Use a Caesar Cipher with digraph coding to encrypt the following plaintext: ‘buffer’ with a key of K=100. (Show your intermediate calculations and formulas used) Hint: Table below help you to find the code equivalent to each letter.
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Use a Caesar Cipher with digraph coding to encrypt the following plaintext: ‘buffer’ with a key of
K=100. (Show your intermediate calculations and formulas used)
Hint: Table below help you to find the code equivalent to each letter.
![a be d ]e]f]g hi j kl m n o p q[r s u v \w\x ]y]z
0 12 3 4 5 6 7 8 9 10 11 12 13 14|15 16 |17|18 19 20 21 22| 23 24 25](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fffb04fd8-2646-4e36-8afb-6615a41002df%2Fb31fd8ff-ad6f-4313-96c1-5dc4d9ea2181%2Fpdp8n0j_processed.png&w=3840&q=75)
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- Plz fix this error I couldn’t Figur out.S ettt JoNa (ede ,{c( odd no . b[u) t 4o I#include <stdio.h> #include <stdlib.h> double f(double x){ return x*x; } double reimannSums(double f(double), double a, double b, int n, double deltax) double sum=0.0, x; int i; printf("n = %d\n",nvoidvoid); for (i=0;i<n;i++){ x= ( (a+i*deltax)+a+(i+1)*deltax))/2; printf("x = %lf\n,x"); sum=sum + f(x); } return sum = (sum * deltax); double integ(double a, double b, double deltax) { double result; int n; n = (b - a) / deltax; return reimannSums(f, a, b, n, deltax); } int main(int argc, char * argv[]) { int a = atoi(argv[1]); int b = atoi(argv[2]); double deltax = atof(argv[3]); printf("result= %lf",integ(a, b, deltax)); } CAN YOU PLEASE CHECK MY WORK I GOT BUILD FAILED ERROR C LANGUAGE I NEED ANSWER IN 30 MIN THANKS<3
- given ASCIIOFSET .FILL x0030 NegASCIIOFSET .FILL xFFD0 ASCIINewline .FILL x000d ; Newline ascii code ASCIISpace .FILL x0020 ; Space ascii code ASCIIComma .FILL x002C ; Comma ascii code Use LC-3 I/O device to print number, commas, and spaces in printCommaSpace subroutine, also check code to see if it follows instructions and modify accordinglyIf the user chooses to generate a Pentagon o This is a combination of a triangular shape and a square shape -- a symmetric triangle sits on top of a rectangle. o The triangular shape must be symmetric about the vertical axis. There must be exactly one character in the first line, and the number of characters increases by 2 in every successive lineo The rectangle shape must have LENGTH lines of height and all lines are 13 characters wide. o Make sure to add delay to view a progressively growing pentagon, otherwise you will just see the final shape. Code for triangle: for (int i = 1; i <= length; i++) { sleep (1); for (int j = length - i; j > 0; j--) cout << " "; for (int b = 1; b <= i; b++) cout << char(rdm) << " "; cout << endl; } Code for square: for (int row = 1; row <= length; row++)…This is the updated code that I have: .data .globl main .textmain: # compute the next state of the LFSR for each input state li $a0, 0x00000001 jal lfsr_next_state move $t0, $v0 li $a0, 0xdeadbeef jal lfsr_next_state move $t1, $v0 li $a0, 0x200214c8 jal lfsr_next_state move $t2, $v0 li $a0, 0x00000000 jal lfsr_next_state move $t3, $v0 # print the output states move $a0, $t0 li $v0, 1 syscall move $a0, $t1 li $v0, 1 syscall move $a0, $t2 li $v0, 1 syscall move $a0, $t3 li $v0, 1 syscall # exit the program li $v0, 10 syscall # Function to compute the next state of an LFSR# Input parameter: $a0 = current state# Output: $v0 = next statelfsr_next_state: # Initialize upper mask with 0x7fff and lower mask with 0x8000 lui $t0, 0x7fff ori $t0, $t0,…
- int a[10] = {0,1,2,3,4,5,6,7,8,9};int *m = &a[0];int *p = &a[5];int *q = &a[1]; p = (int) m + (int) p - (int) q; what is the value of of *pHello I am really really struggling with this problem because i don't know how to do this problem or how to answer this question can you please help me answering this problem. I added the text below. question that I need help with: 1.48 Let Σ = {0,1} and let D = {w|w contains an equal number of occurrences of the substrings 01 and 10}. Thus 101 ∈ D because 101 contains a single 01 and a single 10, but 1010 6∈ D because 1010 contains two 10s and one 01. Show that D is a regular language.include<stdio.h> #include<stdlib.h> int main() { int N; scanf("%d", &N); int input[N], ind, rem; for(ind=0;ind<N;ind++) scanf("%d", &input[ind]); // 8 7 6 4 int two_digits[22], two_digits_ind; unsignedlonglongint res = 0; two_digits[0] = 1; for(two_digits_ind=1;two_digits_ind<22;two_digits_ind++) two_digits[two_digits_ind] = ((two_digits[two_digits_ind-1])*2)%100; for(ind=0;ind<N;ind++) { if(input[ind]>21) { rem = input[ind] % 22; rem += 2; } else rem = input[ind]; res = res + two_digits[rem]; } printf("%llu", res%.
- A = {0,2,4,6} B = {{0},{2}, {4},{6}} C = A ∪ Ø D= A ∪ {Ø} E= { n ∈ ℕ | n² ∈ ℕ } F= { n² ∈ ℕ | n ∈ ℕ } Which are the following are true? 1. A ⊆ E, B ⊆ E, A ⊆ F, B ⊆ F, E ⊆ F, F ⊆ E. 2. O ∈ A, 0 ∈ B, {0} ∈ A, {0} ∈ B.what must be the value of r in a= bq+r ? a.all real number b.always in the form of decimal c.always positve integer d. both choice 1 and 2Can you assist me in finding the error in my code. My output does not equal the expected output as shown in the image. My code is written in C and is as follows #include <stdlib.h>#include <stdio.h>#include <string.h> const int MAX_NAME_SIZE = 50; void swap(char **x, char **y){char *temp;temp = *x;*x = *y;*y = temp;}void AllPermutations(char**permList, int permSize, char**nameList, int nameSize){if(permSize == nameSize - 1){for(int i = 0; i < nameSize; i++){ if(i==nameSize-1) printf("%s", nameList[i]); else printf("%s, ", nameList[i]);}printf("\n");} else{for(int i = permSize; i < nameSize; i++){swap(nameList + permSize, nameList+i);AllPermutations(permList, permSize+1, nameList, nameSize);swap(nameList+permSize, nameList+i);}}} int main(void) { int size; int i = 0; char name[MAX_NAME_SIZE]; scanf("%d", &size); char *nameList[size]; char *permList[size]; for (i = 0; i < size; ++i) { nameList[i] = (char…
