a) Encrypt the plaintext "blotto" using CT R mode. Please enter your answer in hex. (Please do *not** enter an Ox, as this has been done.) Ox b) Encrypt the plaintext "blind" using ECB mode. Please enter your answer in hex. Ox c) Encrypt the plaintext "rango" using CF B mode. Please enter your answer in hex. Ox d) Encrypt the plaintext "rind" using C BC mode. Please enter your answer in hex. Ox e) Encrypt the plaintext "pepper" using OF B mode. Please enter your answer in hex. Ох
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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 cipher(unsigned char block, char key) { return (key+11*block)%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 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 O. 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 "blotto" using CTR mode. Please enter your answer in hex. (Please do **not** enter an Ox, as this has been done.) Ox b) Encrypt the plaintext "blind" using ECB mode. Please enter your answer in hex. Ox c) Encrypt the…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;} 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 encrypt various plaintexts using modes for this cipher. In every case in which the mode requires an IV, the IV will be 0xAA. In the case of CTR mode, we use a (nonce || counter) arrangement in which the nonce is the left 5 bits of 0xAA 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 "?????" using ??? mode. Please enter your answer in hex. (Please do **not** enter an 0x, as this has been done.)0x= * Can you please answer this question // input answer next to the "=" sign.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; } 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 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 Ox, as this has been done.) Ox b) Encrypt the plaintext "chirper" using ECB mode. Please enter your answer in hex. Ox c) Encrypt the…
- 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; } 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 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 "lippo" using CTR mode. Please enter your answer in hex. (Please do **not** enter an Ox, as this has been done.) Ox b) Encrypt the plaintext "lippi" using ECB mode. Please enter your answer in hex. Ox c) Encrypt the…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…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…
- 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 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 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"…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 0xAA. In the case of CTR mode, we use a (nonce || counter) arrangement in which the nonce is the left 5 bits of 0xAA 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…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 0xAA. In the case of CTR mode, we use a (nonce || counter) arrangement in which the nonce is the left 5 bits of 0xAA 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 "2227370922273709"…
- 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…Encryption is commonly used to disguise messageson the internet. A Caesar cipher performs a shift ofall of the characters in a string (based on their ASCIIvalues, see Table 2.1), e.g.h e l l o → m j q q tThe example shows a shift with a distance of 5characters, i.e. h(ASCII:104) → m(ASCII:109)Write a C/C++ program that asks the user to input aline of plaintext and the distance value and outputsan encrypted text using a Caesar cipher, with theASCII values range from 0 through 127. Useunderscores (ASCII: 95) to represent spacecharacters.Underscore characters should not be encrypted,and any character that is encrypted may notbecome an underscore. In this case, the charactershould be changed to the next character in theASCII table.The program should work for any printablecharacters.NB: No strings (datatype) or library functions maybe used.See Figure 2.1 for example output.Using C programming language: A Transposition Cipher A very simple transposition cipher encrypt(S, N) can be described by the following rules: If the length of S is 1 or 2, then encrypt(S, N) is S. If S is a string of N characters s1 s2 s3... sN and k = N/2, then encrypt(S)= encrypt(sk sk−1... s2 s1 ,K)+ encrypt(sN sN−1... sk+1 ,N - K) where + indicates string concatenation. For example, encrypt("Ok", 2) = "Ok" and encrypt("12345678", 8) = "34127856". Write a program to implement this cipher. The input is a file that is guaranteed to have less then 2048 characters. Code structure might look like this: #define MAX_SIZE 2048char text_buffer[MAX_SIZE];int main(){ // read file into text_buffer encrypt(text_buffer, n); // print out text_buffer return 0;}