The Rat

I have started the tale on the event page. Let's see the rest of you tell it from your perspective - and finish the story. I want to send the link to Scryer, among others.

I plan to be there for the big stakeout, but if I'm late, which is a strong possibility, is there a backup rendezvous point? Maybe even a cell phone number someone can e-mail me?

Venona - en garde! Il y a une carte d'un agent federal de securite dans le cache.

quote:Originally posted by georgeandmary:Venona, Are you saying that you will be at the double agent cache at noon P.S.T. to drop off the gold garbage travel bug and not one of your minions? Mary Why would he drop off one of his minions?

quote:Originally posted by georgeandmary:Pepper, Fizzy, and Venona, I'm looking forward to seeing you all on the 31st. George and I have changed our travel plans in order to attend this event. I really hope you all can make it. I was planning on being there, too, but all my decryption came to nought. I DNF the dead drop today. I have sent a secret message to Lamneth via secure method. Maybe I can still make it...

Back in Venona's day OTP's were done by hand using modular arithmetic base 10 (i.e., there was no carrying or borrowing, so 9+6=5, etc.). This is essentially the same as XOR - except XOR is always base 2.

coldwar?age Travel Bug will be placed in Double Agent Cacheuratnefndd.Noon on December 31, 2002. It will be removed atur.600 03 same day if not retrieved.ur It uses ASCII symbols inclusing punctuation and numerals, so you can't use mod 26 to keep everything in the alphabet. Still needs some work, but I'm going to bed.

quote:Originally posted by Mr. Vic:Wow! All this secret code stuff sounds like the book I just finished reading; Cryptonomicon by Neal Stephenson. Secret codes, treasure hunting with GPS's, lot's of good stuff. One of the best books I've read in years... -Mark. Cryptonomicon is a great read. I have a partial solution, but I'm not quite there. THEdaarbag e.Travel.B ug.will.be .placed.in .Double.Ag ent.cacheu ratnefndd. Noon.on.De cember.gez nfddfbn.It .will.be.r emoved.atu renddnjg.s ame.day.if .not.retri evedbur This was obtained by XOR'ing the first text against the other - assuming one to be the OTP left unsecure. The remaining ct was periodic at per. 10, and when I processed that ct with the word REFRACTORY I got an interim ct that decrypted as a Vigenere with key TTTTTTTTTT as above, but all the spaces came out as N's and some of the text was garbled. The capital letters are my edits. Obviously I didn't process the text quite right. I'm not sure whether to xor the interim ct against the key, subtract the key, or what. I have to reduce it mod 26 to get all the numbers in the same range. The fact that there are spaces may make it necessary to process mod 27 (26 letters plus 1 space). Still working on it. C'mon you programmers, help me out here.

I must be getting close, but I'll let someone else be the one to crack it. If you convert the ct to binary then subtract the ASCII value of the key REFRACTORY (repeating every 10 bytes) you end up with following interim ct (in decimal): 432 700 698 686 447 319 429 819 175 296 688 700 571 433 449 447 300 690 688 553 304 443 701 175 833 957 430 177 560 424 431 445 571 433 320 319 173 434 687 554 304 700 572 815 448 704 686 435 431 679 559 445 572 686 705 190 558 305 431 552 688 573 827 687 191 830 173 689 305 552 687 443 571 430 322 702 430 434 430 426 432 444 573 175 704 703 558 562 942 553 688 572 442 815 321 447 686 434 559 680 430 445 698 559 703 445 814 945 430 425 687 445 698 559 321 575 429 434 942 423 688 699 443 559 448 445 430 691 559 427 559 700 442 430 704 319 430 436 815 427 174 572 827 687 577 574 557 690 561 296 560 443 699 943 448 702 558 I have arranged these in columns of 10. As you can see, the frequencies in each column are highly non-random. They are typical of English frequencies in fact, and the frequencies in columns 1, 4, and 9 are essentially the same, indicating they were encrypted with the same column key (i.e. the letter R in REFRACTORY). Now all we have to do is to figure out exactly how the key was originally applied and reverse to decrypt.

quote:Originally posted by georgeandmary:Any reason why he won't send it? george Yes, I asked him not to spoil it for me. He'll send it if I ask him to. I just haven't given up on doing it myself ... yet.

Well, I feel inadequate. I sent VENONA'S cipher to SCRYER, the best solver in the American Cryptogram Association. He broke it in about an hour. However, he has not sent me the solution, just a hint from the plaintext - beware of the spoiler. This is the hint: There will be a one-hour window in which to retrieve it on a particular day that has not yet arrived. I will keep working on the decipherment.

