*Chng 4. Hm bm xc thc v ch k s

Hong Thu Phng - Khoa ATTT

*Ni dung Gii thiu4.1 Cc hm bm v tnh ton vn ca d liu4.2 Trao i v tho thun kho4.3 H mt da trn nh danh4.4 Cc s ch k s khng nn4.5 Cc s ch k s c nn

*Gii thiuMt s khi nim:Xc thc mu tin lin quan n cc kha cnh sau khi truyn tin trn mngBo v tnh ton vn ca mu tin: bo v mu tin khng b thay i hoc c cc bin php pht hin nu mu tin b thay i trn ng truyn.Kim chng danh tnh v ngun gc: xem xt mu tin c ng do ngi xng tn gi khng hay mt k mo danh no khc gi.Khng chi t bn gc: trong trng hp cn thit, bn thn mu tin cha cc thng tin chng t ch c ngi xng danh gi, khng mt ai khc c th lm iu . Nh vy ngi gi khng th t chi hnh ng gi, thi gian gi v ni dung ca mu tin.

*Gii thiuCc yu cu bo mt khi truyn mu tin trn mng:Tm cc bin php cn thit chng i li cc hnh ng ph hoi nh sau: l b mt: gi b mt ni dung mu tin, ch cho ngi c quyn bit.Thm m ng truyn: khng cho theo di hoc lm tr hon vic truyn tin.Gi mo: ly danh ngha ngi khc gi tin. Sa i ni dung: thay i, ct xn, thm bt thng tin.Thay i trnh t cc gi tin nh ca mu tin truyn.

*Gii thiuSa i thi gian: lm tr hon mu tin.T chi gc: khng cho php ngi gi t chi trch nhim ca tc gi mu tin.T chi ch: khng cho php ngi nhn ph nh s tn ti v n ch ca mu tin gi.

*Gii thiuCc hm bm mt m ng vai tr quan trng trong mt m hin i:c dng xc thc tnh nguyn vn d liu c dng trong qu trnh to ch k s trong giao dch in t.Cc hm bm ly mt thng bo u vo v to mt u ra c xem nh l:M bm (hash code), Kt qu bm (hash result), Hoc gi tr bm (hash value).

*Gii thiuVai tr c bn ca cc hm bm mt m l mt gi tr bm coi nh nh i din thu gn, i khi gi l mt du vt (imprint), vn tay s (digital fingerprint), hoc tm lc thng bo (message digest) ca mt xu u vo, v c th c dng nh l mt nh danh duy nht vi xu .Cc hm bm thng c dng cho ton vn d liu kt hp vi cc lc ch k s. Mt lp cc hm bm ring c gi l m xc thc thng bo (MAC) cho php xc thc thng bo bng cc k thut m i xng.

*Gii thiu

*4.1 Cc hm bm v tnh ton vn ca d liuGii thiu hm bmVic s dng cc h mt m v cc s ch k s, thng l m ha v k s trn tng bit ca thng tin, s t l vi thi gian m ha v dung lng ca thng tin. Thm vo c th xy ra trng hp: Vi nhiu bc thng ip u vo khc nhau, s dng h mt m, s k s ging nhau (c th khc nhau) th cho ra kt qu bn m, bn k s ging nhau (nh x N-1: nhiu mt). iu ny s dn n mt s rc ri v sau cho vic xc thc thng tin.

*4.1 Cc hm bm v tnh ton vn ca d liuVi cc s k s, ch cho php k cc bc thng ip (thng tin) c kch thc nh v sau khi k, bn k s c kch thc gp i bn thng ip gcV d: vi s ch k chun DSS ch k trn cc bc thng ip c kch thc 160 bit, bn k s s c kch thc 320 bit. Trong khi trn thc t, ta cn phi k cc thng ip c kch thc ln hn nhiu, chng hn vi chc MB. Hn na, d liu truyn qua mng khng ch l bn thng ip gc, m cn bao gm c bn k s (c dung lng gp i dung lng bn thng ip gc), p ng vic xc thc sau khi thng tin n ngi nhn.

*4.1 Cc hm bm v tnh ton vn ca d liuMt cch n gin gii bi ton (vi thng ip c kch thc vi chc MB) ny l chia thng ip thnh nhiu on 160 bit, sau k ln cc on c lp nhau. Nhng bin php ny c mt s vn trong vic to ra cc ch k s: Th nht: vi mt thng ip c kch thc a, th sau khi k kch thc ca ch k s l 2a (trong trng hp s dng DSS). Th hai: vi cc ch k an ton th tc chm v chng dng nhiu php tnh s hc phc tp nh s m modulo. Th ba: vn nghim trng hn l kt qu sau khi k, ni dung ca thng ip c th b xo trn cc on vi nhau, hoc mt s on trong chng c th b mt mt, trong khi ngi nhn cn phi xc minh li thng ip. Ta cn phi bo v tnh ton vn ca thng ip

