The Krypto K4 hints thread

So suppose you used one alphabet, a shift -1 up to 26 with some tweaks, a shift of +1 from 27-52 with some tweaks...You used different alphabets, etc etc.
Suppose you got something like this (using different results):
(the space alignment doesn't work so good, but the second one should also come till the end)
CIBKRUOXOGHULBSOLIFBBWFLRVQQPRNGKSSOTWTQSJQSSEKZZWATJKLUDIAWINFBNYPVTTMZFPKWGDKZXTJCDIGKUHUAUEKCAR
01:FORMEADYQ YMEXRHHIYXKEASTNORTHEASTPXRZBAREDECODEDBMCIAB UNFNHUAB RJSH GOHBBK JLDUYQ OEOT WEAPONAD
BC:FORMEADEH EMEARHHWEAKEASTNORTHEASTPARTBAREDECODEDBMCWAB UNFNHUAB RLSH GOHBBK LLDUEH OEOT WEAPONAD

CIBKRUOXOGHULBSOLIFBBWFLRVQQPRNGKSSOTWTQSJQSSEKZZWATJKLUDIAWINFBNYPVTTMZFPKWGDKZXTJCDIGKUHUAUEKCAR
01:. RLDA YP XM XRH C KYBSTNORTHEBSTP R CBQ.E.DOE.E BJBA D EI V JB WXVPOW VW F KVX A DTR E
BC:. RLDA EP AM ARH C KEBSTNORTHEBSTP R CBQ.E.DOE.E BJBA D EI V JB WAVPOW VW F KVA A DTR E
1234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567

So what I did was I used an alphabet of 26 letters and an alphabet of 26 letters + '.'. This post is about the weakness of the cipher itself. The clues EAST, NORTHEAST, BERLIN and CLOCK is what you do know what should be there as an end result. Now some of the problems are collisions...You wanted to do something and you noticed...Damn, I should have got an E for example, but that's not possible. Then the next best shot is...You can use the swaps...In other words, you can search eg. the swapped Y to get to E and see if that spot is available. The second weakness then: you don't need to be a genius to see that despite I used 2 different alphabets...If you start from NORTHEAST...It gives you confirmations. One of the giveaways that a +1 or -1 as shift was used in a rotor system was of course the QQprngkSS. But it seems people don't use all weaknesses. In the cipher you have eg. OBKRUOXO. 3 o's in the beginning, the X that doesn't appear offen, the U and BK. At the end we have UHUAUEKCAR. 3 u's again a K and a R etc. So if you form the things you do know or try to form them...And you didn't yet do swapping of letters, use the enigma uhr, etc etc. Then despite what you do...And how you do it... The system tells you that the first 52 letters won't be that hard, except a certain slot. That will get you puzzled. But the second benefit is that if you form those letters, you should rather quickly get confirmation that you are on the right track. In my second alphabet I placed the . in between E and F. So whatever you do...If you want to have the ST of NORTHEAST you would rather quickly also notice that with the correct shift you end up with that same easyness for the SS after it in QSJQSSEKZZ. To understand how you know it should be de on the SS spot or ARE DECODED, you need to understand how your mechanism works. Because the . was on the E spot originally and the alphabet is now 1 letter longer. It all shifts a bit further. In other words. So BQ.E.DOE.E is still obvious to see it is AREDECODED if you used a shift of 1. But are we sure that the first 26 are -1 and the next +1 and then some notches? So weakness 3 in the same 52 first cipher positions. FLRVQQPRNGKSSO -> YBSTNORTHEBSTP. So here you see the interconnection if you get the A of NORTHEAST and did it right, you get EAST or YAST as well. So if you add that all up...What are the chances that OBKRUOXOGHULBSOLIFBBWFLRVQQPRNGKSSOTWTQSJQSSEKZZWAT -> something but not yet completely plain english. What one must do is verify what are the 'never to be swapped' letters. So weakness 4 is about...If we get an O eg in the cipher, you need to look at does the position of what you used make sense? In other words if you got an F. For the first O. Is that a letter than still can be swapped or one that won't be swapped. Is the FLRV -> EAST truely EAST or did it contain swapped letters or letters that were received only after using the enigma uhr. And that's how you need to build everything together to eventually crack it. Also UAUEKCAR -> WEAPONAD. Does this make sense? Was there another notch? So in total you need to look at how enigma works, what it does, etc and get plain english. And if all of that makes sense, you don't need to send in a solution, you realise you cracked it. Or look at this part IAWINFBN -> UNFNHUAB. It doesn't make sense...But again one must look at every detail in total. But you won't solve the riddle by just trying crackers, AI, ... You need to use the logic and your brain.

