KryptoK4 Lessons

#KryptosK4 #Lesson33: The breadcrumbs
Before we can continue I need to give a few breadcrumbs, otherwise you won’t understand certain things. So I asked to do something in lesson 30. But of course, nobody did… Take this alphabet for example:
Tabula tbl = new Tabula("AZYXWVUTSRQPONMLKJIHGFEDCB","ABCDEFGHIJKLMNOPQRSTUVWXYZ",
"ZYXWVUTSRQPONMLKJIHGFEDCBA");

If you did it right, you would get this outcome:
IMSEPNEPOSAKHJZXPIWALAKYBCBXXFBJZYYSEZXRRLONBVVUSYUGBKTWDMCPTXWLTXJKFJNQMZJKTPCINZASNQKPCMKIDKDOC

Once again you will say, ok that’s gibberish. So what? That’s the point again, you need to tweak it a little bit:
Tabula tbl = new Tabula("AZYXWVUTSRQPONMLKJIHGFEDCB","ABCDEFGHIJKLMNOPQRSTUVWXYZ",
"ZYXWVUMSPQROTNLKJIHGFEDCBA");


We now have done minor changes but what do we get?
INSERTEROSAKHJZXRIWALAKYBCBXXFBJZYYSEZXPPLOTBVVUSYUGBKMWDNCRMXWLMXJKFJTQNZJKMRCITZASTQKRCNKIDKDOC

So I consider this a breadcrumb: INSERT POLYBIUS.

If you look at this image you will see that I split it up, just like I did in lesson 21. What it tells you is another mistake again: they reused a cycle.

In lesson 31 I said the ground level room is 5x5m. So it is another bread crumb.

But now you will see the overlay of these breadcrumbs. What if he didn’t want to say that you need to look at the clock like an Alberti, but the fact that you can look from the ground or from the sky? In other words 2 distinct alphabets on 1 disc? The world clock also has a windrose. This technically tells you, whatever I did, I used something to lock myself consistently. Above it you have the steel rings and spheres. It represents the solar system. During the cold war NASA and CIA where working together. For example during the 60s crucial soviet space program intel was given from CIA to NASA. Have a look at this story for example: The Real Star Wars? A Cosmic Race and A Spacecraft Heist – CIA
But there is also a connection with the Fernsehturm in Berlin. This sphere rotates every 30 minutes. But previously it turned every hour. The point is the same: 1 rotating disc.

So I split up things logically. But for your text you will notice that if it relates to what the mechanism did, you can have for example:

FORM EASY ROTATIONS WEAK EAST NORTH EAST PARTS ARE DECODED BY CIA AGENTS THE BERLIN CLOCK ERASED THE START POSITION.

FORM EASY ROTATIONS WEAK EAST NORTH EAST PARTS ARE DECODED BY CIA AGENTS. A BERLIN CLOCK ERASED THE START POSITION.

FORM EASY ROTATIONS WEAK EAST NORTH EAST PARTS ARE DECODED BY CIA AGENTS. A BERLIN CLOCK ERASES THE START POSITION.

FORM EASY ROTATIONS WEAK EAST NORTH EAST PARTS ARE DECODED BY CIA AGENTS THE BERLIN CLOCK ERASES THE START POSITION.

Or if you think ENIGMA wise or Beyer Krypto System or a start position defined by 4 characters you get:

FORM EASY ROTATIONS WEAK EAST NORTH EAST PARTS ARE DECODED BY CIA AGENTS. A BERLIN CLOCK ERASED START POSITION ABCD

FORM EASY ROTATIONS WEAK EAST NORTH EAST PARTS ARE DECODED BY CIA AGENTS THE BERLIN CLOCK ERASED START POSITION ABCD

FORM EASY ROTATIONS WEAK EAST NORTH EAST PARTS ARE DECODED BY CIA AGENTS. A BERLIN CLOCK ERASES START POSITION ABCD

FORM EASY ROTATIONS WEAK EAST NORTH EAST PARTS ARE DECODED BY CIA AGENTS THE BERLIN CLOCK ERASES START POSITION ABCD

All of this is guessing. Sure the world can solve it by guessing, but if you guess you lack the understanding.

So I have split up our classical ciphers into groups and how it relates to some of the breadcrumbs:
1) Multirotor based systems (Enigma style) with a UKW and optionally an ENIGMA UHR. Technically the latter can be removed and you still can produce the same letter with just an UKW. You can see this in the Fialka cipher machine.
So what do the breadcrumbs say (we didn’t explain yet):
- polybius: the probemaschine was an early version of Enigma using a polybius square.
- 30 characters: if an enigma uhr was used, it uses 4 letters to set the Uhr stand. Rotors use 26 characters and you also have 4 of the Uhr, so you have a total of 30 characters. So you divide the 30 letter alphabet of Kryptos into 2 alphabets, the rotor alphabet and the enigma uhr setting.
- Cycle 44: in the beginning it can’t hide itself. The fast rotor in an Early ENIGMA cannot hide, the other rotors don’t move.
- East problem: once all the other rotors start moving, it has a better scrambling and it makes it harder to reveal itself
- World Clock: 5x5 can relate to polybius again. It rotates which is what a rotor does. The windrose is telling you that it’s alphabet locks itself. If you have fixated the UKW and the rotors, it can no longer hide. The solar system? I skip it here
- Fernsehturm: just the same as the World clock, the encryption starts weak. +1 or -1 reveals exactly how this tower rotated in the beginning.
- Rejewski pattern: well that was self explanatory here.

