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Shootinfish

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Shootinfish last won the day on July 30 2016

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  1. @TheHiddenSpectre Nice write up and I understand it.I used these methods to find out if transposition only or subsitution only on TheGiant cipher and worked out neither was possible The undercroft cipher I can rule out Transposition only Vigenere using standard b64 order alphabet and still being in s B64 format ready to decode 3.Mixed case Ascii mono subsubtution " I cant totally rule out or manually plot ascii mixed case and numbers but doesnt seem likely " That's why I was looking into finding a pattern or keylength with xor and trying to automate it. Do you think this method wont work to test if a key (4-10) characters has been xored to encrypt B64
  2. ADFGX cipher

    Had another thought that it could be double transposed using one key word from the list that meets the requirement of the last 2 columns being the first. The reason i think this is the scibble above the cipher could be 2 blocks of text. For instance using "TRIPOD" for 6 column transpose then collapsing the text and making 4 columns and transpose using "TRIP" Will have to check how many words that can have the 2 end letters ommited and still retain being a 4 letter word. I cant do anymore with the 48 combinations so I'm looking at all other possible ways the cipher could be read. EDIT ///////////// I edited the above word I used because I grabbed it from a list on reddit which I though was more comprehensive than my own but little did I realise that it had errors in it so my fault should of double checked
  3. ADFGX cipher

    Ok i know the transpostion keyword is not important on its own but it could be used in the alphabet. Its obvious the alphabet in not primed with a six letter keyword as it makes too many letters static. But it could be shifted with an offset shifting those static letters backwards and forwards. There cant be that many six letter words that fit the 48 transpostions. A quick look for FEDACB FEDACB ZOMBIE I think i got that right but you get the idea. Also im not sure on this idea put you could also try a primed alphabet with longer isograms that may have a connection to the loadscreen or the map. ambidextrous disreputably incomputable "my fav lol" nightwalkers bankruptcies drumbeatings expurgations unprofitable unforgivable EDIT ///////////////////////////////////////// So I computed an alphabet list for zombie and there is 19 start to end if we ceaser shift these that means 19 *25 (I think there is duplicates so its less than that) and I am guessing there is less than 500 transposition keywords that fit meaning there is less than 250,000 alphabets that can made I suppose you could double that if you do descending alphabets as well , so what do you crazy idea or worth spending the time to create the alphabets ? EDIT 2 //////////////////////////////////// I tried to make a cryptool template to generate a wordlist but im not sure if i trust the output but i got 110 words out of 4663 6 word dictionary Here are my results it maybe a start http://pastebin.com/xNdwBMEh EDIT 3 ///////////////////////////////////// Ok I have found a word that should not be on the list "LYRICS" not sure what is going wrong ? Used ZOMBIE for a test vector as it takes words from each half of the alphabet which should mean it will put a small amount of randomness to the ceaser shift. I found 25 duplicates so it takes the number down to 475 per word and there is 109 words at this time. Here is the sequences for zombie tell me if you think its viable for a alphabet or do you think that its not random enough to fit the cipher. http://pastebin.com/7b9EQ8Y6
  4. ADFGX cipher

    Thinking way out side the box here and been looking at other ways that make sense to interperet the ciphertext. What if there are two sets of columns one of 4 columns and one of 2 columns. The reason i say this as it looks like 2 words scribbled out above the ciphertext and its possible the style the way the letters are squeezed in that they have been put there after The only other obvious is its read by columns but the final AF is evidence that the message is at the stage before that part of the cipher. Mind you reading this did make me wonder https://www.dropbox.com/s/q8xjr9u1z4xnm5r/41784789082381.pdf?dl=0 Interesting if you have time to look at it.
  5. ADFGX cipher

    Ok I'm not 100% I did the right thing but took the strand and did a substitution so AA=a etc for the 25 letters of the alphabet and then run ioc test on it and it compares this to the value of 0.0667 which is the friedman ioc value for English. Why I think its right is because it mirrored the results for both strands starting DG and GD So 0.0667 is the friedman ioc value for English text and the other is the value its calculated the ciphertext at. Maybe I have misunderstood the concept but it maybe worth checking Edit ////////////////// This is a screencap of what I have done
  6. ADFGX cipher