Today will be about the problem with BERLINCLOCK or NYPVTTMZFPK. To understand the script one must rule out certain things including what is possible and what isn't. There are ways to change only parts of the encryption by 'zero-ing' the base to where you want to start. Suppose you used a shift of +5/-5 and splitted up the text in 2 parts... You can get at the end to LIBERTY, but you will also get BERLINCLOSK. Now if you look at the encryption, you will notice the P was only used once in the first 65 characters. The second P is what must form BERLIN. And the third P needs to map to a C. So what one can do know is figuring out...Is the P solid, or does the end result lead to a swapped letter. The first P of noRtheast and the second P of beRlin is that actually a swapped letter? In other words....When you want things to make sense. You need to figure out...Is BERLINCLOSK possible and we need to swap the S or do we say the C is the unswapped letter but the R is the swapped letter...And what we get then is BEBLINCLOCK. And is there another way to get to another shift? Well, here is the interesting part...Suppose I took my shift somewhere else. I use a shift of -2, because I wanted to have the B at the N position. The first P and the second P can provide the same outcome (and for the record with different shifts). In other words...I need to have an R for Northeast and the same R for Berlin... And that is possible. Now if I change the outcome to map to CLOCK then the full part is steady. Because my first P will translate to M and my second P too! This makes it more interesting because whatever this letter is, the moment I swap it to R, I get 2 for the price of 1! This time the first was an M and the second was an M. So one can swap the M later on to R. This is the first part... Now also using this weakness provided, and thus exploiting it. TT... If the outcome of TT is not on a swapped letter, then it must match IN. So this implies that if it isn't a swapped letter, the shift needs to be +5 or -21 (if you use a 26 letter alphabet). If that was eg. 28, then +5 is still possible if you just append eg. the ':' and the '.'. But now it is no longer -21, but -23. To understand this better one can use the zero base. So instead of working with a full encryption, you can set the start point at the TT. By using these techniques, you can start to rule out, which encrypted letters are not swapped and which are and why. Now to do all these things quickly, you must play with it. So 3) A simple technique is checking the distance between 2 encrypted letters, based on what you decoded before see what is possible, what is not, what provides collisions, which shifts are possible, ... But also use the weaknesses provided. We don't know for sure what T is connected to. So the shift is still unknown. But what is known is that you can never get to a correct decipherment for BERLINCLOCK at first hand. Here is why? The distance between the second P and the third P is 6. Now zero base again...But this time on the NYP instead of TT. Zero basing means that we don't alter anything up till the improvised starting point. In this case the shift will start at NYPVTTM. That's why I always used the numbers so I can quickly rebase. And now you will understand how this speeds up the cracking... Instead of figuring out everything else figure out the possibilities with BERLIN clock. Now if you don't want to keep changing the code, you can start with a shift at a given point in time, knowing that this isn't the end alphabet. But it will give you the same collisions as expected and the same impossibilities. Then how do you go fast in decryption? Well basically it goes like this. The cipher letter is a B. You place the B somewhere in your array. It returns eg. D. Then you know you need to go 2 spots back to get the B. Easy...If you change the shift, the letters change too. But again you can use this technique to bring back the magic ;). So when you set it's base and a shift of +5, you a collision. What does this mean? Either there is a notch in it, or +5 is a wrong shift. So why does it never work out... The difference between the R and the C is either 10 (upwards) or 14 (downwards). So I hear you coming...What if we use a shift of +2 or -2...Now suppose you had all your letters from +5 and now quickly need to go to +2...You can get it even faster...1) +3, 2) +6, 3) +9 positions, 4) +12 positions, .... Now...Again you will get... a collision. And once again there is either a notch, or it isn't the right shift...But where is the notch...So eventually you can start figuring things out. Suppose you had: BERLIKCLOFJ. Now if I set my notch (+3) at the K...I get BERLIN, but now the rest needs to lower down 3 spots. And the first one (Clock) immediately gives another collision. And that's a weakness you MUST use! You have the B, F, K, P and T also on other places. So try to solve this in a way that everything makes sense. Or other option, split it in 2 parts (easy part, first 52 characters) and more difficult part. And try to figure out how to solve BERLINCLOCK. Now if you re-read the beginning...How did I get to BERLINCLOSK then? Well I started the +5 shift earlier... And because I started earlier, everything else re-arranged and... no collisions. But you can also use another shift and get no collisions...That is the beauty of the process. I hope you see a bit more why you need to play with the alphabets, to get a solid understanding of what things do and how. Because only logic, not AI and other stuff, can solve this. As always...The thing is not about the solution, it's about the process to find things out and learning from it. The solution is what makes people dumb. And unfortunately I see way to many people begging for solutions, but effort you never put into something never comes out. Keep it in mind. Whether it's sports, gaming, work, ... The process is always the same effort in => quality.

