KryptoK4 Lessons

#KryptosK4 #Lesson11: The EAST problem

The goal of this lesson is showing you that if you have to change something in the beginning of your decryption algorithm, that it has an effect on the whole decryption.

So again another short lesson. Let’s look at an example where we tried to fill in the encrypted letters B, T and O. You would have gotten (see first attachment).

So as you can see, just like DECPDED we don’t want to have FPRM but FORM. So let’s see what happens when we change the rotor shift… As you can see, this affects your whole decryption (see second attachment).

This in a nutshell is what I call the east problem. If we don’t want this problem we are forced to start on the left… Or is there another way? Who paid attention noticed I did something systematically.

So try to figure out how you can solve this problem temporarily.

#KryptosK4 #Lesson12: Resetting back to the original shift position
So as one could have imagined, to solve the east problem, one could correct the shift again. In other words if you had done an additional shift of -1 on position 2, then on position 3 we execute an additional shift of +1 to reset back to the original shift and correct it.

By doing so we can temporarily try to advance further finding new letters of the decryption alphabet before we have to do a complete decryption. But there is also a downside with this work method.

Your homework is trying to understand both the benefits as well as the downsides.

#KryptosK4 #Lesson13: Changing the original alphabet
I temporarily undid the previous changes of the alphabet (that’s why I like to work digitally) and now I added something extra to get to this specific outcome after the shifts (see attachment 1).
What we see here is in fact that that we need to form BERLIN but we have a W instead of an I. Now if we just swap the original alphabet then things become more interesting (see attachment 2).
So as you can see each I turned into a W and each W turned into an I. Like steckerbretting in ENIGMA. And this logic of flow of letters is much more interesting as the previous one, because now we can try to form WEAK EAST NORTH EAST PART.
So try to play with the alphabet and see how far you can get with your decryption process.

Many people stop at shifts and think the machine is solved. It isn't. That's just the surface layer.

What you're touching here is where things start to feel alive - not just moving letters, but bending the alphabet itself. A simple swap like I -> W looks trivial, but it changes the entire flow. Suddenly the system isn't static anymore; it's reactive. One small mutation, and the path of decryption drifts into a completely different landscape.

That "almost BERLIN but not quite" moment is the real signal. Not an error - more like the system hinting that the rules you're using are slightly off-axis.

And yeah, the Enigma parallel is hard to ignore. Once you introduce that kind of wiring logic, you're no longer just decoding - you're navigating a network of transformations.

"WEAK EAST NORTH EAST PART" isn't just a phrase to reach; it's a direction. The trick is not forcing the text to match, but letting the transformations guide you there.

Meanwhile, outside the puzzle, the world runs on shallow shortcuts and quick answers - people want instant validation, not depth. But systems like this don't reward that mindset. They reward patience, precision, and the willingness to question the alphabet itself.

Keep twisting the alphabet. The interesting paths are never the clean ones.

#KryptosK4 #Lesson14: When we can’t change the original alphabet

There are plenty of examples to show this. Let’s swap the positions for example for the A and the X in your original alphabet (see attachment).

So as you can see, it is pretty clear that in this case we can’t change the original alphabet for these letters. Because if we would do this, the whole becomes unreadable.

So find ways to deal with this problem.

#KryptosK4 #Lesson14: When we can’t change the original alphabet

There are plenty of examples to show this. Let’s swap the positions for example for the A and the X in your original alphabet (see attachment).

So as you can see, it is pretty clear that in this case we can’t change the original alphabet for these letters. Because if we would do this, the whole becomes unreadable.

So find ways to deal with this problem.

#KryptosK4 #Lesson15: Collisions
When you were searching for the real solution, you probably bumped into collisions. Collisions are typical when you go from a 1 to many alphabet towards a 1 to 1 alphabet. It means that at a given moment you wanted to have on position X eg. an A but on position Y you also wanted an A. But if you followed your rules of decryption logic it turns out the A is only possible on position X but not on position Y. So for example one typical collision I had was that to decrypt the Y of NYPVTT to form BERLIN...The position for the Y in my logic I used then, was already taken by another letter, and thus I could never decode it towards the E.

So the homework is simple, whenever you encounter a collision but you are certain about which word you want to form… Find ways to overcome this problem.

#KryptosK4 #Lesson16: The soul of Kryptos K4
Today we will look at the whole of kryptos and you basically can split it up in 2 parts. An easy part and a more complex part. We still use only 1 alphabet/decryption alphabet.
So in order to let everyone start with the same setting I’ll share an example where I used the following shifts:
- position 1-26: shift -1
- position 26-52: shift +1
- positon 35: additional shift +1 (once)
- position 52 – end: shift -11
- position 60: shift -1 (once)
- as of position 66: shift +2 in addition (so basically shift -9)
The shifts only used once are done to make things readable.
Alphabet used: ABCDEFGHIJKLMNOPQRSTUVWXYZ
Decryption alphabet used: FJANCWOZEUXSTQYPRBVIKHMLDG

This will give you (what you see in the attachment).

If we translate the Y into an E afterwards, we get FOR MEADE at the beginning. Only what comes afterwards doesn’t make sense. So this seems unlogical to follow. So… Let’s try it differently… If the D isn’t a D, it’s far more logical to form FORM EASY. So keep both solutions in mind, because we don’t want to narrow down yet on our implementation. We first want to understand what Kryptos is about, how it teaches us, etc etc. And that’s what you basically do, when you decrypt something, you try to understand it’s soul.

