First look at what the foursquare would look
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Subtracting an additive from the ciphertext would be more in line with traditional decryption methods from a historical standpoint.
-Kryptosfan
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The other side now
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The idea is that we still don’t know which keyword is in which position so this gives us the chance to consider both and not falsely exclude a positive result.
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Unless it’s a 15+ letter word that ends in U, this one won’t work
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I was trying to highlight where the letters we already had were. I was trying to conceptualize various ideas about vulnerabilities in foursquares and then realized none of that matters if you can’t find a good keyword. The problem is in how you fill the table after your keyword. The U in that position means that you cannot correctly fill your foursquare encryption/decryption charts.
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You run into a similar problem (see previous post) with the subtractive L to R
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It just doesn’t work
Even worse with the W
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Oh well
I lost the paperwork for this one.
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The deal breaker for me was that there was an A at the 16th position, you can see it on the earlier post from before. The problem with that is there is no isogram with 15 letters missing an A therefore you would have to find a 17+ letter word missing an A. Probably one exists but it would probably be very long and why would anyone use a weirdo super long word when something 8-12 letters is good enough and easier to use and remember?
Feel free to go look for one but Hippopotomonstrosesquippedaliophobia is the closest I could get in both the literal and figurative sense.
So I get this idea, could K4 be meant to stay attached to K3 during transposition to reorder the letters upon decipherment of K3 to put them in the correct order for K4 decryption. Instead of trying all of the letters at once, I wanted to do a proof of concept exercise.
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Looks like it’s possible.
Not sure doing it this way would give you much for rearranging things. Might need to try an alternative.
This time I tried setting up two different possible PT where one would encrypt while the other decrypted and vice versa.
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This works but it would always be halfway insecure and all you’d have to do is toggle it back and forth. It works but it’s mostly a mental exercise at this point. I suppose if you did an substitution on the 2nd bit and then it would be splattered all through the other one but still doable.
I guess the only way to tell is if you can keep K3 and K4 together and decrypt the one while the other lingers on as an unwanted house guest.
So someone gives you a “coded” message with a ? mark at the beginning, you will very likely assume it’s backwards. Oddly enough, I never considered it for K4 until now. All of this K3 transposition crap is what made me think of it to be honest. So, maybe I’ll play around with a backwards K4 for awhile.
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I’ve had dumber ideas before.
So it turns out that if you take Kryptos K4 and reverse it and then attempt to remove an additive of the letters KRYPTOS, it is exactly what you’ve done before.
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Take the end of K4:
AUEKCAR, I’ve already done the math on trying to remove the additive SOTPYRK
Now take the same and reverse it as I just did trying to figure out how to manipulate K4:
RACKEUA, and the letters I was going to remove were KRYPTOS
Took me a couple minutes to realize those were the exact same operations. I’m going to go ahead and skip the math on this one since it’s already been done.
I find myself temporarily out of ideas…
Maybe I should try transposing K3+K4 after all…
It’s a little bit of a tangent from K3/K4 transposition. Kryptos K1 meaning? Kind of obscure, until you remember he was an artist…
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At least this way it makes sense. “It’s a grey area” is the oft-cited rationalization for morally ambiguous shiftiness.
Maybe not the most significant one the world has ever seen but important enough to Kryptos fans.
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So take the X’s out and put in some stops:
IT WAS TOTALLY INVISIBLE HOWS THAT POSSIBLE ? THEY USED THE EARTHS MAGNETIC FIELD (STOP) THE INFORMATION WAS GATHERED AND TRANSMITTED UNDERGRUUND TO AN UNKNOWN LOCATION (STOP) DOES LANGLEY KNOW ABOUT THIS ? THEY SHOULD ITS BURIED OUT THERE SOMEWHERE (STOP) WHO KNOWS THE EXACT LOCATION ? ONLY WW THIS WAS HIS LAST MESSAGE (STOP) THIRTY EIGHT DEGREES FIFTY SEVEN MINUTES SIX POINT FIVE SECONDS NORTH SEVENTY SEVEN DEGREES EIGHT MINUTES FORTY FOUR SECONDS WEST (STOP) LAYER TWO
So if you need to brush up on how to write a telegram properly, go ahead and then come back.
