Unsolved

Letter to a Death Row Inmate

Here’s Challenge 22 for February 8, 2018.

Death is a way out, but perhaps there is a better escape. I won’t say much about this concealment technique except that it has a celebrated (and perhaps apocryphal) origin story.

Find the hidden message that made it past the prison mail censors.


Dear Marcus.

I am very sorry for you. Ethel too. Do know that your family loves you. Ask the warden if he will extend our visit time next Tuesday.

Chris, young as he is, will not understand what is happening. I pray that he will some day. He eventually will know what you did. The bad deeds cannot be absolved, but you should still pray for God’s forgiveness.

The dreaded day is almost here. Time eats away at my resolve. I do ask God every day to ease your suffering. Doctor Bradford tells us that it will be quick and painless.

You have so many good qualities. Your heart is pure. You only lack empathy.

I know you are in pain. You must know that the families of your victims also hurt. We all hurt. We seek some sort of closure; but a tidy end, all desire, few secure. It could be that you alone will secure it.

Salut my dear son,

Lester.


With CodeAWeek.com, I hope to release one cipher, puzzle, or mystery each week. Anyone can attempt to solve. The winner is the first person to send a correct solution and a description of the solve method to codemaster@codeaweek.com. Once a correct solution is e-mailed, I will publish a follow-up post, congratulating the winner and revealing the secrets of the code.

You may post questions or theories in the comments, but DO NOT POST SOLUTIONS. E-mail them to codemaster@codeaweek.com.

Challenge 18: More Information Than You Require

Here’s Challenge #18 for January 16, 2018. It’s another picture puzzle. Can you read it?

With CodeAWeek.com, I hope to release one cipher, puzzle, or mystery each week. Anyone can attempt to solve. The winner is the first person to send a correct solution and a description of the solve method to codemaster@codeaweek.com. Once a correct solution is e-mailed, I will publish a follow-up post, congratulating the winner and revealing the secrets of the code.

You may post questions or theories in the comments, but DO NOT POST SOLUTIONS. E-mail them to codemaster@codeaweek.com.

Challenge 16: Hollow Pursuits

I’m woefully behind on posting my weekly puzzles, so I decided to play catch-up by posting as many as I can churn out in one week. Here’s Challenge #16 for January 14, 2018.

This cipher has a picture clue!

This image should give you two pieces of information: the cipher type and the key. Each can be attained by chasing down a different piece of history related to the Jefferson Nickel: for the cipher type, a mystery; for the key, an origin story.

Additional hints:

  1. Notice anything strange in the image?
  2. ET AON RISBCDFGHJKLMPQUVWXYZ
  3. The key is a person’s last name.
  4. The key in letters is 6 characters long. Once converted to numbers, it is 10 digits long.

Here’s the ciphertext. Good luck!

04626 96622 85647 35144 54308 36773 34454 49946 78139 97502 15140 54629 33318 40343 94106 38047 29352 60205 71207 00141 92727 02741 77363 79574 84881 53547 03375 19693 27344 36506 75347 16025 17167 59376 86687 47375 24945 21945 58548 50142 13828 3677


With CodeAWeek.com, I hope to release one cipher, puzzle, or mystery each week. Anyone can attempt to solve. The winner is the first person to send a correct solution and a description of the solve method to codemaster@codeaweek.com. Once a correct solution is e-mailed, I will publish a follow-up post, congratulating the winner and revealing the secrets of the code.

You may post questions or theories in the comments, but DO NOT POST SOLUTIONS. E-mail them to codemaster@codeaweek.com.

Challenge 15: June 1918

Here’s Challenge #15 for January 13, 2018.

The year is 1918. You’re a cryptanalyst working for the Allies, and you’ve just intercepted this German Morse-code communication.

--. ..-. -..- --. ..-. .- -.. -..- .- -.. ..-. .- .- -..- / --. .- --. --. --. -.. --. --. -..- -..- --. -.. -..- ..-. / -..- ..-. .- ..-. .- -..- .- -.. --. --. --. .- ..-. -..- / .- -..- --. -.. .- -..- .- -.. -.. -.. .- -..- --. ..-. / .- -..- ..-. -.. .- .- -.. --. ..-. .- .- -.. -..- ..-.

