Cryptography Stack Exchange is a question and answer site for software developers, mathematicians and others interested in cryptography. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am trying to understand MD5 hash algorithm from the link

At first I was not able to understand padding of MD5. Then I asked a question in this link: To understand a fact related to padding in MD5

From this answer I can understand padding. Now I have faced another problem: to understand the second step of MD5.

I am trying to understand the second step from the same link:

But I have failed to understand a sentence and a phrase as before. Here is the phrase:

A 64-bit representation of b

what does 64-bit representation stand for?

Here is the next sentence:

In the unlikely event that b is greater than 2^64, then only the low-order 64 bits of b are used.

what does lower-order bits mean?

Can anyone explain the above phrase and sentence with better clarity?

share|improve this question

what does 64-bit representation stand for?

It's the length (in bits) of the hashed message, expressed as a 64 bit binary value (in little endian order). If you hash a 1 character message, $b=8$ (as 1 byte == 8 bits); this would be represented as the 8 bytes 08 00 00 00 00 00 00 00 (the 08 is first, because in little-endian order, you place the least-significant part first)

what does lower-order bits mean?

The length of the the $b$ field within the padding is 8 bytes long; if you were to hash an extremely long message (one $2^{61}$ bytes or longer), it might be the case that $b$ wouldn't fit in an 8 byte field. In that case, what you place into the field is the lowest 8 bytes into the field; another way to express this is that you place the value $b \bmod 2^{64}$ into the padding field.

On the other hand, $2^{61}$ bytes is an enormous amount of data; it is approximately the amount of data that flows over the internet globally in a day. It is quite unlikely that you (or anyone else) will ever compute an MD5 hash on that much data; you can in practice ignore this sentence.

share|improve this answer

If your message is longer than $2^{64}$ bits, i.e. $b>2^{64}$, which is $2^{61}$ bytes, or 16,777,216 Terabytes(!), this is more than 16 million terabytes, you just take the least significant 64 bits of the number $b$.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.