The point of today is as with mathematics: A²+B²=C².
So on position 2 we have a P, but we are confident, this needs to be a O. On position 46 we aso have a P but again we are sure this needs to be a O. 5 positions later (position 51)… That’s maybe not that obvious. So instead of focusing on the solution, let’s focus on: does the system behave the same?
Place the text as of position 44 underneath the first 44 text positions. If we encounter a problem on position 2 and we have the same on position 46 (which now is also position 2) Then we notice that BM falls underneath AD. The point here is, we would expect again a shift problem .
Now if the D needs to become an S you either advance 15 positions or go 11 backwards (with an alphabet of 26 letters). Now do the same with BM… We are confident that it needs to become BY instead of BM. So let’s change the shift here as well. Mesdames et messieurs, eh bien voilà… BX. Wait...Why am I not surprised? Because you are at position 51… Chances are, you have a third notch already (if you think enigma wise). But again don’t focus yet on implementation.

So why do this? I want you to thoroughly understand Kryptos and let you find ways to gain confidence on what you want to reveal at the end. That you are sure, that your solution is indeed the solution. The path to Rome has many ways, only a few try all of them. The others just take the highway (asking a solution without understanding it fully).

So the homework...Either you restart from scratch or you find ways to deal with this. Remember each time we change a shift for a specific reason, we need to keep note of it. The further you advance to the right, the more we need to undo all our techniques to form words. Because in the end your whole implementation needs to work logically.

#KryptosK4 #Lesson16: The soul of Kryptos K4
Today we will look at the whole of kryptos and you basically can split it up in 2 parts. An easy part and a more complex part. We still use only 1 alphabet/decryption alphabet.
So in order to let everyone start with the same setting I’ll share an example where I used the following shifts:
- position 1-26: shift -1
- position 26-52: shift +1
- positon 35: additional shift +1 (once)
- position 52 – end: shift -11
- position 60: shift -1 (once)
- as of position 66: shift +2 in addition (so basically shift -9)
The shifts only used once are done to make things readable.
Alphabet used: ABCDEFGHIJKLMNOPQRSTUVWXYZ
Decryption alphabet used: FJANCWOZEUXSTQYPRBVIKHMLDG

This will give you (what you see in the attachment).

If we translate the Y into an E afterwards, we get FOR MEADE at the beginning. Only what comes afterwards doesn’t make sense. So this seems unlogical to follow. So… Let’s try it differently… If the D isn’t a D, it’s far more logical to form FORM EASY. So keep both solutions in mind, because we don’t want to narrow down yet on our implementation. We first want to understand what Kryptos is about, how it teaches us, etc etc. And that’s what you basically do, when you decrypt something, you try to understand it’s soul.

The point of today is as with mathematics: A²+B²=C².
So on position 2 we have a P, but we are confident, this needs to be a O. On position 46 we aso have a P but again we are sure this needs to be a O. 5 positions later (position 51)… That’s maybe not that obvious. So instead of focusing on the solution, let’s focus on: does the system behave the same?
Place the text as of position 44 underneath the first 44 text positions. If we encounter a problem on position 2 and we have the same on position 46 (which now is also position 2) Then we notice that BM falls underneath AD. The point here is, we would expect again a shift problem .
Now if the D needs to become an S you either advance 15 positions or go 11 backwards (with an alphabet of 26 letters). Now do the same with BM… We are confident that it needs to become BY instead of BM. So let’s change the shift here as well. Mesdames et messieurs, eh bien voilà… BX. Wait...Why am I not surprised? Because you are at position 51… Chances are, you have a third notch already (if you think enigma wise). But again don’t focus yet on implementation.

So why do this? I want you to thoroughly understand Kryptos and let you find ways to gain confidence on what you want to reveal at the end. That you are sure, that your solution is indeed the solution. The path to Rome has many ways, only a few try all of them. The others just take the highway (asking a solution without understanding it fully).

So the homework...Either you restart from scratch or you find ways to deal with this. Remember each time we change a shift for a specific reason, we need to keep note of it. The further you advance to the right, the more we need to undo all our techniques to form words. Because in the end your whole implementation needs to work logically.

#KryptosK4 #Lesson17: The alphabet length

As said in lesson 5, if we use the wrong alphabet, we will get shift problems. So based on what we did so far, with as little shift changes as possible, we notice more and more that it seems that just an alphabet length of 26 is the correct one. Again we keep our mind open, but if we have a spot where potentially we want to have a space or a dot, it’s position 33. But you’ll notice, based on the shifts used, that we also would have one then on position 5. So we have reasons to believe the alphabet length is just 26. And if not, we are confident that it won’t be an alphabet of 30 characters.

So how do we actually know this? Well...First go back to lesson 1. We wrote down 26 characters.
2. Let’s have a look, with what we know so far, how many letters you think are in the correct position and how many have a certain doubt or need to change into something else.
By doing so, we get an overview for action 3. Which letters do we still need to crack or alter in such a way that it remains logical.

So lesson 1…
A: 4,B: 5,C: 2,D: 3,E: 2,F: 4,G: 4,H: 2,I: 4,J: 3,K: 8,L: 4,M: 1
N: 3,O: 5,P: 3,Q: 4,R: 4,S: 6,T: 6,U: 6,V: 2,W: 5,X: 2,Y: 1,Z: 4

We have 4 A’s… The first A we encounter in the text to decrypt we are confident that it is a correct value. All other A’s...Too soon to tell but probably still incorrect.
The B appears 5 times. The first B (depending whether you did something to form an O instead of a P) seems to be perfectly fine. The second not sure. Well I know what it needs to be but you won’t get everything yet on a sheet of paper. The next 2 B’s, we have a way to know what it needs to be. And the last B...Not sure.
The C only appears twice. But not in the sections we worked on before .

So the homework: complete the rest and see which letters you are confident of, which need to change in some way, which are completely wrong for some reason, …