substr md5 collision

Question

I need a 4-character hash. At the moment I am taking the first 4 characters of a md5() hash. I am hashing a string which is 80 characters long or less. Will this lead to collision? or, what is the chance of collision, assuming I'll hash less than 65,536 (16⁴) different elements?

Answer 1

Well, each character of md5 is a hex bit. That means it can have one of 16 possible values. So if you're only using the first 4 "hex-bits", that means you can have 16 * 16 * 16 * 16 or 16^4 or 65536 or 2^16 possibilities.

So, that means that the total available "space" for results is only 16 bits wide. Now, according to the Birthday Attack/Problem, there are the following chances for collision:

• 50% chance -> 300 entries
• 1% chance -> 36 entries
• 0.0000001% chance -> 2 entries.

So there is quite a high chance for collisions.

Now, you say you need a 4 character hash. Depending on the exact requirements, you can do:

• 4 hex-bits for 16^4 (65,536) possible values
• 4 alpha bits for 26^4 (456,976) possible values
• 4 alpha numeric bits for 36^4 (1,679,616) possible values
• 4 ascii printable bits for about 93^4 (74,805,201) possible values (assuming ASCII 33 -> 126)
• 4 full bytes for 256^4 (4,294,967,296) possible values.

Now, which you choose will depend on the actual use case. Does the hash need to be transmitted to a browser? How are you storing it, etc.

I'll give an example of each (In PHP, but should be easy to translate / see what's going on):

4 Hex-Bits:

$hash = substr(md5($data), 0, 4);

4 Alpha bits:

$hash = substr(base_convert(md5($data), 16, 26)0, 4);
$hash = str_replace(range(0, 9), range('S', 'Z'), $hash);

4 Alpha Numeric bits:

$hash = substr(base_convert(md5($data), 16, 36), 0, 4);

4 Printable Assci Bits:

$hash = hash('md5', $data, true); // We want the raw bytes
$out = '';
for ($i = 0; $i < 4; $i++) {
    $out .= chr((ord($hash[$i]) % 93) + 33);
}

4 full bytes:

$hash = substr(hash('md5', $data, true), 0, 4); // We want the raw bytes