"TheGiant" Cipher

[Guest]

34 minutes ago, Nieno69 said:

Could you please explain your though?

make a text and encipher it and then convert it to base64?

I don't know what more there is to explain. You could decode something one way and then needed to decode it using a different method. There have been ciphers like this before.

[Nieno69]

29 minutes ago, Nightmare Voyager said:

I don't know what more there is to explain. You could decode something one way and then needed to decode it using a different method. There have been ciphers like this before.

Wow... I know what you meant but I thought you have an idea which ways was used... because for now I think that was clear to all of us working on that cipher ?...

 

[Guest]

56 minutes ago, Nieno69 said:

Wow... I know what you meant but I thought you have an idea which ways was used... because for now I think that was clear to all of us working on that cipher ?...

 

I don't understand anything to do with decoding, I was just trying to help. You were working out how to do it without TheGiant as a key, so I was saying you might have to decode what you decoded with the Key. There was literally no need to be like that...

[WaterKH]

I think what needs to be done is either a transposition or a letter substitution based on the key. Since the key does not include multiple letters (T and t are represented differently in B64) it's a possibility it is columnar shifts, but as of right now, that has not yielded any results. So like you were saying @Nightmare Voyager - I think there are two parts, if you will, for this code/ cipher. (:

[Nieno69]

41 minutes ago, Nightmare Voyager said:

I don't understand anything to do with decoding, I was just trying to help. You were working out how to do it without TheGiant as a key, so I was saying you might have to decode what you decoded with the Key. There was literally no need to be like that...

Maybe I dont understand your Post correctly so i'm Sorry to be so hard...

But to go on with the cipher

There are multiple way to work in that way

I already tried many things but no confirmed clues yet...

[Nieno69]

Someone tried to shuffle the base64 alphabet? I tried with TheGiant and "VGhlR2lhbnQ" as transpositionkey... maybe shuffle it with ubchi or just keyed ceasar....?  

 

[TheGoose-]

Some of the unsolved cyphers have to be unsolvable for now. DLC 3 and 4 should provide more information that will help the community. 

[PINNAZ]

So a Reddit user by the name of Munki was able to crack "The Giant" "Automobile Garage" cipher which is a Lorenz cipher & informed @MrRoflWaffles of the plaintext message - 

 

 

WHEN FINISHED WE WILL RETURN TO THE HOUSE AND THE INFINITE

 

 

 

 

 

 

And the hint was right infront of us. And has been since "Der Riese".

If you have a close look at the Teleporter Cogs you will see alot of gears & numbers like this -

 

 

 

Lorenz Rotor Stream Machine

 

 

 

Lorenz Cipher Simulator

http://adamsgames.com/lorenz/index.htm

 

 

@certainpersonio @WaterKH @Nieno69 Do you reckon you could lay it out here in the thread how it's decoded???

 

[Nieno69]

Sadly its not  "the giant" cipher

 

But i wondered how he figured this out so thanks for this

 

 

 Its "this" the giant cipher :

 

NMFU3DNILVAXTPPH9AYHRXUM3PDBVMHN/CANVGPYS+3HGQKH

 

 

Pls correct me if i'm wrong but i was hyped myself when i saw the title@MrRoflWaffles uploaded...

 

I will try to figure it out how exactly this works 

 

 

 

 

[PINNAZ]

Thanks for the correction @Nieno69

 

Damn I thought it was "The Giant" cipher when it is actually the "Automobile Garage" Cipher-

 

[WaterKH]

@PINNAZ As @Nieno69 said it was another cipher on the map, however, those rotors are definitely interesting! I was thinking of using the Enigma for the other one (Not the one you posted though) that hasn't been solved. So I'm going to use that image as a reference. Do you know if those move though? Or are those constant? I think I remember them moving, but wasn't sure if it was the pieces around it or not. Good catch on those rotors! (:

 

EDIT: Ahh, seems that the rotors, while they are numbered for the Enigma, are numbered 01, 02... 26 - instead of single digits.

[PINNAZ]

@WaterKH This image is taken from the "The Giant" Zombies Bonus Map Gameplay Trailer

 

 

 

Around 8 seconds

 

 

But it is also shown on the Map Intro Cinematic at 16 seconds

 

[WaterKH]

@PINNAZ The one right before the rotors you are talking about look like a Jefferson Wheel Cipher... Going to look into it, but right away it seems like I'll have to find the correct way to do it and do it by hand since there doesn't seem to be an online solver. The one I found uses a Java applet, but it gives me an error when I try to run it. I've only tried Safari and Chrome though.

