Cryptography Stack Exchange is a question and answer site for software developers, mathematicians and others interested in cryptography. It's 100% free, no registration required.

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 used TripleDES so many times but never thought about this. In TripleDES it requires 168bit key in option 1. I don't enter three single 56bit keys while implementing instead I only enter 168bit key. So, I'm guessing 168bit breaks into three 56bit keys. What it process of this breaking mechanism? is it linear?

share|improve this question
Yes, the 3DES key is just the concatenation of three keys. In practice the key is often 192 bits, due to including parity bits. – otus Jul 20 '14 at 10:36
up vote 4 down vote accepted

Yes. The keys are indeed used in a linear manner.

In particular, they are used in $E$-$D$-$E$ mode: encrypt using first 56 bits as key, decrypt using next 56 bits as key and then again encrypt using final 56 bits.

This way its possible to use triple DES (which is officially called TDEA) for the DES, 2-DES and 3-DES variations. The first would use $K_1$-$K_1$-$K_1$ as keys, the second would use $K_1$-$K_2$-$K_1$ as keys and the full 3-DES would use $K_3$-$K_2$-$K_1$ as keys. Sometimes the 2-DES and 3-DES keys are also called ABA and ABC keys respectively.

DES uses one bit of each byte as parity, but this bit is not always validated by implementations. So triple DES keys with parity use 64, 128 or 192 bits. This explains why some implementations allow the output of an MD5 hash - which is 128 bits - to be input as 2-DES key without any error.

References :

share|improve this answer
@otus indeed i meant bits thanks will edit it. – Ravindra HV Jul 20 '14 at 11:23

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.