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Cryptography .NET, Avoiding Timing Attack

I was browsing crackstation.net website and came across this code which was commented as following:

Compares two byte arrays in length-constant time. This comparison method is used so that password hashes cannot be extracted from on-line systems using a timing attack and then attacked off-line.

 private static bool SlowEquals(byte[] a, byte[] b)
    {
        uint diff = (uint)a.Length ^ (uint)b.Length;
        for (int i = 0; i < a.Length && i < b.Length; i++)
            diff |= (uint)(a[i] ^ b[i]);
        return diff == 0;
    }

Can anyone please explain how does this function actual works, why do we need to convert the length to an unsigned integer and how this method avoids a timing attack? What does the line diff |= (uint)(a[i] ^ b[i]); do?

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Baahubali Avatar asked Jun 06 '13 03:06

Baahubali


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1 Answers

This sets diff based on whether there's a difference between a and b.

It avoids a timing attack by always walking through the entirety of the shorter of the two of a and b, regardless of whether there's a mismatch sooner than that or not.

The diff |= (uint)(a[i] ^ (uint)b[i]) takes the exclusive-or of a byte of a with a corresponding byte of b. That will be 0 if the two bytes are the same, or non-zero if they're different. It then ors that with diff.

Therefore, diff will be set to non-zero in an iteration if a difference was found between the inputs in that iteration. Once diff is given a non-zero value at any iteration of the loop, it will retain the non-zero value through further iterations.

Therefore, the final result in diff will be non-zero if any difference is found between corresponding bytes of a and b, and 0 only if all bytes (and the lengths) of a and b are equal.

Unlike a typical comparison, however, this will always execute the loop until all the bytes in the shorter of the two inputs have been compared to bytes in the other. A typical comparison would have an early-out where the loop would be broken as soon as a mismatch was found:

bool equal(byte a[], byte b[]) { 
    if (a.length() != b.length())
        return false;

    for (int i=0; i<a.length(); i++)
       if (a[i] != b[i])
           return false;
    return true;
}

With this, based on the amount of time consumed to return false, we can learn (at least an approximation of) the number of bytes that matched between a and b. Let's say the initial test of length takes 10 ns, and each iteration of the loop takes another 10 ns. Based on that, if it returns false in 50 ns, we can quickly guess that we have the right length, and the first four bytes of a and b match.

Even without knowing the exact amounts of time, we can still use the timing differences to determine the correct string. We start with a string of length 1, and increase that one byte at a time until we see an increase in the time taken to return false. Then we run through all the possible values in the first byte until we see another increase, indicating that it has executed another iteration of the loop. Continue with the same for successive bytes until all bytes match and we get a return of true.

The original is still open to a little bit of a timing attack -- although we can't easily determine the contents of the correct string based on timing, we can at least find the string length based on timing. Since it only compares up to the shorter of the two strings, we can start with a string of length 1, then 2, then 3, and so on until the time becomes stable. As long as the time is increasing our proposed string is shorter than the correct string. When we give it longer strings, but the time remains constant, we know our string is longer than the correct string. The correct length of string will be the shortest one that takes that maximum duration to test.

Whether this is useful or not depends on the situation, but it's clearly leaking some information, regardless. For truly maximum security, we'd probably want to append random garbage to the end of the real string to make it the length of the user's input, so the time stays proportional to the length of the input, regardless of whether it's shorter, equal to, or longer than the correct string.

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Jerry Coffin Avatar answered Oct 05 '22 10:10

Jerry Coffin