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 11100010 = xox1X2X3X4X5X6X7 when encrypted by the LFSR produced the ciphertext 11010110 = Yo Yı Y2 Y3 Y3 Y5 Y6 Y7 · 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). 0010

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Q: What are the tap bits of the LFSR? Please enter your answer as unspaced binary digits

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