    Sounds like you got it covered but your video sparked an interest so I looked over some of the old work and I think there was something wrong with the IOC test because I redid them and got different results where I only had 16 out of the 48 showing mono also they where mirrored in the order in equal amounts for ciphertext starting DG and GD which makes sense. GD FF GX GF GF FA GF GG DD FF FF XX FG FD GF FG FD GF FF FD GG AA FG FG DX FX FX GF XF XD FF FA GG AF RNUSSLSTGNNZOMSOMSNMTAOOKPPSXWNLTC IC_analyzed = 0.06952 IC_provided = 0.0667 Mode = monoalphabetic/cleartext ///////////////////////////////////////////////////////////////////////////////////////////////////// GD FF XG GF GF AF GF GG DD FF FF XX FG FD FG FG FD FG FF FD GG AA FG GF DX FX XF GF XF DX FF FA GG AF RNYSSCSTGNNZOMOOMONMTAOSKPXSXKNLTC IC_analyzed = 0.06952 IC_provided = 0.0667 Mode = monoalphabetic/cleartext ///////////////////////////////////////////////////////////////////////////////////////////////////// GD FF GX GF FG FA GF GG DD FF FF XX FG DF GF FG DF GF FF DF GG AA GF FG DX XF FX GF FX XD FF AF GG AF RNUSOLSTGNNZOHSOHSNHTASOKXPSPWNCTC IC_analyzed = 0.06952 IC_provided = 0.0667 Mode = monoalphabetic/cleartext ///////////////////////////////////////////////////////////////////////////////////////////////////// GD FF XG GF FG AF GF GG DD FF FF XX FG DF FG FG DF FG FF DF GG AA GF GF DX XF XF GF FX DX FF AF GG AF RNYSOCSTGNNZOHOOHONHTASSKXXSPKNCTC IC_analyzed = 0.07308 IC_provided = 0.0667 Mode = monoalphabetic/cleartext ///////////////////////////////////////////////////////////////////////////////////////////////////// GD GX FF GF FA GF GF DD GG FF XX FF FG GF FD FG GF FD FF GG FD AA FG FG DX FX FX GF XD XF FF GG FA AF RUNSLSSGTNZNOSMOSMNTMAOOKPPSWXNTLC IC_analyzed = 0.06952 IC_provided = 0.0667 Mode = monoalphabetic/cleartext ///////////////////////////////////////////////////////////////////////////////////////////////////// GD GX FF GF FA FG GF DD GG FF XX FF FG GF DF FG GF DF FF GG DF AA FG GF DX FX XF GF XD FX FF GG AF AF RUNSLOSGTNZNOSHOSHNTHAOSKPXSWPNTCC IC_analyzed = 0.06952 IC_provided = 0.0667 Mode = monoalphabetic/cleartext //////////////////////////////////////////////////////////////////////////////////////////////////// GD XG FF GF AF GF GF DD GG FF XX FF FG FG FD FG FG FD FF GG FD AA GF FG DX XF FX GF DX XF FF GG FA AF RYNSCSSGTNZNOOMOOMNTMASOKXPSKXNTLC IC_analyzed = 0.06952 IC_provided = 0.0667 Mode = monoalphabetic/cleartext ///////////////////////////////////////////////////////////////////////////////////////////////////// GD XG FF GF AF FG GF DD GG FF XX FF FG FG DF FG FG DF FF GG DF AA GF GF DX XF XF GF DX FX FF GG AF AF RYNSCOSGTNZNOOHOOHNTHASSKXXSKPNTCC IC_analyzed = 0.07308 IC_provided = 0.0667 Mode = monoalphabetic/cleartext ///////////////////////////////////////////////////////////////////////////////////////////////////// DG FF GX FG GF FA FG GG DD FF FF XX GF FD GF GF FD GF FF FD GG AA FG FG XD FX FX FG XF XD FF FA GG FA INUOSLOTGNNZSMSSMSNMTAOOWPPOXWNLTL IC_analyzed = 0.07308 IC_provided = 0.0667 Mode = monoalphabetic/cleartext ///////////////////////////////////////////////////////////////////////////////////////////////////// DG FF XG FG GF AF FG GG DD FF