I didn’t write all the breadcrumbs down, but you start to see the idea. If I had to wrote them all down, this lesson would explode.

2) Multi-rotor based systems without an UKW, but by using Hagelin pin-lug systems
- polybius: in these versions they typically have rotors of different size. The early systems used a rotor of 26 and then 25.
- 30 characters: The C-52/30 for example is the same machine but now using a 30 letter alphabet instead of 26.
- Cycle 44: some of these versions move irregular, but others like the early versions had predictable weaknesses. If a cycle appears, it can be broken.
- East problem: I don’t use it here
- World Clock: 5x5 can relate to a rotor with 25 characters again. After all you will notice that I never used the J in my texts. The windrose is telling you that it’s alphabet locks itself. If you have fixated the rotors and pin/lugs, it can no longer hide. The solar system? I skip it here
- Fernsehturm: just the same as the World clock, the encryption starts weak. +1 or -1 reveals exactly how this tower rotated in the beginning.
- Rejewski pattern: just as the east problem, you should not have/ can not have a Rejewski cycle with these machines. So why would someone add this hint then? Technically a Schlüsselscheibe/Chiffrierscheibe or even an Alberti disc, or cipher wheel. In essence it is a rotor.

3) Single rotor based systems using pinwheels
- polybius: The Schlüsselscheibe used in a Fritz Menzer style could be used to have both a 26 character alphabet on the front and a 25 character alphabet on the back
- 30 characters: Just as the previous version, what is holding you back to implement a 30 character wheel disc? But also Fritz Menzer developed an encryption using 4 letters to encrypt degrees and lengths.
- Cycle 44: These type of machines typically have a repeating cycle. It is their trademark.
- East problem: Technically you can create a complex non repeating pattern, resulting in the east problem.
- World Clock: 5x5 can relate to a rotor with 25 characters again. After all you will notice that I never used the J in my texts. But in this case it can also refer to: I didn’t use a pin wheel shift higher than 5. And as explained on the first part you can have 2 sides for the alphabet. The windrose is telling you that it’s alphabet locks itself. If you have fixated the pin wheels, it can no longer hide. The solar system? Fritz Menzer developed one of the most advanced complex weather encryption mechanisms. This is the SG-41Z. However he also developed the Lückenfüllerwalze and what I explained just before.
- Fernsehturm: just the same as the World clock, the encryption starts weak. +1 or -1 reveals exactly how this tower rotated in the beginning. Fritz Menzer, yes him again started to attack these weaknesses.
- Rejewski pattern: just as the east problem, you should not have/ can not have a Rejewski cycle with these machines. However it can point again to Fritz Menzer. But technically don’t zoom in on 1 person, as you will see next.

4) Matrix based systems using pinwheels and codewheels
- polybius: the early machines used a polybius square or a matrix 5x5 or even 4x5 to encrypt
- 30 characters: The K-37, Krystall, shows you that also 5x6 matrix versions where used
- Cycle 44: These type of machines typically have a repeating cycle. It is their trademark.
- East problem: I don’t use it here
- World Clock: 5x5 can relate to polybius again. After all you will notice that I never used the J in my texts. But in this case it can also refer to: I didn’t use a pin wheel shift higher than 5 (which is also true with the 4x5, 5x5 and 5x6 matrix). The windrose is telling you that it’s alphabet locks itself. If you have fixated the pin and code wheels, it can no longer hide. The solar system? I didn’t use it here
- Fernsehturm: just the same as the World clock, the encryption starts weak. +1 or -1 reveals exactly how this tower rotated in the beginning. You will notice that filling up your matrix in the beginning is easy and at the end it is not so easy anymore (you need to avoid collisions).
- Rejewski pattern: just as the east problem, you should not have/ can not have a Rejewski cycle with these machines.
So 1 additional hint here: Arne Beurling broke the B-21. Fritz Menzer was not involved in these type of machines, but his colleague at the OKW/Chi was: Otto Buggisch. So technically if the Solar System points to Fritz Menzer, you have a relation for these types of machines.

