Concept explainers
1. Encrypt the message “Take your bag with clothes” using the Hill cipher with the key !
3 2
3 5 <-it is matrix
Show your calculations and the result.
2.This problem explores the use of a one-time pad version of the Vigenere cipher. In this
scheme, the key is a stream of random numbers between 0 and 26. For example, if the key is 18
9 3…, then the first letter of plaintext is encrypted with a shift of 18 letters, the second with a
shift of 9 letters, the third with a shift of 3 letters, and so on.
Encrypt the plaintext “cryptography” with the key stream
10 22 5 4 1 0 2 9 18 16 16 0
3.Does the set of residue classes (mod3) form a group with respect to modular addition?
4. Determine -11 mod 9
5. Determine gcd(65042, 40902)
6.Using the extended Euclidean
Please write the correct answer. Thank you!
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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…arrow_forward8. Also, in the Caesar cipher you could not encrypt two different letters to be the same letter (meaning if “p" → "S" then nothing else can go to “S"). Is this true of the Vigen'ere cipher too?arrow_forwardThis 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…arrow_forward
- use python pleasearrow_forwardThis 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 "spider" 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 "spelling" using ECB mode. Please enter your answer in hex. Ox c) Encrypt the…arrow_forwardThis 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…arrow_forward
- q5- The Atbash cipher is a type of monoalphabetic cipher formed by taking the alphabet and mapping it to its reverse, so that the first letter becomes the last letter, the second letter becomes the second to last letter, and so on. For example: A becomes Z and Z becomes A, B becomes Y and Y becomes B, and so on. Numeric digits and special symbols will remain the same. What does XHV2HRZ mean?arrow_forwardSubstitution Cipher is an old technique that substitutes one thing for another. For example, alphabet messages can be ciphered by applying the following substitution. plaintext: abcdefghijklmnopqrstuvwxyz | | V Vciphertext: mnbvcxzasdfghjklpoiuytrewq However, this type of ciphering produces patterns since the same plaintext strings produce the same cipher text. Polyalphabetic Cipher, on the other hand mitigates this problem as plaintext characters are not always replaced with the same cipher text character. The replacement of each character dependson a key as well as on the position of the character in the text; for example, by using a formula such as: for (int i=0; i < length of P; i++) C[i] = P[i] + K + (i mod 3) In essence, if the key, K, is choses as K=10, then 10 is added to characters in position 0, 3, 6, …; and 11 is added for those in…arrow_forwardCan you please show the shifts for the encrypt & decrypt in this format? For example, like this : Example:ABCDEFGHIJKLMNOPQRSTUVWXYZ Row 1FGHIJKLMNOPQRSTUVWXYZABCDE Row 2 If the plaintext to encrypt is:THIS IS THE ANSWER The ciphertext becomes:YMNX NX YMJ FSXBJW This is for me to help understand and to redo the problem for studying. Thank you!!arrow_forward
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