It’s one of those ideas that once you’ve thought of it, you can’t help but want to try it. Can K4 be written onto a pyramidal table and then some awesomely useful text read off of it?

Continuing in my new line of solving strategy, I attempted to first determine if it was possible to fit the text to some kind of triangular/pyramid shaped table.

Conceivably there are three types of arrangements: a pyramid of sequentially increasing tiers (n=1, 2, 3), one of even tiers (1, 2, 4) and one of odd (1, 3, 5). I included 1 in each because it’s not much of a pyramid without a pointy top.

So we can now calculate how many letters we’ll need to assemble a complete pyramid. How, you ask? Since we know the first tier is one and the linear increase in tiers of letters, we can calculate the area or total letters needed.

For example, for a 4 tier sequential pyramid, you the number of letters from each tier. 1 from #1, 2 from #2, 3 from #3, and 4 from #4. You will need 12 letters for a complete 4 tier sequential pyramid. Using this simple math, I calculated the increasing areas of complete pyramids of each type. I kept in mind that K4 is 97 letters in total (and that the Morse code is 81 in phrases that make sense and 101 if you count the extra E’s, you can see I haven’t given that one up).

Closest to the number of available letters:
Sequential: 78, 91, 105
Even: 73, 91, 111
Odd: 81, 100, 121

I think you can see why I got interested enough to pursue it. There are no matches to K4 but the morse code can be fitted to an odd-number tiered pyramid.

I started with the old, let’s start writing it in and see if anything pops up. Not much did.

I then tried the alternate method of fitting one of our keywords, I chose palimpsest, into the grid and then tried to fit the rest of the morse code text around it. There are only two sources of the letter p and I started with the phrase “t is your position”. I quickly saw that it was impossible to get to the “a” for the next letter in the keyword. There’s not enough room vertically to fit -ositionsh- which would sandwich the “pa” with t is your position and shadow forces.

So I then moved on to digetal interpretatiu. At first I believed I would run into the same problem so I figured the possible locations of the p (34 to be exact with some repeats). I then examined each for the orientation of the -a- four letters after the -p-. I found 3 orientations that gave PA. The second and the third had to be discarded because U interrupted the keyword. Since PA was on the end of a row, I tried continuing with -LIMPSEST on the next row. The only M’s can be found in “lucid memory” so I fit memory and then was stumped on fitting LUCID in correctly. This was towards the end of my efforts so I made one last ditch effort and shifted -LIMPSEST down a row which enabled me to wrap LUCID MEMORY around it. I then became stuck again and figured I had probably taken it as far as it would go. I’ve included a picture to help you see the success and the failure.

Kryptos Pyramidal Transposition Cipher

I’ve considered the possibility that we are meant to use the E’s but by my count, there are 101 letters the Morse code with the extraneous E’s. I don’t know of any way to know which one to leave off and don’t actually want to spend the time trying to fit another table at this point. The nice thing would be that the E’s would provide a certain amount of null text to maneuver the useful text around with. I may attempt a quick try but unless I get something really good.

It’s an intriguing diversion but nonetheless proved non-applicable to attempting some kind of solution for K4. Perhaps we should count every failure as one less possibile method.

Kryptos Fan