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.
Before going for the key and the proof, take a look at the trailer for 5th part of the book. I hope you will like it.
Muko Part V – Legends of Aivirai
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:
The first key to solution seems to be the word:
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.
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:
and help children solve the mystery!
If you like it, there are other books in the series: Muko Series
Will you be the one that finds the “missing functions”?
Will you be the one that defeats NSA and CIA?
Or maybe you will be the one that wins millions of dollars 🙂
I hope so.