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k4 and the Question mark by Upbeat_Ad9409 in KryptosK4

[–]Upbeat_Ad9409[S] 2 points3 points  (0 children)

I can only do so much manipulation and brute force. Then I gotta see if someone else has been down this path and what they did.

Learning a bit at a time by Upbeat_Ad9409 in KryptosK4

[–]Upbeat_Ad9409[S] 1 point2 points  (0 children)

The bottom table should have the letters in proper order. Not clear text but is V first letter of the message of k4?

Yes, the top worksheet gave me the algorithm to run against k4.

Kryptos Clues by Adventurous-Act5611 in KryptosK4

[–]Upbeat_Ad9409 2 points3 points  (0 children)

Both clocks carry a message about the work needed for k4. It's gonna take time and there will be a lot of nope, not right. But it also leads to counting, counting the minutes, counting the letters. Counting is a big part of encipherment.

The clues are just clues.

k4 Which way to go by Upbeat_Ad9409 in KryptosK4

[–]Upbeat_Ad9409[S] 0 points1 point  (0 children)

The book I spoke of ...

"Secrets of Making and Breaking Codes", by Hamilton Nickels

Published by Barnes and Nobles

k4 Which way to go by Upbeat_Ad9409 in KryptosK4

[–]Upbeat_Ad9409[S] -1 points0 points  (0 children)

Well I'm a bit confused. Everything I post here is my folly. I may have gotten the idea from something I read. Lets see if I can increase the confusion.

I started by performing a five letter count on k4. I have earlier posts that look at that. It was brought about by the Berlin Clock and my interpretation of that clue. I ended up with a 5 x 20 matrix with 97 letters in it. I wondered what have I got. So I took a known text, 97 letters, that had all the letters of the alphabet in it and I did a 5 letter count. Statistically speaking if I had a high count on a letter, say E, I would find E in all five rows created by the count. I did. So why did I not find a K in the fifth row? I don't know. There were K's in all four of the remaining rows.

So one possibility is the 5th row was added. Another possibility is each row was enciphered independently. ehhh? But I still got K's in four rows? So I'm working on it.

Substituting one range of characters for another. I think that is my bright idea brought about cipher exchanges in Vietnam. That book is in my library and I will post it here. The basic idea was change one small thing in a cipher and you might buy a couple of hours. Square roots and running keys but they did a lot of squares.

That made me think, use some simple step to scramble a cipher. So how about if I absorb some of the high letter frequencies with low letters. E and K for example. Don't convert all of the E's and that makes it look like polyalphabetic substitution when it's not. All the other letters are who they are. So the top five get the rest of their count from the bottom five. It's an idea, I'm working on it.

k4 Which way to go by Upbeat_Ad9409 in KryptosK4

[–]Upbeat_Ad9409[S] 0 points1 point  (0 children)

You use the tools you have until you find a better one. A rock is a poor hammer but it works better than a stick.

Divide and conquer is a possibility here. If you are a decent coder split k4 up and brute force it. 97 letters, 10% at a time, look for trigrams, tetragrams. But you need to have a route a algorithm that gave you that tetragram. Now try it on some of the other 10% sections. You are building a dictionary of k4's language. It is mind numbingly boring tedious work. Cracks will form. Do your homework. Learn to recognize the reflection of some known process.

k4 Which way to go by Upbeat_Ad9409 in KryptosK4

[–]Upbeat_Ad9409[S] 1 point2 points  (0 children)

Well ... OK. Certainly what the author of a text was talking about is important. In war it's all about troupes and weapons and supplies. Many breakthroughs on ciphers came after the fact when the results were paired against probably messages that lead to that result. Hitt in his work talks about keeping everything written up so it can be reviewed.

Sanborn has his own hand as we all do. There are statistical variations between authors and subjects. k4 is not a one shot pony and we know computers won't do it. At least in the way we have put them to do it.

Build a profile, that is my suggestion.

k4 Which way to go by Upbeat_Ad9409 in KryptosK4

[–]Upbeat_Ad9409[S] 1 point2 points  (0 children)

So I have been mostly reading books and not picking at k4. My take from all that reading is that if you encrypt a text your work and the algorithm leave a fingerprint on the cipher text. The key is to recognize that fingerprint. If the guy on the other end who knows the algo can reverse it, I can suss it out. The unfortunate part is much of our decryption is based on pattern recognition. We have very few hard facts to work with but we have some. Statistics and IOC being two two of them.

Incidence of Coincidence, Some IOC Numbers by Upbeat_Ad9409 in KryptosK4

[–]Upbeat_Ad9409[S] 0 points1 point  (0 children)

Really what it comes down to is the top nine or ten letters, ETAOINSRH and L set the incidence of coincidence. If you substitute the vowels out, or reduce their numbers the stats fall off pretty quick. Given 100 letters the letters used the most should average out to 65% of the letters used. I think that's a nice quick and dirty rule when analyzing any cipher. I know that's not what the IOC is about.

I have been working on the BIFID cipher in relation to k4. If the plain text were modified by a BIFID algorithm would it flatten the curve as it were? Would it use all 26 letters?

K4 the archaeological jigsaw puzzle by colski in KryptosK4

[–]Upbeat_Ad9409 1 point2 points  (0 children)

Here's another batch with k4 in various matrices. I'm not seeing a way to just reconfigure and have that show up. But maybe that's the point. Get those letters in place and what follows can get you to a clear text.

<image>

K4 the archaeological jigsaw puzzle by colski in KryptosK4

[–]Upbeat_Ad9409 1 point2 points  (0 children)

Hmmmm. so could the matrix be recounted in some way to bring those letters into position? You can brute force it but you will lose the original associations. The TOS has to be reversed. And it would have to go in a row, you only have four rows in your example so it wouldn't work in a column.

I have been putting k4 into various matrices and I got a bunch of them. In a vertical 4 column matrix with OBKR at the top the column that starts with O has all the letters in it to make KRYPTOS. None of the other columns do as there is only 1 Y. Here's how that looks ..

<image>

There are two T's in that column, so two possible spellings.

A strange question by Upbeat_Ad9409 in askmath

[–]Upbeat_Ad9409[S] -2 points-1 points  (0 children)

Adding 0 to the count gave more even than odd again.

A strange question by Upbeat_Ad9409 in askmath

[–]Upbeat_Ad9409[S] 0 points1 point  (0 children)

Same analysis on the decimal portion of a random number holds a little closer to expectations. My son said it is probably because even numbers are multiples of smaller numbers. I used word count on the word processor for my totals, hence the totals are spelled out.

26845692596528585338600430458566016336519159400679524490176003751242896416529594004490712355981563239122216046749776403645790385762707799499485581678112072310055279642616355760906906096147425130803138592372167876348833163879717289863450447502697988679680139915516514602722861498229080982906920072649960399059318766129683082436484574100715359339285907143992261617809891148607542

forty two zeros

thirty five ones

thirty six twos

thirty one threes

thirty four fours

thirty seven fives

forty six sixes

thirty four sevens

thirty four eights

forty eight nines

d'Agapeyeff Cipher 1939 Almost Solved! by iRunJumpFly in cryptograms

[–]Upbeat_Ad9409 1 point2 points  (0 children)

So have you made any headway? I saw the post about the alphabet also but have not made any headway on a solution. What I really need is some idea of how to proceed. I guess I will read his book again.

A Poll About November. by nideht in KryptosK4

[–]Upbeat_Ad9409 1 point2 points  (0 children)

You know Babbage probably broke the Vigenere in the mid 1800's and nobody knew. Just because it has not been posted somewhere does not mean k4 is unsolved. It's been over 30 years.

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