc ht ta m t mc cao ca h thng. 3.1.1.Thut ton DES 1.Vibnrchotrcxvidi64bit,mtxubtx0scxydngbng cchhonvccbtcaxtheophphonvcnhbanuIP.Tavit:x0= IP(X)= L0R0, trong L0 gm 32 bt u v R0 l 32 bt cui. 2.Sautnhton16lnlptheomthmxcnh.TastnhLiRi,1i16theo quy tc sau: Li = Ri-1 Ri = Li-1f(Ri-1,Ki) Trong k hiu php hoc loi tr ca hai xu bt (cng theo modulo 2). f l mt hmm ta smt sau, cn K1,K2, . . . ,K16 l cc xu bt di 48 c tnh nh hm ca kho K. (trn thc t mi Ki l mt php chn hon v bt trong K).K1,...,K16stothnhbngkho.Mtvngcaphpmhocmt trn hnh di. 3.p dng php hon v ngc IP-1 cho xu bt R16L16, ta thu c bn m y. Tc l y=IP-1(R16L16). Hy ch th t o ca L16 v R16. 3.1.2.M t mt vng ca DES 3.1.3.M t hm f Hm f c hai binvo: bin th nht A l xu bt di 32, bin th hai J lmt xu bt di 48. u ra ca f l mt xu bt di 32. Cc bc sau c thc hin: L thuyt mt m v an ton d liu Trang 26 1. BinthnhtAcmrngthnhmtxubtdi48theomthmm rngcnhE.E(A)gm32btcaA(chonvtheocchcnh)vi16bt xut hin hai ln. 2. Tnh E(A) J v vit kt qu thnh mt chui 8 xu 6 bt = B1B2B3B4B5B6B7B8. 3. Bc tip theo dng 8 bng S1, S2, ... ,S8 ( c gi l cc hp S ). Vi mi Si l mt bng 416 c nh c cc hng l cc s nguyn t 0 n 15. Vi xu bt c di 6 (K hiu Bi = b1b2b3b4b5b6), ta tnh Sj(Bj) nh sau: Hai bt b1b6 xc nh biu din nh phn ca hng r caSj ( 0 r 3) v bn bt (b2b3b4b5) xc nh biu din nh phn ca ct c ca Sj ( 0 c 15 ). Khi Sj(Bj) s xc nh phn t Sj(r,c); phn t ny vit di dng nh phn l mt xu bt c di 4. ( Bi vy, mi Sjc th c coi l mt hm m m u vo l mt xu bt c di 2 v mt xu bt c di 4, cn u ra l mt xu bt c di 4). Bng cch tng t tnh cc Cj = Sj(Bj), 1 j 8. 4. Xu bt C = C1C2... C8 c di 32c hon v theo php honv cnhP. Xu kt qu l P(C) c xc nh l f(A,J). Hm f c m t trong hnh di. Ch yu n gm mt php th (s dng hp S), tip sau l php hon v P. 16 php lp ca f s to nn mt h mt tch nu nh phn trn. L thuyt mt m v an ton d liu Trang 27 3.1.4.M t chi tit cc hm trong DES Php hon v ban u IP nh sau: bng ny c ngha l bt th 58 ca x l bt u tin ca IP(x); bt th 50 ca x l bt th hai ca IP(x), .v.v . . . IP 58 50 42 34 26 18 10 2 60 52 44 36 28 20 12 4 62 54 46 38 30 22 14 6 64 56 48 40 32 24 16 8 57 49 41 33 25 17 9 1 59 51 43 35 27 19 11 3 61 53 45 37 29 21 13 5 63 55 47 39 31 23 15 7 Php hon vi ngc IP -1 IP -1 L thuyt mt m v an ton d liu Trang 28 40 8 48 16 56 24 64 32 39 7 47 15 55 23 63 31 38 6 46 14 54 22 62 30 37 5 45 13 53 21 61 29 36 4 44 12 52 20 60 28 35 3 43 11 51 19 59 27 34 2 42 10 50 18 58 26 33 1 41 9 49 17 57 25 Hm m rng E c xc nh theo bng Bng chn E bt 3212345 456789 89101112 13 1213141516 17 1617181920 21 2021222324 25 2425262728 29 28293031321 Tm hp S ln lt nh sau S1 14 4131215118310 312 5 917 11574142131106 1211 9 538 41148136 2 111512 97 3 1050 15 128249 1 7 511314 10 0613 S2 151 8146113 4 9 72 13 12 0 510 3 13 471528 14 12 01 10 6 911 5 0 14 71110413 1 5 812 6 9 32 15 138 1013154 2 11 67120 5 149 S3 100 91463 155 113127 11 42 8 137 093461028514 12 1115 1 136 49853 011 1212 5 1014 7 110 130698 7 415143 11 52 12 L thuyt mt m v an ton d liu Trang 29 S4 713 14 3 0 6 9 10128 5 11 12415 138 11 5 6 15 03 472 12 1 10 149 106 9 0 12 11 7 131513 14 5 2 84 315 0 6 10 1 138 945 11 12 7 2 14 S5 212 4 1710 116 85 315 130149 1411 2 124 7131 5015 10 3986 42 1 111013 78 15912 5 630 14 118 12 71142 13 615 09 10453 S6 121 10 15 926 8 0133414 71511 1015 4 2 7129 5 61 1314 011 3 8 914 15 5 2812 3 704 10 11311 6 43 2 12 95 151011 14 117 608 13 S7 411 12 1415081331297 5106 1 13 011 7 4 9110 14 35 122158 6 1 411 13123714 10 1568 059 2 6 1113 8 1 410 79 50 15 142 3 12 S8 1328 4 6 15 111 109 314 50127 1 1513 8 10 3 74 125 611 0 14 9 2 7 11 41912 142 0610 13 15 3 58 21147410 813 15 12 90 35 611 V php hon v P c dng P 167 20 2912 28 1 15 23 L thuyt mt m v an ton d liu Trang 30 5 18 31 32 273 19 13 30 22114 3.1.5.Tnh ton bng kha t kha K Trn thc t, K l mt xu bt di 64, trong 56 bt l kho v 8 bt kim tra tnh chn l nhm pht hin sai. Cc bt cc v tr 8,16, . . ., 64 c xc nh sao cho mi byte cha mt s l cc s "1". Bi vy mt sai st n l c th pht hin c trong mi nhm 8 bt. Cc bt kim tra b b qua trong qu trnh tnh ton bng kho. 1.Vi mt kho K 64 bt cho trc, ta loi b cc bt kim tra tnh