Can you figure out the message and save lives?

Here are some hints:

1. You need to translate these dots and dashes into letters.
2. Is there anything unique about the resulting letters?
3. Who was the first to break this type of cipher?*
4. You’ll need 2 keywords: One for a mixed alphabet and one for a transposition.
5. Before entering the keywords, delete repeat occurrences of any letters. (Ex. NEBEL would convert to NEBL.)
6. I=J
7. The plaintext is in English, not German.

* For purposes of simplicity, this cipher type is an earlier version of the type he cracked. If you are familiar with the two types, it’ll be obvious which it is based on the ciphertext letters. 

With CodeAWeek.com, I hope to release one cipher, puzzle, or mystery each week. Anyone can attempt to solve. The winner is the first person to send a correct solution and a description of the solve method to codemaster@codeaweek.com. Once a correct solution is e-mailed, I will publish a follow-up post, congratulating the winner and revealing the secrets of the code.

You may post questions or theories in the comments, but DO NOT POST SOLUTIONS. E-mail them to codemaster@codeaweek.com.

Challenge 14: What Futility

Here’s Challenge #14 for December 5, 2017.

Perhaps nothing really matters, but don’t give up hope on this cipher. The method was first used by a political group in the late 1800s to organize terrorism against a regime.

Here’s the ciphertext:

35567034476476485766664594664748705756684736670957649465665776352458564548986557663558749844657733573488382658733566873446583445577738269437474486344658663936896644865435579738476474575780564754567565096456543367648945287773756689444577435834707447854565489865576635395407572476335738703428564358340957349555584888574666443944983437583745467654647645656476662467374744867728545539367057345455587478345685745757805647946478747064456763676686484756473664783428943357387936367743676770363654377546765436946658367074248673676570344677433875884624666467440634268634754767744764377574075636545436348964475476

Hints:

  1. No J.
  2. There is a keyword, and there is a key. Both are last names. One is a real person. The other is a fictional character.
  3. How well do you know your classic Russian literature? I’m thinking of a particular book that popularized a term used in the cipher type’s name.

With CodeAWeek.com, I hope to release one cipher, puzzle, or mystery each week. Anyone can attempt to solve. The winner is the first person to send a correct solution and a description of the solve method to codemaster@codeaweek.com. Once a correct solution is e-mailed, I will publish a follow-up post, congratulating the winner and revealing the secrets of the code.

You may post questions or theories in the comments, but DO NOT POST SOLUTIONS. E-mail them to codemaster@codeaweek.com.

Challenge 12: Nigh Impossible

Here’s Challenge #12 for November 8, 2017.

Today, I wanted to learn about homophonic substitution ciphers. These ciphers aim to thwart frequency analysis by assigning multiple ciphertext symbols to each plaintext symbol. The higher frequency of a letter, the more cipher symbols it is assigned. One of the most famous examples of homophonic ciphers is Rossignols’ Great Cipher.

For my homophonic cipher, I wanted to closely-match the frequency of each letter. For example, the letter E has the highest frequency at 12.7 percent. If my cipher used a pool of 100 numbers, 13 of those would represent the letter E. But if I were to use only 100, what would I do with letters such as Z, which have less than 0.1 percent frequency? In order to represent these low frequency numbers, I had to increase my pool to 1000 numbers. So now, the letter E is represented by 127 different characters!

My symbols-to-letters distribution corresponds fairly closely to the characteristic distribution. The letters Q and Z each have 1 plaintext symbol. The letters M and W each have 24. The letter A has 82.

Such a large pool of symbols means that cracking this cipher without hints will be nigh impossible. I’ll reveal a few hints now and occasionally update the post with further clues. I expect this to be the most challenging cipher I’ve yet posted on CodeAWeek.