*4.1 Cc hm bm v tnh ton vn ca d liuGii php cho cc vn vng mc n ch k s l dng hm bm tr gip cho vic k sCc thut ton bm vi u vo l cc bc thng ip c dung lng, kch thc ty (vi KB n vi chc MB thm ch hn na) cc bc thng ip c th l dng vn bn, hnh nh, m thanh, file ng dng v.v - v vi cc thut ton bm: MD2, MD4, MD5, SHA cho cc bn bm u ra c kch thc c nh: 128 bit vi dng MD, 160 bit vi SHA.Nh vy, bc thng ip kch thc ty sau khi bm s c thu gn thnh nhng bn bm c gi l cc vn bn i din c kch thc c nh (128 bit hoc 160 bit).

*4.1 Cc hm bm v tnh ton vn ca d liuVi mi thng ip u vo ch c th tnh ra c mt vn bn i din gi tr bm tng ng duy nht Hai thng ip khc nhau chc chn c hai vn bn i din khc nhau. Khi c vn bn i din duy nht cho bc thng ip, p dng cc s ch k s k trn vn bn i din

*4.1 Cc hm bm v tnh ton vn ca d liuGi s A mun gi cho B thng ip x. A thc hin cc bc sau:(1) A bm thng ip x, thu c bn i din z = h(x) c kch thc c nh 128 bit hoc 160 bit. (2) A k s trn bn i din z, bng kha b mt ca mnh, thu c bn k s y = sig(z). (3) A gi (x, y) cho B.

*4.1 Cc hm bm v tnh ton vn ca d liu

*4.1 Cc hm bm v tnh ton vn ca d liuKhi B nhn c (x, y). B thc hin cc bc sau:(4) B kim tra ch k s xc minh xem thng ip m mnh nhn c c phi c gi t A hay khng bng cch gii m ch k s y, bng kha cng khai ca A, c z. (5) B dng mt thut ton bm tng ng vi thut ton bm m A dng bm thng ip x i km, nhn c h(x).(6) B so snh 2 gi tr bm z v h(x), nu ging nhau th chc chn rng thng ip x m A mun gi cho B cn nguyn vn, bn cnh cng xc thc c ngi gi thng tin l ai.

*4.1 Cc hm bm v tnh ton vn ca d liu

*4.1 Cc hm bm v tnh ton vn ca d liuHm bm tr gip cho cc s k s nhm gim dung lng ca d liu cn thit truyn qua mngV d: lc ny ch cn bao gm dung lng ca bc thng ip gc v 256 bit (s dng MD) hay 320 bit (s dng SHA) ca bc k s c k trn bn i din ca thng ip gc, tng ng vi vic gim thi gian truyn tin qua mng. Hm bm thng kt hp vi ch k s to ra mt loi ch k in t va an ton hn (khng th ct/dn) va c th dng kim tra tnh ton vn ca thng ip.Hm bm c ng dng rt mnh trong vn an ton thng tin trn ng truyn. Cc ng dng c s dng hm bm khng ch m bo v mt an ton thng tin, m cn to c lng tin ca ngi dng v h c th d dng pht hin c thng tin ca mnh c cn ton vn hay khng, h bit rng thng tin ca mnh chc chn c b mt vi pha cc nh cung cp.

*4.1 Cc hm bm v tnh ton vn ca d liunh ngha hm bm: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 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.

*4.1 Cc hm bm v tnh ton vn ca d liuc trng:Hm bm h l hm bm mt chiu (one-way hash) vi cc c tnh sau:Vi thng ip u vo x thu c bn bm z = h(x) l duy nht.Nu d liu trong thng ip x thay i hay b xa thnh thng ip x th h(x) h(x). Cho d ch l mt s thay i nh hay ch l xa i 1 bit d liu ca thng ip th gi tr bm cng vn thay i. iu ny c ngha l: hai thng ip hon ton khc nhau th gi tr hm bm cng khc nhau. Ni dung ca thng ip gc khng th b suy ra t gi tr hm bm. Ngha l: vi thng ip x th d dng tnh c z = h(x), nhng li khng th (thc cht l kh) suy ngc li c x nu ch bit gi tr hm bm h

*4.1 Cc hm bm v tnh ton vn ca d liuTnh cht:Vic a hm bm h vo dng trong s ch k s khng lm gim s an ton ca s ch k s v n l bn tm lc thng bo bn i din cho thng ip c k ch khng phi l thng ip gc. iu cn thit i vi h l cn tha mn mt s tnh cht sau trnh b gi mo:Tnh cht 1: Hm bm h l khng va chm yu.