chn l v hon v cc bt cn li ca K theo php hon v c nh PC-1. Ta vit: PC-1(K) = C0D0 2.Vi i thay i t 1 n 16: Ci = LSi(Ci-1) Di = LSi(Di-1) Vic tnh bng kho c m t trong hnh sau: Cc hon v PC-1 v PC-2 c dng trong bng kho l: PC-1 L thuyt mt m v an ton d liu Trang 31 5749 41 33 2517 1 58 50 42 3426 10 2 59 514335 1911 3605244 635547 393123 7 6254 463830 14 6 61 534537 2113 5282012 By gi ta s a ra bng kho kt qu. Nh ni trn, mi vng s dng mt kho 48 bt gm 48 bt nm trong K. Cc phn t trong cc bng di y biu th cc bt trong K trong cc vng kho khc nhau. Vng 1 10 51 34 60 49 17 35 57 2 9 19 42 3 35 26 25 44 58 59 1 36 27 18 41 22 28 39 54 37 4 47 30 5 53 23 29 61 21 38 63 15 20 45 1413 62 55 31 Vng 2 2 43 26 52 41 9 25 49 59 1 11 34 60 27 18 17 36 50 51 58 57 19 10 33 14 20 31 46 29 63 39 22 28 45 15 21 53 13 30 55 7 12 37 6 5 54 47 23 Vng 3 51 27 10 36 25 58 9 33 43 50 60 18 44 11 21 49 34 35 42 41 3 59 17 61 4 15 30 13 4723 6 12 29 62 5 37 28 14 39 54 63 21 53 20 38 31 7 Vng 4 35 11 59 49 9 42 58 17 27 34 44 2 57 60 51 50 33 18 19 26 25 52 43 1 45 55 62 14 28 31 7 53 63 13 46 20 21 12 61 23 38 47 5 37 4 22 15 54 Vng 5 19 60 43 33 58 26 42 1 11 18 57 51 L thuyt mt m v an ton d liu Trang 32 41 44 35 34 17 2 3 10 936 27 50 29 39 46 61 12 15 54 37 47 28 30 4 .5 63 45 7 22 31 20 21 55 6 62 38 Vng 6 3 44 27 17 42 10 26 50 60 2 41 35 25 57 19 18 1 51 52 59 58 49 11 34 13 23 30 45 63 62 38 21 31 12 14 55 20 47 29 54 6 15 4 5 39 53 46 22 Vng 7 52 57 11 1 26 5910 34 44 51 25 19 9 41 3 2 50 35 36 43 42 33 60 18 28 7 14 29 47 46 22 5 15 63 6139 431 13 38 53 62 55 20 23 38 30 6 Vng 8 36 41 60 5010 43 59 18 57 35 9 3 58 25 5251 34 19 49 27 26 17 44 2 12 54 61 13 31 30 620 62 47 45 23 55 15 28 22 37 46 39 4 721 14 53 Vng 9 57 33 52 422 35 51 10 49 27 1 60 50 17 44 43 26 11 41 19 18 9 36 59 4 46 53 5 23 22 61 12 54 39 37 15 47 7 20 14 29 38 31 63 6213 6 45 Vng 10 41 17 36 26 511935 59 33 11 50 44 34 1 57 27 10 60 25 3 2 58 49 43 55 30 37 20 7 6 45 63 38 23 21 62 31 54 4 61 13 22 15 47 46 28 53 29 Vng 11 25 1 49 10 35 3 19 43 17 60 34 57 18 50 41 11 59 44 9 52 51 42 33 27 39 14 21 4 54 53 29 47 22 7 5 46 15 38 55 45 28 6 62 31 30 12 37 13 L thuyt mt m v an ton d liu Trang 33 Vng 12 9 50 33 59 19 52 3 27 1 44 18 41 2 34 25 60 43 57 58 36 35 26 17 11 23 61 5 55 38 37 13 31 6 54 20 30 62 22 39 29 12 53 46 15 14 63 21 28 Vng 13 58 34 17 43 3 36 52 11 50 57 2 25 51 189 44 27 41 42 49 19 10 1 60 7 45 20 39 22 21 28 15 53 38 4 14 46 6 23 13 63 37 30 62 61 47 5 12 Vng 14 42 18 1 27 52 49 36 60 34 41 51 9 35 2 58 57 11 25 26 333 59 50 44 54 29 4 23 6 5 12 62 37 22 55 61 30 53 7 28 47 21 14 46 45 31 20 63 Vng 15 26 2 50 11 36 33 49 44 18 25 35 58 19 51 42 41 60 9 10 17 52 43 34 57 38 13 55 7 53 20 63 46 21 639 45 14 37 54 12 315 61 30 29 15 4 47 Vng 16 18 59 42 3 57 25 41 36 10 17 27 50 11 43 34 33 52 1 2 9 44 35 26 49 30 5 47 62 45 12 55 58 13 61 31 37 6 27 46 4 23 28 53 22 21 7 62 39 Php gii m c thc hin nh dng cng thut ton nh php m nu u vo l ynhngdngbngkhotheothtngcliK16,...K1.uracathuttonsl bn r x. Sau khi thay i, han v, , v dch vng, bn c th ngh rng thut ton gii m hontonkhcvphctp,khhiunhthuttonmha.Trili,DESsdng cng thut ton lm vic cho c m ha v gii m. L thuyt mt m v an ton d liu Trang 34 Vi DES, c th s dng cng chc nng gii m hoc m ha mt khi. Ch c skhcnhaulcckhaphicsdngtheothtngcli.Nghal,nu cc kha m ha cho mi vng l k1, k2, k3 , ... , k15, k16 th cc kha gii l k16, k15, ... , k3,k2,k1.Thuttondngsinhkhacsdngchomivngtheokiuvng quanh.Khacdchphi,vsnhng vtrcdchctnhtcuica bng ln, thay v t trn xung. 3.2.V d Sau y l mt v d v php m DES. Gi s ta m bn r ( dng m hexa): 0 1 2 3 4 5 6 7 8 9 A B C D E F Bng cch dng kho K m ha l: 1 2 3 4 5 7 7 9 9 B B C D F F 1 Kho dng nh phn (khng cha cc bt kim tra) l: 00010010011010010101101111001001101101111011011111111000 S dng IP, ta thu c L0 v R0 ( dng nh phn) nh sau: L0 = 1100110000000000110010011111111 L1 =R0 = 11110000101010101111000010101010 Sau thc hin 16 vng ca php m nh sau: E(R0) = 011110100001010101010101011110100001010101010101 K1 = 000110110000001011101111111111000111000001110010 E(R0) K1 = 