Consider a simplest transposition cipher, such as one with block size 8 and encryption key 43725681, the decryption key will be 84215637. As the keys are different, may I say a transposition cipher is an asymmetric algorithm, except for some special values of the keys, such as 87654321?


1 Answer 1


No, it is not an asymmetrical algorithm, becayse the decryption key can be derived from the encrytion key.

Even in a symmetrical algorithm the encryption and decryption algorithms may be different, and the decryption algorithm might include firct calculating a decryption key from the encryption key. However, the same key can be used for both encryption and decryption in the sense that if you know that key you are able to both encrypt and decrypt messages.

In an asymmetrical algorithm, someone who only knows the public key can not decrypt messages. That is not the case for a transposition cipher.

  • 1
    $\begingroup$ Nice explanation, these definitions are far more accurate than those written in many books. :) $\endgroup$
    – GreenPenguin
    Jan 16, 2017 at 18:53
  • 3
    $\begingroup$ @GreenPenguin if you find the answer to be correct, can you please mark it as accepted? $\endgroup$
    – Limit
    Jan 16, 2017 at 21:53
  • 3
    $\begingroup$ "and someone who only knows the private key can not encrypt them. " that's not true in general. You typivally can compute the public key from the private key. $\endgroup$ Jan 17, 2017 at 7:32
  • $\begingroup$ @CodesInChaos What if I know the public key, can I compute the private key? I think in RSA, the choice of public key and private key can be interchanged, right? $\endgroup$ Jan 18, 2017 at 18:29
  • $\begingroup$ @GreenPenguin Depends on how you choose the public key (small fixed values are common) and on how you define the private key (private exponent vs knowledge of the prime factors). And for typical Discrete logarithm based crypto you compute the public key by multiplying the private key with the generator. $\endgroup$ Jan 18, 2017 at 20:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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