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

Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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The answer above is NOT correct.
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 — ХоXјX2XҙX4XsX6X7
when encrypted by the LFSR produced the ciphertext
11010110 — Уo У1 У2 Уз Уз У5 У6 Ут.
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
Transcribed Image Text:The answer above is NOT correct. 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 — ХоXјX2XҙX4XsX6X7 when encrypted by the LFSR produced the ciphertext 11010110 — Уo У1 У2 Уз Уз У5 У6 Ут. 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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