The Krypto K4 hints thread

First Gizmore, if you have a better place for this, feel free to move the thread.

So...As it turns out Kryptos K4 was never solved, which I personally find odd...
The K4 challenge is encrypted: OBKRUOXOGHULBSOLIFBBWFLRVQQPRNGKSSOTWTQSJQSSEKZZWATJKLUDIAWINFBNYPVTTMZFPKWGDKZXTJCDIGKUHUAUEKCAR

The TLDR version: 1) Enigma, 2) check Enigma-Uhr 3) check Rejewski's methods used

Given the fact that it was built in 1990 and we are 2024...+30 years seems to me a very long time, while it really (I'm honest) didn't take long to see 1) the encryption algo used 2) the cycles and 3) that's for later.

First things first, the official hints:
- October 2010 the hint NYPVTT = BERLIN
- October 2014 the hint MZFPK = CLOCK

Turns out, it didn't ring a bell for many...But I'll tell you which clock it is, because it isn't a strange looking clock at all!

- January 2020 the hint QQPRNGKSS = NORTHEAST was given
- April 2020 the hint FLRV = EAST was given

And last time I checked it still wasn't solved...So on 20 August 2023 I wrote Jim Sandborn the following (the reason why I won't provide the solution):
'Whilst everyone is trying to break section 4 (assumption), I was thinking why you would have made it. So I believe that you made it, so we had a better understanding about the difficulty to crack something unknown. So I am going to tell some things and why it's then better not to crack it. Enigma means (in Dutch/Flemish: raadsel) so riddle. You want to let them solve the riddle and gain insight how hard this was for people like Alan Turing, Rewejski and many others. While you provided hints about the fourth section you did it very specific. Northeast for example was related to the line, so you gave away it's a transposition cipher. Also: the structure of the text, gives away that it is enigma encoded. Eg. Two Q's after each other, but still having a different decoded letter.'

Encryption algorithm used?
--> K4 is yet another polyalphabet substition cipher, named Enigma. Now I did make a mistake in that email about the rotors used. And it was because I only cracked another portion of it without cracking the rest.

So technically you can decode it, writing your own enigma system. But then you need to understand how enigma works. The other way to decode it, is by using the technique Rejewski used.

But here is WHY I am not providing solutions:
'You see humanity works this way: they seek a solution for something. They ask the creator for hints, if it's correct, etc. So they actually want shortcuts to solve it quicker. Now that's not how I work. If many people solved an encryption challenge, and I still didn't. Then that's 'frustrating', because you solved many others, they couldn't. But still it tells me, there is some lack of understanding for me. That's the frustration part.
Now suppose I would crack it (and share the result), then the world has the solution. But then this also implies, that the world is no longer going to search for the solution and this would just break the whole idea you have come up with. So that's why I decided: maybe it's best I don't start to decrypt it, even though I know how to do it. After all, I tried to understand your mindset, and to me, that has more value than the solution of the encryption challenge.

In other words, if we provide the solution, we take away the whole benefit of gaining insight on how hard things are. I hope you understand me.'

So why this thread? Because I gave additional hints, and by using this forum, one has it all in one place, instead of cluttered between twitter tweets. After all this place gathers many crypto enthousiasts...

Here are a couple of untold additional hints (which make it easier for you to decrypt Kryptos K4). Because I also noticed there was still a K5, and one needs K4 for it.

