February 6th, 2018, 06:30 PM  #1 
Newbie Joined: Jan 2018 From: Toronto Posts: 12 Thanks: 0  Caesar Cipher help!
Not sure whether this question was answered already, though I saw the same question. Assume that we choose N (that is, K=14) as an encryption key in Caesarâ€™s cipher. What is the result Ci of the following (double) encryption Ci = E (K,E(K, Mi)), where Mi is an arbitrary plaintext element? Any suggestions on how to answer this question will be greatly appreciated. Thank you. Last edited by skipjack; February 16th, 2018 at 02:54 AM. 
February 6th, 2018, 08:16 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 19,866 Thanks: 1833 
Why is the encryption key referred to as a letter? How would the encryption be done using a letter without converting that letter to a number?

February 7th, 2018, 06:44 AM  #3 
Senior Member Joined: Apr 2014 From: UK Posts: 891 Thanks: 328 
The answer is: Ci = E (2K, Mi) Given that 2K in the example is 28, you might prefer the slightly longer answer of: Ci = E (2K(Mod 26), Mi) Any encryption which replaces one fixed set of symbols for another fixed set never benefits from double encryption as you can always decrypt it in a single step (because you can also double encrypt in a single step). 
February 7th, 2018, 09:21 AM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,502 Thanks: 2511 Math Focus: Mainly analysis and algebra 
That's a great example of using notation to obfuscate.

February 7th, 2018, 05:58 PM  #5  
Newbie Joined: Jan 2018 From: Toronto Posts: 12 Thanks: 0  Quote:
If N is used as an encryption key in Caesar cipher and N=Q then should it not be Ci = E (2Q (Mod 26), Mi) but not clear on how you got 2. (The change of each letter to the third letter is how I got the Q) Also, what is meant by where Mi is an arbitrary plaintext element? Thanks for your help. Patricia  
February 8th, 2018, 12:56 AM  #6 
Senior Member Joined: Apr 2014 From: UK Posts: 891 Thanks: 328 
I think I now understand your notation, My equation still works if K is QN (I thought that's what the K=14 was referring to in the original post). So, in the N=Q example, we rotate the alphabet 3 places forward. If we then encrypt the result, we move another 3 places forward, the result is that we move 6 places forward. "Mi is an arbitrary plaintext element" I took this to mean any letter of the alphabet, but this isn't strictly true, plaintext can contain upper and lower case letters, spaces and other punctuation symbols. Computer systems typically allocate a whole byte for each element, so you'd have to use (mod 256) instead of (mod 26) 
February 15th, 2018, 04:58 PM  #7  
Newbie Joined: Jan 2018 From: Toronto Posts: 12 Thanks: 0  Question Quote:
Ci = E (2K(Mod 256), Mi)  
February 15th, 2018, 08:15 PM  #8 
Global Moderator Joined: Dec 2006 Posts: 19,866 Thanks: 1833  If that means that "A" is encrypted as "N" by shifting 13 places along the alphabet (so that "K = 14" was a mistaken assumption), double encryption would leave the original text unchanged (Ci = Mi), assuming that the encryption is applied only to the 26 letters of the alphabet.

February 16th, 2018, 12:10 AM  #9 
Senior Member Joined: Apr 2014 From: UK Posts: 891 Thanks: 328 
Ha, well spotted 

Tags 
caesar, cipher 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Caesar Encryption  mrdumb  Algebra  1  September 29th, 2014 09:24 AM 
PohligHellman Cipher + Affine Hill Cipher, relatively simpl  threesixtify  Applied Math  0  January 22nd, 2013 05:53 AM 
Multiplicative cipher  part 2  purecotton  Number Theory  1  May 28th, 2012 07:30 AM 
a multiplicative cipher  milly2012  Number Theory  7  May 28th, 2012 05:17 AM 
Affine transformation and RSA cipher  nooblet  Number Theory  3  March 12th, 2009 04:46 AM 