Java中的排列而不使用数组

Question

The code below loops through all of the characters submitted, puts them into a string array and then after hashing them, checks against the hashed password originally entered. This logic works fine for up to 5 character passwords however when I get to 6 characters the number of permutations (36^6) is so large that a memory exception is thrown due to the size of the array being too large.

I am therefore trying to modify it too use just string rather than an array to stop this memory issue but cant quite get it to work, any obvious suggestions?

Answer 1

You can just enumerate all the possible passwords containing letters and digits by counting from 0 to infinity and converting the number toString with base 36.

long k = 0;
while (true) {
    String pwd = Long.toString(k++, 36);
    if (SHA1(pwd).equals(hashedPassword)) {
        System.out.println(pwd);
    }
}

Here, toString works for a base up to 36, using [0-9a-z] , ie it works in your case, but if you want to include special characters you will have to create your own number-to-password function (think division and modulo), but the rest remains the same.

This way, the memory requirement is O(1), but of course the complexity is still O(36 ⁿ ) for passwords with up to n characters.

---

One problem with this approach is that -- as with all representations of numbers -- leading zeros will be omitted, thus Long.toString will never produce a password starting with 0 (except 0 itself). To counter this, you could use two nested loops -- the outer loop iterating the number of digits in the password and the inner loop iterating the numbers up to 36 ^d and padding the string with leading zeros, or loop from 36 ^d to 2*36 ^d and cut away the first (non-zero) digit. This may seem like a lot more work than before, but in fact it just produces twice as many passwords.

for (int d = 1; d < 3; d++) {
    long p = pow(36, d);
    for (long n = p; n < 2 * p; n++) {
        String pwd = Long.toString(n, 36).substring(1);
        System.out.println(n + " " + pwd);
    }
}

Output:

0
1
...
z
00
01
...
zz

Answer 2

In addition to tobias' answer.

Here's a way you can use your specified alphabet to recurse through all possible combinations of it, up to a certain length, and operate on each of them without storing them unless you detect that it is a match.

// Depth first combinations
private static void combineRecurse(String base, char[] alphabet, int maxLen,
        Consumer<String> consumer) {
    if (base.length() < maxLen - 1) {
        for (int i = 0; i < alphabet.length; i++) {
            String newPermutation = base + alphabet[i];
            consumer.accept(newPermutation);
            combineRecurse(newPermutation, alphabet, maxLen, consumer);
        }
    } else {
        // the new permutiation will be the maxlength
        for (int i = 0; i < alphabet.length; i++) {
            consumer.accept(base + alphabet[i]);
        }
    }
}

public static void forEachCombination(char[] alphabet, int maxLen, Consumer<String> consumer) {
    if(alphabet == null || consumer == null || maxLen < 1) return;
    combineRecurse("", alphabet, maxLen, consumer);
}

public static void main(String[] args) {

    char[] alphabet = { ... };
    final String hashedPassword = SHA1(password);
    final int maxlen = 16;

    List<String> possiblePasswords = new ArrayList<>();
    forEachCombination(alphabet, maxlen, s -> {
        // System.out.println(s);
        if (SHA1(s).equals(hashedPassword)) {
            possiblePasswords.add(s);
        }
    });
}

Making this iterative rather than recursive might be better, but for now the recursive is easy to implement and is probably more easily understandable for a beginner.