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miskamm [114]
3 years ago
12

Suppose Alice uses the RSA system with a very large modulus n for which the factorization cannot be found in a reasonable amount

of time. Suppose Bob sends Alice a message by representing each letter by its ascii value (A->65,B->66, ...) and then encypting each number separately using RSA with the large modulus and some large key e. Is this method secure? If not, describe an efficient attack.
Computers and Technology
1 answer:
Yuliya22 [10]3 years ago
4 0

Answer:

Encryption method security:-

  • Encryption method representing each alphabetic character as an integer between 0 and 25.
  • Now, apply RSA algorithm for each integer is not an efficient secure encryption method.

Explanation:

  • Let consider alphabetic characters set {A, B... Z} .
  • Each alphabetic character forms a set of message block values M={0,1,2...,25).
  • From encryption method the corresponding cipher text block value C = {0e mod n,1e mod n, 2e mod n,..., 25e mod n}.
  • Then, compute the cipher text with knowledge of Bob's public key;
  • If the public key is generated, then possibility to decrypt the cipher text to plain text.
  • The generated public key is easily calculated by everyone and security is lost.
  • Therefore, the encryption method is not secure.

Encryption method is not secure, Most efficient attack against this method:-

  • Now, compute the encryption message with the function Me mod n for all the possible values of M.
  • This is the most efficient attack against the scheme in encryption method.
  • Then, create a look-up table with cipher text as an index to the corresponding plain text as a value for appropriate location in the table.
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Solution :

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$\text{case '5'}:$

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$\text{case '8'}:$

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case 'A':

case 'a':

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case 'B':

case 'b':

strcat(string, "1011");

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case 'C':

case 'c':

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case 'D':

case 'd':

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case 'E':

case 'e':

strcat(string, "1110");

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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];

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//Now, i have used c library strtol

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Answer:

#include<iostream>//library inclusion

using namespace std;

int main()

{

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Explanation:

The program has been commented for you. The do-while loop enters the first loop regardless of the condition. Then after the first iteration, it checks for the condition. If the condition is being met, it will iterate through, again. Otherwise it will break out of the loop and land on the "return 0;" line. Which also happens to be the termination of the program in this case. The if-else condition is used for the user to see when prompted.

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