0120144 134231567892 15 1183 1010 11 12 13 14 15612 5 9 0 71 0 157S12 4 1144 148 13213 1 10 612 113 15 12826 2 11 15 1249 17 5 119 73 14 10 09 53 10385 06 13(Show your work.) Suppose 110101 is fed into S-box, S1. The resulting 4 bits stringisA) 1000B) 0001C) 0011○ D) 0111

To find the output of the S-box S1 when the input is 110101 , follow these steps: 1. Split the…

Answered: 0 1 2 0 14 4 13 42 31 5 6 7 8 9 2 15 11…

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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6-In the representation Complement to 1, we the number when Code the absolute value thus is negative: A = True B- False 7- The result in base 10' A - 44.25 B-42015 C- 51. 13 45. 75 8- The result AT 40. B + 61 q of the transformation (101001.10) 2 usi o the transformation (52) a in bose 10 is: C² 42 D-41 WWW 30 O
For example: Binary value of character ‘B’ from ASCII table is 01000010Dataword is 1000010 Even Parity Bit is 0 Codeword: 10000100 Even Parity – Complete the following table using Even Parity Character Binary Datawordfrom ASCII table(7-bit value) Parity BitValue TransmittedCodeword ReceivedCodewordat Receiver Check ifthecodewordisrejected t 1010100 10101101 h 1101000 11010000 i 1101001 11010010 s 1110011 11100110
7. The normalized representation of 0.0010110 * 2 ^ 9 is a. 0 10001000 0010110 b. 0 10000101 0110 c. 0 10101010 1110 d. 0 11110100 11100
Given this: A B D E H K L M 1 3 5 9. 10 11 12 13 R U V Y 14 15 16 17 18 19 20 21 22 23 24 25 . Use the encoding formula y = x + 15 (mod 26). 1. Encode LOVE (Use capital letters. No space) G7T
Given the truth table below, where A,B,C,D are inputs and Y is the output: Truth Table ABCDY 0 0000 0 1 00010 2 0010 1 3 0011 1 4 0 1000 5 0 1011 6 0 110 1 7 0 111 1 8 1000 0 9 10011 1 10 10 1 |1 10111 1 11001 11010 3 1 11100 1 11110 5 a) Generate the Karnaugh Map. Use the space to the right of the truth table. b) Draw the largest loops you can. Use different colored loops preferably. c) Give the minimal logic expression. (Hint: Each loop should correspond to one part of the expression.) Show the parts of the expression that correspond to each loop.
Given the ASCII table, Find the Ascii value of the lettere and the symbol { 8.8.8.8. 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111 000 NULL SOH STX ETX EOT ENQ ACK BEL BS HT LF VT SO SI ASCII Table 001 DLE DC1 DC2 DC3 DC4 SYN ETB CAN EM SUB ESC FS GS RS US 8.0.0, 010 011 ornninGAY AN 2 100 101 CADCOEFONICKLMNO А PORSTUVWX>N~~- i 110 CAUCOT CF-Eco m EGUL> 3xx DEL
9. Let the bit representations of sets A and B as following A= 1010101010 B 0000011111 find (A U B) – A O 1010100000 0000001010 1010111111 (000010101
x³ is O(x4) and x4 is O(x³) O True O False
15. Add the binary numbers: (b) 10 + 10 (c) 101 + 11 (d) 111 + 110 (e) 1001 + 101 (f) 1101 + 1011 16. Use direct subtraction on the following binary numbers: (a) 11 – 1 (b) 101 – 100 (c) 110 – 101 - - (d) 1110 – 11 (e) 1100 – 1001 (f) 11010 – 10111
The following is a string ASCII characters whose bit patterns have been converted into hexadecimal for compactness: CD 69 E4 54 E5 72 6D Of the eight bits in each pair of digits, the left most is a parity bit and the remaining bits are the ASCII code. Convert the string to bit form and decode the word in ASCII. What is the parity for each letter of the word?
Consider the intensities of the7-bit image below. Answer the following questions: 52 53 71 42 55 95 0 89 101 60 55 70 69 32 63 0 81 100 85 96 60 41 21 23 0 84 99 60 52 92 0 63 62 0 82 99 62 41 41 40 53 68 0 99 96 85 39 36 32 23 52 0 56 52 62 50 96 41 62 33 0 98 65 0 0 0 0 0 0 0 63 99 0 0 0 0 0 0 0 0 0 (i) A 2-dimensional Laplacian filter shown below is applied to pixel f(6,4), find the output pixel g(6,4) and explain the effect of applying a Laplacian filter to an image. 1 0 1 0 -4 0 1 0 1
Find the representation x of the following signed numbers listed below in the base specified on the right. a.- (01100110)2=( x )10 b.- (10110111)2=( x )10 C.- (207)10=( x )2 d.- (-135)10=(× )2 e.- (2D7)16=( x )2 f.- (10110011)2=(x )16 g.- (-5.5)10=( x )2 e.- (11000001010000000000000000001010)2=(x )10 Note: In part e the binary representation on the left corresponds to a real number representation.

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