Using CBC and IV=101, Encrypt 111100011100? Input Output 000 110 001 111 010 101 011 100 100 011 101 000 110 001 111 010
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Using CBC and IV=101, Encrypt 111100011100?
| Input | Output |
|---|---|
| 000 | 110 |
| 001 | 111 |
| 010 | 101 |
| 011 | 100 |
| 100 | 011 |
| 101 | 000 |
| 110 | 001 |
| 111 | 010 |
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- Perform the following internal operations involved in DES and AES encryption techniques. | Q2: Perform shift left by 2 bits for the below given data 1111111000011001100101010101 1110101010101100110011110001S₁ By using advanced encrption standards encrypt the following messag 87 87 87 87 87 87 87 87" with the key (0E32923E A6D0D73) Note: Use these table below for key generation permutation choice 1 table S₁ PC-1 49 57 41 17 25 Tables below for message eneryption 50 26 34 58 1 10 35 2 59 43 19 44 11 3 52 Initial permutation. 55 31 63 47 23 62 7 54 30 38 14 37 61 6 45 IP 21 13 5 20 12 58 50 42 34 26 18 60 52 44 36 28 20 62 54 46 38 30 22 Iteration number tabl 64 56 48 40 32 24 57 49 41 33 25 17 59 51 43 35 27 19 61 53 45 37 29 21 Number of 63 55 47 39 31 23 Iteration Number Left Shifts 1 1 2 1 3 Expansion table 4 5 6 E BIT-SELECTION TABLE 1 2 3 4 5 6 7 8 9 10 11 13 14 15 17 18 19 21 22 23 25 26 27 29 30 31 ▬▬▬▬▬▬▬▬▬▬ 3 23 16 41 30 44 46 7 52 40 49 42 8 9 33 2 1 50 39 6 3 38 51 39 50 42 51 60 46 53 28 10 11 12 13 2 14 15 16 Prmutation choice 2 tabl PC-2 14 17 11 24 1 28 15 21 19 12 26 7 27 13 2 31 47 55 33 48 34 53 29 32 6 4 20 37 2 2 2 2 2 1 2 45 56 36 2 26579586 5 10 8 18 36 15. 22 32 4 8…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) = 665. 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
- 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].Let 2B7E151628AED2A6ABF7158809CF4F3C be a 128-bit AES key (given in hexadecimal notation). Use the given 128-bit key to compute the encryption of the following plaintext (given in hexadecimal notation) using the 10-round AES: 3243F6A8885A308D313198A2E0370734Given the binary value 11011.01011, what is its normalized value?
- You have intercepted an RSA encrypted message with encoding exponent 7949 and base 44799308449717. Decode the location: 12011932256475,5786692130817, 36685018221928. Enter the decoding exponent: Enter the location (case sensitive):Your professor asked you to encrypt the plaintext bits: 1000 0111 with key bits: 00100 10111(after parity drop) using S- DES, where the block size is 8 bits and round key generator input size 10 bits and output size 8 bits. Use the following S- boxes and P-boxes for encryption. Perform one left shift. 1 IP 2 63 148 5 IP-1 4 572 8 6 0123 0 [1032] SO-1 3 2 10 2021 3 3 313 P10 41019 8 6 0 S1-1 2 3 6 E/P 2 2 4 0123 0 1 2 3] |2 0 13 30 10 103 P8 4 8 3 2 P4 3 5 10 9Given the codeword=01110101111101010, find if it has an error bit or none. If it has an error bit, indicate the bit position. Data message has 12 bits.
- p T Note: The notation from this problem is from Understanding Cryptography by Paar and Pelzl. Suppose you have an LFSR with 6 state bits. The first 12 bits of output produced by this LFSR are 100011011000 = s0 $1 $2 S3 S4 S5 S6 87 Sg S9 S10 S11 - The first bit produced is the leftmost bit and the bit most recently produced is the rightmost bit. a) What is the initial state of the LFSR? Please enter your answer as unspaced binary digits (e.g. 010101 to represent S5 = 0, s4 = 1, 83 = 0, s2 = 1, s1 = 0, so = 1). %3D b) What are the tap bits of the LFSR? Please enter your answer as unspaced binary digits (e.g. 010101 to represent P5 = 0, p4 = 1, pP3 = 0, p2 = 1, p1 = 0, po = 1).If we use the 128-bit binary field extension F 128 to decode the 127-bit binary BCH code at a distance of 6, we get a total of 127 bits. How large is the code, and is there a minimum distance you can guarantee?Calculate the state of PRESENT-80 after the execution of one round.Use the following values (in hexadecimal notation):Plaintext: 0000 0000 0012 3456Key: BBBB 5555 5555 EEEE FFFF