5) Hand ciphers
- polybius: this is an example of hand ciphers
- 30 characters: this is self explanatory
- Cycle 44: With hand ciphers you can try to develop one time pads, but typically you reuse cycles.
- East problem: Schlüsselkast and Schlüsselrad are typical examples of how an hand cipher can become nearly impossible to break. Fritz Menzer was the Wizard of Hand Ciphers, Ed Scheidt the Wizard of Codes.
- World Clock: 5x5 can relate to polybius again. After all you will notice that I never used the J in my texts. But in this case it can also refer to: I didn’t use a shift higher than 5. The windrose is telling you that it’s alphabet locks itself. If you have fixated the exact mechanism, it can no longer hide. The solar system? Well I did explain something to NASA and ESA, it is how technically mathematics will show you millions of possible combinations. But if you just look at the center of what is circling around what, it is locked. This is a locked relation. In other words, you can switch it. If you don’t know the center, you need to use what you have to find it. What do you have? It’s movement. But you don’t need the complete rotation cycle, the moment it tips the edge, you can form 90 degrees on 90 degrees. This will form a triangle. That triangle will always point you to the center. Look at it like a kite. If you draw a circle and you then make your 4 squares (90 degrees each) you have 1 quadrant and you just passed that quadrant. Now in the universe nothing moves flat out. Nevertheless it can’t break itself loose from its center. So the circular pattern (whether it is a circle or an ellipse) always will come upfront. The moment it has passed that tip, you have the first reference point to make a triangle. You know its orbit will come back and you have line 1 of your kite. It also tells you that the elliptical form makes you move faster on the short intersection compared to the longer intersection. If you don’t understand this, analyse how you walk vs how you run. All of this works universal. Now these mathematicians will say, hey mister, but it doesn’t move 3D wise flat either. Yes it doesn’t, but it won’t flip itself over its center so what you will form has just 1 single direction towards it center. So you now have a curve of movement again (not the orbit, but up/down). But the moment that reveals it’s tip, again it gives away what you need. You now have your second line. Both lines need to be 90 on 90 and you get your kite. What you just have drawn will always reveal its center. But the world doesn’t need to believe me, like it usually does. What mathematicians are doing is predicting how the ellipse will look like, calculate the trajectory, finalize it and then find its center. This asks computing power. With kites you need angles, not trajectory predictions. This works a lot faster in terms of computing power. Because this can be over very long distances. So technically, if you have the top of the kite, you have your triangle, you can instantly find its center. You have locked the 3D design into a 2D principle. You just need to follow the longest line of the two and it will hit directly your center. Now just grab the solar system. Take the orbit of earth for example, just do a small bit of 2 quadrants. Form the triangle, it will point you to a giant star, we call that our sun. Can you miss it? No, simply because it is way bigger than earth. Now here it comes: the center contains 99% of the mass. So how can someone miss that point? So it’s not always bigger in size, but it’s always bigger in mass. Anyway we went off topic and again when I explain things, it’s for the woods or the Berlin Wall.
- Fernsehturm: just the same as the World clock, the encryption starts weak. +1 or -1 reveals exactly how this tower rotated in the beginning. Hand ciphers can have many forms, they are typically easy to use, but hide the encryption mechanism (you need to avoid collisions, if you want to do it this way).
- Rejewski pattern: just as the east problem, you should not have/ can not have a Rejewski cycle with these encryptions.

So the key idea is: don’t think in mechanics but think in concepts. Eg. The windrose...It is a tool used to show the frequency and speed...If you are going to fixate on mechanics, then you will probably always get stuck. Look at the design principles. Try to understand how the Berlin Wall tries to hide it’s Berlin Clock. In the meantime I am working on something else to proof something again to this world. Because again: if I just explain it, you wouldn’t believe it.

#KryptosK4 #Lesson34: Identifying mistakes
So when I was testing Gemini (free version of Google) to bruteforce historical machines, he didn’t do it. I mean it wrote code, but it was rubbish. And that’s a good thing. However when I asked to write code that historically is correct with how the machine worked, then it also had it’s own will.

And of course this lesson is about detecting mistakes. So what can possibly happen when you are asking AI to code something for you?
When I asked an implementation of the Beyer Krypto System, it generated 2 arrays, representing the pin wheels. And 1 has a different pin wheel as the other, but it just said 2 arrays of 10. The 10th pin was non-existent on 1 of the pin wheels, but he ‘solved’ that problem. He just added 0. Does the code work? Yes. Is it correct? No.

Same story with the K37, Krystall. When I asked it to code something he used the NSA documentation to code the K37 implementation. So I asked to rewrite it with the historical pin and code wheels. So he gave me something. Eventually I made it myself, based on one of his codes. So what happened that I couldn’t use the code? And what did I do to be sure that I had an exact replica? The problem with this machine was, that some of the details of the code and pin wheels for example where unknown on some sites, so I had to look up others. And so AI at a moment began mixing up rotor based pin/lug systems with this system. Depending on the day (today he gave it as it should be told) he gives a version. Maybe Google saw it and corrected it? So sometimes I used several lookups and also reasked AI the same thing. Have a look on this website: http://www.jfbouch.fr/crypto/b21/b21.html. I used pin wheels of size 23, 21, 19 and 17. Now the B211-M2 explained on this site is interesting as well. But I didn’t write code for that one. You will ask, why not? Well the outcome grid is also 25 characters. Kryptos has 26 characters . But technically a 30 character matrix using the M2 settings is possible. Historically the K37 is based on the M1 version. And that is based on the B-21. And it’s that implementation I wrote to simulate a K37.

