No. MD5 is not encryption (though it may be used as part of some encryption algorithms), it is a one way hash function. Much of the original data is actually “lost” as part of the transformation. Think about this: An MD5 is always 128 bits long. That means that there are 2128 possible MD5 hashes. … Read more
Basically it’s because the output of MD5 contains less information than the input. This is basically what distinguishes a hash algorithm from an encryption algorithm. Here’s a simple example: imagine an algorithm to compute the hash of a 10-digit number. The algorithm is “return the last 2 digits.” If I take the hash of 8023798734, … Read more
Of all bit-operations XOR has the best bit shuffling properties. This truth-table explains why: A B AND 0 0 0 0 1 0 1 0 0 1 1 1 A B OR 0 0 0 0 1 1 1 0 1 1 1 1 A B XOR 0 0 0 0 1 1 1 0 … Read more
Yes. MD5 has been completely broken from a security perspective, but the probability of an accidental collision is still vanishingly small. Just be sure that the files aren’t being created by someone you don’t trust and who might have malicious intent.
Probably because 397 is a prime of sufficient size to cause the result variable to overflow and mix the bits of the hash somewhat, providing a better distribution of hash codes. There’s nothing particularly special about 397 that distinguishes it from other primes of the same magnitude.
You’re correct that it cannot guarantee uniqueness, however there are approximately 3.402823669209387e+38 different values in a 32 digit hex value (16^32). That means that, assuming the math behind the algorithm gives a good distribution, your odds are phenomenally small that there will be a duplicate. You do have to keep in mind that it IS … Read more
For a set of even billions of assets, the chances of random collisions are negligibly small — nothing that you should worry about. Considering the birthday paradox, given a set of 2^64 (or 18,446,744,073,709,551,616) assets, the probability of a single MD5 collision within this set is 50%. At this scale, you’d probably beat Google in … Read more
In your case, breaking the hash algorithm is equivalent to finding a collision in the hash algorithm. That means you don’t need to find the password itself (which would be a preimage attack), you just need to find an output of the hash function that is equal to the hash of a valid password (thus … Read more
CRC works fine for detecting random errors in data that might occur, for example, from network interference, line noise, distortion, etc. CRC is computationally much less complex than MD5 or SHA1. Using a hash function like MD5 is probably overkill for random error detection. However, using CRC for any kind of security check would be … Read more
Yes, MD5 is somewhat less CPU-intensive. On my Intel x86 (Core2 Quad Q6600, 2.4 GHz, using one core), I get this in 32-bit mode: MD5 411 SHA-1 218 SHA-256 118 SHA-512 46 and this in 64-bit mode: MD5 407 SHA-1 312 SHA-256 148 SHA-512 189 Figures are in megabytes per second, for a “long” message … Read more