Chapter 2: Problem 7 Previous Problem Problem List Next Problem Results for this submission Entered Answer Preview Result 0110 0110 incorrect The answer above is NOT correct. (1 point) Note: The notation from this problem is from Understanding Cryptography by Paar and Pelzl. We conduct a known-plaintext attack against an LFSR. Through trial and error we have determined that the number of states is m = 4. The plaintext given by 10111001 = xox1x2x3X4X5X6X7 when encrypted by the LFSR produced the ciphertext 01011010 = yo Y Y2 Y3 Y3 Ys Y6y7. What are the tap bits of the LFSR? Please enter your answer as unspaced binary digits (e.g. 0101 to represent p3 = 0, P2 = 1, p1 = 0, po = 1). 0110 Results for this submission Entered Answer Preview Result 111011 111011 incorrect 10101 10101 incorrect At least one of the answers above is NOT correct. (1 point) Note: The notation from this problem is from Understanding Cryptography by Paar and Pelzi. Suppose you have an LFSR with 6 state bits. The first 12 bits of output produced by this LFSR are 111100011111 = 50 51 52 53 54 55 56 57 58 59810911. 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, 4 = 1, S3 = 0, 2 = 1,51 = 0,5o = 1). 111011 b) What are the tap bits of the LFSR? Please enter your answer as unspaced binary digits (e.g. 010101 to represent ps = 0,4 = 1, p3 = 0, p2 = 1.p1 = 0, po = 1). 10101