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Shootinfish

Loadscreen cipher

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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 by Shootinfish

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@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 by Shootinfish

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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.

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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   

 

 

 

 

 

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