p T 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 are100011011000 = s0 $1 $2 S3 S4 S5 S6 87 Sg 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 representS5 = 0, s4 = 1, 83 = 0, s2 = 1, s1 = 0, so = 1).%3Db) What are the tap bits of the LFSR? Please enter your answer as unspaced binary digits (e.g. 010101 to representP5 = 0, p4 = 1, pP3 = 0, p2 = 1, p1 = 0, po = 1).

LFSR (Linear Feedback Shift Register) A linear feedback shift register (LFSR) is a shift register…

Answered: p T Note: The notation from this…

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Consider a binary code that consists of only four valid code wards as given below: 00000, 01011, 10101, 11110 Let the minimum Hamming distance of the code be p and the maximum number of erroneous bits that can be corrected by the code be q. Then the values of p and q are
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Note: The notation from this problem is from Understanding Cryptography by Paar and Pelzi. Suppose you have an LFSR with 6 state bits. The first 12 bits of output produced by this LFSR are 101011100010 = 80818283848586878889810811- 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 85 = 0, $4 = 1, 83 = 0, 82 = 1, 81 = 0, 80 = 1). b) What are the tap bits of the LFSR? Please enter your answer as unspaced binary digits (e.g. 010101 to represent p = 0, P₁ = 1, P3 = 0, P2 = 1, P₁ = 0, Po = 1).
Draw a FSA which accepts all binary strings which start with 0, the length of the string is even but the number of 1's in the string is odd. 0111, 01 and 000111 should be accepted while 111 and 0011 should not
This is all about BINARY CODES... PLEASE GIVE SHORT SOLUTION ONLY PLEASE! show the process how you are able to arrive with your chosen answer
Assume that D has the value 1010101010 and that the 5-bit generator G=10011 is used. What does R stand for? B. C. Does corruption exist? Publish your work. Repetition of parts a and b with the addition of the following: the numbers 1001010101, 0101101010, and 1010100000
For each of the following binary codes, answer all of the following questions with justification: (a) (b) (c) (d) ● ● Is it linear? What is its block length, message length, and rate? What is its distance? How many errors can it detect, and how many can it correct? (To answer this question you should give the largest number of errors that the code can guarantee to detect or correct. For example a simple parity bit can detect 1 or 3 errors but cannot detect 2 errors so we would say this can detect only a single error. No calculations of probability are required.) The code with seven codewords where seven messages are encoded as follows: 0000000 0010011 0100101 0110110 1001001 101 1010 1101100 The code consisting of thirty-two codewords where each codeword consists of a 5-bit message padded with an additional bit to give ODD parity: that is, each codeword contains an odd number of 1s. For example the word 11111 would be padded with a 0 to give 111110, while 00000 would be padded with a 1…
Perform the following internal operations involved in DES and AES encryption techniques. | Q2: Perform shift left by 2 bits for the below given data 1111111000011001100101010101 1110101010101100110011110001
Provide the correct answers to each question
implement bitMask(x,y) Generate a mask consisting of all 1’s from lowbit to highbit /* * bitMask - Generate a mask consisting of all 1's * lowbit and highbit* Examples: bitMask(5,3) = 0x38* Assume 0 <= lowbit <= 31, and 0 <= highbit <= 31* If lowbit > highbit, then mask should be all 0's* Legal ops: ! ~ & ^ | + << >>* Max ops: 16* Rating: 3*/int bitMask(int highbit, int lowbit) {return 2;}
You can determine the maximum and minimum possible values for a datatype based on the number of bits that it uses (I cover this in lecture, note it differs for unsigned compared with signed). The idea is to consider which bits should be 0 or 1 to achieve the maximum and minimum values, respectively. 1. For each type in the table, enter the minimum and maximum Base 10 values you can represent. Minimum (Base 10) Maximum (Base 10) Type unsigned byte byte unsigned int int Note: You can express your answers as powers of 2 if you want.
Let's consider a Student ID 171014026 as in Decimal number. Converting the last 4-digit of the ID 4026 will generate equivalent 16-bit binary 0000111110111010. Now, convert the last 4-digit of your own student ID into 16-bit binary number, and store the High-significant 8-bits of the binary number into register R1, and Least-significant 8-bits of the binary number into register R2. Compute the register value R2 after an arithmetic shift right followed by a circular shift left followed by an arithmetic shift left. Determine whether there is an Overflow for the register R2.

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