Java generate wordlist

Question

I'm stuck at this subject about 3 days, I can't think a proper algorithm. So could you please help me guys.

I want to generate wordlist with given charset, minimum & maximum length

Given charset:abcdef min:2 max:5

Result:

aa
ab
ac
...
ffffd
ffffe
fffff

Like linux command line crunch.

But there's something, I need to make this in a for loop doesn't matter how many loops but matters without calling any user-made functions*.

Like:

for (...) {

   // Password is ready!
   pass = ...;

}

And NOT like:

pass = get_pass(...);

Thank you.

Answer 1

One way could be to use a libraray which creates strings from a given regex like Generex. Using Generex your task is as simple as:

import com.mifmif.common.regex.Generex;

public class Example {

    public static void main(String[] args) {
        Generex gen = new Generex("[abcdef]{2,5}");
        gen.getAllMatchedStrings().forEach(System.out::println);
    }
} 

output:

aa
aaa
aaaa
aaaaa
aaaab
aaaac
aaaad
aaaae
aaaaf
...
...
ffffa
ffffb
ffffc
ffffd
ffffe
fffff

See this post for other libs like generex using-regex-to-generate-strings-rather-than-match-them

Answer 2

I'm pretty sure that it's not the best possible solution, but it works.

public static void main(String[] args) {
    List<Character> charset = new ArrayList<>();
    charset.add('a');
    charset.add('b');
    charset.add('c');
    charset.add('d');
    charset.add('e');
    int min = 2;
    int max = 5;

    List<Integer> word = new ArrayList<>();
    for (int len = min; len <= max; len++) {
        word.clear();
        for (int i = 0; i < len; i++) {
            word.add(0);
        }
        while (word.get(0) < charset.size()) {
            System.out.println(word.stream()
                    .map(e -> charset.get(e).toString())
                    .collect(Collectors.joining()));
            word.set(len - 1, word.get(len - 1) + 1);
            for (int i = len - 1; i > 0; i--) {
                if (word.get(i).equals(charset.size())) {
                    word.set(i - 1, word.get(i - 1) + 1);
                    word.set(i, 0);
                } else {
                    break;
                }
            }
        }
    }
}

Answer 3

Another approach is to interpret your charset as digits of a numbersystem and your desired output as all n digit numbers. For example, in decimal system we have the digits 0 to 9 and all two-digit numbers (with leading zero) would be the numbers

00 - 99

three-digit

000 - 999 

and so on. In octal system, however, your numbers would go from

00 - 77 or from

000 - 777

and in binary

00 - 11 or

000 - 111

If we now replace in your charset the letters with digits, i.e. a with 0, b with 1, c with 2 ...

aaa in your output is the same as 000 and fff the same as 555. That means the task is to create all two-digit, three-digit, ...., 5-digit numbers in the base six number system (charset length) and convert the digits back to the letters from the given charset. A starting point for such an algorithm could be something like:

public static void main(String[] args) {
    String str = "abcdef";

    //create a map which looks like {0=a, 1=b, 2=c, 3=d, 4=e, 5=f}
    Map<Character,Character> map = new HashMap<>();
    for(int i = 0; i <str.length(); i ++){
        map.put((char)(i+'0'), str.charAt(i));
    } 

    //convert numbers to string using //Integer.toString(int i, int radix)
    //use String#format & String#replace to have a string representation with leading zeros
    for(int n = 2; n <= 5; n++){
        int maxValue = (int)Math.pow(str.length(), n);
        for(int i = 0; i < maxValue; i++){
            String temp = String.format("%"+n+"s", Integer.toString(i, str.length())).replace(' ', '0');
            for(char c: map.keySet()){
                temp = temp.replace(c, map.get(c));
            }                
            System.out.println(temp);
        }
    }
}