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Below is an example of key generation, encryption, and decryption using RSA. Look at the example, then fill in the blanks to indicate what each part is or answer the question. You do not have to do any math for this. Here are the words to select from to fill in the blanks. Modulus, encrypt key, decrypt key, factor, plaintext, ciphertext Public key is (23, 11) What is 23 called?_ Private key is (23, 13) What is 23 called? 23 can be part of the public key because it is very hard to ENCRYPT (m) = m^e mod n What is m? DECRYPT (c) = c^d mod n What is c? What is 11 called?_ What is 13 called?_ what is n? large prime numbers. (generic name, not a number)

Below is an example of key generation, encryption, and decryption using RSA. Look at the example, then fill in the blanks to indicate what each part is or answer the question. You do not have to do any math for this. Here are the words to select from to fill in the blanks. Modulus, encrypt key, decrypt key, factor, plaintext, ciphertext Public key is (23, 11) What is 23 called?_ Private key is (23, 13) What is 23 called? 23 can be part of the public key because it is very hard to ENCRYPT (m) = m^e mod n What is m? DECRYPT (c) = c^d mod n What is c? What is 11 called?_ What is 13 called?_ what is n? large prime numbers. (generic name, not a number)

Database System Concepts
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
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Below is an example of key generation, encryption, and decryption using RSA. Look at the example, then fill in the blanks to indicate what each part is or answer the question. You do not have to do any math for this. Here are the words to select from to fill in the blanks. Modulus, encrypt key, decrypt key, factor, plaintext, ciphertext Public key is (23, 11) What is 23 called? Private key is (23, 13) What is 23 called? 23 can be part of the public key because it is very hard to ENCRYPT (m) = m^e mod n What is m? DECRYPT (c) = c^d mod n Search for the following. What does "mod" mean? What is c? What is 11 called? What is 13 called? what is n? large prime numbers. (generic name, not a number)
Transcribed Image Text:Below is an example of key generation, encryption, and decryption using RSA. Look at the example, then fill in the blanks to indicate what each part is or answer the question. You do not have to do any math for this. Here are the words to select from to fill in the blanks. Modulus, encrypt key, decrypt key, factor, plaintext, ciphertext Public key is (23, 11) What is 23 called? Private key is (23, 13) What is 23 called? 23 can be part of the public key because it is very hard to ENCRYPT (m) = m^e mod n What is m? DECRYPT (c) = c^d mod n Search for the following. What does "mod" mean? What is c? What is 11 called? What is 13 called? what is n? large prime numbers. (generic name, not a number)
EXAMPLE An RSA public-key / private-key pair can be generated by the following steps: 1. Generate a pair of large, random primes p and q. NOTE for examples we use small prime numbers due to the size of the number generated. p=3, q=5 2. Compute the modulus n as n = pq. n = p*q, 15=3*5 3. Select an odd public exponent e between 3 and n-1 that is relatively prime to p-1 and q-1. e is between 3 and 14, e is relatively prime to 2 (3-1) and 4 (5-1) (n)=2*4=8 e=7, could also be 9 or 11 or 13 4. Compute the private exponent d from e, p and q. (See below.) e^-1d (mod o(n)) 7^-1d (mod 8) 7* d = 1 (mod 8) What number multiplied by 7 and then divided by 8 equals 1? Multiples of 8 8 16 24 32 48 56 64 72 80 88 96 104 Multiples of 7 7 14 21 28 35 42 49 56 63 70 77 84 91 98 105 7*15= 1 (mod 8) = 105/8= 13 remainder 1 d = 15 5. Output (n, e) as the public key and (n, d) as the private key. PUBLIC KEY IS (15, 7) (Modulus, e), PRIVATE KEY IS (15, 15) (Modulus,d) The encryption operation in the RSA cryptosystem is exponentiation to the e th power modulo n: Plaintext = 8 ciphertext = ENCRYPT (m) = m^e mod n c = 8^7 mod 15 = 2,097,152 mod 15 = 2 (ciphertext) The decryption operation is exponentiation to the dth power modulo n: m = DECRYPT (c) = c^d mod n. m= 2^15 mod 15 = 32,768 mod 15 = 8 (plaintext)
Transcribed Image Text:EXAMPLE An RSA public-key / private-key pair can be generated by the following steps: 1. Generate a pair of large, random primes p and q. NOTE for examples we use small prime numbers due to the size of the number generated. p=3, q=5 2. Compute the modulus n as n = pq. n = p*q, 15=3*5 3. Select an odd public exponent e between 3 and n-1 that is relatively prime to p-1 and q-1. e is between 3 and 14, e is relatively prime to 2 (3-1) and 4 (5-1) (n)=2*4=8 e=7, could also be 9 or 11 or 13 4. Compute the private exponent d from e, p and q. (See below.) e^-1d (mod o(n)) 7^-1d (mod 8) 7* d = 1 (mod 8) What number multiplied by 7 and then divided by 8 equals 1? Multiples of 8 8 16 24 32 48 56 64 72 80 88 96 104 Multiples of 7 7 14 21 28 35 42 49 56 63 70 77 84 91 98 105 7*15= 1 (mod 8) = 105/8= 13 remainder 1 d = 15 5. Output (n, e) as the public key and (n, d) as the private key. PUBLIC KEY IS (15, 7) (Modulus, e), PRIVATE KEY IS (15, 15) (Modulus,d) The encryption operation in the RSA cryptosystem is exponentiation to the e th power modulo n: Plaintext = 8 ciphertext = ENCRYPT (m) = m^e mod n c = 8^7 mod 15 = 2,097,152 mod 15 = 2 (ciphertext) The decryption operation is exponentiation to the dth power modulo n: m = DECRYPT (c) = c^d mod n. m= 2^15 mod 15 = 32,768 mod 15 = 8 (plaintext)
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Suppose we have 1G x 16 RAM chips that make up a 32G x 64 memory that uses high-order interleaving. (This means that each word is 64 bits in size and there are 32G of these words.) a) How many RAM chips are necessary? b) Assuming four chips per bank, how many banks are required? c) How many lines must go to each chip? d) How many bits are needed for a memory address, assuming it is word addressable? e) For the bits in part d, draw a diagram indicating many and which bits are used for chip select, and how many and which bits are used for the address on the chip.
Transcribed Image Text:Suppose we have 1G x 16 RAM chips that make up a 32G x 64 memory that uses high-order interleaving. (This means that each word is 64 bits in size and there are 32G of these words.) a) How many RAM chips are necessary? b) Assuming four chips per bank, how many banks are required? c) How many lines must go to each chip? d) How many bits are needed for a memory address, assuming it is word addressable? e) For the bits in part d, draw a diagram indicating many and which bits are used for chip select, and how many and which bits are used for the address on the chip.
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