I heard he was going to sell the solution, but I haven't heard of it since...So here is something that might be valuable for your quest. Have a look at the beaufort cipher, the Hagelin M-209 and the DIANA crypto system. Happy New Year.

Maybe I need to add this too. The full text is 97 characters. This isn't random. It will make sense once you see the full picture.

Little did they know, it was always there. ABCD just meant 1234. If you change the I for 1, you will meet a Belgian King. Unfortunately he was the bastard son of an Italian noble. If you understand what you need to do, you will understand Kryptos too.

You might find out that the HILL at the end doesn't refer to the hill cipher, but to hill climbing. A technique very effective against enigma. Try to decode the first 52 characters first, because they are the easiest to form the alphabet. You will find out that a shift of -1 for the first 26 characters and a shift of +1 for the next 26 characters will help you a lot with the full encrypted text. Did you also spot that in the first 52 you see often an O. And after that O in the enciphered text you get rubbish. But the second part doesn't have an O...Figure out the relationship between swapping eg. S and D and P and O in your alphabet to understand what is going on. Why will these shifts be very effective?

Well since someone earned a billion and asked 50 dollars if you want an email reply, here is the next kryptos hint. Rest assured, he'll soon won't make any more money of it, just to earn money. East North East refers to another enigma code breaking technique: Rodding. The idea behind it is that if you develop a rod, which is basically a tabula recta, and you don't know which rotor alphabet is used, the alphabet reveals itself. Just like Alberti used 2 alphabets, you can develop an imaginary fixed alphabet and a fixed entry alphabet. Since the right rotor advances one, one can detect the used alphabet by looking east and then north east. In other words diagonal. The first 52 characters will give you enough cribs to play with. And you can be confident about at least 4 of them.

So let's get back to the beginning... I started these hints on 22 august 2023, after I know that mathematicians, who usually have a theoretical background, overcomplicate the understanding of our universe. They will come back on their black holes one day, when they understand that you need observation. Not formulas based on theories that even go completely wild. But who am I. So today's hint will again be about what you can see. Why I started simple with eg. 1 rotor technique etc etc. You got 4 hints from the creator. We are now almost 3 years further...So there are only two options: people really don't know what to do. Or don't see it, when they are on the right track. Or people really can't solve it. I'm an optimist: unsolvable isn't in my dictionary. Persistence, determination, ... Are. So before we dig into enigma rotors or other rotor based systems, time to focus on basics... As you will see here (copy paste it and use eg. font consolas):
private char[] alfabet = new char[]{'A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z'};
private char[] decrypt = new char[]{'F','J','A','N','C','W','O','Z','E','U','X','S','T','Q','H','P','R','B','V','Y','K','I','D','L','M','G'};
12345678901234567890123456
OBKRUOXOGHULBSOLIFBBWFLRVQ
FPRMEADYQEYLEXRHEIYXKEASTN
12345678901234567890123456
QPRNGKSSOTWTQSJQSSEKZZWATJ
ORTHEASTPXRZBAREDECODEDBMC
12345678901234567890123456
KLUDIAWINFBNYPVTTMZFPKWGDK
IABDRNFKHUABGRJSHIGOHBBKWJ
12345678901234567890123456
ZXTJCDIGKUHUAUEKCAR
LDUYQXLEOTMWEAPONAD