FF XX GF FD FG GF FD FG FF FD GG AA FG GF XD FX XF FG XF DX FF FA GG FA INYOSCOTGNNZSMOSMONMTAOSWPXOXKNLTL IC_analyzed = 0.06952 IC_provided = 0.0667 Mode = monoalphabetic/cleartext ///////////////////////////////////////////////////////////////////////////////////////////////////// DG FF GX FG FG FA FG GG DD FF FF XX GF DF GF GF DF GF FF DF GG AA GF FG XD XF FX FG FX XD FF AF GG FA INUOOLOTGNNZSHSSHSNHTASOWXPOPWNCTL IC_analyzed = 0.06952 IC_provided = 0.0667 Mode = monoalphabetic/cleartext ////////////////////////////////////////////////////////////////////////////////////////////////////// DG FF XG FG FG AF FG GG DD FF FF XX GF DF FG GF DF FG FF DF GG AA GF GF XD XF XF FG FX DX FF AF GG FA INYOOCOTGNNZSHOSHONHTASSWXXOPKNCTL IC_analyzed = 0.06952 IC_provided = 0.0667 Mode = monoalphabetic/cleartext ///////////////////////////////////////////////////////////////////////////////////////////////////// DG GX FF FG FA GF FG DD GG FF XX FF GF GF FD GF GF FD FF GG FD AA FG FG XD FX FX FG XD XF FF GG FA FA IUNOLSOGTNZNSSMSSMNTMAOOWPPOWXNTLL IC_analyzed = 0.07308 IC_provided = 0.0667 Mode = monoalphabetic/cleartext ///////////////////////////////////////////////////////////////////////////////////////////////////// DG GX FF FG FA FG FG DD GG FF XX FF GF GF DF GF GF DF FF GG DF AA FG GF XD FX XF FG XD FX FF GG AF FA IUNOLOOGTNZNSSHSSHNTHAOSWPXOWPNTCL IC_analyzed = 0.06952 IC_provided = 0.0667 Mode = monoalphabetic/cleartext /////////////////////////////////////////////////////////////////////////////////////////////////////// DG XG FF FG AF GF FG DD GG FF XX FF GF FG FD GF FG FD FF GG FD AA GF FG XD XF FX FG DX XF FF GG FA FA IYNOCSOGTNZNSOMSOMNTMASOWXPOKXNTLL IC_analyzed = 0.06952 IC_provided = 0.0667 Mode = monoalphabetic/cleartext //////////////////////////////////////////////////////////////////////////////////////////////////////// DG XG FF FG AF FG FG DD GG FF XX FF GF FG DF GF FG DF FF GG DF AA GF GF XD XF XF FG DX FX FF GG AF FA IYNOCOOGTNZNSOHSOHNTHASSWXXOKPNTCL IC_analyzed = 0.06952 IC_provided = 0.0667 Mode = monoalphabetic/cleartext /////////////////////////////////////////////////////////////////////////////////////////////////////////// This might be worth investigating again as it could cut your workload down if its right I used cryptool2 and converted AA to a etc
  7. @oxin8 I know you are using your own software but have you investigated cryptool2 IDP analyser for ideas in your own software the source is available and I believe it is written in c# I can test but I'm having problems with page navigation on the site with IE and FF only see the first and last page of the topic so I cant see the ciphertext easily can someone paste it as I'm still working on "theGiant" cipher which btw I seem to have exhausted vigenere to base64 conversion I just cant seem to get the starting blocks to 01 and kinda out run out of ideas for it EDIT ///////////// I got it I think r rh tgeRbt bnad ee shisfptFtn et eo enebe lesdepj n .a cih,eepa t gv g o,aiM Av sT aevtH esoe.nS eaeat , w,arhaeohjemuo s,detibR nhtkphaetaid ygoTt.irc"tma"rocnrihr mapa oapIthhboyst.pr eear heuaAtm et . net 'o clms e dat sl drwht hedyaehsnrdoeot oihtrvire Ci m euh emtethtr mp cn
  8. "TheGiant" Cipher