011000010001011110111010100001100110010100100111 S-box outputs01011100100000101011010110010111 f(R0,K1) = 00100011010010101010100110111011 L2 = R1 = 11101111010010100110010101000100 E(R1) = 011101011110101001010100001100001010101000001001 K2 = 011110011010111011011001110110111100100111100101 E(R1) K2 = 000011000100010010001101111010110110001111101100 S-box outputs11111000110100000011101010101110 f(R1,K2) = 00111100101010111000011110100011 L3 = R2 = 11001100000000010111011100001001 E(R2) = 111001011000000000000010101110101110100001010011 K3 = 010101011111110010001010010000101100111110011001 E(R2) K3 = 101100000111110010001000111110000010011111001010 L thuyt mt m v an ton d liu Trang 35 S-box outputs00100111000100001110000101101111 f(R2,K3) = 01001101000101100110111010110000 L4 =R3 = 10100010010111000000101111110100 E(R3) =01010000010000101111100000000101011111111010100 K4 = 011100101010110111010110110110110011010100011101 E(R3) K4 = 001000101110111100101110110111100100101010110100 S-box outputs00100001111011011001111100111010 f(R3,K4) = 10111011001000110111011101001100 L5 = R4 = 01110111001000100000000001000101 E(R4) = 101110101110100100000100000000000000001000001010 K5 = 011111001110110000000111111010110101001110101000 E(R4) K5= 110001100000010100000011111010110101000110100010 S-box outputs01010000110010000011000111101011 f(R4,K5) = 00101000000100111010110111000011 L6 = R5 =10001010010011111010011000110111 E(R5) = 110001010100001001011111110100001100000110101111 K6 = 011000111010010100111110010100000111101100101111 E(R5) K6 =101001101110011101100001100000001011101010000000 S-box outputs01000001111100110100110000111101 f(R5,K6) = 10011110010001011100110100101100 L7 = R6 = 11101001011001111100110101101001 E(R6) = 111101010010101100001111111001011010101101010011 K7 = 111011001000010010110111111101100001100010111100 E(R6) K7 = 000110011010111110111000000100111011001111101111 S- box outputs00010000011101010100000010101101 f(R6,K7) = 10001100000001010001110000100111 L8 = R7 = 00000110010010101011101000010000 E(R7) = 000000001100001001010101010111110100000010100000 K8 = 111101111000101000111010110000010011101111111011 E(R7) K8 = 111101110100100001101111100111100111101101011011 S-box outputs01101100000110000111110010101110 f(R7,K8) = 00111100000011101000011011111001 L9 = R8 = 11010101011010010100101110010000 E(R8) = 011010101010101101010010101001010111110010100001 L thuyt mt m v an ton d liu Trang 36 K9 = 111000001101101111101011111011011110011110000001 E(R8) K9 = 100010100111000010111001010010001001101100100000 S-box outputs00010001000011000101011101110111 f(R8,K9) = 00100010001101100111110001101010 L10 = R9 = 00100100011111001100011001111010 E(R9) = 000100001000001111111001011000001100001111110100 K10 = 101100011111001101000111101110100100011001001111 E(R9) K10 = 101000010111000010111110110110101000010110111011 S-box outputs11011010000001000101001001110101 f(R9,K10) = 01100010101111001001110000100010 L11 = R10 = 10110111110101011101011110110010 E(R10) = 010110101111111010101011111010101111110110100101 K11 = 001000010101111111010011110111101101001110000110 E(R10) K11 = 011110111010000101111000001101000010111000100011 S-box outputs 01110011000001011101000100000001 f(R10,K11) = 11100001000001001111101000000010 L12 = R11 = 11000101011110000011110001111000 E(R11) = 011000001010101111110000000111111000001111110001 K12 = 011101010111000111110101100101000110011111101001 E(R11) K12 = 000101011101101000000101100010111110010000011000 S-box outputs01110011000001011101000100000001 f(R11,K12) = 11000010011010001100111111101010 L13 = R12 = 01110101101111010001100001011000 E(R12) = 001110101011110111111010100011110000001011110000 K13 = 100101111100010111010001111110101011101001000001 E(R12) K13 = 101011010111100000101011011101011011100010110001 Sbox outputs10011010110100011000101101001111 f(R12,K13) = 11011101101110110010100100100010 L14 = R13 = 00011000110000110001010101011010 E(R13) = 000011110001011000000110100010101010101011110100 K13 = 010111110100001110110111111100101110011100111010 E(R13) K14 = 