The palimpsest

Note: If you haven’t already made yourself familiar with the premises that guide this work, then you should do that now. Follow this link and then return to begin the journey to the “APEX” of Kryptos.

Motivation:

Ever since Jim Gillogly’s crack of K1, the keyword “Palimpsest” has inspired some would-be crackers to postulate that this clue might indicate that we are supposed to form a Palimpsest by somehow aligning the characters of the “Code” side of Kryptos with those of the “Tableau” side.

While I am not the first person to consider this particular interpretation of the keyword “Palimpsest”, I am unsure who is. Quite likely, Mr. Gillogly himself considered the possibility shortly after obtaining the key. Furthermore, it is likely that numerous individuals independently had this notion, for it is quite obvious. There doesn’t seem to be any way to “credit” anybody in particular as the founder of this idea.

Pre-Step Observations:

The first obvious challenge to forming a Palimpsest between the two sides of Kryptos is the fact that they have a slightly different number of characters. (The fact that they are even close, though, is what makes this concept somewhat tantalizing.)

Regarding the number of characters on each side, I make the following observations:

1)Whereas the “Code” side of Kryptos is quite irregular in shape and poses interesting questions regarding whether we are to delete some characters (such as punctuation) or add others (such as missing letters from K2 & K3) the “Tableau” side seems quite deliberate. That is, in spite of the fact that only the inner “core” of the Tableau was needed for encryption of K1 & K2, Mr. Sanborn included alphabetic rows on top and bottom, extra columns on the left and right, and a dangling “L” in a column all to itself.

2)The number of characters on the “Tableau” side is 867, which has only two prime divisors: 3 and 17 (with multiplicity 2). Therefore, there are only a few rectangular matrices that we may need to consider:
• 17 x 51 (or its transpose)
• 3 x 289 (or its transpose)
• 3 different square (17 x 17) matrices

I decided to start with a premise that the Tableau side is to remain unaltered, because the above observations seemed to suggest that. Indeed, it is fairly easy to come up with a way to modify the Code side so that it has 867 characters. Note that the punctuation (i.e. question marks) never played a role in encryption, so they were added by Mr. Sanborn after-the-fact. This may have been done for two purposes: (1) provide interrogative context of the dialog, and (2) obscure the “Palimpsest” nature of these two layers. So I deleted all of the question marks from the Code side. That left me two characters short. Coincidentally, that’s exactly how many “missing” characters there are on the Code side of Kryptos: one near the end of K2 that was removed after the encryption process, and another near the beginning of the K3 plaintext that was deleted prior to encryption.

Step Process:

Write out the entirety of the Kryptos plaintext (K1, K2, then K3) with the two missing letters reinserted, the question marks deleted, and with the K4 cipher text appended to the end. If you form this into a matrix that is 51 rows tall by 17 columns wide, then you will obtain the array labeled “Layer One” in the figure below. The other array, labelled “Layer Two”, is simply the tableau side of Kryptos formatted identically. Note also that I have assigned unique colors to the different sections (K1=red, K2=green, K3=blue, K4=black) and that I have indicated the locations of the misspelled characters in K1, K2, and K3, by placing a border around the cells containing each.