*4.1 Cc hm bm v tnh ton vn ca d liuV d xt mt kiu tn cng sau:ng l: thng tin phi c truyn ng t A n B

Nhng: trn ng truyn, thng tin b ly trm v b thay i

*4.1 Cc hm bm v tnh ton vn ca d liuNgi A gi cho B (x, y) vi y = sigK(h(x)). Nhng trn ng truyn, tin b ly trm. Tn trm, bng cch no tm c mt bn thng ip x c h(x) = h(x) m x x. Sau , hn a x thay th x ri truyn tip cho ngi B. Ngi B nhn c v vn xc thc c thng tin ng n. Do , trnh kiu tn cng nh trn, hm h phi tha mn tnh khng va chm yu: 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).

*4.1 Cc hm bm v tnh ton vn ca d liuTnh cht 2: Hm bm h l khng va chm mnhXt mt kiu tn cng nh sau: 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, hn 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: 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).

*4.1 Cc hm bm v tnh ton vn ca d liuTnh cht 3: Hm bm h l hm mt chiu:Xt mt kiu tn cng nh sau: Vic gi mo cc ch k trn bn tm lc z thng xy ta vi cc s ch k s. Gi s tn gi mo tnh ch k trn bn tm lc z, sau hn tm mt bn thng ip x c tnh ngc t bn i din z, z = h(x). Tn trm thay th bn thng ip x hp l bng bn thng ip x gi mo, nhng li c z = h(x). V hn k s trn bn i din cho x bng ng ch k hp l. Nu lm c nh vy th (x, y) l bc in gi mo nhng hp l. trnh c kiu tn cng ny, h cn tha mn tnh cht mt chiu: Hm bm h l mt chiu nu khi cho trc mt bn tm lc thng bo z th khng th thc hin v mt tnh ton tm ra thng ip ban u x sao cho h(x) = z .

*4.1 Cc hm bm v tnh ton vn ca d liuHm bm gm 2 loi:Hm bm khng c kha: cc hm bm da trn mt m khi. Hm bm c kha (MAC) dng xc thc thng bo

*4.1 Cc hm bm v tnh ton vn ca d liuCc hm bm khng c khanh ngha 1: Mt m khi (n, r) l mt m khi xc nh mt hm kh nghch t cc bn r n bit sang cc bn m n bit bng cch s dng mt kho r bit. Nu E l mt php m ho nh vy th Ek(x) k hiu cho php m ho x bng kho k.nh ngha 2: Cho h l mt hm bm c lp c xy dng t mt mt m khi vi hm nn thc hin s php m ho khi x l tng khi bn tin n bit. Khi tc ca h l 1/s.

*4.1 Cc hm bm v tnh ton vn ca d liuTnh cht:Tnh cht nnTnh d dng tnh tonTnh kh tnh ton nghch nhKh tm nghch nh th hai: vi x cho trc th khng c kh nng tm x x sao cho: h(x) = h(x)Tnh khng va chm: khng c kh nng v tnh ton tm hai u vo khc nhau bt k x v x h(x) = h(x)Hm bm tha mn tnh cht trn c gi l hm bm mt m hay hm bm an ton.

*4.1 Cc hm bm v tnh ton vn ca d liuMDC ( Manipulating detection codes): m pht hin sa iCc hm bm dng MD: MD2, MD4, MD5 c Rivest a ra thu c kt qu u ra vi di l 128 bit. Hm bm MD4 a ra vo nm 1990. Mt nm sau phin bn mnh MD5 cng c a ra thay th cho MD4.Mc ch ca MDC l cung cp mt biu din nh hoc bm ca thng bo, n l lp con ca cc hm bm khng c kha. Cc lp c bit ca MDC l:Cc hm bm mt chiu (OWHF): l cc hm bm m vic tm mt u vo bm thnh mt gi tr bm c xc nh trc l rt khCc hm bm khng va chm (CRHF): l cc hm bm m vic tm hai u vo c cng gi tr bm l kh

*4.1 Cc hm bm v tnh ton vn ca d liuMDC di n.Ba s di y c lin quan cht ch vi cc hm bm di n, xy dng trn cc mt m khi. Cc s ny c s dng cc thnh phn c xc nh trc nh sau:Mt mt m khi n bit khi sinh Ek c tham s ho bng mt kho i xng k.Mt hm g nh x n bit vo thnh kho k s dng cho E (Nu cc kho cho E cng c di n th g c th l hm ng nht)Mt gi tr ban u c nh IV thch hp dng vi E.