Well, I feel inadequate. I sent VENONA'S cipher to SCRYER, the best solver in the American Cryptogram Association. He broke it in about an hour. However, he has not sent me the solution, just a hint from the plaintext - beware of the spoiler. This is the hint: There will be a one-hour window in which to retrieve it on a particular day that has not yet arrived. I will keep working on the decipherment.

quote:Originally posted by The Rat:More analysis shows that when you convert both the first and second hex ciphers to binary and slide them against each other, they show an elevated IC at offset 35 and even higher at offset 350. This suggests that the hex code could be a compression of a 5-bit or 7-bit based cipher (the divisors of 35) and more importantly that the two ciphers used the same key (like the reused one-time pad in the real VENONA) at an offset of 350 bits (either 70 characters or 50 characters, depending on whether it is the 5-bit or 7-bit system used). It also suggest that there could be a 10-character key (e.g. REFRACTORY). This is by no means certain, but I think there is something there. I did the 5-bit conversion and the distribtuion is less flat than the the hex distribution, so that might be suggestive. Unfortunately all 32 characters appear. I was hoping that no more than 26 would appear (i.e. the number of letters in the alphabet). I'm not a computer professional, so let's see some of you C++ or Perl whizzes work on those two binary ciphertexts. I think this is promising. Oops - I made a programming error in creating the second binary file. I told you I wasn't a computer professional. Ignore the numbers 35 and 350 above. I still think the approach is worth pursuing, though. Convert the first and second ciphertexts from hex to binary, then slide them against each other to see if a high index of coincidence appears to indicate the two sections were encrypted using the same key.

More analysis shows that when you convert both the first and second hex ciphers to binary and slide them against each other, they show an elevated IC at offset 35 and even higher at offset 350. This suggests that the hex code could be a compression of a 5-bit or 7-bit based cipher (the divisors of 35) and more importantly that the two ciphers used the same key (like the reused one-time pad in the real VENONA) at an offset of 350 bits (either 70 characters or 50 characters, depending on whether it is the 5-bit or 7-bit system used). It also suggest that there could be a 10-character key (e.g. REFRACTORY). This is by no means certain, but I think there is something there. I did the 5-bit conversion and the distribtuion is less flat than the the hex distribution, so that might be suggestive. Unfortunately all 32 characters appear. I was hoping that no more than 26 would appear (i.e. the number of letters in the alphabet). I'm not a computer professional, so let's see some of you C++ or Perl whizzes work on those two binary ciphertexts. I think this is promising.

quote:Originally posted by Marty Fouts:it should be pretty easy to write a decoding program. i'll do that now. not a simple henrietta-16 code Marty Fouts ae6ip I agree. Converting the hexadecimal to base 4 yields an interim ciphertext that contains strings where each of the four digits (0 - 3) appears at least once in a string of three repeats. Thus none of them can represent the end of word or end of character symbol needed for the next (Morse) step of H16. However, that does not mean H16 can be eliminated. This could be a variant. The penultimate step of H16 is a slight variation of a Fractionated Morse. In a FM there is only one symbol 'x' to represent both end of char and end of word (but the x is doubled at the end of the word). For the same reason as above, none of the four symbols can be an 'x' though. There are never three adjacent x's in a FM. The ci-text is aperiodic, so if REFRACTORY is the keyword (not certain - maybe it just appears in the plaintext) of a normal periodic there should be an elevated Index of Coincidence at period 10. There is none. Hexadecimal ciphers in general only began to appear when computer cryptology did, and there is no chance this group could break a true computer-based encryption scheme unless we have someone here with pretty sophisticated tools. So I suspect that it is just a variant on a common paper-and-pencil cipher. I don't know of any that use sixteen symbols, though. The character distribution is relatively smooth, but by no means flat. It is also not the same for the first ciphertext and the second. In both cases the most frequent character (hexadigit) appears 25 times but in the first section it is the 2,3, and 8, but in the second it is the E. Similarly the D is least frequent in the second section (13 appearances) while the 9 is least frequent in the first. The fact that the two parts differ suggests either a different key for the two sections, or possibly the same key but a different period length. This type of distribution usually indicates a cipher that encrypts pairs or trigrams, or those that involve fractionation (like the FM). I would not rule out variants on Autokey, Grandpre, or Franctionated Morse. Even a Baconian should be considered, although we'd have to figure out a way to get to a ciphertext count that is divisible by 5. I'll have to do more analysis, but this might help someone get started.

I'm like Marty - I've been too busy working to do much geocaching lately, and I don't read the board much, so I thought I'd ask two questions at once. Someone will know the answer to both questions. 1. I sent my check in for Charter Member two weeks ago and still haven't received the e-mail. I'd like to do Amazing Race. Can anyone tell me how long it takes for that to come through? 2. I'm ready to upgrade. My eTrex doesn't do well in the woods and ravines. I know this isn't the gear board, but you guys know the terrain around here. What model(s) do you recommend for my Christmas stocking? Any particular features to look for - or that aren't worth the trouble?

There are two or three caches out in the Alviso Slough area (e.g., Alviso Slow, Rebecca Cat) that recommend bicycle access. Can anyone tell me if a mountain bike is necessary or will regular street tires do?

I notice that there are several caches out in the Alviso slough area that recommended for bikes. Can anyone tell me if they require mountain bikes, or can a bike with ordinary street tires do those?