    Sorry for the double post again but i thought this was interesting. I dont know what to think about and cant be sure if its correct or not. If you take "TheGiant" cipher text and input it as ASCII and then convert to UTF-16 you get this. 䍫汭䙧㙩啇乊歧汎儱ㄴ扦祦潌䙃䍔汶歱楚佬汋塁穁ㅐㅕ祵䈱㑅晕䉐灦测䱭扏奪兮剎慂瑐楋坖捺䄵瘴睏砳敬䘸桏䝁婊朷椴佮湷䩤䵸癏㍏捤䴱㈸瑡吲㤶㔳潲煔坹杄䝴⽄睨剷㍆䡯䙱㕍捖ㅷ瑊乬獢坧浒漴⼴畱䑅婫砷䈱㜲戵㍘䘯ㅯ What is strange that it seems from a first glance it seems the glyphs are from the same language and all the symbols are there and none are missing so it seems like it isnt just a random load of bytes converted to unicode. If it where surely there would be glyphs from all different codebases. Im not sure what to think of it can someone take a look at it. I have no idea how to go about translating it or working out its not what the cipher converts to. I actually doubt this is what its supposed to be and maybe printable ASCII is in this character sets byte range but would be good to hear what everyone thinks EDIT////////////////////////////////////////////////////////////////////////////// I have thought about it and I think this is just a coincidence that 2 printable 8 bit ascii characters line up with 1 utf-16 character and its in the same codebase The reason I think this is, why only use the B64 alphabet and also "TheGiant" keyword isn't used so I think its a false positive I also tried to translate it and the first character is for a mystical goat like creature so that doesn't make sense
  9. "TheGiant" Cipher

