Talk:RSA code

Draft

There's a lot to do to clean this up to the level of being a full task. For one thing, it's not at all clear what this code is trying to do! It also lacks links to resources (e.g., wikipedia) that describe the algorithm involved, and the Python solution needs much work too (especially in its surrounding descriptive text, which looks to me more like text that belongs here on the talk page). –Donal Fellows 06:31, 24 March 2011 (UTC)

+1 on Dkf's comments. How to perform the task needs to be in the task description in a more language neutral form. --Paddy3118 13:12, 24 March 2011 (UTC)

Sorry about the shoddy state of my code, it's one of the first programs i have written on my own, and there are probably far more efficient ways of performing many functions that i just brute forced my way past. But it works, and i am rather proud of it, ugly as it may be. I have added detail about the function of the code and the way the RSA algorithm works. Please let me know if there is anything else i can clear up! --Erasmus 03:00, 26 March 2011 (UTC)

I believe my code is now annotated enough so that it makes sense to read it. I am streamlining (at least as much as i can) a program to generate new keys to use, i will upload that when i am done --Erasmus 02:21, 29 March 2011 (UTC)

Blocking?

"This yields two blocks of numbers ..." I can see that a series of numbers are produced, but the method of splitting into blocks is not given. --Paddy3118 02:45, 26 March 2011 (UTC)

The blocking code is more complicated than the encryption code. But:

When converting letters to numbers, the numbers should be non-zero.
If if X is the largest number value, then pick the largest K such that (1+X)^K < N (where N is from the key). K is the number of letters represented in a block.
If v is an array of numbers repesenting the letters in a block, the numeric value of the block itself is can be computed (using psuedo-C or maybe psuedo-javascript):

   block= 0;
   for (int i= 0; i<v.length; i++) {
       block= v[i]+(X+1)*block;
   }

To go the other direction:

   for (int i= v.length-1; i >= 0; i--) {
       v[i]= block % (X+1);  /* remainder function corresponding to division on next line */
       block= block / (X+1); /* integer division */
   }

--Rdm 03:31, 26 March 2011 (UTC)

RSA in J

I may simply be naive, but it seems to me that the example doesn't code "hi there" correctly. The output is given as "695 153 2377 260". I tried to decrypt those blocks, but i can only recover gibberish from them. Even abandoning my python code and simply doing the modular exponentiation yields the same results. I get "575, 1230, 2387, 205" if i split "hi there" into 4 blocks. Can you explain how exactly you are encoding and decoding? I am not familiar with J, so i cant really tell whats going on in your code. --Erasmus 03:04, 30 March 2011 (UTC)

I believe the pseudo code I posted, above, matches the process I used in J, though I cannot actually run the pseudo code, and it might be buggy. But I have updated the J entry with a blocking example. So, for example: 'h' is letter 8 and 'i' is letter 9 and 'hi' is thus (32*8)+9 or 265. But perhaps we should be using a different blocking algorithm? --Rdm 12:26, 30 March 2011 (UTC)

I should perhaps also note that I cannot get the python example to work. It fails for me with the message ImportError: No module named tkinter. But these work for me:

<lang python>>>> import _tkinter

>>> import Tkinter >>> Tkinter._test()</lang>

And the python implementation does not supply any canned examples.

So, anyways, I have not tested against the python code. --Rdm 17:35, 30 March 2011 (UTC)

Ah...that explains it. Our blocking methods were just entirely different. I recovered 265, but i didnt know how to turn that back to letters. My code blocks "hi" as "0809", since "h" is "08" and "i" is "09". If "f" is the number of digits in "n", my python code only places (f/2)-1 letters per block. I realize this doesn't make sense for a small n, but at larger n it isn't noticeable, and it prevents my blocks from being larger than "n".

I have uploaded a version of my code that doesn't use the Tkinter window. I find running it from python IDLE seems to work best, since i can copy paste plaintext or ciphertext to and from the window. --Erasmus 01:18, 1 April 2011 (UTC)