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Computer Security Set of slides 4 Dr Alexei Vernitski.

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1 Computer Security Set of slides 4 Dr Alexei Vernitski

2 Public-key cipher We consider a scenario when Alice wants to send a confidential message to Bob Alice and Bob use two different keys Alices key is the public key: it is publicly known Bobs key is the private key: only Bob knows it Also called asymmetric cipher

3 Public-key cryptography Public-key cryptography is called public-key cryptography because it uses two types of keys: – Public keys, which are known to everyone and used to encrypt messages – Private keys, which are known only to the person who has received the message and wants to decrypt it.

4 Public-key cryptography Suppose Bob wants other people to send messages to him confidentially He chooses (but does not tell anyone) a private key. This is the key he shall use for decrypting messages arriving to him. At the same time, he chooses and published a public key. This is the key other people will use to encrypt messages to send them to Bob.

5 Keys and blocks In ciphers like DES, keys are just arrays of bits. In public-key cryptography, keys are parameters of some complicated calculations, and they are not necessarily arrays of bits. In ciphers like DES, a message is treated as a long array of bits, and is split in blocks. In public-key cryptography, blocks are not necessarily arrays of bits.

6 RSA RSA is a public-key cipher invented in the 1970s. It is still considered secure and is used in many applications

7 Modular arithmetic This example is modulo 7 The numbers allowed are 0 to 6 After 6, numbers wrap around 0 = 7 (mod 7) 3+3 = 6 (mod 7) 4+4 = 1 (mod 7)

8 Mock RSA This is a simplified version of RSA Bob finds three numbers e, d, n such that ed = 1 (mod n) e is for encryption, d is for decryption For example, e = 2, d = 3, n = 5 Each block m in a message is a number between 0 and n-1

9 Mock RSA For example, e = 2, d = 3, n = 5 m is a number between 0 and n-1 To encrypt, calculate c = em modulo n To decrypt, calculate dc = dem = 1m = m modulo n Alices (public) key is the pair e and n Bobs (private) key is the pair d and n Both keys are prepared by Bob

10 RSA For example, e = 3, d = 7, n = 33 m is a number between 0 and n-1 To encrypt, calculate c = m e modulo n To decrypt, calculate c d = m ed = m 1 = m modulo n Alices (public) key is the pair e and n Bobs (private) key is the pair d and n Both keys are prepared by Bob

11 Now say we want to encrypt the message m = 7 c = m e (mod n) = 7 3 (mod 33) = 343 (mod 33) = 13. Hence the ciphertext c = 13. To decrypt, we compute m = c d (mod n) = 13 7 (mod 33) = 7.

12 RSA RSA is secure because it is difficult to find d when n and e are known Of course, n, e and d should be larger than in our example (say, )

13 Large integers We need to perform arithmetic with large integers, say, numbers occupying 1000 bits in memory. Is the standard implementation of integer suitable for this?

14 Raising into large powers We need to raise into large powers For the sake of an example, we can say that we need to calculate m 100 How can we do this efficiently? – Using the modular arithmetic – Re-using smaller powers, where possible

15 Encoding data Blocks of RSA have an exotic format How do you prepare data for being encrypted by RSA? Homework: where can you find the standard describing the recommended scheme for data encryption and decryption with RSA?

16 Using RSA with other ciphers How can RSA and, say, AES work together as parts of a cryptographic protocol of a software system? We want to use the best of each of them

17 RSA – Problem 1 Recall how the RSA works: The public key is a pair e and n Bobs private key is a pair d and n To encrypt, calculate c = m e (mod n) To decrypt, calculate c d = m ed = m 1 = m (mod n) Problem 1: Bob has published the public key e = 7, n = 247. Use this public key to encrypt a message m = 100.

18 RSA – Problem 2 Problem 2: Bob has published the public key e = 317, n = 851. Alice has encrypted a message m = 111 using this key and obtained an encrypted message c = 148. Use this information to find the private key. Recall how the RSA works: The public key is a pair e and n Bobs private key is a pair d and n To encrypt, calculate c = me (mod n) To decrypt, calculate cd = med = m1 = m (mod n)

19 Stream ciphers What is the simplest implementation of a cipher based on a key stream? What is the difference between a one-time pad cipher and a stream cipher? What are the ways of obtaining a random key stream for a one-time pad cipher? What are the ways of obtaining a pseudorandom key stream for a stream cipher?

20 Linear feedback shift register At each step, each bit is shifted by one position to the right The new value of the leftmost bit is calculated as an XOR of the bits that stood at so-called tap positions XOR

21 Linear feedback shift register For example, populate the register as follows: Use the rightmost bit (1) as the first bit of the key stream Find the bits in the tap positions and XOR their values: Shift the register: ? Provide a new value for the leftmost bit (as the XOR of the bits that were in tap positions):

22 Linear feedback shift register LFSRs can be used to produce a pseudorandom key stream The length of the register and the choice of the tap positions are important If they are chosen correctly, the LFSR will get back to its original value only after it has taken all other possible values Such an LFSR is called maximum-length

23 Sample exam questions Explain the difference between symmetric and asymmetric ciphers. What are the relative advantages of each of these types of cipher? Give an example of a public key cipher Show exactly (with formulas) how a message is encrypted and decrypted in RSA

24 Sample exam questions Explain the difference between block ciphers and stream ciphers Compare one-time pad ciphers and stream ciphers. What are the relative advantages of each of these types of cipher? Explain briefly how a pseudorandom key stream can be produced for a stream cipher


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