I read in a book that: let $f(K,R)$ is cryptographic transformation of $R$ using $K$, $K\{R\}$ means $R$ is encrypted using $K$ and $h\{R\}$ is the hash of $R$ and then the example follows.

I want to know what is difference between cryptographic transformation of $R$ and encryption of $R$ using the same key? Aren't they same thing?


Usually a cryptographic transformation can mean anything. It is just a cryptographic function whose output is based on the input, $K$ and $R$ in this case. You could have a $MAC(K, R)$ where the transformation is a message authentication code or MAC for instance. A MAC also takes a key and a message, but its purpose is rather different.

So $K\{R\}$ is a specialization of $f$. Often $K\{R\}$ it is written down as $E(K, R)$ which would probably make this more clear (although $E_k(R)$ is also often used).

It could also be that $K\{R\}$ is the keyed encryption function, taking just the input $R$ while $f(K, R)$ is the uninitialized or stateless cipher taking both the key $K$ and the input message $R$. If the math notation is not precisely specified you may have to rely on context.

| improve this answer | |
  • $\begingroup$ I'm not sure about the term "specialization" though, maybe there is a better mathematical term for this. $\endgroup$ – Maarten Bodewes Jun 16 '15 at 19:54
  • $\begingroup$ maybe: "special case"? But "specialization" is also fine according to wikipedia. $\endgroup$ – SEJPM Jun 16 '15 at 19:58
  • $\begingroup$ @SOJPM Do you agree on the edit below the horizontal line? $\endgroup$ – Maarten Bodewes Jun 16 '15 at 20:33
  • $\begingroup$ As the OP didn't (yet) reply, stating this addition is absolutely reasonable as it gives the OP a second way of interpretation if the first (more likely) one doesn't make sense. If your "uninitailized" indicates that you look for a better word, I may suggest "stateless". $\endgroup$ – SEJPM Jun 16 '15 at 20:41

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