Using the key leg with vigenere cipher 2 a decrypt the...

Name: using the key leg with vigenere cipher 2 a decrypt the ciphertext c0 c1 ldots c31 0101 1111 1111 0100 0001 0101 0111 1010 from the following lfsr 6 co3 k1 k0 0 1 2 3 4 5 6 7 tion techniqu 25237
Uploaded: 2022-05-17T08:08:52-08:00
Duration: 4 min 22 s
Channel: Madhur L
Description: using the key leg with vigenere cipher 2 a decrypt the ciphertext c0 c1 ldots c31 0101 1111 1111 0100 0001 0101 0111 1010 from the following lfsr 6 co3 k1 k0 0 1 2 3 4 5 6 7 tion techniqu 25237

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We conduct a known-plaintext attack on an LFSR-based stream cipher. We know that the plaintext sent was: 1001001001101101100100100110. By tapping the channel carrying the ciphertext, we observe the following stream: 1011110000110001001010110001. (a) What is the length m of the key stream generator (assume the length is no more than 6 and you can use any tool to compute the determinant. But you must display the matrices used)? (b) What is the initialization vector? (c) Determine the feedback coefficients of the LFSR.

Madhur L.

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 "2227370922273709" using CTR mode. Please enter your answer in ASCII characters (aka words). b) Decrypt the ciphertext "3EAC33803EAC3380" using ECB mode. Please enter your answer in ASCII characters (aka words). c) Decrypt the ciphertext "314C23FAA3314C23FAA3" using CFB mode. Please enter your answer in ASCII characters (aka words). d) Decrypt the ciphertext "8ADF9821F48ADF9821F4" using CBC mode. Please enter your answer in ASCII characters (aka words). e) Decrypt the ciphertext "3BD56939533BD5693953" using OFB mode. Please enter your answer in ASCII characters (aka words)

Akash M.

All of the results of this section can be generalized to the case where the plaintext is a binary message; that is, it is a sequence of 0's and 1's. In this case we do all of our modular arithmetic using modulus 2 rather than modulus $26 .$ Thus, for example, $1+1=0(\bmod 2) .$ Suppose we want to encrypt the message $110101111 .$ Let us first break it into triplets to form the three and let us take $\left[\begin{array}{lll}1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 1 & 1\end{array}\right]$ as our $\left[\begin{array}{l}1 \\ 1 \\ 0\end{array}\right],\left[\begin{array}{l}1 \\ 0 \\ 1\end{array}\right],\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]$ vectors enciphering matrix. (a) Find the encoded message. (b) Find the inverse modulo 2 of the enciphering matrix, and verify that it decodes your encoded message.

Elementary Linear Algebra: Applications Version

Applications of Linear Algebra

Cryptography

Transcript

00:01 Hi, in this question we can say that by zeroing the cyporetext stream with the key stream, we get the following, 0 .0111, 1110, 11101, 11001, 1 -0111, 1 -0111, 1001 and 01111.

00:34 Now if we arrange this into 7 bit blocks then we can easily see that the key represents every 7 bit.

00:43 The period of the so the period of the lf sr is therefore 7.

00:55 So 0 -1 -001 -1 -1 -1 -1 -0 -1 -1 -1 -1 -1 -0 -1 -1 -1 -0 -1 -1 -1 -1 therefore the period of an lf sr is defined by we have p is equal to 2 raise to m minus 1.

01:35 M will be equal to log of p plus 1 to the base 2...

Question

using the key leg with Vigenere cipher 2 a) Decrypt the Ciphertext ($$C_0, C_1, \ldots C_{31}$$) = (0101 1111 1111 0100 0001 0101 0111 1010) from the following LFSR 6 CO3 k₁ k₀ 0 1 2 3 4 5 6 7 tion technique. Apply Rail fence 4 CO2

Question

using the key leg with Vigenere cipher 2 a) Decrypt the Ciphertext ($$C_0, C_1, \ldots C_{31}$$) = (0101 1111 1111 0100 0001 0101 0111 1010) from the following LFSR 6 CO3 k₁ k₀ 0 1 2 3 4 5 6 7 tion technique. Apply Rail fence 4 CO2

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