Each 4 digit number represents a letter in the English alphabet. Here’s the ciphertext:

0385 0376 0275 0591 0106 0856 0957 0894
0808 0997 0830 0801 0511 0556 0648 0995
0295 0587 0756 0686 0983 0169 0207 0353
0111 0447 0545 0168 0162 0294 0632 0475
0937 0951 0996 0444 0549 0410 0652 0939
0701 0936 0312 0231 0770 0186 0898 0458
0374 0507 0479 0423 0017 0198 0323 0550
0306 0233 0460 0702 0625 0583 0708 0004
0524 0205 0305 0037 0038 0677 0351 0465
0299 0092 0753 0293 0018 0775 0100 0654
0311 0938 0108 0612 0496 0118 0495 0698
0665 0201

Notice that no number repeats, which tells you that the plaintext has one or no occurrences of Q and Z.

Here’s the first big hint, which will hopefully take this cipher from nigh impossible to nearly nigh impossible. Here are all 90 symbols for the letter T:

0268 0930 0367 0715 0294 0709 0688 0010 0704 0858 0266 0306 0886 0438 0502 0655 0595 0885 0673 0995 0708 0468 0142 0040 0078 0863 0558 0399 0210 0436 0091 0090 0952 0029 0701 0417 0264 0092 0175 0861 0250 0170 0605 0729 0681 0318 0221 0263 0077 0597 0197 0686 0430 0650 0542 0864 0805 0693 0505 0017 0924 0488 0625 0543 0362 0099 0849 0682 0946 0192 0422 0012 0239 0312 0194 0184 0347 0277 0064 0425 0557 0753 0156 0663 0877 0481 0830 0434 0410 0827

With CodeAWeek.com, I hope to release one cipher, puzzle, or mystery each week. Anyone can attempt to solve. The winner is the first person to send a correct solution and a description of the solve method to codemaster@codeaweek.com. Once a correct solution is e-mailed, I will publish a follow-up post, congratulating the winner and revealing the secrets of the code.

You may post questions or theories in the comments, but DO NOT POST SOLUTIONS. E-mail them to codemaster@codeaweek.com.

Challenge 1: The Expert’s Curse

With CodeAWeek.com, I hope to release one cipher, puzzle, or mystery every Wednesday evening. Anyone can attempt to solve. The winner is the first person to send a correct solution and a description of the solve method to codemaster@codeaweek.com. Once a correct solution is e-mailed, I will publish a follow-up post, congratulating the winner and revealing the secrets of the code. At this stage in the game, there are no prizes except the thrill of the solve.

You may post questions or theories in the comments, but DO NOT POST SOLUTIONS. E-mail them to codemaster@codeaweek.com.

Here’s Challenge #1 for August 9, 2017:

THE EXPERT’S CURSE

While attempting to solve the fourth part of Jim Sanborn’s CIA headquarters sculpture Kryptos, I theorized what seemed to me like a novel method.* It didn’t work. But I liked it so much that I decided to make a cipher of my own. And that cipher is “The Expert’s Curse.”

Here are a few clues:

  1. The solution (aka plaintext) is fewer characters than the code (aka ciphertext).
  2. The numbers 2 and 13 are significant.
  3. It is not a substitution cipher. Every character in the plaintext is in the ciphertext, unsubstituted.
  4. I present the ciphertext to you in five rows. The first thing you need to do is break up the text into a different number of rows. I won’t say how many, but it is more than five.
  5. Solving does not involve any of the following manipulations to the ciphertext: reversing, turning upside down, rotating right or left, switching rows or columns, or following a pattern (i.e. moving diagonally or moving down two rows and over three columns, etc).
  6. Some characters are the treasure map, some are the treasure, and some are both. The rest fill the gaps.
  7. The solve method bears some relation to a play mechanic in a popular, two-player guessing game created over 30 years ago. Perhaps it’s even older.
  8. The end is at the beginning of the end.

Here’s the code. Happy hunting!

GQNEZNBPARCMWTQSDZLRRIPHQYJGDPKRFX
SEZACCNJTHVOPUHFFREAZXSKFGOIRVJKPE
IXBOXUORQDZIRNWENDLCBGXFMKRADGSJJU
QCIUDHOIZLLYAACFKNTWLWGVXJENBHYLBO
CUMIFORVCZGPPABMWFDEYXHLNDTSNOLQT

* I thought the method was novel. However, after further research, it appears to be related to a known technique, although there are features I have not seen before.