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 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 Ð E(Ro) = We now apply the S-box transformation. S(k1 Ð E(Ro)) = Finally we apply the permutation box to complete the function f. f(Ro) = P(S(k¡ O E(Ro))) = We can now compute the state of DES going into the next round. L = R =

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 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 Ð E(Ro) = We now apply the S-box transformation. S(k1 Ð E(Ro)) = Finally we apply the permutation box to complete the function f. f(Ro) = P(S(k¡ O E(Ro))) = We can now compute the state of DES going into the next round. L = R =

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

See similar textbooks

Similar questions

2.) This question builds somewhat on top of the previous question. Instead of adding single bits at a time, let's scale this up to two bits. This is visually represented below: + Cout R1 RO For this problem, you are given a carry-in bit `Cin`, as well as the operands `A` and `B`, just as before. However, `A` and `B` are now two bits apiece, with A0 holding the bit at position 0 and `Al` holding the bit at position 1, and so on. This still produces `R` as a result, though the result is now two bits large, with `R0` holding the result value at position 0, and `R1` holding the result value at position 1. `Cout holds the carry-out bit, as before. As before, fill in the table below. Note the ordering of the inputs. The first few have already been completed for you. 2a) 0 2b) 0 2c) 0 2d) 0 Cin Al B1 A0 B0 | Cout R1 R0 2e) 0 2f) 0 2h) 0 21) 0 2k) 0 2g) 0 0 1 0 0 A1 A0 В1 ВО 2r) 1 2s) 1 2j) 0 1 0 2t) 1 2u) 1 2v) 2w) 1 2x) 1 2y) 1 2z) 1 21) 0 2m) 2n) 0 1 O O O O O O OOHHddd□□□ O O O O O O…
Geometric Sequence main.c Test Case 1-input 3, expected output 3 9 27 Test Case 2-input 6, expected output 3 9 27 81 243 729 Test Case 3-input 9, expected output 3 9 27 81 243 729 2187 6561 19683
Divide 5 by 4 in unsigned binary numbers, where each number is represented using 4 bits. The flowchart of restoring division algorithm is shown below. START M– Divisor QE Dividend Count – n Shift left A, Q AEA – M No Yes A< 0? Qo + 0 A A + M Qo–1 Count Count – 1 No Yes Count = 0? END
HAS TO BE SOLVED BY USING MICROSHIP STUDIO with the ATmega2560 device
PROGRAAM PYTHON EXPRESSION G.
Given a positive integer, check whether it has alternating bits: namely, if two adjacent bits will always have different values. For example: Input: 5 Output: True because the binary representation of 5 is: 101. Input: 7 Output: False because the binary representation of 7 is: 111. Input: 11 Output: False because the binary representation of 11 is: 1011. Input: 10 Output: True because The binary representation of 10 is: 1010. """ # Time Complexity - O(number of bits in n) def has_alternative_bit(n): first_bit=0 second_bit=0 whilen: first_bit=n&1 ifn>>1:.
Write a program that swaps 5th~11th bits in data_a with 25th~31th bits in data_bYour program must work for any data given, not just the example belowIn this question, we assume that the positions of bits count from right to left.That is, the first bit is the least significant bit.data_a DCD 0x77FFD1D1data_b DCD 0x12345678
Use C++ coding language and provide a code that works properly input.txt 4 51 0 1 0 01 0 1 1 11 1 1 1 11 0 0 1 05 71 0 1 0 0 1 11 0 1 1 1 1 11 1 1 1 1 0 11 0 1 1 1 0 11 0 1 0 1 0 10
this is COmputer machine architecture ! Help me fix the following code as theres an error. what were trying to accoplish; The assignment is to create a MIPS program that corrects bad data using Hamming codes. The program is to request the user to enter a 12-bit Hamming code and determine if it is correct or not. If correct, it is to display a message to that effect. If incorrect, it is to display a message saying it was incorrect and what the correct data is (the 12-bit Hamming code) again in hex. I will be testing only with single bit errors, so the program should be able to correct my tests just fine. You do not need to worry about multiple bit errors. # This program corrects bad data using Hamming codes # It requests the user to enter a 12-bit Hamming code and determines if it is correct or not # If correct, it displays a message to that effect. If incorrect, it displays a message # saying it was incorrect and what the correct data is (the 12-bit Hamming code) again in hex. # This…
The result is negative when the dividend and divisor have not same sign. * O False O True In BCD Addition, if and only a 4-bit sum is greater than 9, it is an invalid result. * False True The negative numbers are stored in a computer as 1's * complement form plus 1. O False O True
3. Implement a 32-bit subtraction of two numbers X and Y that are stored in memory as: X = (DS: 503H) MS byte of MS word (DS: 0502H) (DS: 0501H) (DS: 0500H) LS byte of MS byte of LS byte of MS word LS word LS word Y = (DS:0507H) (DS: 0506H) (DS: 0505H) (DS: 0504H) The result of X - Y is to be saved in the place where X is stored in memory.
Which of the following statements about the "count leading zeros" operation is false? It cannot be accomplished in less than O(n) time for n bits It is useful for normalization in floating-point arithmetic An efficient implementation could be written as a recursive Verilog modula