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 256return (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 E24110012" using CTR mode. Please enter your answer in ASCII characters (aka words).b) Decrypt the ciphertext "80338 BEEF9" using EC B mode. Please enter your answer in ASCII characters (aka words).c) Decrypt the ciphertext "3 BF8D72D83" using CFB mode. Please enter your answer in ASCII characters (aka words).d) Decrypt the ciphertext "5BE1595AOD3B" using CBC mode. Please enter your answer in ASCII characters (aka words).e) Decrypt the ciphertext "37C D6733* using OFB mode. Please enter your answer in ASCII characters (aka words).

Step 1: a) Decrypt the ciphertext "303 E241 10012" using CTR mode. Please enter your answer in…

Answered: ar cipher(unsigned char block return…

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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write python code Example 2: key length = block size Input data:Message: “Hello” → x48656C6C 6FSecret Key: “4q72JHgX89z3BkFMt6cwQxL1rD28jpN5UfVhIZYPbCSeuGovRaWmA0sD9ECtX7Jf” → x34713732 4A486758 38397A33 426B464D 74366377 51784C31 72443238 6A704E35 55665668 495A5950 62435365 75476F76 5261576D 41307344 39454374 58374A66 Constants for our case hash algorithm (SHA1) with block size 64 bytes:ipad (in HEX): x36363636 36363636 36363636 36363636 36363636 36363636 36363636 36363636 36363636 36363636 36363636 36363636 36363636 36363636 36363636 36363636 (x36 repeated 64 time)opad (in HEX): x5C5C5C5C 5C5C5C5C 5C5C5C5C 5C5C5C5C 5C5C5C5C 5C5C5C5C 5C5C5C5C 5C5C5C5C 5C5C5C5C 5C5C5C5C 5C5C5C5C 5C5C5C5C 5C5C5C5C 5C5C5C5C 5C5C5C5C 5C5C5C5C (x5C repeated 64 time) Steps of creating HMAC (you could use the calculator to verify the result): If the key length = block size, then we do nothing. In our case, with a key length of 64 bytes, we proceed with no additional action.So new key is K0 = 34713732…
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(2) {w = {0, 1]* : w corresponds to the binary encoding, without leading O's, of natural numbers that are powers of 4} Testing strings: €, 0, 1, 01, 10, 11, 100, 101, 1000, 1010, 10000 The accepted strings are: L [ E 01 11 100 101 1000 1010 10000
(2) Encode the letters 'A' to 'Z' as the numbers 10 to 35, respec- tively. Taking p = 47, q = 83, use the Rabin cryptosystem to encode the message consisting of the first two letters of your family name (in the unlikely event that your name only has one letter, add a second letter). Note that the two letters yield a single, 4-digit block for encryption, not two 2-digit blocks. Decrypt the encrypted message. As part of your solution, present in full all computations, including both encoding and decoding.
Its a python programming Based on the following Sorted List, show the steps of a Binary Search if the search key was -3: [ -34, -23, -11, -5, 1, 3, 7, 11, 14, 19, 20, 25, 33, 42 ]
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…
Suppose that we have the following 128-bit AES key, given in hexadecimalrepresentation:54 77 6F 20 4F 6E 65 20 4E 69 6E 65 20 54 77 6F Construct the round key for the first round.
write an encoder and a decoder for a modified "book cipher." A book cipher uses a document or book as the cipher key, and the cipher itself uses numbers that reference the words within the text. For example, one of the Beale ciphers used an edition of The Declaration of Independence as the cipher key. The cipher you will write will use a pair of numbers corresponding to each letter in the text. The first number denotes the position of a word in the key text (starting at 0), and the second number denotes the position of the letter in the word (also starting at 0). For instance, given the following key text (the numbers correspond to the index of the first word in the line) [0] 'Twas brillig, and the slithy toves Did gyre and gimble in the wabe; [13] All mimsy were the borogoves, And the mome raths outgrabe. [23] "Beware the Jabberwock, my son! The jaws that bite, the claws that catch! [36] Beware the Jubjub bird, and shun The frumious Bandersnatch!" [45] He took his vorpal sword in…
choose the correct answer.
21. RSA problem: given plaintext P, e, prime numbers p, q, compute ciphertext C. Decrypt C and verify you get plaintext P back.
(c) The following table provides all possible block pairs of plaintext/ciphertext when using some block cipher E with a particular key K. Plaintext Ciphertext Plaintext 0000 1100 1000 1111 1001 0111 1010 1110 1011 1001 0010 1101 0001 0010 0011 0100 0101 0110 0111 1011 1100 1101 1110 1111 Ciphertext 0001 0000 0101 0100 0011 1000 0110 1010 Given that a plaintext P was encrypted using this block cipher operating in a counter mode starting with the counter value of 0, i.e. 0, 1, 2, ..., decrypt the ciphertext C = 010001110111.
AES [ Choose ] obsolete from the 1970's; uses just one 56-bit key symmetric algorithm; supports 128, 192, and 256-bit keys Is easily defeated by a Kasiski examination Not the worst one on this list, but you still shouldn't use it anymore asymmetric algorithm DES 3DES [ Choose ] RSA [ Choose ]

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ar cipher(unsigned char block return (key+11"block)%256; e inverse of this cipher is sho ar inv_cipher(unsigned char b 163 is the inverse of 11 mod return (163*(block-key+256))9 ote that the block size is 8 bits