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Aleksandr [31]
3 years ago
13

R6. Suppose N people want to communicate with each of N - 1 other peo- ple using symmetric key encryption. All communication bet

ween any two people, i and j, is visible to all other people in this group of N, and no other person in this group should be able to decode their communication. How many keys are required in the system as a whole
Computers and Technology
1 answer:
evablogger [386]3 years ago
5 0

Answer:

N(N-1)/2

Explanation:

You would need a symmetric key for each pair of people, so you need the number of pairs.

Mathematically this can be written as

_NC_2 = \frac{N!}{2!(N-2)!} = \frac{N(N-1)}{2}

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When driving, your attention is __________.
maks197457 [2]
It is c hope I helped
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Hey, Another question. I'm sure it's possible in the future, but I wanted to ask if HIE would be possible. I'm sure it would be,
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Answer:

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Explanation:

please mark this answer as brainliest

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3 years ago
Write a C program that reads two hexadecimal values from the keyboard and then stores the two values into two variables of type
sattari [20]

Solution :

#include  $$

#include $$

#include $$

//Converts $\text{hex string}$ to binary string.

$\text{char}$ * hexadecimal$\text{To}$Binary(char* hexdec)

{

 

long $\text{int i}$ = 0;

char *string = $(\text{char}^ *) \ \text{malloc}$(sizeof(char) * 9);

while (hexdec[i]) {

//Simply assign binary string for each hex char.

switch (hexdec[i]) {

$\text{case '0'}:$

strcat(string, "0000");

break;

$\text{case '1'}:$

strcat(string, "0001");

break;

$\text{case '2'}:$

strcat(string, "0010");

break;

$\text{case '3'}:$

strcat(string, "0011");

break;

$\text{case '4'}:$

strcat(string, "0100");

break;

$\text{case '5'}:$

strcat(string, "0101");

break;

$\text{case '6'}:$

strcat(string, "0110");

break;

$\text{case '7'}:$

strcat(string, "0111");

break;

$\text{case '8'}:$

strcat(string, "1000");

break;

$\text{case '9'}:$

strcat(string, "1001");

break;

case 'A':

case 'a':

strcat(string, "1010");

break;

case 'B':

case 'b':

strcat(string, "1011");

break;

case 'C':

case 'c':

strcat(string, "1100");

break;

case 'D':

case 'd':

strcat(string, "1101");

break;

case 'E':

case 'e':

strcat(string, "1110");

break;

case 'F':

case 'f':

strcat(string, "1111");

break;

default:

printf("\nInvalid hexadecimal digit %c",

hexdec[i]);

string="-1" ;

}

i++;

}

return string;

}

 

int main()

{ //Take 2 strings

char *str1 =hexadecimalToBinary("FA") ;

char *str2 =hexadecimalToBinary("12") ;

//Input 2 numbers p and n.

int p,n;

scanf("%d",&p);

scanf("%d",&n);

//keep j as length of str2

int j=strlen(str2),i;

//Now replace n digits after p of str1

for(i=0;i<n;i++){

str1[p+i]=str2[j-1-i];

}

//Now, i have used c library strtol

long ans = strtol(str1, NULL, 2);

//print result.

printf("%lx",ans);

return 0;

}

4 0
3 years ago
Write a program that replaces words in a sentence. The input begins with an integer indicating the number of word replacement pa
kramer

Answer: provided in the explanation segment

Explanation:

This is the code to run the program:

CODE:  

//Input: 3 automobile car manufacturer maker children kids

  //15 The automobile manufacturer recommends car seats for children if the automobile doesn't already have one.

  import java.util.Scanner;

  public class LabProgram{

  public static int findWordInWordList(String[] wordList, String wordToFind, int numInList) {

          for (int i = 0; i < numInList; i++) {

              if (wordList[i].equals(wordToFind)) {

                  return i;

              }

          }

          return -1;

      }

      public static void main(String[] args) {

          Scanner scan = new Scanner(System.in);

          String[] original_String = new String[20];

              String[] modified_String = new String[20];

          int numInList = scan.nextInt();

          for (int i = 0; i < numInList; i++) {

              original_String[i] = scan.next();

              modified_String[i] = scan.next();

          }

          int num_Of_Words = scan.nextInt();

          String[] words_To_Replace = new String[num_Of_Words];

          for (int i = 0; i < num_Of_Words; i++) {

              words_To_Replace[i] = scan.next();

          }

          int indx;

          for (int i = 0; i < num_Of_Words; i++) {

              indx = findWordInWordList(original_String, words_To_Replace[i], numInList);

              if (indx != -1)

                  words_To_Replace[i] = modified_String[indx];

          }

          for (int i = 0; i < num_Of_Words; i++)

              System.out.print(words_To_Replace[i] + " ");

          System.out.println();

      scan.close();

      }

  }

cheers i hope this helps!!!!1

3 0
4 years ago
The following program segment is designed to compute the product of two nonnegative integers X and Y by accumulating the sum of
mote1985 [20]

The program is correct: at the beginning, product = 0. Then, we start summing Y to that variable, and we sum Y exactly X times, because with each iteration we increase Count by 1, and check if Count=X so that we can exit the loop.

5 0
3 years ago
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