Do something first. Based on the tips and understanding the weakness of the right rotor, one can use a bit of shift tweaking to form text. The tabula recta on the right of Kryptos can be seen as a reference to Alberti. Look up the Italian and his cipher and the history of western cryptology. The second thing then. If you look up Berlin clock for Wikipedia... It's not a strange clock. It refers to the enigma Uhr or Stecker Uhr, as said on 10 april 2024. So it's time to connect the dots. That's why AI can't solve this, because it doesn't understand re-encryption. It needs to be teached re-encryption before he recognises it. We humans are far more better in recognising patterns. Don't forget that, but you need to train your brain. Nothing comes the easy way. So why would I divert people to eg. Lorenz cipher, the combined cipher machine, Hagelin, Diana, ... Because it all completes your understanding. You try to understand what you see. If the crypto K4 is about sharing knowledge then chances are that he used things in a historic way. So go back to 10 July 1944. Who used the enigma Uhr? The Luftwaffe. What was typical about the Luftwaffe using enigma? 3 rotors in conjunction with the Uhr. What else can we look up to downsize, what could be possibly used? 1939 5 possible rotors. The Uhr makes the crypto non-reciprocal. The Kriegsmarine used far more complex encryption than the Luftwaffe. So it means that you are lucky if it was the Luftwaffe or Wehrmacht. The Luftwaffe grouped in 5, whereas the Kriegsmarine grouped in 4. The Luftwaffe aligned with the Wehrmacht. They used Enigma I. So when moving towards enigma, chances are that it was encrypted using an Enigma I and the Enigma Uhr. Now you have narrowed down your possibilities. But as always: keep your mind open. Don't get locked into a tunnel vision. So we come back as to: why did I spread things out in groups of 26 (the extract above)? Because all rotor systems have something in common: the correct start position. If you look at the second 26 characters, there is a key weakness in it and it allows you to gain confidence into what is not bretted, could be bretted, or is bretted. Why do you want to do this? Because you gain confidence about what is not bretted and then can gain confidence is on position X this letter fine? For starters: if you look at the encrypted letter B then at first glance in every occasion it's always rubbish. Look at the T... Now let's look at F,L,V,R,Q,P,N,F,K,S,O,W,J,E... Whole bunch of letters but if those are not bretted...Aha... Then it means you can pinpoint on them to find eg. Notches, settings, ... What I mean by that is: if we just use a simple rotor shift. What are the chances that the first eight form: Form Easy. Or For Meade. Is just before east: weak? Or did we enter a tunnel vision and do we need to rethink everything. If we made sure that we used a shift aligning with northeast towards Berlin clock. Is that possible? What did we get and why did we get it. That's stuff AI or brute force can't come up with. That's human intel. And it's that intel that you need to use. Suppose we follow the 3 rotor track. The second one is the weakest because we figured out something and the weakest rotor is the right rotor. So that's a shift + 1 each time until you arrive at the notch of the second or the third rotor. That's how thinking in terms of rotors work. It's that weakness that you need to exploit. You don't enter computer systems by thinking you can just randomly use scripts in the hope to exploit it. As a cyber security specialist you write your own code, develop your own methods. It's you versus you, day in, day out. The more you learn, the more you improve. The more you know, the better you recognise patterns.

So if you have an enigma machine, you have the uhr. You used all ordinary rotors, didn't use the uhr, didn't use steckerbretts and did something to show the world: do they really understand everything about enigma? What did I change, and how can someone do this on an existing enigma machine. The enigma machine couldn't do, what I do here. The text is for information purpose. Not everything on the internet you will find is how you solve things. You need to understand fundamental things, basics, in order to advance and change things in your favor. To proof that it works consistent...If you understand this, then you can open pandora's box. What I show here, is with just a minor change to existing data, you encode to the same letter. Impossible with standard wirings.
IN: FORMNOTSOEASYSTUFFCODEEASTNORTHEASTPARTSAREDECODED
OUT: RQWZPSDFIVQCUSOBPOPWDYJAPIEQWDQEVFJOQRBSWDAQEVKWJO

IN: FORMEASYCRACKSTODECODEEASTNORTHEASTPARTSAREDECODED
OUT: RQWZLQMBPRQSXSOOXMPWDYJAPIEQWDQEVFJOQRBSWDAQEVKWJO

So in the first (because it isn't so easy to spot on this font) it is the E (dEcoded), in the second it is the S. (crackS, partS)

In the coming weeks I let you focus on the east problem and how it is related to Berlin. If you follow the orders of the provided hints (by Jim Sandborn) there is logic in it. And after the last hint he gave, you should understand how the east problem is one of the biggest hints you could get. And if you think that letting you research Alberti was a waste of time... I don't think so. Just try to understand the east problem and what happens and find the correlation with Berlin.