3
$\begingroup$

Let $n=11, d=3, e=7$ and $M=3$.

Encryption:

  M = BSK
  B -> 1 , S -> 18 -> k -> 10
  M1 => C1 = 1^7 mod 11 = 1
  M2 => C2 = 18^7 mod 11 = 6
  M3 => C3 = 10^7 mod 11 = 10

  C = 1 6 10 
  <=> 
  1=>B, 6=>G 10=>K

Decryption:

  C1 => M1 = 1^3 mod 11 = 1
  C2 => M2 = 6^3 mod 11 = 18
  C3 => M3 = 10^3 mod 11 = 10

  M = BSK

I would like to know that I am correctly doing encryption and decryption?

$\endgroup$
3
$\begingroup$

Using the Pohlig-Hellman cipher, you encrypt your message $M = 3$ using an encryption key $(e,n) = (7,11)$. The encryption is $C = M^e \mod n = 3^7 \mod 11=9$.

You decrypt with a decryption key $(d,n) = (3,11)$, such that $e\cdot d = 1 \mod (n-1)$. Since $3 \cdot 7 \mod (11-1) \equiv 21 \mod 10 \equiv 1$, this is satisfied.

So, to decrypt, compute $9^3 \mod 11 \equiv 3$, which was your $M$. You're done.

It looks like you are confusing the Pohlig-Hellman exponentiation cipher with the Pohlig-Hellman theorem used to solve a discrete logarithm problem. These are related problems.

You use the Pohlig-Hellman algorithm when you want to solve $g^x = h \mod p$, where $g,h$ are in a finite field $\mathbb{F}^*_p$, $g$ is a primitive root and $p$ is a prime. When $p-1$ factors into a bunch of small primes, you can solve several smaller discrete logarithm problems and reassemble $x$ from the solutions to those smaller problems via the Chinese remainder theorem.

Also, make sure you know why you're using this cipher in the first place.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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