010100000101010110110001011110000100110111001110 S-box outputs01100100011110011001101011110001 f(R13,K14) = 10110111001100011000111001010101 L15 = R14 = 11000010100011001001011000001101 L thuyt mt m v an ton d liu Trang 37 E(R14) = 111000000101010001011001010010101100000001011011 K15 = 101111111001000110001101001111010011111100001010 E(R14) K15 = 010111111100010111010100011101111111111101010001 S-box outputs10110010111010001000110100111100 f(R14,K15) = 01011011100000010010011101101110 R15 = 01000011010000100011001000110100 E(R15) = 001000000110101000000100000110100100000110101000 K16 = 110010110011110110001011000011100001011111110101 E(R15) K16 = 111010110101011110001111000101000101011001011101 S-box outputs10100111100000110010010000101001 f(R15,K16) = 11001000110000000100111110011000 R16 = 00001010010011001101100110010101 Cui cng p dng IP-1 vo L16,R16 ta nhn c bn m hexa l: 8 5 E 8 1 3 5 4 0 F 0 A B 4 0 5 3.3.Tranh lun v DES Khi DES c xut nh mt chun mt m, c rt nhiu kin ph phn. Mt ldophniDESclinquanncchpS.MitnhtonlinquannDES ngoi tr cc hp S u tuyn tnh, tc vic tnh php hoc loi tr ca hai u ra cng gingnhphphocloitrcahaiuvoritnhtanura.CchpS-cha ng thnh phn phi tuyn ca h mt l yu t quan trong nht i vi mt ca h thng( Ta thytrong chng 1 l cc hmt tuyn tnh - chng hn nh Hill - c th d dng b m thm khi b tn cng bng bn r bit). Tuy nhin tiu chun xy dng cc hp S khng c bit y . Mt s ngi gi l cc hp S phi cha cc "ca sp" c du kn, cho php Cc An ninh Quc gia M (NSA) gii m c cc thng bo nhng vn gi c mc an ton ca DES. D nhin ta khng th bc b c khng nh ny, tuy nhin khng c mt chng c no c a ra chng t rng trong thc t c cc ca sp nh vy. Nm 1976 NSA khng nh rng,cc tnh cht sau ca hp S l tiu chun thit k: P0 Mi hng trong mi hp S l mt hon v ca cc s nguyn 0, 1, . . . , 15. P1 Khng mt hp S no l mt hm Affine hoc tuyn tnh cc u vo ca n. P2 Vic thay i mt bt vo ca S phi to nn s thay i t nht l hai bt ra. L thuyt mt m v an ton d liu Trang 38 P3ivihpSbtkvviuvoxbtkS(x)vS(x001100)phi khc nhau ti thiu l hai bt ( trong x l xu bt di 6 ). Hai tnh cht khc nhau sau y ca cc hp S c th coi l c rt ra t tiu chun thit k ca NSA. P4 Vi hp S bt k, u vo x bt k v vi e, f {0,1}: S(x) S(x 11ef00). P5 Vi hp S bt k , nu c nh mt bt vo v xem xt gi tr ca mt bt u ra c nh th cc mu vo bt ra ny bng 0 s xp x bng s mu ra bt bng 1. Ch rng, nu c nh gi tr bt vo th nht hoc bt vo th 6 th c 16 mu vo lm cho mt bt ra c th bng 0 v c 16 mu vo lm cho bt ny bng 1. Vi cc bt vo t bt th hai n bt th 5 th iu ny khng cn ng na. Tuy nhin phn b kt qu vn gn vi phn b u. Chnh xc hn, vi mt hp S bt k, nu ta c nh gi tr ca mt bt vo bt k th s mu vo lm cho mt bt ra c nh no c gi tr 0 (hoc 1) lun nm trong khong t 13 n 19. Ngi ta khng bit r l liu c cn mt chun thit k no y hn c dng trong vic xy dng hp S hay khng. S phn i xc ng nht v DES chnh l kch thc ca khng gian kho: 256 l qu nh m bo an ton thc s. Nhiu thit bi chuyn dng c xut nhm phc v cho vic tn cng vi bn r bit. Php tn cng ny ch yu thc hin tm kho theo phng php vt cn. Tc vi bn r x 64 bt v bn m y tng ng, mi kho u c th c kim tra cho ti khi tm c mtkhoKthomneK(x)=y.CnchlcthcnhiuhnmtkhoKnh vy). Ngay t nm 1977, Diffie v Hellman gi rng c th xy dng mt chp VLSI (mch tch hp mt ln) c kh nng kim tra c 106kho/giy. Mt my c th tm ton b khng gian kho c 106 trong khong 1 ngy. H c tnh chi ph to mt my nh vy khong 2.107$. Trong cuc hi tho ti hi ngh CRYPTO'93, Michael Wiener a ra mt thit k rt c th v my tm kho. My ny xy dng trn mt chp tm kho, c kh nng thchinngthi16phpmvtcti5107kho/giy.Vicngnghhin nay,chiphchtokhong10,5$/chp.Gicamtkhungmycha5760chpvo khong 100.000$ v nh vy n c kh nng tm ra mt kho ca DES trong khong 1,5ngy.Mtthitbdng10khungmynhvycgichng106$sgimthi gian tm kim kho trng bnh xung cn 3,5 gi. 3.4.DES trong thc t L thuyt mt m v an ton d liu Trang 39 Mc d vic m t DES kh di dng song ngi ta c th thc hin DESrt ha hiu bng c phn cng ln phn mn. Cc php ton duy nht cn c thc hin l php hoc loi tr cc xu bt. Hm m rng E, cc hp S, cc hon v IP v P v vic tnhtonccgitriK1,...,K16ucththchinccnglcbngtrabng( trong phn mn ) hoc bng cch ni cng chng thnh mt mch. Cc ng dng phn cng hin