You have an originating plain text message (in hexadecimal): 41 72 61 62 Use your Student ID number to obtain a 10-bit key as follows: Key (in decimal) = (ID mod 1024) +1.You have to manually cipher the first block of the following plain text message (first 8 bits). The encrypting process will be done using Simplified DES (S-DES). you must explain all steps in details [mandatory: write down all the calculations].
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You have an originating plain text message (in hexadecimal): 41 72 61 62 Use your Student ID number to obtain a 10-bit key as follows: Key (in decimal) = (ID mod 1024) +1.You have to manually cipher the first block of the following plain text message (first 8 bits). The encrypting process will be done using Simplified DES (S-DES). you must explain all steps in details [mandatory: write down all the calculations].
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- To prevent the tapping and use of information on a wireless network which has brought untold mistrust amongst the directors of Okonko Systems, you suggested that every message that is transmitted should go through the following: A systematic two left bit shift; After “a” above, the algorithm f[(6m +2) + k] mode 13 should be applied to the cipher text, where m is the message and k =9 is a key; After “b” above, perform a systematic right bit shift to the new cipher; The message should then be transmitted. If on a trial basis, the message is “genomics” Use the mono – alphabetic letters where a = 0, b = 1, ….z = 25 to find the cipher that would be transmittedAssume you are using any correct plaintext padding method, such as those described in lecture, with a 128-bit (16-byte) block cipher. If you are sending a message that is 184 bytes long, how many padding bytes would you need to add? Justify your answer. For the remaining questions, consider a 4-bit block cipher, described in hexadecimal by the following table: Plaintext Ciphertext Plaintext Ciphertext 0 a 8 e 1 c 9 d 2 f a 0 3 6 b 7 4 3 c 5 5 8 d b 6 4 e 9 7 2 f 1 You can think of this as a simple substitution cipher for hexadecimal digits. There is no “key” other than the table itself.Suppose we will use Advanced Encryption Standard (AES) to encrypt a block of plaintext and the known round key for round 7 is: B5 8D BA D2 31 2B F5 60 7F 8D 29 2F EA D2 73 21 What is the third word (i.e. the third 4 bytes) of the round key for round 8?
- Compute CBC-MAC for a message of 16 bits, “ABCD” (in Hexa). Assume a block size of 8 bits with an IV=C9 (in hexa). For simplicity, assume the encryption to be a simple XOR of the key with the plaintext. Let the encryption key be D8 (in Hexa). (Hint: Divide the message into blocks of 8 bits each; XOR each block with the previous cipher output; then encrypt this with the key. For the first block, XOR it with IV.There are other modes of block cipher besides the ones (OFB,CFB,CTR). One of these modes is named Plaintext Block Chaining (PBC) Mode. On the encryption side, the following is executed to obtain the nth ciphertext: Cn := Ek(Mn)XOR Mn-1. Suppose that we need to encrypt M1; : : : ;M5 using the PBC mode. Show the explicit formulas to obtain C1; : : : ;C5. What do you need to use for M0? Also, show the steps on the decryption side to obtain M1; : : : ;M5.Suppose we will use Advanced Encryption Standard (AES) to encrypt a block of plaintext andthe known round key for round 7 is:B5 8D BA D2 31 2B F5 60 7F 8D 29 2F EA D2 73 21Please calculate the first four words of the round key for round 8 by following the steps below:(1) What is the ‘temp’ value after RotWord? (2) What is the value after SubWord? (3) What is the value after XOR with Rcon? (4) What is the first word (i.e. the first 4 bytes) of the round key for round 8? (5) What is the third word (i.e. the third 4 bytes) of the round key for round 8?
- N blocks of plaintext (P1 to PN) are being encrypted in CFB mode into ciphertext (C1 to CN). Assume that two bits of C3 are erroneous (their binary values have flipped). What bits of C3, C4, and C20 are erroneous?The following is a message encrypted using affine cipher and you know the multiplicative key is 7 and the additive key is 3. Decrypt the message and show all your work. RFS DEY DCL CFA XXV HDE YFO REV WVO VEX VHD EYF ORE VAD BQO BOW EYD WVF ORF XVK XLP YBO BEL QDB XEV WDH BOW EYF OTF ORA DBQ OFO ROB ELQ BAA NQD XLA EXa. Compute CBC-MAC for a message of 16 bits, “8642” (in Hexa). Assume a block size of 8 bits with an IV=F1 (in hexa). For simplicity, assume the encryption to be a simple XOR of the key with the plaintext. Let the encryption key be B4 (in Hexa). (Hint: Divide the message into blocks of 8 bits each; XOR each block with the previous cipher output; then encrypt this with the key. For the first block, XOR it with IV. Details in pages 325-326 Ch.12 of the textbook) b. Suppose Alice computes the Secret prefix MAC (page 322: secret prefix MAC(x) = h(key || x)) for the message ”AM” (in ASCII) with key “G” (in ASCII) that both Alice and Bob know. The hash function that is used is h(x1x2x3)= g(g(x1 XOR x2) XOR x3 ) where each xi is a character represented as 8 bits, and g(x) is a 8-bit string that is equal to the complement of bits in x. For example, g(10110011) = 01001100. The MAC is 8 bits. (8-bit ASCII representation of the characters is given below.) What is the Secret prefix MAC…
- 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…We use DES in cipher feedback mode (CFB) to encrypt a plaintext m = m1m2 ...m100 into a ciphertext c1c2 ...c100, where each mi is 8-bit long. The ciphertext is sent to Bob. If c15 and c25 are missing and c8 and c88 are received as c8' and c88' wrongly, what mi’s can B compute correctly from the received ciphertext?Just comment in my code text and explaining each function or block of code that all i need help on In cryptography, a Caesar cipher is a very simple encryption technique in which each letter in the plain text is replaced by a letter some fixed number of positions down the alphabet. For example, with a shift of 3, A would be replaced by D, B would become E, and so on. The method is named after Julius Caesar, who used it to communicate with his generals. ROT-13 (“rotate by 13 places”) is a widely used example of a Caesar cipher where the shift is 13. In Python, the key for ROT-13 may be represented by means of the following dictionary: key = {‘a’:’n’, ‘b’:’o’, ‘c’:’p’, ‘d’:’q’, ‘e’:’r’, ‘f’:’s’, ‘g’:’t’, ‘h’:’u’, ‘i’:’v’, ‘j’:’w’, ‘k’:’x’, ‘l’:’y’, ‘m’:’z’, ‘n’:’a’, ‘o’:’b’, ‘p’:’c’, ‘q’:’d’, ‘r’:’e’, ‘s’:’f’, ‘t’:’g’, ‘u’:’h’, ‘v’:’i’, ‘w’:’j’, ‘x’:’k’, ‘y’:’l’, ‘z’:’m’, ‘A’:’N’, ‘B’:’O’, ‘C’:’P’, ‘D’:’Q’, ‘E’:’R’, ‘F’:’S’, ‘G’:’T’, ‘H’:’U’, ‘I’:’V’, ‘J’:’W’, ‘K’:’X’, ‘L’:’Y’,…