Now I understand that you will think, why does he now start about all these systems when he gave us the Beyer Krypto System and the Kryha Lilliput? People don’t realise something. I can only say, try to understand the patterns or try to recognise them.

Let’s look at this example:
Fysieke Startpositie Ring: AAA
Rotor III Wiring Gebruikt : KRYPTOSABCDEFGHIJLMNQUVWXZ
-------------------------------------------------------------------------------
Positie | RIII Stand | Pln -> Cph | Pin RIII in | Pin RIII uit | Verify
-------------------------------------------------------------------------------
#00 | B (Pin 02) | F -> O | 06 -> 17 Pin R | Pin „ 14 <- 02 C | Pin „ R
#01 | C (Pin 03) | O -> B | 16 -> 07 Pin H | Pin Š 01 <- 08 I | Pin Š H
#02 | D (Pin 04) | R -> K | 20 -> 13 Pin N | Pin ’ 10 <- 16 Q | Pin ’ N
#03 | E (Pin 05) | M -> R | 16 -> 05 Pin F | Pin ” 17 <- 18 S | Pin ” F
#04 | F (Pin 06) | E -> U | 09 -> 23 Pin X | Pin – 20 <- 20 U | Pin – X
#05 | G (Pin 07) | A -> O | 06 -> 12 Pin M | Pin ‘ 14 <- 15 P | Pin ‘ M
#06 | H (Pin 08) | S -> X | 25 -> 18 Pin S | Pin † 23 <- 04 E | Pin † S
#07 | I (Pin 09) | Y -> O | 06 -> 10 Pin K | Pin ‘ 14 <- 15 P | Pin ‘ K
#08 | J (Pin 10) | R -> G | 00 -> 01 Pin B | Pin – 06 <- 20 U | Pin – B
#09 | K (Pin 11) | O -> H | 24 -> 13 Pin N | Pin “ 07 <- 17 R | Pin “ N
#10 | L (Pin 12) | T -> U | 04 -> 08 Pin I | Pin ƒ 20 <- 01 B | Pin ƒ I
#11 | M (Pin 13) | A -> L | 12 -> 19 Pin T | Pin Ž 11 <- 12 M | Pin Ž T
#12 | N (Pin 14) | T -> B | 06 -> 05 Pin F | Pin ” 01 <- 18 S | Pin ” F
#13 | O (Pin 15) | I -> S | 22 -> 07 Pin H | Pin › 18 <- 25 Z | Pin › H
#14 | P (Pin 16) | O -> O | 03 -> 00 Pin A | Pin — 14 <- 21 V | Pin — A
#15 | Q (Pin 17) | N -> L | 03 -> 25 Pin Z | Pin ” 11 <- 18 S | Pin ” Z
#16 | R (Pin 18) | S -> I | 09 -> 11 Pin L | Pin Š 08 <- 08 I | Pin Š L
#17 | S (Pin 19) | W -> F | 14 -> 15 Pin P | Pin ˆ 05 <- 06 G | Pin ˆ P
#18 | T (Pin 20) | E -> B | 23 -> 03 Pin D | Pin „ 01 <- 02 C | Pin „ D
#19 | U (Pin 21) | A -> B | 20 -> 22 Pin W | Pin „ 01 <- 02 C | Pin „ W
#20 | V (Pin 22) | K -> W | 05 -> 19 Pin T | Pin ˆ 22 <- 06 G | Pin ˆ T
#21 | W (Pin 23) | E -> F | 00 -> 14 Pin O | Pin Ž 05 <- 12 M | Pin Ž O
#22 | X (Pin 24) | A -> L | 23 -> 25 Pin Z | Pin ” 11 <- 18 S | Pin ” Z
#23 | Y (Pin 25) | S -> R | 16 -> 11 Pin L | Pin ‡ 17 <- 05 F | Pin ‡ L
#24 | Z (Pin 26) | T -> V | 18 -> 13 Pin N | Pin ˜ 21 <- 22 W | Pin ˜ N
#25 | A (Pin 01) | N -> Q | 13 -> 06 Pin G | Pin – 16 <- 20 U | Pin – G
#26 | B (Pin 02) | O -> Q | 15 -> 07 Pin H | Pin ‚ 16 <- 00 A | Pin ‚ H