thi c th t c tc m ho cc nhanh. Cng tyDigitalEquipmentthngbotihinghCRUPTO'92rnghschtomt chp c 50 ngn tranzistor c th m ho vi tc 1 Gbt/s bng cch dng nhp c tc 250MHz. Gi ca chp ny vo khong 300$. Ti nm 1991 c 45 ng dng phncngvchngtrnhcscaDEScUbantiuChunqucgiaM (NBS) chp thun.Mt ng dng quan trng ca DES l trong giao dch ngn hng M - (ABA) DES c dng m ho cc s nh danh c nhn (PIN) v vic chuyn ti khon bng my th qu t ng (ATM). DES cng c H thng chi tr gia cc nh bng ca Ngn hng hi oi (CHIPS) dng xc thc cc giao dch vo khon trn 1,51012 USA/tun.DEScncsdngrngritrongcctchcchnhph.Chnghn nh b nng lng, B T php v H thng d tr lin bang. Cc ch hot ng ca DES: C 4 ch lm vic c pht trin cho DES: Ch chuyn m in t (ECB), ch phn hi m (CFB), ch lin kt khi m (CBC) v ch phn hi u ra (OFB). Ch ECB tng ng vi cch dng thng thng ca m khi: vi mt dy cc khi bn r cho trc x1,x2,. . .( mi khi c 64 bt),mixiscmhobngcngmtkhoKtothnhmtchuicckhi bn m y1y2 ... theo quy tc yi = eK(yi-1xi) i 1. Vic s dng ch CBC c m t trn hnh 3.4. L thuyt mt m v an ton d liu Trang 40 Trong cc ch OFB v CFB dng kho c to ra s c cng mod 2 vi bn r (tc l n hot ng nh mt h m dng, xem phn 1.1.7). OFB thc s l mt h mdng ng b: dng kho c to bi vic m lp vc t khi to 64 bt (vc t IV). Ta xc nh z0 =IV v ri tnh dng kho z1z2 . . . theo quy tc zi = eK(zi-1), i1. Dy bn r x1x2 . . . sau s c m ho bng cch tnh yi = xi zi,i 1. Trong ch CFB, ta bt u vi y0 = IV (l mt vc t khi to 64 bt) v to phn tzicadngkhobngcchmhokhibnmtrc.Tczi=eK(yi-1),i1. Cng nh trong ch OFB: yi = xi zi,i 1. Vic s dng CFB c m t trn hnh 3.5 (ch rng hm m DES eK c dng cho c php m v php gii m cc ch CFB v OFB). Ch CBC. x1x2 ++ eKeK y1y2 IV=y0 . . .M ho Encrypt y1y2 dKdK ++ x1x2 IV=y0 . . .Gii m Decrypt L thuyt mt m v an ton d liu Trang 41 Cng cn mt s bin tu ca OFB v CFB c gi l cc ch phn hi K bt (1 < K < 64 ). y ta m t cc ch phn hi 64 bt. Cc ch phn hi 1 bt v 8 bt thng c dng trong thc t cho php m ho ng thi 1 bit (hoc byte) s liu. Bn ch cng tc c nhng u, nhc im khc nhau. ch ECB v OFB, sthayicamtkhibnrxi64btslmthayikhibnmyitngng, nhng cc khi bn m khc khng b nh hng. Trong mt s tnh hung y l mt tnh cht ng mong mun. V d, ch OFB thng c dng m khi truyn v tinh. Mt khc cc ch CBC v CFB, nu mt khi bn r xi b thay i th yi v tt c cc khi bn m tip theo s bi nh hng. Nh vy cc ch CBC v CFB c th c s dng rt hiu qu cho mc ch xc thc. c bit hn, cc ch ny c th cdngtomxcthcbntin(MAC-messageauthenticationcode).MAC c gn thm vo cc khi bn r thuyt phc Bob tin rng, dy bn r thc s l ca Alice m khng b Oscar gi mo. Nh vy MAC m bo tnh ton vn (hay tnh xc thc) ca mt bn tin ( nhng tt nhin l MAC khng m bo mt). Ta s m t cch s dng ch BCB to ra mt MAC. Ta bt u bng vc t khi t IV cha ton s 0. Sau dng ch CBC to cc khi bn m y1,. . . ,yn theo kho K. Cui cng ta xc nh MAC l yn. Alice s pht i dycc khi bn r Ch CFBx1 x2 ++ y1 y2 IV=y0 . . .M ho Encrypt eKeK y1 y2 ++ x1 x2 IV=y0 . . .Gii m Decrypt eKeK L thuyt mt m v an ton d liu Trang 42 x1,x2,. . . ,xn cng vi MAC. Khi Bob thu c x1. . .xn anh ta s khi phc li y1. . .yn bng kho K b mt v xc minh xem liu yn c ging vi MACm mnh thu c hay khng. NhnthyOscarkhngthtoramtMAChpldoanhtakhngbitkhoK m Alice v Bob ang dng. Hn na Oscar thu chn c dy khi bn r x1. . .xn v thay i t nhiu ni dung th th chc chn l Oscar khng th thay i MAC c Bob chp nhn. Thng thng ta mun kt hp c tnh xc thc ln bo mt. iu c th thc hin nh sau: Trc tin Alice dng kho K1 to MAC cho x1. . . xn . Sau Alice xc nh xn+1 l MAC ri m ho dy x1. . .xn+1 bng kho th hai K2 to ra bn m y1. . .yn+1 . Khi Bob thu c y1. . .yn+1 , trc tin Bob s gii m ( bng K2) v kim tra xem xn+1 c phi l MAC i vi dy x1. . .xn dng K1 hay khng. Ngc li, Alice c th dng K1 mhox1. . .xn v tora cy1...yn , sau dng K2 to MAC yn+1 i vi dy y1. . .yn. Bob s dng K2 xc minh MAC v dung K1 gii m y1. . .yn. 3.5.ng dng ca DES Mc d vic m t DES kh di dng song ngi ta c th thc hin DESrt hu hiu bng c phn cng ln phn mm. Cc php ton duy nht cn c thc hin l php hoc loi tr cc xu bit. Hm m rng E, cc hp S, cc han v IP v P v vic tnhtonccgitriK1,...,K16ucththchinccnglcbngtrabng( trong phn mm ) hoc bng cch ni cng chng thnh mt mch. Cc ng dng