Marian Rejewski is the third person I would like to dedicate my book to. He was a great Polish mathematician and cryptologist. He was born in 1905. Today we celebrate his birth. Marian Rejewski – together with Henryk Zygalski and Jerzy Różycki – prepared techniques to decrypt the German Enigma ciphers.
The 5th part of Muko ends with words: “The key to the fourth part seems to be xxxxxxxx.”.
I will spoil it a bit more in a second.
Since the beginning lots of us – kryptos geeks – probably assumed that THIS is the word, the key to solving K4. Unfortunately I haven`t seen any proof that would give us a slight hope that it really is the key and how it could be used.
The series of Muko books (Legends of Aivirai) were created because I wanted to help anyone to find the final solution and put it in decrypted, publicly available form. I wanted the K4 to be left in a state where anyone (with just a minor knowledge about simple ciphers) will be able to move it forward.
The key: To the point. Last sentence in the book is:
“The key to the fourth part seems to be KRYPTOS.”
A bit geek stuff for more advanced readers:
Now part for more interested readers, geeks like me who spent a lot of time on K4. Everything started with a message for which hash was put in book 1 – “Muko and the Secret”. The 2nd message from at the end of the book (with hashes) begins with:
BEGIN The first key to solution seems to be the word: KRYPTOS (…)
This keyword combined with clues gives us something interesting. With BERLIN clue – it could be an incident With CLOCK clue – it could be a coincidence With NE clue – it seems to be a pattern
All three clues and the keyword show the same characteristics. The calculations behind the pattern also shows some additional insight about the methods used as masking and encryption.
Simple definition needed:
What is the pattern and the crack in Kryptos K4? If we get a random key and a random not-so-easy-cipher than the average distance between decrypted and encrypted letter on a specific position should be something around 6,5. why? dist(A, A) = 0, dist(A, B) = dist(A, Z) = 1, dist(A, C) = dist(A, Y) = 2, … dist(A, N) = 13 = dist(B, O) = …
average distance between A and all letters in alphabet is (0 + 1 + 2 + 3 + … + 12 + 13 + 12 + 11 + … + 1) / 26 = 6,5 The same calculations work for all other letters.
Calculations for K1, K2, K3:
Now average distances in Kryptos between decrypted and encrypted texts are: Part 1: around 6,7 Part 2: around 8,9 Part 3: around 6,7 Seem to be close to 6,5
if we split letters of decrypted messages to two sets:
and calculate distances separately, then the distances are: Part 1 for (IS_NOT_KR vs IS_KR) = (7,1 vs 5,5) – similar (7,1/5,5 = 1,29) Part 2 for (IS_NOT_KR vs IS_KR) = (8,3 vs 9,8) – similar (8,3/9,8 = 0,84) Part 3 for (IS_NOT_KR vs IS_KR) = (6,4 vs 7,4) – similar (6,4/7,4 = 0,86) everything behaves as it should behave in a correct cipher.
Calculations for clues:
Now lets take a look into clues BERLIN, CLOCK and NORTHEAST
BERLIN clue for (IS_NOT_KR vs IS_KR) = (9 vs 2) – incident (9/2 = 4,5) CLOCK clue for (IS_NOT_KR vs IS_KR) = (11,6 vs 4,5) – coincidence (11,6/4,5 = 2,6) NORTHEAST clue for (IS_NOT_KR vs IS_KR) = (5,25 vs 1,4) – pattern (5,25/1,4 = 3,75) These are much larger values than they should be for a random word as a keyword.
Of course there are other words for which we could find larger quotient, but THIS word is the name of the scuplture!
What is more – like I`ve written – this gives a lot of insight about the methods used as masking and encryption. Let`s think. Why KRYPTOS letters encrypt to other “near” letters?
Are you convinced?
If not than you have to wait a bit more (Muko Part VI) for other “incidents”. Now, take a rest, grab first part of the book for free from:
Jerzy Różycki is the second person I would like to dedicate my book to. Today we celebrate his birth. He was born 111 years ago in 1909. Jerzy Różycki is one of the three cryptologists from Poland who had major impact on cracking german Enigma.
Like I`ve written before I hope that 5th part of the book will be a special one. A little spoiler alert. One of the main characters ends the book with the phrase “The key to the fourth part seems to be xxxxxxxx.” 🙂
I hope that soon it will change the approach to solving k4 in the sculpture for everyone.
The 5th part of the Muko series book is almost finished and the first person I would like to dedicate it to is Henryk Zygalski. Today we celebrate Henryk Zygalski`s birth. He was born 112 years ago in 1908. Henryk Zygalski is one of the three Polish cryptographers who had major impact on cracking German Enigma. Thanks to him and others, duration of WWII may have shortened by years.
5th part of the book is a special one. I hope it changes the approach to solving mysterious cipher that remains uncrackable for almost 30 years. Preparing Muko series I wanted to create someting that will lead one of you to the final solution and will allow you to defeat institutions like CIA or NSA. I hope I managed to do so. I have something interesting that I would like to share with you. Soon…
I would like to share some information about #Kryptos that I haven`t seen anywhere yet, and I strongly believe will lead someone to final solution (as always):)
First one that I wanted to share is connected to the idea that Mr. Sanborn could easily deceive us and make decryption a bit more difficult, even without the knowledge of any advanced math or crypto algorithms by himself.
The idea is following (in pseudo-code): Input: 1) zeros_and_ones: string of 0 and 1 2) plain_text: plain text to encrypt 2) key: key 3) basic_algorithm: selected cryptography algorithm which uses key. algorithm has two methods: 3.1) basic_algorithm.encrypt(position, letter, key) 3.2) basic_algorithm.decrypt(position, letter, key)
The idea for new_algorithm is following:
new_algorithm.encrypt(zeros_and_ones, position, plain_text, key): letter = plain_text[position] if (zeros_and_ones[position] == 0): return basic_algorithm.encrypt(position, letter, key) else: return basic_algorithm.decrypt(position, letter, key)
similar approach to decryption:
new_algorithm.decrypt(zeros_and_ones, position, encrypted_text, key): letter = encrypted_text[position] if (zeros_and_ones[position] == 1): return basic_algorithm.encrypt(position, letter, key) else: return basic_algorithm.decrypt(position, letter, key)
To sum up in one sentence: it says that in order to produce encrypted letter you either must encrypt it or decrypt – which approach to use is based on the zeros_and_ones string
Why did I write about this idea? If you get the top rows from morse code: E DIGETAL EEE T IS YOUR EE VIRTUALLY E EE SHADOW EE LUCID EEE
remove the debris (Es) DIGETAL T IS YOUR VIRTUALLY SHADOW LUCID
correct the mistake (DIGETAL -> DIGITAL) and you have exactly 97 zeros and ones (dashes and dots in morse code). This leads us to Part II (soon).
still working on this pattern, but not found anything interesting
before/after AGENCIES and SYMBOL until now. The Creator is not accepting
partial solutions, so no constructive response acquired. I`ve tried to
look for something interesting near BLETCHLEY PARK and … UNITED
KINGDOM is found.