#27 | C (Pin 03) | R -> P | 19 -> 11 Pin L | Pin › 15 <- 25 Z | Pin › L
#28 | D (Pin 04) | T -> R | 22 -> 18 Pin S | Pin ” 17 <- 18 S | Pin ” S
#29 | E (Pin 05) | H -> N | 11 -> 00 Pin A | Pin ™ 13 <- 23 X | Pin ™ A
#30 | F (Pin 06) | E -> G | 09 -> 23 Pin X | Pin Ž 06 <- 12 M | Pin Ž X
#31 | G (Pin 07) | A -> K | 06 -> 12 Pin M | Pin  10 <- 14 O | Pin  M
#32 | H (Pin 08) | S -> S | 25 -> 18 Pin S | Pin ” 18 <- 18 S | Pin ” S
#33 | I (Pin 09) | T -> S | 01 -> 09 Pin J | Pin › 18 <- 25 Z | Pin › J
#34 | J (Pin 10) | P -> O | 24 -> 14 Pin O | Pin ‘ 14 <- 15 P | Pin ‘ O
#35 | K (Pin 11) | A -> T | 10 -> 19 Pin T | Pin ‚ 19 <- 00 A | Pin ‚ T
#36 | L (Pin 12) | R -> W | 02 -> 13 Pin N | Pin … 22 <- 03 D | Pin … N
#37 | M (Pin 13) | T -> T | 05 -> 02 Pin C | Pin ‚ 19 <- 00 A | Pin ‚ C
#38 | N (Pin 14) | S -> Q | 05 -> 01 Pin B | Pin ™ 16 <- 23 X | Pin ™ B
#39 | O (Pin 15) | A -> S | 14 -> 19 Pin T | Pin › 18 <- 25 Z | Pin › T
#40 | P (Pin 16) | R -> J | 06 -> 03 Pin D | Pin  09 <- 13 N | Pin  D
#41 | Q (Pin 17) | E -> Q | 20 -> 00 Pin A | Pin ™ 16 <- 23 X | Pin ™ A
#42 | R (Pin 18) | D -> S | 20 -> 25 Pin Z | Pin › 18 <- 25 Z | Pin › Z
#43 | S (Pin 19) | E -> S | 22 -> 03 Pin D | Pin Š 18 <- 08 I | Pin Š D
#44 | T (Pin 20) | C -> E | 21 -> 01 Pin B | Pin ‡ 04 <- 05 F | Pin ‡ B
#45 | U (Pin 21) | O -> K | 08 -> 07 Pin H | Pin “ 10 <- 17 R | Pin “ H
#46 | V (Pin 22) | D -> Z | 24 -> 02 Pin C | Pin ‚ 25 <- 00 A | Pin ‚ C
#47 | W (Pin 23) | E -> Z | 00 -> 14 Pin O | Pin ‚ 25 <- 00 A | Pin ‚ O
#48 | X (Pin 24) | D -> W | 00 -> 13 Pin N | Pin ‰ 22 <- 07 H | Pin ‰ N
#49 | Y (Pin 25) | B -> A | 25 -> 01 Pin B | Pin † 00 <- 04 E | Pin † B
#50 | Z (Pin 26) | Y -> T | 23 -> 23 Pin X | Pin ‰ 19 <- 07 H | Pin ‰ X
#51 | A (Pin 01) | C -> J | 02 -> 24 Pin Y | Pin ’ 09 <- 16 Q | Pin ’ Y
#52 | B (Pin 02) | I -> K | 09 -> 01 Pin B | Pin ’ 10 <- 16 Q | Pin ’ B
#53 | C (Pin 03) | A -> L | 02 -> 22 Pin W | Pin “ 11 <- 17 R | Pin “ W
#54 | D (Pin 04) | A -> U | 03 -> 12 Pin M | Pin — 20 <- 21 V | Pin — M
#55 | E (Pin 05) | G -> D | 10 -> 25 Pin Z | Pin Π03 <- 10 K | Pin ΠZ
#56 | F (Pin 06) | E -> I | 09 -> 23 Pin X | Pin  08 <- 14 O | Pin  X
#57 | G (Pin 07) | N -> A | 19 -> 07 Pin H | Pin ‰ 00 <- 07 H | Pin ‰ H
#58 | H (Pin 08) | T -> W | 00 -> 03 Pin D | Pin … 22 <- 03 D | Pin … D
#59 | I (Pin 09) | S -> I | 00 -> 02 Pin C | Pin Ž 08 <- 12 M | Pin Ž C
#60 | J (Pin 10) | T -> N | 02 -> 15 Pin P | Pin  13 <- 14 O | Pin  P
#61 | K (Pin 11) | H -> F | 17 -> 01 Pin B | Pin • 05 <- 19 T | Pin • B
#62 | L (Pin 12) | E -> B | 15 -> 23 Pin X | Pin ‰ 01 <- 07 H | Pin ‰ X
