(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=4m=4. The plaintext given by

11001101=x0x1x2x3x4x5x6x711001101=x0x1x2x3x4x5x6x7

when encrypted by the LFSR produced the ciphertext

01000011=y0y1y2y3y3y5y6y701000011=y0y1y2y3y3y5y6y7.

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,p0=1p3=0,p2=1,p1=0,p0=1).

Chapter 2: Problem 7 Previous Problem Problem List Next Problem (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 11001101 = 2021222324252627 when encrypted by the LFSR produced the ciphertext 01000011 = yoy1Y2Y3Y3 Y5 Y6Y7 - 0, P2 1, P1 = 0, Po = 1). What are the tap bits of the LFSR? Please enter your answer as unspaced binary digits (e.g. 0101 to represent P3 0101 Preview My Answers Submit Answers