Td corrigé EXAMEN DE CRYPTOGRAPHIE pdf

EXAMEN DE CRYPTOGRAPHIE

by using Euclid's Algorithm. Hint: you don't need to simplify x. Corrigé : Exercice 1 : 1. On a aa' = 1 + k(p ? 1) donc maa' = m.mk(p?1) m mod p, puisque m(p?1) ...




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Examen de Sécurité Informatique
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Exercice 1 (7pts)
1. On considère le crypto système (sans clé) suivant :
Un grand nombre premier p est public et les unités de message sont des entiers m, 1 d" m 5lhmH sH hE-dhE-dhmH sH hÛghmH sH hE-dhmH sH hÛgh‘V­H*hmH sH h‘V­h‘V­H*hmH sH  jºðh‘V­hmH sH h‘V­h‘V­hmH sH h‘V­hmH sH 0‚N¾` ¾MNOPQR¼U’ññàÏϾ°°°°°°°ŸŸŸ$
& F1$7$8$H$a$gd¤,E
$1$7$8$H$a$gd¤,E$
& F1$7$8$H$a$gd>5l$
& F1$7$8$H$a$gdE-d$
& F1$7$8$H$a$gd‘V­
$1$7$8$H$a$gd‘V­ TU‘’ž-QTUZ\j’“˜œž ¦§¨©ª¸Ñã*,1678£¥Ó"óêóêóêóêáÕÉÕÉÕÉÕÉÕÉáÉáÉáÉá긬óáóáóá󠔈shî~hî~hmH sH hî~hmH sH hISYhISYhmH sH hISY5\hmH sH h¤,Eh°s§hmH sH h¤,Ehû]ÅhmH sH  jÕ hû]Åhû]ÅUhmH sH hû]Åh¤,EhmH sH hû]Åhû]ÅhmH sH hû]ÅhmH sH h¤,EhmH sH h¤,Eh¤,EhmH sH )¨ª78ëL„Ó#…¤¥ý`º U r îØʾµµµµµµµµµ©©©©©© $„¤ÿ]„¤ÿa$gdî~„¤ÿ]„¤ÿgdISY $„¤ÿ]„¤ÿa$gd˜GX
$1$7$8$H$a$gd¤,E$„8„h1$7$8$H$^„8`„ha$gdû]Å$
& F1$7$8$H$a$gdû]År ¥ !U!f!¼!!"‚"Ô"Õ"#Y#£#î#:$„$›$Ð$%e%©%ð%x@ú@ü@AóóóóóóóóóåååååååååååååÙÙÙ $„¤ÿ]„¤ÿa$gd¯h€
$1$7$8$H$a$gd¯h€ $„¤ÿ]„¤ÿa$gdî~Ó"Õ""#$#(#*#;#>#A#E#L#O#R#V#_#a#t#u#{#|#¯#°#¶#·#Ô#×#Ø#Ù#Ú#Ý#$$:$
hj>
5U\hmH sH hj>
5\hmH sH h¯h€hmH sH Uh¯h€h¯h€hmH sH h¯h€h¯h€6]hmH sH  2 = 7717, y = 12667 and z = 14702. In this case, compute c1
by using Euclid s Algorithm. Hint: you don t need to simplify x.


Corrigé :


Exercice 1:
1. On a aa = 1 + k(p " 1) donc maa = m.mk(p"1) ( m mod p, puisque m(p"1) ( 1 mod p.
2.
(a) 215(1 mod 31, 415(1 mod 31
(b) #A = '(30) = '(2)'(3)'(5) = 8. A = {1, 7, 11, 13, 17, 19, 23, 29, }, éléments
dont les inverses modulo 30 sont respectivement 1, 13, 11, 7, 23, 19, 17, 29.
(c) On doit avoir b " 1 multiple de 5, d'où b = 11.
(d) De 4b = 4 le pirate déduit b = 1 ou b = 11.
Si b = 1, b0 = 1 et m = Eb0 = 8.
Si b = 11, b0 = 11 et m = Eb0 _ 811 _ 233 _ 23 = 8 mod 31.

Exercice 2:


Exercice 3:
(a) The integer k must be coprime to p"1 for the inverse k"1 mod p"1 to exist.
(b) Bob computes u ( yrrs mod p and v ( gm mod p; show that if the signature is valid, then u = v.
yrrs ( gar(gk)k"1(m"ar)( gar+m"ar ( gm mod p.
(c) she can choose integers u and v with v coprime to p " 1, compute r ( guyv mod p and s ( "rv"1 mod p"1; verify that (r, s) is a valid signatuAAAA2AÞAæA&BÊBfCÎC.DpDæDèDEEEE¾EóçÜÑÃñÃÃÃÃÃÃÃÑÜÜܨ„¤ÿ]„¤ÿgdç„$„Ð1$7$8$H$`„Ða$gdžK
$1$7$8$H$a$gd¥-ð
1$7$8$H$gd¥-ð
1$7$8$H$gd¸,ú $„¤ÿ]„¤ÿa$gd¯h€ $„¤ÿ]„¤ÿa$gdÃ/$¼AÆAÈAÊAäAæAîAòAöAøABBB$BäDèDúDüDEEEE&E(E”E˜E¨EªEäEæEêEìEîEðE
F FòæÚæÑæÑòÅÑòÅÑæ»±ª±» ’„y„i„y„[„i„i„[ jºðhç„OJQJ^Jhhç„h>5lH*OJQJ^Jhhç„OJQJ^Jhh>5lh>5lOJQJ^Jhh>5lhç„OJQJ^Jhh¸,úh¸,ú5h h¥-ð5hh¸,úh¥-ð5hh¥-ðh¥-ð\h j¹ðhžKhmH sH h¥-ðhmH sH  jºðh¥-ðhmH sH h¥-ðh¥-ðhmH sH h¥-ðh¥-ðH*hmH sH # FFFBFDF†FˆFŠFŒFŽFF”F˜FœFžF F²F´F¸FÆFÈFÊFÎFÐFhGjGnGpGtGvGxGzG”G–GðGòGHPHPJPLPNPPPRPVPXP\P^P`PbPdPfPhPjPnPrPñáñÖñáñáñÈñáñáñáÈñáñÈñáñÖñÈñáñáñÈñ½ñ»ñ«ñ«ññ«ñ«ñ«ñ«ññ« jºðho3fOJQJ^Jhho3fh>5lH*OJQJ^JhUho3fOJQJ^Jh jºðhç„OJQJ^Jhhç„OJQJ^Jhhç„h>5lH*OJQJ^Jhh>5lh>5lOJQJ^Jh7¾E„FàF6PüP_R`RaR~R€RRƒR„R†R‡R‰RŠR‹RŒRööêáÕɾ°®®®®®®®®¬®
$1$7$8$H$a$gd¢
~
1$7$8$H$gdS{¿ $„¤ÿ]„¤ÿa$gd>5l $„¤ÿ]„¤ÿa$gdK,„¤ÿ]„¤ÿgdK, $„¤ÿ]„¤ÿa$gdo3f„¤ÿ]„¤ÿgdç„re for the message m = us.
We find yrrs ( yr(guyv)s ( gusyry"r ( gus mod p, so (r, s) is the signature of the message m = us.
(d) Here (r, s) is a valid signature if yrrs ( gh(m) mod p. In order to find a message with signature (r,s), Malice now has to look for an m with h(m) = us; but this is next impossible for pre-image resistant hash functions.


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Département informatique Le 7/06/2011
Faculté Electronique et Informatique Master RSD 1ère année
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