Additional hints:
1) The world didn't know until 22 august 2023 that is was Enigma. What gives it away are the QQ, SS positions of NORTHEAST. QQ = NO, SS = ST. Clear evidence that this cipher is ENIGMA.
2) On 23 august 2023 I gave an additional hint: SJQSSEKZZW is decrypted AREDECODED. Here is yet again the giveaway. SS -> ZZ -> SS = DE, ZZ - DE. This thing you witness is what we technically call cycles. Have a look at what Rejeweski did and what is meant by cycles. So in order to crack unswapped letters and/or swapped letters you need to turn your alphabet to the right. Eg.When the decryption alphabet is used the first time the S was on the D of the alphabet and the second turn on the E. Afterwards the same happened with the Z. So it turns. You should now be able to crack more and eventually have 15 out of 26 letters in a not so long time...
3) On that same day I also told the world, that he did something, K = K of BERLINCLOCK and the C cycles show that the cycles aren't following all the time. There is a reason for this all. How to break it: notepad, brains, logic.
4) On 10 April 2024 I gave additional hints:BERLINCLOCK refers not to what is on Wikipedia on this day, it refers to Enigma-Uhr, this was an addon for the Enigma I cipher machine. It was used by the Luftwaffe to improve the cipher security. It was manufactured by Konski and Kruger in Berlin. Makes more sense than that strange Wiki clock, doesn't it ;). So the Steckeruhr, plug-clock or Enigma-Uhr was introduced in July 1944. The difference: The Steckerbrett is self-reciprocal. This means that it if the configuration is exact, it can be both used to encrypt and decrypt.
5) The Enigma Uhr gives away implicitely that 16 non-swapped letters are used and 10 swapped letters. Now with the hint of AREDECODED, one should already have 15 out of 26 letters.
6) QQ, SS, ZZ tell you something, but the TT of BERLIN also tells you something...One needs to swap letters.

Now I personally believe that this can't last for another couple of years, if you know all of this. But nevertheless...I am planning my next hint at the end of August 2024. But maybe it's time I also crack it completely. By doing so, I have a deal with the world...Suppose each one of you follows my thumb rule: don't provide the solution. The solution will never be made public. So when I cracked it completely, I will print it out, together with a date and hide it behind something at my desk. My ancestors can thus still provide the solution.

But my personal belief...I believe with all those hints, many should now be able already to crack it anyway. At the end of August I probably provide the last hint, because otherwise it would be TOO easy to crack it. You already should have 15 out of 26 letters. 11 left...10 are probably swapped.

Happy cracking folks ;).

People always have a choice: if you pay Jim Sandborn and provide the solution, you will get your name on the history plate. If you don't, you give others the chance to crack it too, without knowing the solution.

The text is English of course.

I hope this was informative, and I'm rather confident, many of us can crack it by now.

For those of you trying to solve it. You might want to code a scritchmus or clock method. This will speed up your journey a bit, but it will also rule out human error mistakes. Which I sometimes made.

And it might be you will need to understand not only how the enigma Uhr or steckerbrett works, but Banburismus might come in handy as well.

It is best to start first without swapped letters, then you can see when you need a rotor notch and when there is an actually letter swap.

Hello XStreamke!
Maybe people can't solve this riddle because some of your hints are wrong.
You say that:
1. MZFPK = CLOCK, and
2. QQPRNGKSS = NORTHEAST

Which implies that there are two letters, K in the first hint and S in the second, that encrypt to themselves.
But we know that in a rotor-based encrypting machine like the Enigma was, this is not possible.
Cheers.

@FranzT: the enigma uhr does something that changes your Enigma. In a normal enigma system everything is reciprocal. Meaning that an R can become an A and as such an A can become an R. But when using an enigma uhr, this changes. Meaning that an R doesn't necessarily mean that it is always an A. And if it is an A it isn't always an R. So yes Jim said that some letters decrypt as themselves, but the basic rules apply. Now first things first: I added some more tips on how to break it on X.
So can you find the spot of the next shift...So the last rotor? Let's use Jim Sandborn's hints: FLRVQQPRNGKSS -> EASTNORTHEAST, and NYPVTTMZFPK -> BERLINCLOCK. Here we see that PRN became RTH and NYP became BER. Now the N and the P can be a weak spot to break it. Because you know what needs to come and if the P and the N are not swapped letters, you can match it accordingly. Thus knowing what shift you need to use again.
And then the final hint:
So suppose you did some substitition and you got something eg. OBKRUOXOGHULBSOLIFBBWFLRVQQPRNGKSSOTWTQSJQSSEKZZWATJKL became FORMEADYQUYLEXRHEIYXKEASTNORTHEASTPXRZBAREDECODEDBMBJB. This sounds like gibberish but is there another weakness that can be flaw. What can FBBW mean...It can basically mean several things. It can become FORM, WEAK, ... But the idea behind all of this is that if you replace the characters consequently you can get to less gibberish. So what looks at first like gibberish, can become more plane english. But it must be applied on your full first sequence 'decrypted' text. Meaning if you replace every X with an A and every Y with an A then you have to do this on the full sequence. So you can also change Q to H, U to Q, Z to T etc etc. So you look for the swaps to get more plain english. So this can become:
FOR MEADE HQ ELEARHE WEAK EASTNORTHEAST PART B ARE DECODED BY CIA
Or this can also become:
FOR MEADE HQ: A CHANGE, WEAK EAST NORTHEAST PARTS ARE DECODED BY THE USE OF A NEW BERLIN CLOCK.

