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Problem Session CMPE553 29122010

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The document discusses asymmetric key cryptography techniques like RSA and Diffie-Hellman key exchange as well as hash functions.

In little-endian, the bytes are interpreted with the least significant byte first while in big-endian it is the other way around.

To encrypt, compute the ciphertext as the plaintext raised to the power of the public key modulo n. To decrypt, raise the ciphertext to the power of the private key modulo n.

Problem Session CMPE-553 Cryptography and Network Security 29.12.

2010
Chapter 9. RSA 9.2. Perform encryption and decryption using RSA algorithm for the following a. p=3, q=11, e=7, M=5 b. p=5, q=11, e=3, M=9 c. p=7, q=11, e=17, M=8 d. p=17, q=31, e=11, M=7 e. p=17, q=31, e=7, M=2

9.3. In a public-key system using RSA, you intercept the ciphertext C=10 sent for a user whose public key is e=5, n=35. What is the plaintext M? M=5 9.4. In an RSA system, the public key of a given user is e=31, n=3599. What is the private key of the user? D=3031 Chapter 10. Key Management 10.1. Users A and B use the Diffie-Hellman key exchange technique a common prime q=71 and a primitive root a=7. a. if user A has private key XA=5 what is As public key YA? b. if user B has private key XB=12 what is As public key YB?

10.2. Consider a Diffie-Hellman scheme with a common prime q=11 and a primitive root a=2. a. show that 2 is a primitive root of 11 b. if user A has public key YA=9, what is As private key XA? c. If user B has public key YB=3, what is the shared secret key K?

10.4. In 1985, T. ElGamal announced a public-key scheme based on discrete logarithms, closely related to the Diffie-Hellman technique. As with Diffie-Hellman, the global elements of the ElGamal scheme are a prime number q and a,a primitive root of q. A user A selects a private key XA and calculates a public key YA as in Diffie-Hellman. User A encrypts a plaintext M<q intended for user B as follows: 1. Choose a random integer k such that 1<=k<q 2. Compute K=(YB)k(modq) 3. Encrypt M as the pair of integers (C1, C2) where C1=ak(modq), C2=KM(modq) User B recovers the plaintext as follows: 1. Compute K=(C1)XB(modq) 2. Compute M=(C2*K-1)modq Show that the system works; that is, show that the decryption process does recover the plaintext.

10.5. Consider an ElGamal scheme with a common prime q=71 and a primitive root a=7. a. If B has public key YB=3 and A chose the random integer k=2, what is the ciphertext of M=30? b. If now A chooses a different value of k, so that the encoding of M=30 is C=(59, C2), what is the integer C2?

Example 1 Consider an El Gamal scheme with a common prime q=71 and a primitive root =7. (a) If Bob has public key YB=3 and Alice chose the random integer k=2, what is the ciphertext of M=30? (b) If Alice now chooses a different value of k, so that the encoding of M=30 is C=(59, C2), what is the integer C2? Solution (a) K=(YB)k mod q=32 mod 71=9 C1=k mod q=72 mod 71=49 C2=KM mod q=930 mod 71 =270 mod 7157 (b) In this case we have C1=59=k mod q=7k mod 71 We need to solve a discrete logarithm to find k. It can be shown that k=3 because: 73 mod 71=343 mod 71=59=C1 K=Y mod q=3kB mod 71 =27 C2=KM mod 71=2730 mod 71 =810 mod 7129
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Chapter 11. 11.6. Using an encryption algorithm, construct a one-way hash function. Consider using RSA with a known key. Then process a message consisting of a sequence of blocks as follows: Encrypt the first

block, XOR the result with the second block and encrypt again, and so on. Show that this scheme is not secure by solving thee following problem. Given a two-block message B1, B2, and its hash RSAH(B1,B2)=RSA(RSA(B1)+B2) And given arbitrary block C1, choose C2 so that RSAH(C1, C2)=RSAH(B1,B2)

Chapter 12. 12.4 How 4-byte word a1a2a3a4 is interpreted in computers with little-endian and big-endian architectures? Big-endian: a1*2^24+a2*2^16+a3*2^8+a4 Little-endian: a4*2^24+a3*2^16+a2*2^8+a1 How MD5 (assuming a little-endian architecture) addition operation (X+Y) would be carried out on a big-endian machine?

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