*4.1 Cc hm bm v tnh ton vn ca d liu

*4.1 Cc hm bm v tnh ton vn ca d liuThut ton bm Matyas - Meyer OseasVo: Xu bit nRa: M bm n bit ca x(1) u vo x c phn chia thnh cc khi n bit v c n nu cn thit nhm to khi cui cng hon chnh. Ta c t khi n bit: x1 x2 xt. Xc nh trc mt gi tr ban u n bit (k hiu IV)(2) u ra l Ht c xc nh nh sau: H0 = IV, Hi = Eg(xi) xi, 1 i t

*4.1 Cc hm bm v tnh ton vn ca d liuThut ton bm Davies - Meyer Vo: Xu bit nRa: M bm n bit ca x(1) u vo x c phn chia thnh cc khi n bit v c n nu cn thit nhm to khi cui cng hon chnh. Ta c t khi n bit: x1 x2 xt. Xc nh trc mt gi tr ban u n bit (k hiu IV)(2) u ra l Ht c xc nh nh sau: H0 = IV, Hi = Exi(Hi-1) Hi-1, 1 i t

*4.1 Cc hm bm v tnh ton vn ca d liuThut ton bm Miyaguchi - PreneelS ny tng t nh thut ton M-M-O ngoi tr Hi-1 (u ra giai on trc) c cng mod 2 vi tn hiu ra giai on hin thi. Nh vy: H0 = IV, Hi = Eg(Hi-1)(xi) xi Hi-1; 1 i t

*4.1 Cc hm bm v tnh ton vn ca d liuMDC di kp:MDC 2 v MDC 4 l cc m pht hin s sa i yu cu tng ng l 2 v 4 php ton m ho khi trn mi khi u vo hm bm. MDC-2 v MDC- 4 s dng cc thnh phn xc nh nh sau:DES c dng lm mt m khi Ek c u vo/ ra 64 bit v c tham s ho bng kho k 56 bit.Hai hm g v nh x cc gi tr 64 bit U thnh cc kho DES 56 bit nh sau:Cho U = u1 u2 u64, xo mi bit th 8 bt u t u8 v t cc bit th 2 v th 3 v "10" i vi g v "01" i vi iu ny m bo rng chng khng phi l cc kho DES yu hoc na yu. ng thi iu ny cng m bo yu cu bo mt l g(IV)

*4.1 Cc hm bm v tnh ton vn ca d liuThut ton MD2VO: Xu bit x c di r = 64t, t 2RA: Bn bm, i din cho thng ip gc, di c nh 128 bit M t thut ton(1) Phn x thnh cc khi 64 bit xi: x1, , xt(2) Chn IV v nh sau: IV = 0x5252525252525252; (3) K hiu l php ghp v CiL, CiR l cc na 32 bit phi v tri ca Ci u ra c xc nh nh sau (vi 1 i t):

*4.1 Cc hm bm v tnh ton vn ca d liu

*4.1 Cc hm bm v tnh ton vn ca d liuThut ton MD4:

*4.1 Cc hm bm v tnh ton vn ca d liuThut ton MD5M t thut tonu vo: l mt thng ip c di tu u ra l mt chui c di c nh l 128 bit. Thut ton c thit k chy trn cc my tnh 32 bit.Thut ton?

*4.1 Cc hm bm v tnh ton vn ca d liuThut ton MD5M t thut tonu vo: l mt thng ip c di tu u ra l mt chui c di c nh l 128 bit. Thut ton c thit k chy trn cc my tnh 32 bit.Thut ton:Thng ip u vo c di b bit bt k. Biu din cc bit di dng nh sau: m[0] m[1] m[2] ... m[b-1]Bc1: Cc bit gn thm : Thng ip c m rng, thm bit vo pha sau sao cho di ca n (bit) ng d vi 448 theo mun 512. Ngha l thng ip c m rng sao cho n cn thiu 64 bit na th s c mt di chia ht cho 512. Vic thm bit ny c thc hin nh sau: Mt bit 1 c thm vo sau thng ipSau cc bit 0 c thm vo c mt di ng d vi 448 mun 512.

*4.1 Cc hm bm v tnh ton vn ca d liuBc 2: Gn thm di: Dng biu din 64 bit di b ca chui ban u c thm vo pha sau kt qu ca bc 1. Bc 3: Khi to b m MD: Mt b m 4 t (A,B,C,D) c dng tnh m s thng ip. y mi A,B,C,D l mt thanh ghi 32 bit. Nhng thanh ghi ny c khi to theo nhng gi tr hex sau : A=0x01234567 B=0x89abcdef C=0xfedcba98 D=0x76543210 Bc 4 : X l thng ip theo tng khi 16 t. nh ngha cc hm ph, cc hm ny nhn gi tr u vo l 3 t 32 bit v to to ra mt word 32 bit. F(X,Y,Z) = XY v not(X) Z G(X,Y,Z)= XZ v Y not(Z) H(X,Y,Z) = X xor Y xor Z I(X,Y,Z) = Y xor (X v not(Z)). Bc ny s dng mt bng 64 gi tr T[1 .. 64] c to ra t hm sin. Gi T l phn t th i ca bng, th T l phn nguyn ca 4294967296*|sin(i)| , i c tnh theo radian

*4.1 Cc hm bm v tnh ton vn ca d liunh gi thut ton MD5V tc sinh ra chui ct yu th MD5 chm hn so vi MD4 nhng n li an ton hn rt nhiu so vi MD4. Thut ton s ha thng ip MD5 kh n gin thc hin, cung cp mt gi tr bm ca thng ip vi di tu .