    I have been looking at B64 character "/" bit pattern "111111". There are 3 occurances of "/" in the whole of the ciphertext. The only ASCII character i can map that to is "?" bit pattern "00111111". That would mean it would only fit into the 4th slot (bit 18-24) of the 24 bit block. I dont think that "?" question mark would be in the plaintext so that would mean a transposition only on the ciphertext and then decode from B64 cannot be possible. What do you people think have i got this right and do you think that assuming a question mark in the plaintext is so unlikely that i can write off transposition only encryption entirely ? As i explained in the post above and my work is correct substitution only encryption wont work because there are too many unique B64 characters that are in the first B64 character block of the 24 bit block. I know tried bruteforcing it before with both individual methods but as i said last post i made mistakes and i wanted to rule these methods out a second time a more methodical way. So I'm really puzzled how the keyword "TheGiant" is used Also you must think I mad with my previous Lorenz posts as I didn't realise it was confirmed that it was simplified Lorenz cipher and the wheel positions had been posted by DragonGJY I'm kinda glad in way because I didn't understand how the full machine worked or the cribs involved :)
  10. "TheGiant" Cipher

    Sorry to triple post but i have changed subject. I have been looking at "TheGiant" cipher again and have made a fresh start to see if i have missed anything also i found i had made some mistakes in my charts so this is a recap and rebuild of data. I started by using this Cipher text lowercase l for lima and uppercase O for Oscar. kCmlgFi6GUJNgkNl1Q41fbfyLoCFTCvlqkZilOKlAXAzP1U1uy1BE4U fPBfpKmmLObjYnQNRBaPtKiVWzc5A4vOw3xle8FOhAGJZ7g4inOwn dJxMOvO3dc1M82at2T6935roTqyWDgtGD/hwwRF3oHqFM5Vcw1 JtlNbsgWRm4o4/quEDkZ7x1B275bX3/Fo1 Firstly i have converted each character to B64 6 bit binary The chart is here http://pastebin.com/c2kXcPFD I have also indented and "*" the start of each B64 block If it was a unspaced string of ASCII characters using just A-Z and a-z with no spaces then at every indent "*" needs to be a binary number starting "01". These are B64 characters Q-Z and a-e. Even if there are spaces and numbers involved then the binary pattern at the start of the block must start with zero. If its a space this can only be a B64 "I" bitpattern 001000 (there is no I in the ciphertext as its been read as l). B64 "M" 001100 can be used to start the ASCII Character numbers 0,1,2,3 B64 "N" 001101 can be used to start the ASCII Character numbers 4,5,6,7 B64 "O" for Oscar 001110 can be used to start the ASCII Character numbers 8,9,0 So if im right encoded B64 will only have these characters at the start of each 24 bit block and there is 19 of these B64 characters that can be used at the start of each 24bit block. Problem i am having is there is 30 unique B64 characters over all of the 48 starting blocks ???? If someone has time and can check this out and confirm this it would be good as i think it would mean we can rule out some possibilties.
  11. "TheGiant" Cipher

    Ok I got it I was overcomplicating things and I should have kept things simple but I had fun working out the full Lorenz machine and now have a small insight into the cribs they used on the real cipher even though that wasn't required I will leave it for others to find out as its enjoyable to work out
  12. "TheGiant" Cipher

    I know its been a while but i would like also to know how the lorenz cipher was solved Im not sure how the method to program all 501 pin locations was found I tried to reverse engineer the first letter to start with I may be wrong but to turn a W anto an N you need the sum of xor the key to be 11111 11001 W 00110 N I can only think of twelve sums to get the bit pattern 11111 XOR SUM 11111 USE 00000 and 11111 / and 8 USE 11111 and 00000 8 and / USE 10101 and 01010 Y and R USE 01010 and 10101 R and Y USE 11100 and 00011 U and M USE 00011 and 11100 M and U USE 01001 and 10110 L and F USE 10110 and 01001 F and L USE 00001 and 11110 T and K USE 11110 and 00001 K and T USE 00100 and 11011 9 and + USE 11011 and 00100 + and 9 Im not sure where i can go with this as i can plot an xor course reverse engineering each letter from one of these starting patterns but i dont know how to tell which is correct so there must be some guideline as you can plot many diffrent patterns so there must be some other clue to let you know the cipher is correct we are missing ? EDIT ///////////// Firstly above there is more combinations I am currently checking it out and will update when I worked it out I have found this PDF useful with a binary conversion table and also an addition table http://www.bletchleypark.org.uk/edu/maths/codes_u19_text.pdf I have calculated the total sum value of the XOR and converted the plaintext and ciphertext to binary to see if there are any clues and it just seems to be a random stream of letters apart from the plaintext is exposed twice in a row twice shown by the "/" "00000"
  13. CG_Spawn: no free fake entities

    I dont know what it is but it could be a generalised error code meaning its not always the same problem generating the error code I not tech savy or know anything about the toolset but i do believe that the devs work hard to constantly improve and maintain the toolset Its really not as simple as saying lets dump the engine and get a new one There not going to move to a different engine when they have the gtkradiant based toolset already developed and working successfully Even if they did it wouldnt be a error free zone and you would still need to customise the toolset It takes a long time to develop toolsets and its a difficult job to develop a tool set in parallel with the software you develop it with
  14. What are you listening right now?

    Liking this at the moment Bit of mix but I just wanted show a little of the different genres I like
  15. Apothicon and Keeper Language Revealed *Updated

    Alpha i totally understand that waiting for him to release the info is fustrating and i am sorry if my manner was very blunt in the posts above.I have tried to be fair and see everyones point of view but i can be inarticulate at times when getting my point across i hope this situation is solved soon without anymore conflict.
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