phn cng hin thi c th t c tc m ha cc nhanh. Nm 1991 c 45 ng dng phn cng v chng trnh c s ca DES c U ban tiu Chun quc gia M (NBS) chp thun.Mt ng dng quan trng ca DES l trong giao dch ngn hng M - (ABA) DES c dng m ha cc s nh danh c nhn (PIN) v vic chuyn ti khon bng my th qu t ng (ATM). DES cng c H thng chi tr gia cc nh bng ca Ngn hng hi oi (CHIPS) dng xc thc cc giao dch. DES cn c s dng rng ritrong cc t chc chnh ph. Chnghn nh b nng lng, B Tphp v H thng d tr lin bang. L thuyt mt m v an ton d liu Trang 43 4.M HA KHA CNG KHAI Nh nhngtnh cht u vit cam hakha cng khai, dn n s pht trin rt ln ca hm hanyv c l chnh n to ra cuc cchmng trong ton b lch s ca m ha. Trong thc t c rt nhiu loi m ha kha cng khai nhng trong phm vi phn ny ta ch xt mt s thut ton rt ph bin l Elgamal v RSA 4.1.Bi ton Logarit ri rc (DL) H m ha Elgamal c xut t nm 1985, da trn c s bi ton logarith ri rc. Chng ta s bt u bng vic m t bi ton khi thit lp mi trng hu hn Zp, p l s nguyn t (Nhm nhn Zp* l nhm cyclic v phn t sinh ca Zp*c gi l phn t nguyn thy). Bi ton logarith ri rc trong Zp l i tng trong nhiu cng trnh nghin cu v c xem l bi ton kh nu p c chn cn thn. C th khng c mt thut ton thi gian a thc no cho bi ton logarith ri rc. gy kh khn cho cc phng php tn cng bit, p phi c t nht 150 ch s v (p-1) phi c t nht mt tha s nguyn t ln. Li th ca bi ton logarith ri rc trong xy dngh m ha l kh tm c cc logarith ri rc, song bi ton ngc ly ly tha li c th tnh ton hiu qu theo thut ton bnh phng v nhn. Ni cch khc, ly tha theo moun p l hm mt chiu vi cc s nguyn t p thch hp. 4.2.Cc thut ton cho bi ton Logarit ri rc Trongphnnytaxemrngplsnguynt,lphntnguynthutheo moun p. Ta thy rng p v l cc s c nh. Khi bi ton logarith ri rc c th cphtbiudidngsau:tmmtsmaduynht,0ap-2saochoa (mod p), vi Zp* cho trc. c trng ca bi ton: I = (p,,) trong p l s nguyn t, Zp l phn t nguyn thy, Zp*

Mc tiu: Hy tm mt s nguyn duy nht a, 0 a p-2 sao cho: a (mod p) Ta s xc nh s nguyn a bng log Bi ton logarith ri rc trong Zp L thuyt mt m v an ton d liu Trang 44 Rrnglbitonlogarithrirc(DiscreteLogarith-DL)cthgiibngmt php tm kim vt cn vi thi gian c O(p) v khng gian c O(1) (b qua cc tha s logarith). Bng cch tnh ton tt c cc gi tr a c th v sp xp cc cp c th t (a, a mod p) c lu n cc to th hai ca chng, ta c th gii bi ton DL vi thi gian c O(1) bng O(p) php tnh ton trc v O(p) b nh (vn b qua cc thaslogarith).Cmtsthuttonchobitonlogarithrircnh:Shanks, Pohlig-Hellman, phng php tnh ton ch s Chng ta s m t mt thut ton c tn l Shanks, mt thut ton ti u ha thi gian - b nh ca Shanks. Nucn,ccbc1v2cthtnhtontrc(tuynhin,iunykhngnh hng ti thi gian chy tim cn) Tip theo cn l nu (j,y) L1 v (i,y) L2 th mj = y = -i Bi vy mj+i = nh mong mun.Ngc li, i vi bt k ta c th vit: log = mj+i trong 0 j,i m-1. V th php tm kim bc 5 chc chn thnh cng. C th p dng thut ton ny chy vi thi gian O(m) v vi b nh c O(m) (b qua cc tha s logarith). Ch l bc 5 c th thc hin mt cch (ng thi) qua tng danh sch L1 v L2. Sau y l mt v d nh minh ho: Gi s p=809 v ta phi tm log3525. Ta c = 3, = 525 v m = 808( = 29.Khi : 29 mod 809 = 99 Trc tin tnh cc cp c sp (j,99j mod 809) vi 0 j28. Ta nhn c danh sch sau: (0,1)(1,99)(2,93)(3,308)(4,559) 1.t m =p-1(2.Tnh mj mod p, 0 j m-1 3.Sp xp m cp th t (j,mj mod p) c lu ti cc to th hai ca cc cp ny, ta s thu c mt danh sch L1 4.Tnh -i mod p, 0 i m-15.Sp xp m cp th t (i, -i mod p) c lu ti cc ta th hai ca cc cp c sp ny, ta s thu c mt danh sch L2 6.Tm mt cp (j,y) L1 v mt cp (i,y) L2 (tc l mt cp c to th hai nh nhau) Thut ton Shanks cho bi ton DL L thuyt mt m v an ton d liu Trang 45 (5,329)(6,211)(7,664)(8,207)(9,268) (10,644)(11,654)(12,26)(13,147)(14,800) (15,727)(16,781)(17,464)(18,314)(19,275) (20,582)(21,496)(22,564)(23,15)(24,676) (25,586)(26,575)(27,295)(28,81) Danh sch ny s c sp xp to L1. Danh sch th hai cha cc cp c sp (i,525(3i)-1 mod 809), vi 0 i 28. Danh sch ny gm: (0,525)(1,175)(2,328)(3,379)(4,396) (5,132)(6,44)(7,554)(8,724)(9,511) (10,440)(11,686)(12,768)(13,256)(14,,355) (15,388)(16,399)(17,133)(18,314)(19,644) (20,754)(21,496)(22,564)(23,15)(24,676) (25,356)(26,658)(27,489)(28,163) Saukhispxpdanhschny,tacL2 .Byginuxlngthiquachai danhsch,tastmc(10,644)trongL1v(19,644)trongL2.Bygitacth tnh: log3525 = 2910+19 = 309 C th kim tra thy rng qu thc 3309 525 (mod 809). 