Cs 506 (Week 4)

Front 1

Block Ciphers

Back 1

sequence of permutation and substitution ciphers

Front 2

Electronic Codebook Mode (ECB)

Back 2

Identical blocks result in identical ciphertext; blocks are encrypted independently; no error propagation

Front 3

Cipherblock Chaining Mode (CBC)

Back 3

same plaintexts = same ciphertexts if same key and IV used. Each section depends on the one before it.

Front 4

Cipher Feedback Mode (CFB)

Back 4

same plaintexts = same ciphertexts if same key and IV used; chaining happens. idk how the errors work

Front 5

Output Feedback Mode (OFB)

Back 5

same plaintexts = same ciphertexts if same key and IV used; plaintext-independent preceding blocks; error in one bit affects only its decryption

Front 6

Counter Mode (CTR)

Back 6

same plaintexts = same ciphertexts if same key and nonce used; no chaining; system recovers from ciphertext bit errors

Front 7

Public-Key Encryption

Back 7

Incoming messages would have been encrypted with the recipient's public key and can only be decrypted with his corresponding private key.

Front 8

Merkle Hellman Cryptosystem

Back 8

public private key, uses knapsacks

Front 9

merkle hellman key generation

Back 9

make an easy knapsack as a private key; make public key by turning it into a difficult knapsack

Front 10

merkle hellman encryption

Back 10

pack knapsack; comput the ssum of knapsack using public key.

Front 11

merkle hellman decryption

Back 11

Transform the sum of the knapsack and solve the respective knapsack problem using the private key.

Front 12

RSA

Back 12

generate 2 large prime, p,q; compute n=pq and o=(p-1)(q-1); select random int e .: 1 < e < o .: gcd(e, o) = 1; compute d .: 1 < d < o .: = 1 mod (o); A's public key is (n,e); private is d.

Front 13

RSA encryption

Back 13

B encrypts message M for A by taking A's public key and running computations on m based on A, then sends ciphertex to A

Front 14

RSA decryption

Back 14

Use private key d to find m = c^d mod n.

Front 15

Symmetric Key Cryptography

Back 15

keys for decryption and encryption are exactly the same shared secret.

Front 16

TTP

Back 16

Trusted Third Party

Front 17

Properties of acryptographic hash function

Back 17

pre-image resistance; second pre-image resistance; collision resistance

Front 18

How to encrypt with RSA (n=33, e=3, d=7)

Back 18

m=5 -> c=5^3 mod 33 = (25*5) mod 33 = 26

Front 19

How to compute RSA signature (n=33, d=7)

Back 19

m=4 -> s = 4^7 mod 33 = 16