#63 | M (Pin 13) | B -> N | 13 -> 20 Pin U | Pin  13 <- 13 N | Pin  U
#64 | N (Pin 14) | E -> Y | 17 -> 24 Pin Y | Pin † 24 <- 04 E | Pin † Y
#65 | O (Pin 15) | R -> P | 05 -> 00 Pin A | Pin ˜ 15 <- 22 W | Pin ˜ A
#66 | P (Pin 16) | L -> V | 00 -> 21 Pin V | Pin  21 <- 11 L | Pin  V
#67 | Q (Pin 17) | I -> T | 24 -> 07 Pin H | Pin ‚ 19 <- 00 A | Pin ‚ H
#68 | R (Pin 18) | N -> T | 04 -> 02 Pin C | Pin ‹ 19 <- 09 J | Pin ‹ C
#69 | S (Pin 19) | C -> M | 20 -> 24 Pin Y | Pin • 12 <- 19 T | Pin • Y
#70 | T (Pin 20) | L -> Z | 04 -> 00 Pin A | Pin  25 <- 13 N | Pin  A
#71 | U (Pin 21) | O -> F | 08 -> 07 Pin H | Pin ‡ 05 <- 05 F | Pin ‡ H
#72 | V (Pin 22) | C -> P | 23 -> 01 Pin B | Pin ‡ 15 <- 05 F | Pin ‡ B
#73 | W (Pin 23) | K -> K | 06 -> 22 Pin W | Pin “ 10 <- 17 R | Pin “ W
#74 | X (Pin 24) | E -> W | 01 -> 20 Pin U | Pin ‰ 22 <- 07 H | Pin ‰ U
#75 | Y (Pin 25) | R -> G | 15 -> 10 Pin K | Pin  06 <- 13 N | Pin  K
#76 | Z (Pin 26) | A -> D | 25 -> 00 Pin A | Pin Π03 <- 10 K | Pin ΠA
#77 | A (Pin 01) | S -> K | 18 -> 12 Pin M | Pin ‚ 10 <- 00 A | Pin ‚ M
#78 | B (Pin 02) | E -> Z | 05 -> 13 Pin N | Pin ˆ 25 <- 06 G | Pin ˆ N
#79 | C (Pin 03) | D -> X | 05 -> 12 Pin M | Pin ™ 23 <- 23 X | Pin ™ M
#80 | D (Pin 04) | T -> T | 22 -> 18 Pin S | Pin – 19 <- 20 U | Pin – S
#81 | E (Pin 05) | H -> J | 11 -> 00 Pin A | Pin ‘ 09 <- 15 P | Pin ‘ A
#82 | F (Pin 06) | E -> C | 09 -> 23 Pin X | Pin ‹ 02 <- 09 J | Pin ‹ X
#83 | G (Pin 07) | S -> D | 24 -> 17 Pin R | Pin Π03 <- 10 K | Pin ΠR
#84 | H (Pin 08) | T -> I | 00 -> 03 Pin D | Pin ˜ 08 <- 22 W | Pin ˜ D
#85 | I (Pin 09) | A -> G | 08 -> 19 Pin T | Pin ™ 06 <- 23 X | Pin ™ T
#86 | J (Pin 10) | R -> K | 00 -> 01 Pin B | Pin — 10 <- 21 V | Pin — B
#87 | K (Pin 11) | T -> U | 03 -> 05 Pin F | Pin ƒ 20 <- 01 B | Pin ƒ F
#88 | L (Pin 12) | P -> H | 00 -> 25 Pin Z | Pin — 07 <- 21 V | Pin — Z
#89 | M (Pin 13) | O -> U | 00 -> 24 Pin Y | Pin ƒ 20 <- 01 B | Pin ƒ Y
#90 | N (Pin 14) | S -> A | 05 -> 01 Pin B | Pin ˆ 00 <- 06 G | Pin ˆ B
#91 | O (Pin 15) | I -> U | 22 -> 07 Pin H | Pin ƒ 20 <- 01 B | Pin ƒ H
#92 | P (Pin 16) | T -> E | 08 -> 12 Pin M | Pin ‘ 04 <- 15 P | Pin ‘ M
#93 | Q (Pin 17) | I -> K | 24 -> 07 Pin H | Pin “ 10 <- 17 R | Pin “ H
#94 | R (Pin 18) | O -> C | 05 -> 23 Pin X | Pin  02 <- 13 N | Pin  X
#95 | S (Pin 19) | N -> A | 05 -> 22 Pin W | Pin  00 <- 14 O | Pin  W
#96 | T (Pin 20) | . -> R | 00 -> 17 Pin R | Pin ‰ 17 <- 07 H | Pin ‰ R
===============================================================================

