Shootinfish Report post Posted November 3, 2015 (edited) This is regarding the MOTD loadscreen cipher Now im not even going to attempt to decode it as cryptography is all new to me But i am interested on the complexity calculation of the cipher and wondered if anyone has any insight in getting a actual number if we fix some variables We can debate the variables make them constants and try calculate total number of combinations Assuming a 5*5 polybius for the ADFGX (im guessing there is 625 combinations of alphabets but im proberly wrong) Assuming the loadscreen column width is key length which is 6 which i have squared for transposition Assuming the key is an word in the english dictionary (Needs more research but 15222 is a quick and dodgy web number) 625 * (6*6) * 15222 I came up with 342495000 possible combinations Now i know some people will laugh as i proberly got this totally wrong but please correct me and add your input on the complexity of this cipher Hopefully we can get an agreed formula to start with and then debate the numbers EDIT These calculations are totally useless and I need to go back to school to learn basic maths like factorials Hopefully the cipher breakdown below is correct if not please let me know Edited November 3, 2015 by Shootinfish 0 Share this post Link to post

PINNAZ Report post Posted November 3, 2015 It's way over my head. @MrRoflWafflesis on it though. I reckon he has started decoding it. 0 Share this post Link to post

Shootinfish Report post Posted November 3, 2015 (edited) @PINNAZ Yes hopefully he will have it cracked soon Its just i thought there was some maths gurus on the site and they could crunch the numbers while waiting for launch Anyway Let see if i understand this cipher right Before you read anymore this isnt me trying to crack the code its just to get the understanding of how it works and if i got my calculations right (which i proberly havent) FFGXGD GFFAGF GGDDGF FFXXFF FDGFFG FDGFFG FDGGFF FGFGAA FXFXDX XFXDGF FAGGFF AF Ok im assuming it hasnt been concacated its still in its grid but columnar transposition has been performed and its a basic order of alphabet So i take a 6 letter keyword im going to be unoriginal and use ISLAND ADLINS // Island in alphabtical order FFGXGD GFFAGF GGDDGF FFXXFF FDGFFG FDGFFG FDGGFF FGFGAA FXFXDX XFXDGF FAGGFF AF ISLAND // columns rearranged GDXFGF FFAGGF DFDGGG XFXFFF GGFFFD GGFFFD GFGFGD FADFAG FXXFDF XFGXGX GFDFFA F A Which would refractioned to GD XF GF FF AG GF DF DG GG XF XF FF GG FF FD GG FF FD GF GF GD FA DF AG FX XF DF XF GX GX GF DF FA FA Now the 5*5 polybius square i could have looked at frequencies but just used numbers instead of letters for simplicity A D F G X A 1 2 3 4 5 D 6 7 8 9 10 F 11 12 13 14 15 G 16 17 18 19 20 X 21 22 23 24 25 So the decode output would be 17,3,18,13,4,18,8,9,19,23,23,13,19,13,12,19,13,12, 18,18,17,11,8,4,15,23,8,23,20,20,18,8,11,11 Hopefully i understand it correctly Im looking at some c sourcecode but its for ADFGVX cipher ,if i was more proficient in programming it could be altered to ADFGX and batched output Then i dont know what to do with the output maybe i could grep it somehow Anyway i had my fun with it hopefully it will be decoded soon BTW This what happens when your waiting for a Black ops to drop you go a little insane "Well i have anyway" Edited November 3, 2015 by Shootinfish 0 Share this post Link to post

Cunambula Report post Posted November 3, 2015 Figured this out on my own a while back, though i shared it with everyone, maybe they didnt believe me? Anywhoo....if you want to do it yourself I can save you a ton of time by giving you the key-word Spoiler Portals It is deciphered using the standard adfgx method. I will refrain from posting the revealed message , although I have allready in prev. post. Its neat but, it is something you may have heard since...sam says something very similar. 0 Share this post Link to post

Shootinfish Report post Posted November 3, 2015 Not sure why im replying but you havent supplied a fractionation key to go with the transposition key you suggested "Because portals is a severn letter word" The total number of transposition keys for 7 columns is (7!) = 5400 This is a guess BTW to end this thread The total number of polibuis square fractionation keys is (25!) = 15511210043330985984000000 I do believe it has not been concacated and there are six columns but this still would mean (6!) = 720 transposition keys Still insane figures for pen and paper cipher And if I was to try generate all the fractionation keys on computer I would be well dead before it finished 0 Share this post Link to post

Shootinfish Report post Posted November 4, 2015 (edited) Sorry to double post but Firstly I should say that I should not just learn maths but also how to search the forum If I had I would have found the answer to the calculations the thread below http://www.callofdutyzombies.com/index.php?/topic/154218-adfgx-cipher/&page=1 Edited November 4, 2015 by Shootinfish 0 Share this post Link to post