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C++ 1 Time Pad Optimization
I wrote a code in C++ to try to brute force the 1-time pad method. Basically the issue is that it goes through the loops about 5000 times in about two seconds then it gets unbearably slow. I know I would never crack it with the current code, but I have no clue how I can further optimize it. Any ideas would be great.
#include <windows.h>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <string.h>
using namespace std;
int main(){
string binary[26]; // for binaryary letterbet
string letter[26]; // ASCII letterbet
string text[3]; // 1 Time Pad Packets
string key; // binaryary Key To decryptpt 1 Time Pad
string out; // Hold the "decryptpted" binaryary
string decrypt[4]; // Holds the 1 byte string to decrpyt
binary[0] = "01100001";
binary[1] = "01100010";
binary[2] = "01100011";
binary[3] = "01100100";
binary[4] = "01100101";
binary[5] = "01100110";
binary[6] = "01100111";
binary[7] = "01101000";
binary[8] = "01101001";
binary[9] = "01101010";
binary[10] = "01101011";
binary[11] = "01101100";
binary[12] = "01101101";
binary[13] = "01101110";
binary[14] = "01101111";
binary[15] = "01110000";
binary[16] = "01110001";
binary[17] = "01110010";
binary[18] = "01110011";
binary[19] = "01110100";
binary[20] = "01110101";
binary[21] = "01110110";
binary[22] = "01110111";
binary[23] = "01111000";
binary[24] = "01111001";
binary[25] = "01111010";
letter[0] = "a";
letter[1] = "b";
letter[2] = "c";
letter[3] = "d";
letter[4] = "e";
letter[5] = "f";
letter[6] = "g";
letter[7] = "h";
letter[8] = "i";
letter[9] = "j";
letter[10] = "k";
letter[11] = "l";
letter[12] = "m";
letter[13] = "n";
letter[14] = "o";
letter[15] = "p";
letter[16] = "q";
letter[17] = "r";
letter[18] = "s";
letter[19] = "t";
letter[20] = "u";
letter[21] = "v";
letter[22] = "w";
letter[23] = "x";
letter[24] = "y";
letter[25] = "z";
text[0] = "00001001000010010000110000010110";
text[1] = "00000011000001000001101000011000";
text[2] = "00011011000001000001111000010110";
int count=0;
cout<<"\t\t\tBruteForcing Initiated \n\t\t";
for (int a=0;a<26;a++){
for (int b=0;b<26;b++){
for (int c=0;c<26;c++){
for (int d=0;d<26;d++){
key = binary[a]+binary[b]+binary[c]+binary[d]; // Sets the key to the 4 part binaryary key
for (int e=0;e<32;e++){ // 32 part loop for the 32 bit enc
if (key[e]==text[0][e]){ // if bit "a" in key == bit "b" in text
out = out + "0"; // add a "0" to out
} //
if (key[e]!=text[0][e]){ // else
out = out + "1"; // add a "1"
}
}
for (int t=0;t<8;t++){
decrypt[0] =decrypt[0]+out[t];
}
for (int g=0;g<26;g++){ // 1-8 part loop for the 32 bit enc
if(decrypt[0] == binary[g]){
cout<<"\n -- Packet 1 = "<<letter[g];
g=26;
}
}
for (int t=8;t<16;t++){
decrypt[1] =decrypt[1]+out[t];
}
for (int g=0;g<26;g++){ // 9-16 part of the 32 bit enc
if(decrypt[0] == binary[g]){
cout<<letter[g];
g=26;
}
}
for (int t=16;t<24;t++){
decrypt[2] =decrypt[2]+out[t];
}
