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(1 point) 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 "303E24110012" using CTR mode. Please enter your answer in ASCII characters (aka words). purple b) Decrypt the ciphertext "80338BEEF9" using ECB mode. Please enter your answer in ASCII characters (aka words). c) Decrypt the ciphertext "3BF8D72D83" using CFB mode. Please enter your answer in ASCII characters (aka words). d) Decrypt the ciphertext "5BE1595 A0D3B" using CBC mode. Please enter your answer in ASCII characters (aka words). e) Decrypt the ciphertext "37CD6733" using OFB mode. Please enter your answer in ASCII characters (aka words).

(1 point) 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 "303E24110012" using CTR mode. Please enter your answer in ASCII characters (aka words). purple b) Decrypt the ciphertext "80338BEEF9" using ECB mode. Please enter your answer in ASCII characters (aka words). c) Decrypt the ciphertext "3BF8D72D83" using CFB mode. Please enter your answer in ASCII characters (aka words). d) Decrypt the ciphertext "5BE1595 A0D3B" using CBC mode. Please enter your answer in ASCII characters (aka words). e) Decrypt the ciphertext "37CD6733" using OFB mode. Please enter your answer in ASCII characters (aka words).

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
Problem 1PE
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Give the answer to each question. Find the answer my coding using python

(1 point) 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 "303E24110012" using CTR mode. Please enter your answer in ASCII characters (aka words). purple b) Decrypt the ciphertext "80338BEEF9" using ECB mode. Please enter your answer in ASCII characters (aka words). c) Decrypt the ciphertext "3BF8D72D83" using CFB mode. Please enter your answer in ASCII characters (aka words). d) Decrypt the ciphertext "5BE1595 A0D3B" using CBC mode. Please enter your answer in ASCII characters (aka words). e) Decrypt the ciphertext "37CD6733" using OFB mode. Please enter your answer in ASCII characters (aka words).
Transcribed Image Text:(1 point) 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 "303E24110012" using CTR mode. Please enter your answer in ASCII characters (aka words). purple b) Decrypt the ciphertext "80338BEEF9" using ECB mode. Please enter your answer in ASCII characters (aka words). c) Decrypt the ciphertext "3BF8D72D83" using CFB mode. Please enter your answer in ASCII characters (aka words). d) Decrypt the ciphertext "5BE1595 A0D3B" using CBC mode. Please enter your answer in ASCII characters (aka words). e) Decrypt the ciphertext "37CD6733" using OFB mode. Please enter your answer in ASCII characters (aka words).
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