Brute Force

[Giorgos Mourtasagas]

Hi my name is George.
I want to make a program generate all the possible characters of length 8 using all the characters in charSet
char[] charSet = {0,1,2,3,4,5,6,7,8,9}.
I made that program to understand the algorithm , then I want to make the same with 8 char length.

import java.io.*;
class Brutus {
	public static void main(String args[]) {
		char charSet[] = {'0','1','2','3','4','5','6','7','8','9','a', 'b', 'c', 'd', 'e', 
                'f', 'g', 'h', 'i', 'j','k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't','u', 'v', 'w', 'x', 'y', 'z'};
                String x = "zzzz";
		 for (int i = 0; i < charSet.length; i++) {
		 	for (int j = 0; j < charSet.length; j++) {
                            for (int k = 0; k < charSet.length; k++) {
                                for (int l = 0; l < charSet.length; l++) {
				String s = charSet[i] + "" + charSet[j]+ "" + charSet[k]+ "" + charSet[l];
                                System.out.println(s);
                                if (x.equals(s)){
                                    System.out.println("the paswd found");
                                }else{
                                    System.out.println("the pswd isn't found");
                                }
		 	}
		 }
 
    	
	}
}

But as you see thai is onle 4 characters and I want 8.
I believe that there is another way to make without to make 8

for

.
If anyone have an idea help me.

[Nathan Leniz]

I would ask why you'd want to brute force something like this, because choosing 4 out of a set of 36 produces only 82,251 combinations while choosing 8 produces a much more respectable 145,008,513. All that is assuming you'll be matching a string in which the order doesn't matter. I see in your example you are using it as a "password" which would mean that given the same set of 36 characters and trying to brute force a password (which should, but not always, result in the order mattering) consisting of 8 characters, it could take you set^subset attempts, in this case a staggering 2,821,109,907,456 total permutations.

My math is a bit rusty and the numbers might be off, but Brute Forcing something like that is astronomical. If it's still something you want to give a shot at, the Google bot returned this link for me. http://www.merriampark.com/perm.htm

[Giorgos Mourtasagas]

I want to make it because it's my homework.
I have a zip file with AES encryption and I want to find out the password with brute force and dictionary attack.
I have made it with dictionary attack but with brute force I have a problem.
I understand that is useless but I must done it.

[Giorgos Mourtasagas]

I wrote this program but I have a little problem.
I don't know how to convert the elements of an integer array to a string array to make the comparison and I found the password.
But I don't know how.I found examples but it isn't with arrays.It is only convert an integer to a string.
Here is the code.

import java.io.*;
import com.javamex.arcmexer.*;
import java.*;
class Brutus4 {
public static void main(String args[]){
char charSet[] = {'0','1','2','3','4','5','6','7','8','9','a', 'b', 'c', 'd', 'e', 
                'f', 'g', 'h', 'i', 'j','k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't','u', 'v', 'w', 'x', 'y', 'z'};   
try {
FileInputStream f = new FileInputStream("C://ReadZip.zip");
ArchiveReader r = ArchiveReader.getReader(f);
ArchiveEntry entry = r.nextEntry() ; 
int index[] = new int[8];
for (int c0 = 0; c0 <charSet.length; c0++) {
index[0] = c0;
for (int c1 = 0; c1 <charSet.length; c1++) {
index[1] = c1;
for (int c2 = 0; c2 <charSet.length; c2++) {
index[2] = c2;
for (int c3 = 0; c3 <charSet.length; c3++) {
index[3] = c3;
for (int c4 = 0; c4 <charSet.length; c4++) {
index[4] = c4;
for (int c5 = 0; c5 <charSet.length; c5++) {
index[5] = c5;
for (int c6 = 0; c6 <charSet.length; c6++) {
index[6] = c6;
for (int c7 = 0; c7 <charSet.length; c7++) {
index[7] = c7;
String s = Integer(index).toString();     /*here is the problem*/
System.out.println(s);
if (entry.isProbablyCorrectPassword(s)){
System.out.println("the paswd found");
}else{
System.out.println("the pswd isn't found");
}
}
}
}
}
}
}
}
}
}catch (Exception e){
System.out.println("Exception raised!");
e.printStackTrace();
}
}
}

[Wouter Oet]

The key word is permutations. Here is an example.