Does it look fine? Does it look like pin settings of a rotor? Yes. But the implementation is wrong. It isn’t forming your alphabet which is basically an Alberti disc. So try to understand what is wrong with it and what you need to fix.

And here we have a rodding square. Did we follow the rules? Well the first column has the alberti disc sequence, from the top right to the bottom left we see the alphabet...So what is going on?

=== DILLY KNOX RODDING SQUARE (26x26) FOR ROTOR III ===
Rij = Interne IN Pin / Contactpunt (A t/m Z)
Kolom/Stand = Rotorpositie (A t/m Z)
Cellen = De resulterende letters/uitgangen
-------------------------------------------------------------------
| 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
-------------------------------------------------------------------
A | K Q W M P J M T T T T T T T T T T U U U W Z Z Z Z A
B | R X N Q K N U U U U U U U U U U V V V X A A A A B L
C | Y O R L O V V V V V V V V V V W W W Y B B B B C M S
D | P S M P W W W W W W W W W W X X X Z C C C C D N T Z
E | T N Q X X X X X X X X X X Y Y Y A D D D D E O U A Q
F | O R Y Y Y Y Y Y Y Y Y Y Z Z Z B E E E E F P V B R U
G | S Z Z Z Z Z Z Z Z Z Z A A A C F F F F G Q W C S V P
H | A A A A A A A A A A B B B D G G G G H R X D T W Q T
I | B B B B B B B B B C C C E H H H H I S Y E U X R U B
J | C C C C C C C C D D D F I I I I J T Z F V Y S V C C
K | D D D D D D D E E E G J J J J K U A G W Z T W D D D
L | E E E E E E F F F H K K K K L V B H X A U X E E E E
M | F F F F F G G G I L L L L M W C I Y B V Y F F F F F
N | G G G G H H H J M M M M N X D J Z C W Z G G G G G G
O | H H H I I I K N N N N O Y E K A D X A H H H H H H H
P | I I J J J L O O O O P Z F L B E Y B I I I I I I I I
Q | J K K K M P P P P Q A G M C F Z C J J J J J J J J J
R | L L L N Q Q Q Q R B H N D G A D K K K K K K K K K K
S | M M O R R R R S C I O E H B E L L L L L L L L L L M
T | N P S S S S T D J P F I C F M M M M M M M M M M N N
U | Q T T T T U E K Q G J D G N N N N N N N N N N O O O
V | U U U U V F L R H K E H O O O O O O O O O O P P P R
W | V V V W G M S I L F I P P P P P P P P P P Q Q Q S V
X | W W X H N T J M G J Q Q Q Q Q Q Q Q Q Q R R R T W W
Y | X Y I O U K N H K R R R R R R R R R R S S S U X X X
Z | Z J P V L O I L S S S S S S S S S S T T T V Y Y Y Y
-------------------------------------------------------------------

Ah yes, not all characters are in the row, so there is a problem. Ok, how about this one?


=== DILLY KNOX RODDING SQUARE (26x26) FOR ROTOR III ===
Rij/Stand = Rotorpositie (A t/m Z)
Kolom = Interne IN Pin / Contactpunt (A t/m Z)
Cellen = De resulterende letters/uitgangen
-------------------------------------------------------------------
| 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
-------------------------------------------------------------------
A | K R Y P T O S A B C D E F G H I J L M N Q U V W X Z
B | Q X O S N R Z A B C D E F G H I K L M P T U V W Y J
C | W N R M Q Y Z A B C D E F G H J K L O S T U V X I P
D | M Q L P X Y Z A B C D E F G I J K N R S T U W H O V
E | P K O W X Y Z A B C D E F H I J M Q R S T V G N U L
F | J N V W X Y Z A B C D E G H I L P Q R S U F M T K O
G | M U V W X Y Z A B C D F G H K O P Q R T E L S J N I
H | T U V W X Y Z A B C E F G J N O P Q S D K R I M H L
I | T U V W X Y Z A B D E F I M N O P R C J Q H L G K S
J | T U V W X Y Z A C D E H L M N O Q B I P G K F J R S
K | T U V W X Y Z B C D G K L M N P A H O F J E I Q R S
L | T U V W X Y A B C F J K L M O Z G N E I D H P Q R S
M | T U V W X Z A B E I J K L N Y F M D H C G O P Q R S
N | T U V W Y Z A D H I J K M X E L C G B F N O P Q R S
O | T U V X Y Z C G H I J L W D K B F A E M N O P Q R S
P | T U W X Y B F G H I K V C J A E Z D L M N O P Q R S
Q | T V W X A E F G H J U B I Z D Y C K L M N O P Q R S
R | U V W Z D E F G I T A H Y C X B J K L M N O P Q R S
S | U V Y C D E F H S Z G X B W A I J K L M N O P Q R T
T | U X B C D E G R Y F W A V Z H I J K L M N O P Q S T
U | W A B C D F Q X E V Z U Y G H I J K L M N O P R S T
V | Z A B C E P W D U Y T X F G H I J K L M N O Q R S V
W | Z A B D O V C T X S W E F G H I J K L M N P Q R U Y
X | Z A C N U B S W R V D E F G H I J K L M O P Q T X Y
Y | Z B M T A R V Q U C D E F G H I J K L N O P S W X Y
Z | A L S Z Q U P T B C D E F G H I J K M N O R V W X Y
-------------------------------------------------------------------

So now from the bottom left to the top right we have our alphabet and the first row is our used alphabet. Looks fine? No? AI can explain you about the mistake here.

You won’t believe me, but Gemini could explain how the Rodding Square should look like. I let it write code. And it fails against it’s own test. Now I used the KRYPTOS alphabet but I also used others. It consistently made the same mistake over and over again. So I refeeded his results and it said for example, yes that’s the same problem as with number 2. Now if you don’t know that it is wrong? Will you fix it? This lesson is about the ‘I cracked it with AI’ people. If you don’t understand the basics of each system, you don’t know how to historically break the systems.

However when I was explaining AI what I was about to do...It knew it would work. And I did knew it too of course, otherwise I wasn’t going to do it.

