Monday, July 20, 2009

Lecture 2 : Authentication & Cryptography

Posted by gina at 3:38 AM

Some of people is very familiar with authentication, as you know it is something that verification of our identity as security to some data...Such as login for some application which we need to insert our username and password..Besides, we use smart card, passport, biometrics system, signature, retina or use voice to verification. All this kind of technique only for one purpose, protection from atacker.
Below is an example that we need to protect our password..


  1. 1st thing, do not show password to anybody. Example : cannot write password anywhere and login the password only at a secure places.
  2. Important to find a password that is easy for us to remember but difficult to guess by others.
  3. Create password more that six character(especially using combination of number and alphabets)
  4. Do not use familiar information such as names, birthday,phone number, plate numbers or family names
  5. Others

What best part of lecture today, we learned how to guess other passwords.

  • Use default passwords to guess.
  • Use short word and easy character to guess
  • Use all the words in an electronic dictionary(60,000).
  • Hm how about collect information about the user name, family names, birthday, so on.
  • Guess user’s phone number, social security number, street address.
  • Guess license plate numbers
  • Use a Key Looger or Trojan horse so we can keep track what had been type by users.
  • Lastly, tap the line between a remote user and the host system.
I'm leaning a new thing about CRYPTOGRAPHY today..

General meaning of cryptography is about how to get secret writing..It also about the plaintext and ciphertext.

The solution of RSA calculation :
Solution :
n = p x q
= 3 x 11
= 33
Ø(n)=(q-1)(p-1)
=(10)(2)
= 20
e = 7
d = e ¯¹ mod Ø(n)
= 7 ¯¹ mod 20
a = 7 , b = 20
b = x(a) + y
20= 2(7) + 6 …………………………………(1)
7 = 1(6) + 1………………………………….(2)


From (1)
1 = 7-1(6)…………………………………….(3)
From (2)
6 = 20 – 2(7)…………………………………(4)
From (3)
1 = 7-1(6)
= 7-1 [20-2(7)]
= 7 – 20 + 2(7)
= 3(7) – 20…………………………………(5)

Multiple (5) with mod 20
1 = 3(7) – 20
mod 20 = 3(7) mod 20 – 20 mod 20
mod 20 = 3(7) mod 20
1/7 mod 20 = 3 mod 20
7 ¯¹ mod 20 = 3 mod 20
d = 3

Here is example of calculation to find the plaintext:
In a public key-stream using RSA, you intercept the ciphertext c=10 sent to whose public key is e=5, n=35. What is the plaintext m?
p = 5, q = 7
n = p * q
= 5 * 7 = 35
Ø(n) =(q-1)(p-1)
=(6)(4)
=24
d = e ¯¹ mod Ø(n)
= 5 ¯¹ mod 24
a = 5 , b = 24
b = x (a) + y
24 = 4(5) + 4…………………………………………………(1)
5 = 1(4) + 1…………………………………………………(2)

From (1)
1 = 5 – 1(4)…………………………………………………..(3)
From (2)
4 = 24 – 4(5)………………………………………………...(4)
From (3)
1 = 5 -1(4)
= 5-1[24-4(5)]
= 5 – 24 + 4(5)
= 5(5) – 24…………………………………………………(5)

Multiple (5) with mod 24
1 = 5(5) – 24
mod 24 = 5(5) mod 24 – 24 mod 24
1/5 mod 24 = 5 mod 24
5 ¯¹ mod 24 = 5 mod 24
d = 5
m = Cˆd mod n
= 10ˆ5 mod 35
= 5


 

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