4.3.H mt RSA KhinimhmtmRSAcrainm1976bicctcgiR.Rivets, A.Shamir, v L.Adleman. H m ha ny da trn c s ca hai bi ton : Bi ton Logarith ri rc Bi ton phn tch thnh tha s. TronghmhaRSAccbnr,ccbnmvcckha(publickeyvprivate key) l thuc tp s nguyn ZN = {1, . . . , N-1}. Trong tp ZN vi N=pq l cc s nguyn t khc nhau cng vi php cng v php nhn moun N to ra moun s hc N. Kham ha EKB l cps nguyn (N,KB) v khagiim DkBl cp snguyn (N,kB),ccslrtln,sNcthlnti hngtrmchs.Ccphngphpm ha v gii m l rt d dng.Cng vicm ha l s bin i bn r P (Plaintext) thnh bnm C (Ciphertext) da trn cp kha cng khai KB v bn r P theo cng thc sau y : C = EKB(P) = PKB (mod N) (1) L thuyt mt m v an ton d liu Trang 46 CngvicgiimlsbiningclibnmCthnhbnrPdatrncp kha b mt kB , moun N theo cng thc sau : P = DkB(C) = CkB (mod N)(2) D thy rng, bn r ban u cn c bin i mt cch thch hp thnh bn m, sau c th ti to li bn r ban u t chnh bn m : P = DkB(EKB(P))(3) Thay th(1) vo (2) ta c : (PKB)kB = P (mod N)(4) Ta thy N=pq vi p, q l s nguyn t. Trong ton hc chng minh c rng, nu N l s nguyn t th cng thc (4) s c li gii khi v ch khi: KB.kB 1 (mod (N))(5) trong (N) = LCM(p-1,q-1) .LCM (Lest Common Multiple) l bi s chung nh nht. 4.3.1.nh ngha h mt RSA Chun b Cho n = p*q vi p,q l s nguyn t ln.t P = C = Zn t (n) = (p-1)*(q-1) Chn b nguyn t vi (n) nh ngha kha K = {(n,a,b): a*b1 (mod (n))} Mt cch k hiu khc: K = {(n,Kb,kb): Kb*kb1 (mod (n))} Xc nh h mt RSA Hai gi tr n, b cng khai; cc gi tr a l b mt Vi mi gi tr K=(n, a, b) v xP; yC ta xc nh hai hm sau Hm m ha: y = ek(x) = xb mod n Hm gii m: x = dk(y) = ya mod n L thuyt mt m v an ton d liu Trang 47 Nimtcchkhc,utinnginhnBlachnmtkhacngkhaiKBmt cchngunhin.KhikhabmtkBctnhrabngcngthc(5).iuny hon ton tnh c v khi B bit c cp s nguyn t (p,q) th s tnh c (N). V d: N=11413=101113,(N)=100112=11200=26527.KBphichnsaochokhng chia ht cho 2,5,7. Chn, chng hn KB=3533 khi kB = KB-1 = 6579 mod 11200. V ta c kha cng khai l (N,KB) = (11413,3533), kha b mt l 6579. Php lp m v gii m l: EKB(P) = PKB (mod N) = P3533 (mod 11413) DkB(C) = CkB (mod N) = C6579 (mod 11413) Chng hn vi P = 9726, ta c C=5761. Chn p v q Tnh N=pq Tnh (N) Chn kha KB C = PKB (mod N) P = CkB ( mod N ) Chn kha kB KB kB Bn r P Bn mC Bn r gc P S cc bc thc hin m ha theo thut ton RSA L thuyt mt m v an ton d liu Trang 48 4.3.2. an ton ca h RSA Mt nhn nh chung l tt c cc cuc tn cnggii m u mang mc ch khng tt. Tnh bomt caRSA chyu da vovic gi bmt kha giim hay gi b mt cc tha s p,q ca N. Ta th xt mt vi phng thc tn cng in hnh ca k ch nhm gii m trong thut ton ny (nhm xm phm ti cc yu t b mt ).Trng hp 1: Chng ta xt n trng hp khi k ch no bit c moun N,khacngkhaiKBvbntinmhaC,khikchstmrabntingc (Plaintext)nhthno.lmciukchthngtncngvohthng mt m bng hai phng thc sau y: Phngthcthnht:TrctindavophntchthasmounN.Tip theo sau chng s tm cch tnh ton ra hai s nguyn t p v q, v c kh nng thnh cng khi s tnh c (N) = (p-1)(q-1) v kha b mt kB. Ta thy N cn phi l tch ca hai s nguyn t, v nu N l tch ca hai s nguyn t th thuttonphntchthasngincntiaN1/2bc,bivcmts nguyntnhhnN1/2.Mtkhc,nuNltchcansnguynt,ththut ton phn tch tha s n gin cn ti a N1/n bc.Phng thc th hai:Phng thc tn cngth hai vo hm ha RSA l c th khi u bng cch gii quyt trng hp thch hp ca bi ton logarit ri rc.TrnghpnykchctrongtaybnmCvkhacngkhaiKB tc l c cp (KB, C) Trng hp 2: Chng ta xt trng hp khi k ch no bit c moun N v (N), khi k ch s tm ra bn tin gc (Plaintext) bng cch sau: Bit (N) th c th tnh p,q theo h phng trnh: pq = N, (p-1)(q-1) = (N) do p v q l nghim ca phng trnh bc hai: x2 - (n - (N) +1) + n = 0 V d: n=84773093, v bit (N) = 84754668. Gii phng trnh bc hai tng ng ta s c hai nghim p=9539 v q=8887 4.3.3.Mt s tnh cht ca h RSATrongcchmtmRSA,mtbntincthcmhatrongthigiantuyn tnh. L thuyt mt m v an ton d liu Trang 49 i vi cc bn tin di, di ca cc s c dng cho cc kha c th c coinhlhng.Tngtnhvy,nngmtslnlythacthchin trong thi gian hng, cc s khng c php di hn mt di hng. Thc ra tham s ny che du nhiu chi tit ci t c lin quan n vic tnh ton vi cc