*4.1 Cc hm bm v tnh ton vn ca d liuCc mc tiu ca i phng vi cc thut ton MDC:Mc tiu ca i phng mun tn cng mt MDC l nh sau:(a) tn cng mt OWHF: cho trc gi tr bm y, tm mt tin nh x sao cho y =h(x) hoc mt cp (x, h(x)), tm mt tin nh th hai x sao cho h(x) = h(x)(b) tn cng mt CRHF: tm hai u vo bt k x, x sao cho h(x) = h(x). Mt CRHF phi c thit k chng li cc tn cng ngy sinh nht

*4.1 Cc hm bm v tnh ton vn ca d liuTn cng ngy sinh nht C th ngh hash 64 bit l an ton, c ngha l kh tm c bn tin c cng hash. Nhng khng phi vy v nghch l ngy sinh nht nh sau: Trong lp c t nht bao nhiu sinh vin, xc sut c t nht 2 sinh vin trng ngy sinh nht l ln hn 0.5?Theo l thuyt xc sut thng k gi s sinh vin t nht trong lp l k, khi xc sut q khng c 2 ngi no trng ngy sinh l t s gia cch chn k ngy khc nhau trong 365 ngy trn s cch chn k ngy bt k trong 365 ngy. Vy: q = Ck365/ 365k Do , xc sut p c t nht 2 ngi trng ngy sinh l:p = 1 q = 1 Ck365/ 365k

*4.1 Cc hm bm v tnh ton vn ca d liu p > 0.5 th k > 22 hay k =23, c th khi p = 0.5073 Khi cha tnh ton chi tit chng ta ngh l trong lp phi c t nht khong 365/2 tc l 184 sinh vin. Nhng trn thc t con s t hn rt nhiu ch cn 23 sinh vin, chnh v vy ta gi y l nghch l ngy sinh nht. iu mun ni ln rng, trong nhiu trng hp xc sut hai mu tin c cng bn Hash l khng nh nh chng ta tng.

SHA1SHA1 (Secure Hash Function) c NSA (M) thit k nm 1995 thay th cho SHA0;Chui u ra ca SHA1 c kch thc 160 bt v thng c biu din thnh 40 s hexa;H hm bm SHA: SHA-0, SHA-1, SHA-2, SHA-3:SHA0 t c s dng trn thc t;SHA1 tng t SHA0, nhng khc phc mt s li;SHA2 ra i nm 2001 khc phc li ca SHA1 v c nhiu thay i. Kch thc chui u ra c th l 224, 256, 384 v 512 bt;SHA3 ra i nm 2012, cho php chui u ra c kch thc khng c nh.SHA1 c s dng rng ri m bo tnh xc thc v ton vn thng ip.

SHA1Qu trnh x l thng ip ca SHA1:SHA1 s dng th tc x l thng ip tng t MD5;Thng ip c chia thnh cc khi 512 bt. Nu kch thc thng ip khng l bi s ca 512 ni thm s bt thiu;Phn x l chnh ca SHA1 lm vic trn state 160 bt, chia thnh 5 t 32 bt (A, B, C, D, E);Cc t A, B, C, D, E c khi tr bng mt hng c nh;Tng phn 32 bt ca khi u vo 512 bt c a dn vo thay i state;Qu trnh x l gm 80 vng, mi vng gm cc thao tc: add, and, or, xor, rotate, mod.

SHA1Lu x l mt vng ca SHA1:A, B, C, D, E: cc t 32 bitWt: khi 32 bit thng ip u vo;Kt: 32 bit hng. Mi s dng mt hng khc nhau;

*4.1 Cc hm bm v tnh ton vn ca d liuCc hm bm c kha (MAC):MAC l mt lp con ca hm bm c kha. Mc ch ca MAC l bo m c ti nguyn ca thng bo v tnh ton vn ca n. Gm cc tnh cht sau:D dng tnh ton: vi hk bit, gi tr k cho trc v mt u vo x, hk(x) c th c tnh d dng (hk(x) c gi l gi tr MAC)Nn: nh x mt u vo x c di bit hu hn tu ti mt u ra hk(x) c di bit n c nh.Khng tnh ton: Vi cc cp gi tr (xi, h(xi)) khng c kh nng tnh mt cp (x, h(x)) vi x xi (k c c kh nng hk(x) = hk(xi) vi mt i no ).

