The notation from this problem is from Understanding Cryptography by Paar and Pelzl. Sappose you have an LFSR with 6 state bits. The first 12 bits of output produced by this LFSR are 010010000110 = S0S1 S2 S3 S4 S5 S6 S7 S8 S9 S10 S11. ( 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, $3 = 0, $2 = 1, $₁ = 0, so = 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, Po = 1).
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- A Gray code is a sequence of binary numbers with the property that no more than 1 bit changes in going from one element of the sequence to another. For example, here is a 3-bit binary Gray code: 000, 001, 011, 010, 110, 111, 101, and 100. Using three D flip-flops and a PLA, construct a 3-bit Gray code counter that has two inputs: reset, which sets the counter to 000, and inc, which makes the counter go to the next value in the sequence. Note that the code is cyclic, so that the value after 100 in the sequence is 000.We can perform logical operations on strings of bits by considering each pair of corresponding bits separately (called bitwise operation). Given two eight-bit strings A = 10110001 and B = 10101100 , evaluate the eight-bit result after the following logical operations: XORWe can perform logical operations on strings of bits by considering each pair of corresponding bits separately (called bitwise operation). Given two eight-bit strings A = 10110001 and B = 10101100 , evaluate the eight-bit result after the following logical operations: OR
- In the Hamming coding system (7.3), its representation is expressed as 3 control (test) bits and 7 total bits (data+control bits).In the literature, control bits are represented as (cl, c2, c3, or (pl, p2, p3, .) The bit string sent by encoding with Hamming coding (15,4) is on the receiving side, It is retrieved as 001100101100 (pl.p2.dll). According to this bit sequence;Which option is given according to the control bits (p1.p2.p3.p4) sequence of the received bit sequence?A 0110B 1100C 1001D 0011E 0010In the steps of “Append the padding bits” in SHA-512 system (shown in Figure 1), the message is padded so that its length is congruent to 896 mod 1024 or (length ≅ 896 mod 1024). State the value of the following padding field is the message length is: a) 1919 b) 1920 c) 2942In this problem we will work through a round of DES. For notational simplicity, assume it is the first round. Please enter all answers as strings of 0's and 1's. The input is the 64 bit block 0000001000100000000000000000000000000000000000000100000000010010 Suppose that the subkey for the current round is this 48 bit number: 000000000000000000000000000001000000000000000000 What does the 64 bit state look like after the IP transformation is applied to the input? Now find Lo and Ro, the left and right halves of the state. Lo Ro = What is the result of applying the expansion box to Ro? E(Ro) = What is the result of XORing the subkey with E(Ro)? ki O E(Ro) = We now apply the S-box transformation. S(k, O E(R)) Finally we apply the permutation box to complete the function f. f(Ro) = P(S(k1 O E(Ro))) = We can now compute the state of DES going into the next round. L, = Rj =
- In this problem we will work through a round of DES. For notational simplicity, assume it is the first round. Please enter all answers as strings of 0's and 1's. The input is the 64 bit block 0000000000000010000000000000001000000100000001000100000000000000 Suppose that the subkey for the current round is this 48 bit number: 000000000000000001000000000000000000000000000000 What does the 64 bit state look like after the IP transformation is applied to the input? 00000000000110000100000000000000 Now find Lo and Ro, the left and right halves of the state. Lo 01000000010000000000000000010000 = Ro=00000000000000100000000000000000 What is the result of applying the expansion box to R₁? E(Ro) = 00000000000000100000000000000000 What is the result of XORing the subkey with E(Ro)? k₁ E(Ro) = 00000000000000100000000000000000 We now apply the S-box transformation. S(k₁ E(Ro)) = 11011000110110001101111110111100 Finally we apply the permutation box to complete the function f. f(Ro) = P(S(k₁ + E(Ro)))…You have been hired by a company to design a circuit to encrypt passwords. Assumepasswords are 3 bits in size and that the encrypted password is the original password inverted(0's become 1's and 1's become 0's) and then add 3 (modulo 8 as we only have 3 bits). Forexample if the password is 000, we invert it to give us 111. We then add 3 (7 + 3 = 10 mod8 = 2) which makes the encrypted password 010. Design and build a sequential circuit (usingthe procedure from your notes) using 3 T flip-flops to perform the above function. You needto show the State Table, K-maps and draw the sequential circuit.Cs The redundant bit offers a simple error correction. Example The sixth digit in the ISBN 0 – 7923 519 - X has - faded out. We want to find the missing digit.
- Simplify the function F represented by the following Kmap into the simplest SOP expression: AB 00 CD 01 11 10 00 01 1 11 1 10 Select one: O a. F= B'D' ABC A CD Ob. F= CD ABC + A'CD AB CD O c. F= CD ABC + A CD - AB'D' Od. F= CD ABC A CDScenario: In a biased N-bit binary number system with bias B, positive and negative numbers are represented as their value plus the bias B. For example, for 5-bit numbers with a bias of 15, the number 0 is represented as 01111, 1 as 10000, and so forth. Biased number systems are sometimes used in floating point mathematics. Consider a biased 8-bit binary number system with a bias of 12710 Question: What is the representation and value of the most negative number?In this problem we will work through a round of DES. For notational simplicity, assume it is the first round. Please enter all answers as strings of O's and 1's. The input is the 64 bit block 0000000000000100000000000100001000100000000000000000100000000000 Suppose that the subkey for the current round is this 48 bit number: 000000000000000010000000000000000000000000000000 What does the 64 bit state lok like after the IP transformation is applied to the input? Now find Ln and Ro, the left and right halves of the state. Lo =0 Ro = What is the result of applying the expansion box to Ro? E(Ro) = 0 What is the result of XORing the subkey with E(Ro)? k1 e E(Ro) = We now apply the S-box transformation. S(kı O E(Ro)) = Finally we apply the permutation box to complete the function f. f(Ro) = P(S(kı O E(Ro))) = We can now compute the state of DES going into the next round. L, =|| Rj =