315HW7 - (2) In order to obtain d, we should have ed mod z...

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(1) (a) M = “ help ” = 01000 00101 01100 10000 G = 101011 M * 2 5 mod G = R = 10010 = 18 (10) So the digital signature = P A - (R) = 18 3 mod 33 = 24 = 11000 (2) (b) message numeric representation P B + (M) = c=m^e mod n h 01000=8 8^13 mod 33 = 17 = 010001 e 00101=5 5^13 mod 33 = 26 = 011010 l 01100=12 12^13 mod 33 = 12 = 001100 p 10000=16 16^13 mod 33 = 4 = 000100 P A - (R) 011000=24 24^13 mod 33 = 30 = 011110 So the entire stream is 010001 011010 001100 000100 011110 (c) received message P B - (RM) = m = c^ (P b - ) mod n message 17 = 010001 17^17 mod 33 = 8 h 26 = 011010 26^17 mod 33 = 5 e 12 = 001100 12^17 mod 33 = 12 l 4 = 000100 4^17 mod 33 = 16 p 30=011110 => 30^17 mod 33 = 24 => P A - (R) So Bob can recover the whole message and the signature with his private key.
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Unformatted text preview: (2) In order to obtain d, we should have ed mod z = 1; Next, we want to show that when e and z are not relatively prime, we can not get such z: Since e and z are not relatively prime, they should have a common factor, say a, Then, e = i*a, z = j*a, then ed = k*z + 1, means i*d*a-k*j*a=1, that is (id-kj)a=1, which is not possible. (3) alice and bob need send in a message containing two parts : 1) solution 2) random number (only alice and bob knows the random number. Of course, both parts need to be hashed with hd5 or other hashing function. (4) we can use ssl (transport layer) security to do authentication. A wep code will be assigned during the authentication process....
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