for (int g=0;g<26;g++){ // 17-24 part of the 32 bit enc
if(decrypt[0] == binary[g]){
cout<<letter[g];
g=26;
}
}
for (int t=24;t<32;t++){
decrypt[3] =decrypt[3]+out[t];
}
for (int g=0;g<26;g++){ // 25-32 part of the 32 bit enc
if(decrypt[0] == binary[g]){
cout<<letter[g]<<"\nKey: "<<letter[a]<<letter[b]<<letter[c]<<letter[d]<<"\n";
g=26;
}
}for (int e=0;e<32;e++){ // 32 part loop for the 32 bit enc
if (key[e]==text[1][e]){ // if bit "a" in key == bit "b" in text
out = out + "0"; // add a "0" to out
} //
if (key[e]!=text[1][e]){ // else
out = out + "1"; // add a "1"
}
}
for (int t=0;t<8;t++){
decrypt[0] =decrypt[0]+out[t];
}
for (int g=0;g<26;g++){ // 1-8 part loop for the 32 bit enc
if(decrypt[0] == binary[g]){
cout<<"\n -- Packet 2 = "<<letter[g];
g=26;
}
}
for (int t=8;t<16;t++){
decrypt[1] =decrypt[1]+out[t];
}
for (int g=0;g<26;g++){ // 9-16 part of the 32 bit enc
if(decrypt[0] == binary[g]){
cout<<letter[g];
g=26;
}
}
for (int t=16;t<24;t++){
decrypt[2] =decrypt[2]+out[t];
}
for (int g=0;g<26;g++){ // 17-24 part of the 32 bit enc
if(decrypt[0] == binary[g]){
cout<<letter[g];
g=26;
}
}
for (int t=24;t<32;t++){
decrypt[3] =decrypt[3]+out[t];
}
for (int g=0;g<26;g++){ // 25-32 part of the 32 bit enc
if(decrypt[0] == binary[g]){
cout<<letter[g]<<"\nKey: "<<letter[a]<<letter[b]<<letter[c]<<letter[d]<<"\n";
g=26;
}
}
for (int e=0;e<32;e++){ // 32 part loop for the 32 bit enc
if (key[e]==text[2][e]){ // if bit "a" in key == bit "b" in text
out = out + "0"; // add a "0" to out
} //
if (key[e]!=text[2][e]){ // else
out = out + "1"; // add a "1"
}
}
for (int t=0;t<8;t++){
decrypt[0] =decrypt[0]+out[t];
}
for (int g=0;g<26;g++){ // 1-8 part loop for the 32 bit enc
if(decrypt[0] == binary[g]){
cout<<"\n -- Packet 3 = "<<letter[g];
g=26;
}
}
for (int t=8;t<16;t++){
decrypt[1] =decrypt[1]+out[t];
}
for (int g=0;g<26;g++){ // 9-16 part of the 32 bit enc
if(decrypt[0] == binary[g]){
cout<<letter[g];
g=26;
}
}
for (int t=16;t<24;t++){
decrypt[2] =decrypt[2]+out[t];
}
for (int g=0;g<26;g++){ // 17-24 part of the 32 bit enc
if(decrypt[0] == binary[g]){
cout<<letter[g];
g=26;
}
}
for (int t=24;t<32;t++){
decrypt[3] =decrypt[3]+out[t];
}
for (int g=0;g<26;g++){ // 25-32 part of the 32 bit enc
if(decrypt[0] == binary[g]){
cout<<letter[g]<<"\nKey: "<<letter[a]<<letter[b]<<letter[c]<<letter[d]<<"\n";
g=26;
}
}
out="";
count=count+1;
if (count%1000==0){
cout<<".";
}
}
}
}
}
cout<<"\n\n\t\t\t"<<count<<" Keys tried\n";
cout<<"\t\t\tNo Matches Found\n\t\t\t";
system("pause");
}
If you're talking about a pure brute force attack against a securely implemented one time pad, it's actually been proven that it is mathematically impossible.
The only things that can lead to a compromise in the security of a OTP is if the pad used is not totally random, or if the same key is used for more than one different message.
You're familiar with python, no? Check out http://www.hellboundhackers.org/articles/691-cryptanalysis-case-study:-the-one-time-pad.html