This question concerns block cipher padding. Suppose the block cipher has a block size of 16 bytes. A certain message ends with a partial block which is 13 bytes. We must reversibly pad out the message to use some block cipher modes, such as CBC (even if the last block is full). Consider the following padding strategy. Determine the number of padding bytes required. This is a number n which satisfies 1

This question concerns block cipher padding. Suppose the block cipher has a block size of 16 bytes. A certain message ends with a partial block which is 13 bytes. We must reversibly pad out the message to use some block cipher modes, such as CBC (even if the last block is full). Consider the following padding strategy. Determine the number of padding bytes required. This is a number n which satisfies 1

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

Problem 1PE

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RSA cryptosystem Let p = 29 and q = 31. Then the RSA modulus n = pq = 899. We set the length of plaintext blocks to be 2 and the length of ciphertext blocks 3. Let the encryption exponent b = 29. If, for example, we want to encrypt the word live, we first divide the word into blocks of length 2, li and ve, and then find its corresponding number blocks 11 8 and 21 4. We consider these number blocks as numbers in base 26. Then we find the corresponding integers in base 10, that is 11 8 = 11 × 26 + 8 = 294 and 21 4 = 21 × 26 + 4 = 550. The integers 294 and 550 are regarded as modular numbers in Z899 and encrypted as 29429 (mod 899) = 91 and 5502⁹ (mod 899) = 492. Now write 91 and 492 in base 26 of length 3 we get 91 = 0 x 26² +3 x 26 +13 and 492 = 0 x 262 +18 × 26 +24 and the corresponding number blocks of length 3 are 0 3 13 and 0 18 24. Hen the ciphertext for live is ADNASY as 0 3 13 = ADN and 0 18 24 = ASY.
In the case of Public Key Authority It is necessary to make a request for B's public key, and the Public Key Authority must prepare a message (E(PRauth[PUB] || Request || Time1]) before sending it to A. What key A should use to decipher the message is up for debate.
In this problem we will work through a round of DES. For notational simplicity, assume it is the first round. Please enter all answers as strings of 0's and 1's. The input is the 64 bit block 0000000000000010000000000000001000000100000001000100000000000000 Suppose that the subkey for the current round is this 48 bit number: 000000000000000001000000000000000000000000000000 What does the 64 bit state look like after the IP transformation is applied to the input? 00000000000110000100000000000000 Now find Lo and Ro, the left and right halves of the state. Lo 01000000010000000000000000010000 = Ro=00000000000000100000000000000000 What is the result of applying the expansion box to R₁? E(Ro) = 00000000000000100000000000000000 What is the result of XORing the subkey with E(Ro)? k₁ E(Ro) = 00000000000000100000000000000000 We now apply the S-box transformation. S(k₁ E(Ro)) = 11011000110110001101111110111100 Finally we apply the permutation box to complete the function f. f(Ro) = P(S(k₁ + E(Ro)))…
Public Key Authority is one such instance. The Public Key Authority must prepare a request for B's public key before delivering a message to A (E(PRauth[PUB || Request || Time 1]). It is debated if the letter A should be used to decipher the message.
Given the RSA public key (3,70747) and private key (46811,70747). a. Encrypt the letter ‘z’ using its ASCII value and implement in python. b. Decrypt the result from the subtask above and generate the original text and implement in python. c. Encrypt the word ‘Cyber’ two characters at a time padding with a space and implement in python. d. Decrypt the following three blocks 14520 66071 27623. The blocks are the encryption of a five-letter word, encrypted two characters at a time and padded with a space. Recover the word and implement in python. d. The blocks are the encryption of a five letter word, encrypted two characters at a time and padded with a space. Recover the word and implement in python. e. Determine the maximum block size, in bits, that this modulus can encrypt and decrypt.
Please explain the problem in detail so that I can understand.
This question concems block cipher modes. We will use a simple affine cipher, which can be expressed in C as follows. char cipher(unsigned char block, char key) return (key+11"block)%6256, Note that the block size is 8 bits, which is one byte (and one ASCII character). We will work with the fixed key Ox08. We now encrypt various plaintexts using modes for this cipher In every case in which the mode requires an IV, the IV will be OXAA. In the case of CTR mode, we use a (nonce || counter) arrangement in which the nonce is the left 5 bits of OXAA and the counter is a 3 bit counter that begins at 0. In all of the problems given below, one character is one block. Each character of the plaintext should be regarded as its corresponding ASCII code a) Encrypt the plaintext "cheery using CTR mode. Please enter your answer in hex. (Please do *"not"" enter an 0x, as this has been done) Ox b) Encrypt the plaintext "chirper" using ECB mode Please enter your answer in hex Ox C) Encrypt the plaintext…
Plz don't use chat gpt
The hacker group Desdeia tries to break our old weak cipher which we used for the communication with the Duchess of Ligovia. It's generally known that an actual breaking of the cipher needs the reviewing a huge amount of variants which probably lasts a couple of months. Additionally we know that the reviewing of different variants needs always the same server resource. Using our intelligence sources we tried to get some information about the actual progress of breaking, but we got only some fragments: • The hacker group Desdeia uses for breaking their powerful four-core server which resources are fully dedicated to breaking process. . The actual breaking process began exactly at noon, but we don't know, on which day. . By the noon of 2nd of April, already 648,467,673 variants were successfully reviewed. Exactly at this noon, when the breaking process was already lasted for 19 days, it was attempted to double the server speed. The doubling itself was failed, but instead of it there was…
This question concerns block cipher modes. We will use a simple affine cipher, which can be expressed in C as follows. char cipher(unsigned char block, char key) { return (key+11"block)%256; } The inverse of this cipher is shown below. char inv_cipher(unsigned char block, char key) { // 163 is the inverse of 11 mod 256 return (163*(block-key+256))%256; } Note that the block size is 8 bits, which is one byte (and one ASCII character). We will work with the fixed key 0x08. We now decrypt various ciphertexts using modes for this cipher. In every case in which the mode requires an IV, the IV will be OXAA. In the case of CTR mode, we use a (nonce || counter) arrangement in which the nonce is the left 5 bits of OXAA and the counter is a 3 bit counter that begins at 0. In all of the problems given below, one character is one block. Each character of the plaintext should be regarded as its corresponding ASCII code. The ciphertext is given in hexadecimal. a) Decrypt the ciphertext "303…
use python please
Explain what the following lines of code do pad = len(byteblock)%16 * (-1)byteblock_trimmed = byteblock[64:pad]ciphertext = cipher.encrypt(byteblock_trimmed)ciphertext = byteblock[0:64] + ciphertext + byteblock[pad:]