*4.1 Cc hm bm v tnh ton vn ca d liuCc hm bm c kho c s dng xc thc thng bo v thng c gi l cc thut ton to m xc thc thng bo (MAC).MAC da trn cc mt m khi.Thut ton VO: D liu x, mt m khi E, kho MAC b mt k ca E.RA : n bit MAC trn x (n l di khi ca E)(1) n v chia khi: n thm cc bit vo x nu cn. Chia d liu n thnh tng khi n bit : x1, , xt

*4.1 Cc hm bm v tnh ton vn ca d liu(2) X l theo ch CBC. K hiu Ek l php m ho E vi kho k.Tnh khi Ht nh sau: H1 Ek(x1) Hi Ek(Hi-1 xi) 2 i t(3) X l thm tng sc mnh ca MAC. Dng mt kho b mt th hai k k. Tnh: Ht Ek-1(Ht)Ht Ek(Ht)(4) X l thm tng sc mnh ca MAC(5) Kt thc: MAC l khi n bit Ht

*4.1 Cc hm bm v tnh ton vn ca d liu

*4.1 Cc hm bm v tnh ton vn ca d liuCc m xc thc mu tin MAC cung cp s tin cy cho ngi nhn l mu tin khng b thay i v t ch danh ngi gi. Cng c th s dng m xc thc MAC km theo vi vic m ho bo mt. Ni chung ngi ta s dng cc kho ring bit cho mi MAC v c th tnh MAC trc hoc sau m ho, tt hn l thc hin MAC trc v m ho sau. S dng MAC c nhc im l MAC ph thuc vo c mu tin v c ngi gi, nhng i khi ch cn xc thc mu tin v thng tin xc thc ch ph thuc mu tin lu tr lm bng chng cho tnh ton vn ca n. Khi ngi ta s dng hm Hash thay v MAC. Cn lu rng MAC khng phi l ch k in t, v c ngi gi v ngi nhn u bit thng tin v kho.

*4.1 Cc hm bm v tnh ton vn ca d liu

*4.1 Cc hm bm v tnh ton vn ca d liuTnh ton vn ca d liu v xc thc thng boC ba phng php cung cp tnh ton vn ca d liu bng cch dng cc hm bm.Ch dng MAC

*4.1 Cc hm bm v tnh ton vn ca d liuDng MDC v m ha:

S dng MDC v knh tin cy:

*4.1 Cc hm bm v tnh ton vn ca d liuCc mc tiu ca i phng i vi cc thut ton MAC:Tn cng vi vn bn bit: mt hoc nhiu cp (xi, hk(xi)) l c gi tr.Tn cng vn bn chn lc: mt hoc nhiu cp (xi, hk(xi)) l c gi tr vi xi c chn bi i phng.Tn cng vi vn bn chn lc thch ng: xi c th c chn bi i phng nh trn, by gi cho php la chn thnh cng da trn cc kt qu truy vn c u tin.

*4.2 Trao i v tho thun khoGi s A v B mun lin lc s dng h mt kho b mt. tho thun mt kho K chung cho c hai bn qua mt knh khng an ton m khng ai khc c th bit c, A v B c th dng th tc tho thun kho Diffie -Hellman sau:(1) Chn trc mt s nguyn t p thch hp v mt phn t sinh ca Zp* (2 p 2) . Cc gi tr p v c cng khai.(2) A gi cho B gi tr (2.1) B gi cho A gi tr (2.2)

*4.2 Trao i v tho thun khoThc hin cc bc sau mi khi cn c kha chung:(a) A chn mt s nguyn b mt x: 1 x p 2 v gi cho B thng bo x mod p (2.1)(b) B chn mt s nguyn b mt y: 1 y p 2 v gi cho A thng bo y mod p (2.2).(c) B thu c x v tnh kho chung k: k = (x)y mod p(d) A thu c y v tnh kho chung k: k = (y)x mod p

*4.2 Trao i v tho thun khoV d:Gi s A v B chn p = 11 v = 2. Nhm nhn xyclic sinh bi : {i, i = 0, , 9 } = {1, 2, 4, 8, 5,10, 9, 7, 3, 6}. Cc phn t sinh ca nhm ny bao gm cc phn t sau: = 2, 3 = 8, 7 = 7, 9 = 6. Gi s A chn gi tr ngu nhin x = 4 v gi cho B gi tr 24 mod 11 = 5.Gi s B chn gi tr ngu nhin y = 7 v gi cho A gi tr 27 mod 11 = 7.B nhn c 5 v tnh kho chung k = 57 mod 11 = 3A nhn c 7 v tnh kho chung k = 74 mod 11 = 3

*4.3 H mt da trn nh danh tng c bn:H mt da trn nh danh do Shamir xut l mt h mt bt i xng trong :Thng tin nh danh ca thc th (tn ring) ng vai tr kho cng khai ca n. Trung tm xc thc T c s dng tnh kho ring tng ng ca thc th ny

*4.3 H mt da trn nh danhS trao i kho Okamoto-Tanaka: gm 3 pha(1) Pha chun b: Trung tm xc thc tin cy chn 2 s nguyn t p v q v a cng khai cc gi tr n, g v e, trong : n = p.qg l phn t sinh ca c Zp* v Zq*e Z*(n). y, hm Carmichael ca n c xc nh nh sau:(n) = BCNN(p 1, q 1)Tnh kho b mt ca trung tm d = e-1 mod (n) vi d Z*(n).