The whole idea is that when having this semi gibberish, that the cycles must fit and eventually the whole gets broken. Now you can say: this isn't enigma. But it does use a form of the rotor system, with the fact that decrypted can lead to the same encrypted letter.
Just make sure, that if you see plain english, you don't focus too hard on it, and make sure that every rule you apply still needs to work out on the whole. What I mean by that is that if you are in doubt if something is really decoded that, use the cycles to check whether it applies completely.

That was the final tip (I was thinking on giving it at the end of august, but no one knows when he dies).

And if you want a real braintwist: FOR MEADE HQ: A CHANGE, WEAK EAST NORTHEAST PARTS ARE DECODED BY THE USE OF A NEW BERLIN CLOCK, WHICH IS OUTDATED SECURITY. Is this the solution? I won't tell. But it suits all letters. There are several flaws that can be used. Another one is for example. You decrypt something, you got FF...You use another shift, now it is no longer FF, but a bit further in the sequence something now is MM. No one needs to believe what I say, but if you apply the basics, you will eventually see that there is a sequence that fits all. A CHANGE could also be ExCHANGEd if it is a typical Jim Sandborn error. The whole point is, that inside the first part, there is a lot to aid for the latter part.

For those who got stuck on just 15 or 16 out of 26 letters...If you couldn't find another weak spot with all the hints...Here's how you can go to 20 out of 26...How? Basically by using flaws...But that's easy said when you don't see it. So here is another weak pot: UAUEKC. As you will see this is at the end. If you did things correctly, you will see the easyness to form a word...This will get you to 20 out of 26 letter. You just need another 6! Without providing the clear text...Have a look at (encrypted sections) OBK (beginning), QQP and EKZ...Normally if you did the previous things you will say...What the hell? Why? Because of the flaws, to make sense in clear text...There aren't that many possibilities. Going beyond 20 is like providing a solution. So I won't prodive the word neither...But you'll understand once decrypted correctly. You'll need this, because this will break the latter part.

Who followed it on X. I wrote that as of July 2025 I will add solutions so everyone can step up to 20 out of 26 letters. But assume you don't have understanding of programming...I already provided a mechanism that provides a rotor system.
The code hereunder is Java code. The whole idea is you have an alphabet, a rotor implementation (or more than 1) a decryption alphabet, the ciphertext and the shift (schuif). Sorry the code is in Dutch. You start with S equals S. Hint NORTHEAST = QQPRNGKSS. So it is this S you start with and the second S becomes T. So the thing you need to do is add letters in the decrypt array in such a way, you get more. But also keep in mind that you have more rotors probably. So schuif can be different and let's say start with 26. Alfabet could also be rewritten to eg. letters + numbers. etc etc. But assume we stick to 26. So you will need to add more logic in here, to get more out of it. But this will give you a head set and perhaps, you can have a bit of looking up how the Java programming language works and have play with it.

I hope this helps for many. I didn't provide the enigma uhr changes, the additional steps needed, but the first say 50 characters can be solved in less than 3 hours (if you use the code). If you want more decryptions you can add more decrypt arrays. So I kept it minimalistic for everyone, but here you have freedom. You don't need everything to solve parts. You can sort your way on how to do it progressively.