[PINNAZ]

Opps, seemed I got abit carried away. hahah

 

 

Difference engine - https://en.wikipedia.org/wiki/Difference_engine

 

[WaterKH]

Haha, Oh well - Looks like Jefferson cipher is a no-go for the other one as well. However, their IoC's are extremely similar - Cipher: 0.036 - Jefferson: 0.038 - @certainpersonio you seem to have a better understanding of what these mean. But do these just mean that they are random? Or since they are so similar should we be looking for rotor based substitutions, like the Enigma?

[certainpersonio]

@WaterKH My understanding of the IOC (value of 0.0367) is that it just suggests whether it's either random text or polyalphabetic. Probably polyalphabetic, but if it actually is random it could be a key for a different cipher. Remember too that the smaller the length of the cipehrtext the less sensitive the test becomes because sample viability will have a stronger effect. I think that at 46 characters it's a large enough sample size to get the idea, but certainly on the smaller size.

 

I also don't think you use it to say whether a rotor system was used or not, but I'll try to confirm it. It's really good to figure out if you're looking at poly-alphabetic system or a mono-alphabetic system. 

[Nieno69]

0,036 or 0,038 means that the letters are that "random" that it should/could be polyalphabetic - 

Plain is in englisch about 0,064

 

But sure some plain could have 0,049

 

Further you could check with this test how many alphabets was used 

 

For more infos:

ioc

Or check out kasiski

[WaterKH]

4 minutes ago, certainpersonio said:

@WaterKH My understanding of the IOC (value of 0.0367) is that it just suggests whether it's either random text or polyalphabetic. Probably polyalphabetic, but if it actually is random it could be a key for a different cipher. Remember too that the smaller the length of the cipehrtext the less sensitive the test becomes because sample viability will have a stronger effect. I think that at 46 characters it's a large enough sample size to get the idea, but certainly on the smaller size.

 

I also don't think you use it to say whether a rotor system was used or not, but I'll try to confirm it. It's really good to figure out if you're looking at poly-alphabetic system or a mono-alphabetic system. 

 

Okay, so perhaps like a private/ public key for PGP. And the rotor speculation was based off of the images I was looking at and the fact that a Jefferson Wheel Cipher had almost the same IoC, so I was thinking lower IoC would correlate to a "random" alphabet which could be just the usage of rotors. 

[DragonGJY]

After going over @MrRoflWaffles's video, i go and have a first-step study of the Lorenz Cipher.

And this is actually a simplified form of the cipher, you can find the link here

The ciphertext, NMFU3DNILVAXTPPH9AYHRXUM3PDBVMHN/CANVGPYS+3HGQKH, is the Z mentioned in the link.

You will have to use K-wheel start position of 9 and S-wheel start position of 4 to crack this code.

Letter addition follows this rule here

So far, the method suggested by the site where "the ΔZ+ΔK sequence where "/" appears the most often corresponds to the correct K-wheel start position." do not apply to this ciphertext.

But you can get it by putting all those into the blank above, so i think more observations or calculations will have to be put in there to find out thos start position.

[MixMasterNut]

 

Wow, and I've been wondering all along what these numbered gears are.  They are the one thing from the intro cinematic that don't seem to appear anywhere within the map.

 

Great find,

- Mix

[certainpersonio]

@DragonGJY I think I have an explanation for why the method proposed by the simulator website doesn't work. 

 

If you go the Cryptanalysis of the Lorenz Cipher wikipedia page, they explain the proper set-up of the machine to be as follows:

1. Chi wheel= moves every time new letter of plaintext is used
2. Psi wheel= does not move every time a new letter is used
3. Mu wheel= controls when Psi wheel moves

The overall effect of these three wheels is to create a pseudo-random stream of letters with which to encrypt/decrypt the message. The goal of this machine is to create cipher-text that is as random as possible. The thing that Bletchley park discovered though, was that if you create the difference of the stream the randomness of the sequence will decrease. This is because the the Psi wheel doesn't move every turn. Thus when you perform the XOR function on the sequence itself to find the difference, you get "/" appearing at these "skip points". 

