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( 1 point ) Note: The notation from this problem is from Understanding Cryptography by Paar and Pelzl. Consider the LFSR represented by the polynomial

(1 point) Note: The notation from this problem is from Understanding Cryptography by Paar and Pelzl.
Consider the LFSR represented by the polynomial x4+x2+x+1.
What are the tap bits of the LFSR? Please enter your answer as unspaced binary digits (e.g.010101 to represent
p5=0,p4=1,p3=0,p2=1,p1=0,p0=1.(1 point) Note: The notation from this problem is from Understanding Cryptography by Paar and Pelzl.
Suppose you have an LFSR with 6 state bits. The first 12 bits of output produced by this LFSR are
001100011111=s0s1s2s3s4s5s6s7s8s9s10s11.
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,s4=1,s3=0,s2=1,s1=0,s0=1).
b) What are the tap bits of the LFSR? Please enter your answer as unspaced binary digits (e.g.010101 to represent
{:p5=0,p4=1,p3=0,p2=1,p1=0,p0=1).(1 point) Note: The notation from this problem is from Understanding Cryptography by Paar and Pelzl.
Consider the LFSR represented by the polynomial x4+x3+x2+x+1.
What are the tap bits of the LFSR? Please enter your answer as unspaced binary digits (e.g.010101 to represent
p5=0,p4=1,p3=0,p7=1,p1=0,p0=1.(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
00110001=x0x1x2x3x4x5x6x7
when encrypted by the LFSR produced the ciphertext
01001000=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=1).
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