Pohlig-Hellman Cipher + Affine Hill Cipher, relatively simple (not for me!) questions

Hi guys,

How you all doing?

Would be a great help if you could help me solve these equations. I guess more than the answers I need the process by which you get to them.

Here are the questions:

---------------------------------------------------------------------------------------------------------------------------------------------------------

1) The cryptotext 3276 is a result of the Pohlig-Hellman Cipher with p = 7823 and e = 3129. Find the Plaintext.

Encryption algo: Ee(x) =xe (mod p)

Decryption algo: Dd(y) = yd (mod p)

---------------------------------------------------------------------------------------------------------------------------------------------------------

2) The Affine Hill Ciper has the encryption function

E(x^1,...,x^m) = (x^1,...,x^m) A+(b^1,...,b^m) (mod 26)

where A is invertible m x m matrix. Encrypt the message "tavern" using the encryption key m=2, A= (see matrix below)

( 4 11

3 19 )

and (b^1, b^2) = (13, 5).

Check your result by decrypting.

---------------------------------------------------------------------------------------------------------------------------------------------------------

Ok. That's it!

Any and all help would be appreciated... Thankss(Nerd)

Re: Pohlig-Hellman Cipher + Affine Hill Cipher, relatively simple (not for me!) quest

Hey threesixtify.

Can you show us what you have tried? Also do you have a software package that can do the matrix and mod operations? (A free package is Octave that you can download with GUIOctave as a GUI front end).

Re: Pohlig-Hellman Cipher + Affine Hill Cipher, relatively simple (not for me!) quest

Moi threesixtify,

Mita kuuluu ?

you can find "d" by using e.d = 1(mod p-1) [where '=' means congruent]

Then apply the decryption formula which you have mentioned.

And yes you are right, Steps are required if Tomi asks to solve (this 25% of the questions) on the board :P

Which batch are you in ? 10-12 or 12-2 ?

nähdään huomenna

Terveisin

AliceBobEve