con s di, chi ph ca cc php ton thc s l mt yu t ngn cn s ph bin ng dng ca phng php ny. Phn quan trng nht ca vic tnh ton c lin quannvicmhabntin.Nhngchcchnlskhngchmhano ht nu khng tnh ra c cc kha ca chng l cc s ln. CckhachohmhaRSActhctoramkhngphitnhtonqu nhiu.Mtlnna,talininccphngphpkimtrasnguynt.Mis nguyn t ln c th c pht sinh bng cch u tin to ra mt s ngu nhin ln, sau kim tra cc s k tipcho ti khi tmcmt snguyn t. Mt phng php n gin thc hin mt php tnh trn mt con s ngu nhin, vi xcsut1/2schngminhrngsckimtrakhngphinguynt.Bc cui cng l tnh p da vo thut ton Euclid. Nhphntrntrnhbytronghmhacngkhaithkhagiim (privatekey)kBvccthasp,qlcgibmtvsthnhcngca phng php l tu thuc vo k ch c kh nng tm ra c gi tr ca kB hay khng nu cho trc N v KB. Rt kh c th tm ra c kB t KB, cn bit v p v q. Nh vy cn phn tch N ra thnh tha s tnh p v q. Nhng vic phn tch ra tha s l mt vic lm tn rt nhiu thi gian, vi k thut hin i ngy nay th cn ti hng triu nm phn tch mt s c 200 ch s ra tha s. antoncathuttonRSAdatrncsnhngkhkhncavicxc nh cc tha s nguyn t ca mt s ln. Bng di y cho bit cc thi gian d on, gi s rng mi php ton thc hin trong mt micro giy. S cc ch s trongs c phn tch Thi gian phn tch 504 gi 75104 gi 10074 nm 2004.000.000 nm 30051015 nm 50041025 nm L thuyt mt m v an ton d liu Trang 50 4.3.4.ng dng ca RSA HmhaRSAcngdngrngrichyuchoWebvccchngtrnh email.Ngynay,RSAcncsdngrngritrongcccngnghbomts dng cho thng mi in t (v d nh cng ngh bo mt SSL). 4.4.H mt Elgamal Elgamal pht trin mt h mt kha cng khai da trn bi ton logarith ri rc. H thng ny c trnh by di y Trong h mt ny, bn m ph thuc vo c bn r x ln gi tr ngu nhin k. Bi vy s c nhiu bnm c m t cng mt bn r. Sau y sm t s lc cch lm vic ca hmt Elgamal. Bn r x c che du bng cchnhn n vik to ra y2 . Gi tr k cng c gi i nh mt phn ca bn m. Nu mt ngi bit s m b mt a c th tnh c k t k . Sau anh ta s tho mt n bng cch chia y2 cho k thu c x. V d:Cho p = 2579, = 2, a = 765. Khi = 2765 mod 2579 = 949 ChoplsnguyntsaochobitonlogarithrirctrongZp lkh gii. Cho Zp* l phn t nguyn thy. Gi s P = Zp* C = Zp* Zp* . Ta nh ngha: K= {(p, ,a,): a (mod p)} Cc gi tr p, , c cng khai, cn a gi kn Vi K = (p, ,a,) v mt s ngu nhin b mt k Zp-1 , ta xc nh: ek (x,k) = (y1 ,y2 ) trong y1 = k mod p y2 = xk mod p vi y1 ,y2 Zp* ta xc nh: dk(y1 ,y2 ) = y2 (y1a )-1 mod p H mt kha cng khai Elgamal trong Zp* L thuyt mt m v an ton d liu Trang 51 By gi ta gi s Alice mun gi thng bo x = 1299 ti Bob. Gi s s ngu nhin k m c chn l k = 853. Sau c ta tnh y1 = 2853 mod 2579 = 435 y2 = 1299 949853 mod 2579 = 2396 Sau Bob thu c bn m y = (435,2396), anh ta tnhx = 2396 (435765)-1 mod 2579 =1299 chnh l bn r m Alice m ha. 4.5.CC PHNG PHP KIM TRA S NGUYN T LN 4.5.1.Kim tra Miller-Rabin Kim tra Miller-Rabin l mt thut ton xc sut kim tra tnh nguyn t cng nhccthuttonkimtratnhnguynt:KimtraFermatvKimtraSolovay-Strassen. N c xut u tin bi Gary L. Miller nh mt thut ton tt nh, da trngithitRiemanntngqut;MichaelO.Rabinsachanthnhmtthut ton xc sut. KhisdngkimtraMiller-RabinchngtacncvomtmnhQ(p,a)ng vi cc s nguyn t p v mi s t nhinv kim tra xem chng c ng vi s n mun kim tra v mt sc chn ngu nhin hay khng? Nu mnh Q(n,a) khng ng, tt yu n khng phi l s nguyn t, cn nu Q(n,a) ng, s n c th l s nguyn t vi mt xc sut no . Khi tng s ln th, xc sut n l s nguyn t tng ln. Tiu chun kim tra Q(n,a) Cn bc hai ca 1 trong Trc ht l mt b v cn bc hai ca n v trong trng hu hn , trong p l s nguyn t. Chc chn rng 1 v -1 lun l cc cn bc hai ca 1 theo moun p. Chng l hai cn bc hai duy nht ca 1. Tht vy, gi s rng x l mt cn bc hai ca 1 theo moun p. Khi : T , x 1 hoc x + 1 l chia ht cho p. Tiu chun Miler-Rabin L thuyt mt m v an ton d liu Trang 52 By gi gi s p l mt s nguyn t l, khi p - 1 l s chn v ta c th vit p 1 di dng, trong s l mt s t nhin >=1 v m' l s l - iu ny ngha l ta rt ht cc tha s 2 khi p 1. Ly s a bt k trong tp {1,2,..,p-1}. Xt dy s vi k=0,1,2,...,s. Khi xk = (xk 1)2, vi k=1,2,...,s v xs = p &minus 1. T nh l Fermat nh: hay hay . Do,hochoc. Nuta dng li, cn nu ngc li ta tip tc vi xs 2. Sau mt s hu hn bc hoc ta c mt ch s k,sao cho, hoc ti k=0 ta vn c. Ta c mnh Q(p,a) nh sau: Nu p l s nguyn t l v p - 1 =th vi mi a: 0