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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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…
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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