 

That may or may not make 100% to you (I certainly don't understand the full implications of this myself) but I do understand one thing. By differencing the sequence, it should become less random; in other words, the IOC should increase!

 

If you test this out on a full Lorenz Cipher Simulator, you will find this:

• I took our plaintext "WHENFINISHEDWEWILLRETURNTOTHEHOUSEANDTHEINFINITE" and ran it through the simulator on a random setting.
• This created the cipher text: +vsc9gw89azdtuzhg9y4ij48z8blzm3f3gxtxgppxo8fal4
  • IOC: 0.0357
• I then created the differenced text: S+JRVDN+ULOBQPSCVZQ9FDXCCIJAFHSSLUFFUN/JSULCZT
  • IOC: 0.0453

If you do this for our original cipher on TG, these are the results:

• Original cipher text: NMFU3DNILVAXTPPH9AYHRXUM3PDBVMHN/CANVGPYS+3HGQKH
  • IOC: 0.0418
• Differenced text: TZRKESRHNXVFI/4TUPEVQG+HV8TU43ONCFKL9NATVWMCFO+
  • IOC: 0.0337

In the full Lorenz cipher, the IOC got larger with differencing; however, in the Simplified Lorenz cipher, the IOC got smaller with differencing. I hypothesize that the simplified lorenz cipher doesn't actually work using the difference method.

 

From this, this is my hypothesis as to how Munki solved it:

1. On a hunch, researched the Lorenz cipher
2. Recognized that the alphabet used in the Lorenz cipher matches the alphabet of the cipher text.
3. In an attempt how to solve the cipher, worked through the Simplified cipher we've found.
4. Couldn't get the Simplified cipher's "decryption process" to work (just like us)
5. Probably was so confident that this was the right type of cipher, he tried all 56 possible solutions

I definitely applaud Munki because this took some dedication and confidence to solve. If Munki ever wants to tell us how it was actually solved, I would love to hear it. 

 

[Shootinfish]

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"

 

 

 

Spoiler

/////////      ///////////////           /////////////
BINARY        CIPHER TO BINARY           XOR SUM
////////      ////////////////          /////////////
                             
W 11001       N  00110                   8  11111
H 00101       M  00111                   3  00010     
E 10000       F  10110                   N  00110
N 00110       U  11100                   J  11010
F 10110       3  00010                   S  10100
I 01100       D  10010                   K  11110
N 00110       N  00110                   /  00000
I 01100       I  01100                   /  00000
S 10100       L  01001                   Q  11101
H 00101       V  01111                   R  01010
E 10000       A  11000                   4  01000
D 10010       X  10111                   H  00101
W 11001       T  00001                   A  11000
E 10000       P  01101                   Y  10101
W 11001       P  01101                   S  10100
I 01100       H  00101                   L  01001
L 01001       9  00100                   P  01101
L 01001       A  11000                   Z  10001
R 01010       Y  10101                   8  11111
E 10000       H  00101                   Y  10101  
T 00001       R  01010                   G  01011
U 11100       X  10111                   G  01011
R 01010       U  11100                   F  10110
N 00110       M  00111                   T  00001  
T 00001       3  00010                   3  00011
O 00011       P  01101                   C  01110 
T 00001       D  10010                   B  10011
H 00101       B  10011                   F  10110     
E 10000       V  01111                   8  11111
H 00101       M  00111                   3  00010
O 00011       H  00101                   N  00110 
U 11100       N  00110                   J  11010
S 10100       /  00000                   S  10100
E 10000       C  01110                   K  11110
A 11000       A  11000                   /  00000
N 00110       N  00110                   /  00000
D 10010       V  01111                   Q  11101
T 00001       G  01011                   R  01010
H 00101       P  01101                   4  01000
E 10000       Y  10101                   H  00101
I 01100       S  10100                   A  11000
N 00110       +  11011                   Q  11101
F 10110       3  00010                   S  10100
I 01100       H  00101                   L  01001
N 00110       G  01011                   P  01101
I 01100       Q  11101                   Z  10001 
T 00001       K  11110                   8  11111
E 10000       H  00101                   Y  10101

 

[Shootinfish]

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

[Shootinfish]

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.

[Nieno69]

I still think we get to far with this - remember the ciphers are there to be broken so i think we are overcomplicating stuff... but still i don't know what to do with "the giant".... ?