*4.3 H mt da trn nh danh(2) Pha tham gia ca ngi dngCho IDi l thng tin nh danh ca ngi dng th i (I = A, B, C, ). Cho ai l kho b mt ca ngi dng i tho mn: si Idi-d mod n. Sau trung tm s cng b (e, n, g, Idi) v phn pht si ti mi ngi dng i qua mt knh an ton (hoc bng cch dng th)(3) Pha to kha chungTa gi s y rng hai ngi dng Alice v Bob mun chia s mt kho chung (chn hn dng cho mt h mt kho b mt). Trc tin Alice to mt s ngu nhin rA v tnh: v gi n cho Bob.Tng t, Bob to mt s ngu nhin rB v tnh:v gi n cho Alice.

*4.3 H mt da trn nh danhTip theo, Alice tnh: Tng t, Bob tnh:WKAB v WKBA s dng lm kho chung v:

*4.3 H mt da trn nh danh

*4.3 H mt da trn nh danhV d: p = 11, q = 13, n = p.q = 143, (143) = 60. Z*() = {1, 7, 11, 13, 17, 19, 23, 29, 37, 41, 43, 47, 49, 53, 59}. Gi s e = 43. M phng qu trnh trao i kha?

*4.3 H mt da trn nh danhTnh d = e-1 mod (n) = 7Vi IDi = 2 v IDj = 3 ta c: si = 2-7 mod 143 = 19; sj = 3-7 mod 143 = 126. pha to kha chung: Gi s A chn ri = 3, khi Xi = 19.23 mod 143 = 9. A gi Xi cho BGi s B chn rj = 2, khi Xj = 126.22 mod 143 = 75. B gi Xj cho A.A tnh B tnh

*4.4 Cc s ch k s khng nnCh k s:Ta s nghin cu mt ng dng in hnh trong my tnh th hin mt nhu cu thng thng ca con ngi: lnh chuyn tin t mt ngi ny ti mt ngi khc.V vn bn y l mt dng sc c my tnh ha.Giao dch dng giy t c thc hin nh sau:Sc l mt i tng xc nh c t cch giao dch thng miCh k trn sc xc nhn tnh xc thc bi v chc chn rng ch c ngi k hp php mi c th to c ch k nyTrong trng hp bt hp php th s c mt bn th 3 c th c gi vo phn xt tnh xc thc.Sc b hy n khng c s dng liSc giy khng th thay i c, hay hu ht cc kiu thay i u c th d dng pht hin c

*4.4 Cc s ch k s khng nnGiao dch trn my tnh i hi mt m hnh khc. Xt m hnh sau y:Sandy gi cho ngn hng ca mnh mt thng bo y quyn ngn hng chuyn 100$ cho Tim.Ngn hng ca Sandy phi lm nhng vic sau:Kim tra v chng t c rng thng bo ny thc s ti t Sandy, nu sau c ta khng nhn l mnh gi nPhi bit chc rng ton b thng bo ny l ca Sandy v n khng b sa iSandy cng mun bit chc rng ngn hng ca mnh khng th gi mo nhng thng bo tng t.C hai bn u mun m bo rng thng bo l thng bo mi, khng phi l mt thng bo trc c s dng li v n khng b sa i trong khi truyn

*4.4 Cc s ch k s khng nnCh k s l mt giao thc to ra mt hiu qu tng t nh ch k thc: N l mt du hiu m ch c ngi gi mi c th to ra nhng nhng ngi khc c th nhn thy c rng n l ca ngi gi.Ging nh ch k thc, ch k s dng xc nhn ni dung thng bo

*4.4 Cc s ch k s khng nnCh k s phi tha mn iu kin sau y:Khng th gi mo: Nu P k thng bo M bng ch k S(P, M) th khng mt ai c th to c cp [M, S(M,P)]Xc thc: Nu R nhn c cp [M, S(M,P)] c coi l ca P th R c th kim tra c rng ch k c thc s l ca P hay khng. Ch P mi c th to c ch k ny v ch k c gn cht vi M.Khng th thay i: sau khi c pht M khng th b thay i bi S, R hoc bi mt k thu trm noKhng th s dng li: Mt thng bo trc c a ra s ngay lp tc b R pht hin

*4.4 Cc s ch k s khng nnTo ch k s

*Thm nh ch k s 4.4 Cc s ch k s khng nn

*4.4 Cc s ch k s khng nnCh k s Shamir:Chui bt thng bo trc ht c tch thnh cc vct k bt M. Gi s M [0, n 1], M = (m1, , mk)Mt ma trn nh phn b mt k x 2k (ma trn H) c chn ngu nhin cng vi mt gi tr modulo n, trong n l mt s nguyn t ngu nhin k bit. Mt vecto A 2K bit (c dng lm kha cng khai) c chn trn c s gii h phng trnh tuyn tnh sau:

*4.4 Cc s ch k s khng nn

*4.4 Cc s ch k s khng nnXc thc thng bo dng s Shamir Ngi gi A c th chng t cho B tnh xc thc ca thng bo M bng cch dng kha ring ca mnh (HA, nA) i vi thng bo M:

Trong :Cc bt ca thng bo k l: , vi 1 j 2k, si [0, k] Ch c A c th to ra 2Kbt {si} t k bt ca thng bo {mi} v ch c A mi to c 2.k2 phn t ca ma trn {hij}

*4.4 Cc s ch k s khng nnKim tra thng boMi ngi dng trn mng c th kim tra tnh xc thc ca thng bo do A gi bng cch dng thng tin cng khai (AA, nA):

Tc l:

*4.4 Cc s ch k s khng nnV d p dng: Cho k = 3, n = 5 v ma trn

Tm kha cng khai AA khi cho trc cc gi tr a1 = 1, a2 = 3, a3 = 4?Hy xc thc thng bo M = 3 v kim tra tnh xc thc ca thng bo M ?

*4.4 Cc s ch k s khng nn trnh nguy c thm m c th xc nh c ma trn H vi mt cp bn r m thch hp. Ta s tm cch ngu nhin ha thng bo M trc khi k. Ta lm nh sau:Vecto A s c nhn vi mt vecto ngu nhin R c K 2K bit: R = (r1, r2, , r2k)Thc hin php bin i: M = (M R x A) mod n hay M = (M + R x A) mod n k thng bo bin i M ta tnh theo cng thc sau: S = Mr x H + RKhi xc thc, bn nhn tnh: S x A mod n = MV d: Tr li v d trc, ta chn ngu nhin R = (1, 1, 0, 0, 0, 1)Hy xc thc thng bo M = 3 v kim tra tnh xc thc ca thng bo M ?

*4.4 Cc s ch k s khng nnS xc thc Ong-Schnorr-Shamir (1) Ngi gi A chn mt s nguyn ln nA.(2) Sau A chn mt s ngu nhin kA nguyn t cng nhau vi nA(3) Kha cng khai kA c tnh nh sau: Cp (KA, nA) c cng khai cho mi ngi dng trong mng

*4.4 Cc s ch k s khng nn(4) xc thc mt thng bo M (gcd(M, nA) = 1), ngi gi s chn mt s ngu nhin RA (gcd(RA, nA) = 1) ri tnh thng bo c xc thc l cp (S1, S2) sau:

(5) Sau A gi S cho bn thu qua mng (6) Vic kim tra tnh xc thc bn thu c thc hin nh sau:

*4.4 Cc s ch k s khng nnV d:Cho nA = 23, kA = 7Tnh kha cng khai KA?Chn RA = 13, vi M = 25, xc thc M v kim tra tnh xc thc ca M?

*4.4 Cc s ch k s c nnNn ch k

*4.4 Cc s ch k s c nnS ch k Diffie Lamport(1) Chn n cp kha ngu nhin (chng hn nh kha 56 bt ca DES) c gi b mt:

(2) Chn mt dy S gm n cp vct ngu nhin (chng hn nh cc khi u vo 64 bt ca DES), dy ny c a ra cng khai:

(3) Tnh R l dy cc kha m (chng hn l cc dy ra ca DES). Dy R cng c a cng khai, trong , 1 i n, j = 0, 1

*4.4 Cc s ch k s c nnCh k SG(M) ca mt bn tin n bt M = (m1, m2, ..., mn) chnh l dy kha sau:

Trong ij = mjV d: Thng bo

th SG(M) l:

Bn tin M v ch k SG(M) u c gi ti ni thu

*4.5 Cc s ch k s c nn

*4.5 Cc s ch k s c nnBn tin c th kim tra tnh xc thc ca thng bo bng cch:M ha cc vct tng ng ca dy S bit vi ch k SG(M) nhn So snh bn m to ra vi dy R bit

Nu dy n vct ny bng nhau th ch k c xem l xc thc

*4.5 Cc s ch k s c nn

*4.5 Cc s ch k s c nnS ch k RSA

*4.5 Cc s ch k s c nn

*4.5 Cc s ch k s c nnV d: s k s RSAn = p.q vi p, q l cc s nguyn t ln c kch thc tng ngVi K = {(n, e, d): d Zp*, ed 1 mod (n)}Ta c D = d l kha b mt, E = (n, e) l kha cng khai, m l bn tin cn kTo ch k: S = sigD(m) = md mod n Kim tra ch k: verE(m, S) = TRUE m Se mod n

*4.5 Cc s ch k s c nnTrng hp bn tin r m khng cn b mt:A k bn tin m v gi cho B.B kim tra ch k ca A

*4.5 Cc s ch k s c nnTrng hp bn tin r m cn gi b mt:A k bn tin r m c ch k SA. Sau A dng kho m cng khai EB ca B lp bn m M = EB(m, SA) ri gi n B