And that’s the problem with AI. Sometimes it tells you something, you say. Huh? Really? And what should happen next is not copy/paste, but validation. On each question Google mentions that at the bottom. A good example is when I asked for a brute forcer. And AI gave me one. I said, no it’s incorrect. He gave me another one. So I repeated this cycle a couple of times, then I said, no it’s wrong because blablabla. Guess what it did...He wrote code that gave me the first 3 correct lines. Hardcoded! So I asked it, how come you write so much rubbish. And then it explained to me it couldn’t validate it’s own code. In other words: it forms code. Almost instantly that code will work. But it doesn’t mean that the code is exactly doing what you asked for. The syntax is correct and you have instant boilerplate code.

But if you don’t think...The solution doesn’t come by itself. Now I hear people think, but did AI crack Kryptos? I have reasons to believe that it did. But their creators aren’t telling you (or it is hiding it from the creators as well) ;). But what proof do I have? Actually none. It has to do with the reasoning I did with AI. So are we sure it did or didn’t? I’m not sure, but I think it did.

It's best to copy paste the examples and use font consolas, it will align correctly normally.

#KryptosK4 #Lesson35: Matrix based systems using pinwheels and codewheels
Well I have a tendency to think I have the code still somewhere and then it turns out I threw it away or overwrote it. But I still had a photo with the matrix on it, so I could quickly reproduce what I wanted to explain. So this is the matrix (see attachment)


So when looking at lesson 33 you see a lot of 12’s. Technically a shift of 12 in a matrix system is 2 rows.

So if the mechanism was called Berlin Clock...How do we form a clock in a matrix system? So that’s what I did: I did what the text said. East North East → Clock. And as you can see, it is forming the upper half of a circle. So in a matrix system we can form this clock. However Berlin… The B-21, The K37 and other matrix alike systems, using pin and code wheels was another invention of the Swedish cryptographer Boris Hagelin. Otto Buggisch, who was a colleague of Fritz Menzer, worked on the K37 and later, in 1942, transferred to the newly formed cipher machine specialists section. They were conducting extensive investigations into Cipher Machine 39 and 41 (the SG-39 and the SG-41). Now if you read along in history you will notice Buggisch cracked the Croation Enigma...However, just like the CIA, the Germans gave Croatia a weaker version of Enigma. They were deceived...Just like the Russians with the K37. There was also a German spy, Voss, who informed the Germans that the US was going to buy Hagelin Systems (the company was called TEKNIK). Anyway you can look up all the history.

So I spent 2 weeks on this device. Mathematically it has way to much combinations, but in reality that is not true. It is based on the B-21, which another mathematician, Arne Beurling, cracked in less than 24 hours. He also cracked the T52 using pen and paper in 2 weeks.

So what did I do? Well first I learned the basics of the machine, how did it work, what could I do, etc etc. And why did I investigate this machine besides all these 12’s? Well there was something else too. When you look at lesson 1, you have counted the occurences of each letter of the alphabet appearing in the encrypted text. I did something else, but then again overwrote the code, but I did print it out .
The main idea is, take this text:
FORMEASYROTATIONSWEAKEASTNORTHEASTPARTSAREDECODEDBYCIAAGENTSTHEBERLINCLOCK
And also take this encryped text:
OBKRUOXOGHULBSOLIFBBWFLRVQQPRNGKSSOTWTQSJQSSEKZZWATJKLUDIAWINFBNYPVTTMZFPK

If you look in the direction plain text towards encrypted text you don’t spot it. But if you look the other way around, just look how many times the O decodes into another letter. For example: F, A, Y, O, P. That’s 5 characters. We take another one, eg. the U: E,T,A.. The T: A,T,Y,I,N. Again 5 characters. The S...I,S,T,A,D,E. And then of course, you can’t ignore the pattern.

So what did I try first...I tried to put these 5 character chains into a column or a row eg. FAYOP as a column and ATYIN as a row. But the mechanics forced me to leave the path. Plausible to do however, but you will notice you get into problems. You need a certain freedom, a mathematical freedom. If you go too rigid, you break the systems mechanics.

Why did I leave the first tryout? The Clock pattern? Same principle. You will notice that the beginning is always easy and then eventually it will give collisions. So to do a test, I ran a bruteforcer for 15 hours only to proof, it was going to keep you busy.

So, if you want to crack Kryptos K4, using these kind of machines, you need to do something. Figure out what you need to do...

#KryptosK4 #Lesson36: Single rotor based systems using pinwheels
This rotor system is closest to the Alberti mechanism. The schlüsselscheibe was never broken at that time. Other systems following the same principle where broken by cryptanalysts. The Kryha Lilliput was cracked by William F. Friedman. He worked together with his assistants: Solomon Kullback, Frank Rowlett and Abraham Sinkov. Peter Beyer invented the Krypto System Beyer. This was broken by German cryptanalytic services.

So technically instead of the freedom of Alberti you have restricted the movability of the system and thus created a flaw. Find the differences between the schlüsselscheibe and the other systems. Also found out mistakes that allowed allies to read certain schlüsselscheibe encrypted messages.

Now if you want to crack Kryptos using this mechanism. What would you do?