**THE FOUR MESSAGE (K4) OF KRYPTOS**

**+ The fourth coded message has yet to be solved. It is the hardest, arguably, because it is the shortest of the four messages, containing only 97 characters, so it is harder to find patterns. Recently, Sanborn revealed a clue to the New York Times:**

*“The characters that are the 64th through 69th in the final series on the sculpture read NYPVTT. When deciphered, they read BERLIN.”*

**K4 = “OBKRUOXOGHULBSOLIFBBWFLRVQQPRNGKSSOTWTQSJQSS**

**EKZZWATJKLUDIAWINFB NYPVTT MZFPKWGDKZXTJCDIGKUHUAUEKCAR”
**

**Though 6 doubled letters (BB QQ SS SS ZZ TT) is slightly above chance for a 97 character text, that’s not hellishly improbable but If NYPVTT = BERLIN, this means … T= I but also T= N**

**I seriously think this is important. The same letters in sequence, representing different letters.**

**I suppose that when a digraph appears there is a change in the encryption, some type of asymmetry happens**

+ The K4 is divided into the following groups, thus

**OBKRUOXOGHULBSOLIFB**

**-BWFLRVQ**

**-QPRNGKS**

**-SOTWTQSJQS**

**-SEKZ**

**-ZWATJKLUDIAWINFBNYPVT**

**-TMZFPKWGDKZXTJCDIGKUHUAUEKCAR**

**Maybe some kind of function f(x)?**

**EEC ?**

+ The Elliptic curve cryptography (ECC) is an approach to public-key cryptography based on the algebraic structure of elliptic curves over finite fields. An elliptic curve cryptosystem is an asymmetric cryptosystem relying on the hardness of the discrete logarithm problem in elliptic curve groups.

**In mathematics, an elliptic curve (EC) is a smooth, projective algebraic curve of genus one, on which there is a specified point O. An elliptic curve is in fact an abelian variety – that is, it has a multiplication defined algebraically, with respect to which it is a (necessarily commutative) group – and O serves as the identity element. Often the curve itself, without O specified, is called an elliptic curve.**

**Any elliptic curve can be written as a plane algebraic curve defined by an equation of the form:**

**y^2 = x^3 + ax + b**

The figure show a simply elliptic curve.

The

**If you’re interested in the details, read on:**

**“A (Relatively Easy To Understand) Primer on Elliptic Curve Cryptography”**

**http://blog.cloudflare.com/a-relatively-easy-to-understand-primer-on-elliptic-curve-cryptography/**

**BY THE WAY**

** I suggest that K4 solution is based in the place of BERLIN:**

** “PEOPLE TO CREATE A SAFER FREER WORLD AND SURELY THERE IS NO BETTER PLACE THAN BERLIN THE MEETING PLACE OF EAST AND WEST”**

**Barcelona November 02, 2014.**

**Bye**

Hello from Spain

Double Letters are K4 Demarcations.

‘It is a capital mistake to theorize before one has data. Insensibly one begins to twist facts to suit theories, instead of theories to suit facts.’

_Sherlock Holmes Quote from “A Scandal in Bohemia”

The following page explains some of the clues that can be used to deduce a word or a letter in a piece of ciphertext:

http://www.simonsingh.net/The_Black_Chamber/hintsandtips.html

One is Identify Common Pairs Of Letters: If the ciphertext appears to encode a message in English then focus on pairs of repeated letters. In English the most common repeated letters are ss, ee, tt, ff, ll, mm and oo. If the ciphertext contains any repeated characters, you can assume that they represent one of these…… BUT

Jim Sanborn has provided The New York Times with the answers to six letters in the sculpture’s final passage. The characters that are the 64th through 69th in the final series on the sculpture read NYPVTT. When deciphered, they read BERLIN.

Thus If NYPVTT = BERLIN, this means … T= I but also T= N

This is a fact. Now we have to find an explanation to this fact.

The solution must be some old code forgotten

‘There is nothing new under the sun. It has all been done before.’

Sherlock Holmes Quote from “A Study in Scarlet”

Bye.

By the way I keep thinking that the solution is this:

PEOPLE TO CREATE A SAFER, FREER WORLD AND SURELY THERE IS NO BETTER PLACE THAN BERLIN THE MEETING PLACE OF EAST AND WEST (from a Ronald Reagan speech)

TO CONTINUE

I think I’ve found the original text but Mr. Sanborn admitted to introducing misspellings to add a degree of difficulty. Thus the plaintext can be different for this reason.

But, Now what you really want to find the K4 code?

The characters that are the 64th through 69th in the final series on the sculpture read NYPVTT. When deciphered, they read BERLIN. “If that’s not the same letter, ‘Game over, and you didn’t crack the code.’ JS says,

If you pursue the anomaly, you will find the treasure.

‘I ought to know by this time that when a fact appears to be opposed to a long train of deductions it invariably proves to be capable of bearing some other interpretation.’

_Sherlock Holmes Quote -A Study in Scarlet Chapter 3: “Light in the Darkness”

And so we are.

TO CONTINUE

A_My inspiration is the work of commander Joseph J. Rochefort

Since the early spring of 1942, the US had been decoding messages stating that there would soon be an operation at objective “AF”. It was not known where “AF” was, but Commander Joseph J. Rochefort and his team at Station HYPO were able to confirm that it was Midway by telling the basethere by secure undersea cable to radio an uncoded false message stating that the water purification system it depended upon had broken down and that the base needed fresh water. The code breakers then picked up a Japanese message that “AF was short on water.”

In the beggining, a partial decrypt in which the words “koryaku butai”, invasion force, were followed by the geographical designator AF. Koryaku butai had appeared in orders for the invasions of Rabaul, Java, Sumatra, and Bali that Hypo had already read. AF had been tentatively identified as Midway. Rochefort argued that the clincher was an order that air base

equipment was to be shipped to Saipan to be in position for the “AF ground crews.” AF was obviously an island air base; it was, Rochefort insisted, a matter of simple deduction to see that it had to be Midway.

B_Sanborn launched Kryptos Clue to provide an automated way for people to submit what they think the first 10 characters are. I wrote the 10 words on that page but I do not know if these are the words written by Sanborn because my solution is “clean” and free of grammatical errors ( the sculptor has made mistakes before)….. and the site was closed two or three weeks later.

Sanborn says that so far only two of the 97 characters have been deciphered. He told the New York Times he had become so fed up with Kryptos obsessives contacting him to claim success that he has set up his own website to filter out approaches…..but it seems that now he is a poor man and JS could not afford the site ….. or not?.

And so we are.

MY METHOD

Easy ? Yes, I’m not a cryptographer and neither is Mr Sanborn.

This is a game which must be revealed, not hidden for ever and ever

Jim Sanborn said:

Wired News 20 Jan 2005

WN: Do you want the puzzle of Kryptos to be solved?

Sanborn: “Uhhh … I certainly want it to be considered. I had figured that the parts that have been solved already would have been done a lot quicker than they were. But that might just have been a question of focus of the cryptography community. ”

I just apply the Occam’s razor or in latin “lex parsimoniae”

The principle states that among competing hypotheses, the one with the fewest assumptions should be selected. Other, more complicated solutions may ultimately prove correct, but—in the absence of certainty—the fewer assumptions that are made, the better.

This is a problem-solving principle devised by William of Ockham (c. 1287–1347), who was an English Franciscan friar and scholastic philosopher and theologian

So I searched for an original text that fulfills the requirements expressed by the sculptor.