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This question concerns block cipher modes. We will use a simple affine cipher, which can be expressed in C as follows. char cipherfunsigned 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) (1 163 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 charactor). We will work with the fixed key Ox08. 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 OKAA and the counter is a 3 bit counter that begins at 0. In all of th- 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 '212C330F184EBB using CTR mode. Please enter your answer in ASCII characters (aka worda).

This question concerns block cipher modes. We will use a simple affine cipher, which can be expressed in C as follows. char cipherfunsigned 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) (1 163 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 charactor). We will work with the fixed key Ox08. 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 OKAA and the counter is a 3 bit counter that begins at 0. In all of th- 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 '212C330F184EBB using CTR mode. Please enter your answer in ASCII characters (aka worda).

Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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This question concerns block cipher modes. We will use a simple affine cipher, which can be expressed in C as follows. char cipherfunsigned char block, char key) return (key+11'block)%256; The inverse of this cipher is shown below. char inv_cipherfunsigned 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 Ox08. 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 OKAA 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 n hexadecimal. a) Decrypt the ciphertext "212C330F184EBB* using CTR mode. Please enter your answer in ASCII characters (aka words). b) Decrypt the ciphertext "ACCDF904' using ECB mode. Please enter your answer in ASCIl characters (aka words). c) Decrypt the ciphertext *324 F1 F363623' using CFB mode. Please enter your answer in ASCil characters (aka words). d) Decrypt the ciphertext '5BSA91C9274170* using CBC mode. Piease enter your answer in ASCII charactors (aka words). e) Decrypt the ciphertext "3EDF673842* using OF B mode. Please enter your answer in ASCII characters (aka words). Note: You can earn partial credit on this pmhiem
Transcribed Image Text:This question concerns block cipher modes. We will use a simple affine cipher, which can be expressed in C as follows. char cipherfunsigned char block, char key) return (key+11'block)%256; The inverse of this cipher is shown below. char inv_cipherfunsigned 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 Ox08. 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 OKAA 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 n hexadecimal. a) Decrypt the ciphertext "212C330F184EBB* using CTR mode. Please enter your answer in ASCII characters (aka words). b) Decrypt the ciphertext "ACCDF904' using ECB mode. Please enter your answer in ASCIl characters (aka words). c) Decrypt the ciphertext *324 F1 F363623' using CFB mode. Please enter your answer in ASCil characters (aka words). d) Decrypt the ciphertext '5BSA91C9274170* using CBC mode. Piease enter your answer in ASCII charactors (aka words). e) Decrypt the ciphertext "3EDF673842* using OF B mode. Please enter your answer in ASCII characters (aka words). Note: You can earn partial credit on this pmhiem
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