Nothing about this message actually makes much sense. We don’t know the subject (what was invisible), we don’t know who “they” are, what the information is, how something is transmitted underground (besides burying it), was the message to or from WW, and whether LAYER TWO is meant as a noun or verb.
Layer two could be the second layer at the location where “it” was buried, the literal 2nd layer of Kryptos or a cryptographic directive to layer two.
The problem with K2 is that it lacks a lot of context which results in a lot of potential meaning being lost. Perhaps it isn’t quite finished being solved?
UPDATE: How they used the earth’s magnetic field is through ionosphere hops (DXing)
So part 3 of Kryptos reads:
SLOWLY DESPARATLY SLOWLY THE REMAINS OF PASSAGE DEBRIS THAT ENCUMBERED THE LOWER PART OF THE DOORWAY WAS REMOVED WITH TREMBLING HANDS I MADE A TINY BREACH IN THE UPPER LEFT HAND CORNER AND THEN WIDENING THE HOLE A LITTLE I INSERTED THE CANDLE AND PEERED IN THE HOT AIR ESCAPING FROM THE CHAMBER CAUSED THE FLAME TO FLICKER BUT PRESENTLY DETAILS OF THE ROOM WITHIN EMERGED FROM THE MIST X CAN YOU SEE ANYTHING Q?
A fairly pertinent part of Howard Carter’s journal reads:
“There was naturally short suspense for those present who could not see, when Lord Carnarvon said to me `Can you see anything’.”
Here are X and Q:
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X = Howard Carter
Q = Lord Carnarvon
p.s. Q could also stand for “query” which was the telegram convention for a question mark. This would mean X was a stop. Not quite as fun as the above but equally possible… (ref: Nelson Ross, 1928)
So in Kryptos K4, there are probably 98 characters but only 97 letters with the ? being either the end of K3 or part of both K3 and K4. If you take 98 letters and check the multiples, you’ll find that 7×14 works out alright. This originally seemed significant to me because KRYPTOS is 7 letters long. It didn’t really work out. After a lot of frustration and some mulling over, I realized that it would just as easily work as two 7×7 grids. And the really interesting thing with that is you can see the double letters line up on the ends of the rows.
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So if you want to manipulate the 4th part of Kryptos numerically, you need to replace the letters with 1-26 for A-Z. Since I’m working on an idea about how the ciphertext is arranged, I’ve put the numbers into two 7×7 grids.
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Well, Kryptos had two fathers…
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Hard to know which would be first though…
I’m not even interested anymore but for posterity, here is my modular arithmetic of the keywords SANBORN and SCHEIDT into K4 of Kryptos once it has been split into two 7×7 grids.
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I did addition and subtraction because we have no way of knowing which is necessary for decryption.
p.s. Don’t you have a hard time caring about a big blob of calculations like this? I can’t be the only one…
People have different priorities on 1st and 2nd authors…
How can we know what the preferences were for the Kryptos creators?
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Are there? Of course there are! In this iteration of Kryptos modular arithmetic, I kept SANBORN-SCHEIDT as the keywords but changed how you fill the two 7×7 grids. Instead of filling one with the first 49 characters and then filling another, I filled them side-by-side, row-by-row. I skipped some of the introductory stuff for this style of analysis but remember that you can always check my easy notes at the bottom of the pages to see which keywords, which order of keywords, and how I filled the 7×7 matrices for each particular example.
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Don’t worry, I do a little summary at the end.
So this is the new rectangular matrix with SCHEIDT-SANBORN as the keywords for the 4th part of Kryptos
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Yay