public class Kryptos2 {
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[]{' ',' ',' ',' ',' ',' ',' ',' ',' ',' ',' ',' ',' ',' ',' ',' ',' ',' ','S',' ',' ',' ',' ',' ',' ',' '};
private char[] cipher = new char[] {'O','B','K','R','U','O','X','O','G','H','U','L','B','S','O','L','I','F','B','B','W','F','L','R','V','Q','Q','P','R','N','G','K','S','S','O','T','W','T','Q','S','J','Q','S','S','E','K','Z','Z','W','A','T','J','K','L','U','D','I','A','W','I','N','F','B','N','Y','P','V','T','T','M','Z','F','P','K','W','G','D','K','Z','X','T','J','C','D','I','G','K','U','H','U','A','U','E','K','C','A','R'};

public Kryptos2() {
int schuif = 1;
char[] tekst = new char[97];
for (int i=0; i < cipher.length; i++) {
schuif++;

for (int j=0; j < decrypt.length; j++) {
if (cipher == decrypt[j]) {
if (schuif > 25) {
schuif = 0;
}
int positie = j+schuif;
while (positie > 25) {
positie -= 26;
}
while (positie < 0) {
positie += 26;
}
tekst = alfabet[positie];
}
}
}
System.out.print("CI:");
System.out.println(cipher);
System.out.print("01:");
System.out.println(tekst);
System.out.println(" 1234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567");
}

public static void main(String[] arg)
{
Kryptos2 k = new Kryptos2();

}

PS: if you say that S is at the wrong position...Indeed. But you will then also see how things work and what you need to do to keep NORTHEAST.

So if someone says, why don't we use eg. this alphabet:WXZKYXZKRYPTOSABCDEFGHIJLMNQUVWXZKRZZKRYPTOSABCDEFGHIJLMNQUVWXZKRYABCDEFGHIJKLMNOPQRSTUVWXYZABCD to start with. Well if you just use a single rotor, you will almost immediately find out, that with the standard alphabet, you get further than with this one. One of the nobrainers as a give away is the EASTNORTHEAST. If you use the kryptos wise alphabet, then the first thing you need to do because of the rotary is either change the shift or swap the letters. If you used a shift+1 on the standard alphabet you got the N and after the N comes the Q. The kryptos alphabet had NQ...It means with a shift 1 you always get NQ. To get NO from NORTHEAST with a shift 1 you need to swap Q into O. But feel free to toy around with the kryptos alphabet, you'll quite fastly understand that just using the standard alphabet is getting you faster to plain english. I didn't say that some additional steps are needed, but the first part gives you a lot of indications of whether you are on the right track or not. So first stick with just shifting as if you are using only 1 rotor. Now let's say we use the alphabet and the '.' as element 27. As you can see..NORTHEAST is truely a gift to get confident on which alphabet to use. OK.OK...How about the '.' then as the first element...You'll spot the problem. So this can give you confidence on which alphabet to use, to crack it...

Since it's almost Christmas . Let's take a look at the weaknesses...
Weakness 1: the sculpture is actually telling you, that it is using a rotor based mechanism. So if you look at it, it's like it tells you crack the fast rotor...So start with the simple alphabet from A - Z and only crack the first 52 characters of the enciphered text.
Weakness 2: Another weakness in this text is that you have 22 letters out of 26...In other words, something weakly encrypted and if you do it right you will get almost all letters. So here is the decryption alphabet I used...
'J','A','N','C','W','O','Z','E','U','X','S','T','Q',' ','P','R','B','V',' ','K',' ',' ','L',' ','G','F'
Weakness 3: Start with just a shift -1 and a shift +1. And you will notice how weak this part really is. Now you will need to add the 'notches' too. So I did -1 again on position 1 (offset zero based), +1 again on position 2. And also +1 on 35 and 46 and -1 on 45.
This reads: FORMEADYQ YLEXRH IYXKEASTNORTHEASTPXRZBAREDECODEDBMC
Weakness 4: If you pay attention you only have 2 gaps left...The H and the I. So with only D,I,H,M,Y missing you end up with only 3 missing afterwards.
So now comes another easy part...For every X -> A, for every Y -> E, for Every Z -> T and you might consider B -> S.
So what you get now is For MEADE (HQ?) (ELEARHx) weak east north east parts are decoded SMC.
OK. So what is this rubbish ELEARHx? Well if you look at the sculpture...KRYPTOS...ELEARHx is rubbish but is the same size as KRYPTOS.
So where does part 1 end? Well let's assume 97 / 2 -> rounded up... 49. Exactly at are decoded...

Merry christmas and a happy new year folks! Try to understand all this, how it's done. How you code it. The weaknesses, what details you need to look at...Because unfortunately part 2...That isn't this easy ;).