Month: January 2018

Challenge 21: The Cheesening!

Update 2018/02/01: Puzzling.Stackexchange user DqwertyC solved this cipher in 5 days. He gets five points on the Leaderboard. I will post the solution soon.


Here’s Challenge 21 for January 27, 2018: THE CHEESENING!

Greetings fellow agents,

I’m agent Sly Happershat of the Department of Unlikely Threats. One of our undercover operatives, whom we’ll call “Snuggles,” has infiltrated the infamous prankster spy network known as the T.I.N.K.L.E.R.S.*

Snuggles has uncovered their diabolical plan to fly hundreds of drones over a famous United States sculpture. Each drone will be armed with a can of off-brand Easy Cheese. The TINKLERS fully intend to SMOTHER THE MONUMENT WITH AEROSOL-PROPELLED STRINGS OF PROCESSED CHEESE FOOD!

Obviously, we can’t let that happen.

We know WHEN the Cheesening will happen. Exactly ten days from today. But we don’t know WHERE it will happen.

That’s where you come in.

Snuggles has managed to sneak out a TINKLERS document that reveals the target monument and time of attack. Unfortunately, it is enciphered in a code we’ve never seen before.

Can you help us crack the code and SAVE AMERICA?!

Our experts report that the code is too complex to break without clues. We’ve asked Snuggles to gather more intel. But with every suspicious question he asks, he risks exposure as a government spy. And no one wants to be on the receiving end of the TINKLERS’ dreaded tickle torture!

Each day, Snuggles will add one additional clue to help us solve the mystery … until he is exposed or time runs out.

Here is the message. Will you help us?


RLLIR ESLKS TLETS TKLSS SKKNK KSSEE SLIRL SNKLS LLTLS STRES EKESS TELNS EIRNS LETLT ESLSE IRESS LILEE SKNKN KNLSS KTKSN ENNSL TEELE SSKLK KLKKS SREKI SSTNL SNKSR LKSEN SLIRE SSKRN SLIRL RESNS SSELS LILSS TIELS SKIEI NSNIL ISREL IEISS KKNNK SSNTI STTIS RKLIR ESTEI SSLSS KRNLI STLSN SSEKS ELISS NNNSS TTESL SSRLL ISTTN TTTTS RKRLK ISSLI RKRTS SKEKL KSRTT ILISK KSTET SSSEN ESESK ILKIS TINLI SKTNS RKLI


Hints from Snuggles:

Day 1 (2018-01-27): The message is enciphered in 8 unique letters, which when combined, spell TINKLERS. I caught a glimpse of someone decoding from afar, and they seemed to be adding things together. Because of this, I suspect the plaintext is shorter than the ciphertext and that is it a homophonic substitution cipher.

Day 2 (2018-01-28): I managed to smuggle out the coding alphabet with each character’s corresponding number value. My theory is that each letter in “TINKLERS” represents a number, and you add them together to get a number for a corresponding alphabet letter. But how do we know when one plaintext letter ends and another begins? Perhaps there is something I am missing.

Click on the alphabet to enlarge.

Day 3 (2018-01-29):  I think they’re getting suspicious. I got Thompkins tipsy on wine coolers and was able to extract the values of 5 of the 8 code letters. The problem is, I just don’t know WHICH letters. At any rate, I’m getting funny looks from some of the guys. I better lie low for a while.

1 2 3 5 10

The ciphertext has a space after every five letters, but I think that’s just a ruse. There must be some way to demarcate one plaintext letter and the next. Maybe not all of the 8 code letters represent numerals?

Day 4 (2018-01-30): I have to keep this short. They expect me back any minute. The “S” letter in the TINKLERS code alphabet represents a character escape. It does not matter if there are one, two, or three. Whenever you see an S or multiple S’s, the preceding character code ends and the next begins.

Day 5 (2018-01-31): For a second, I thought I had the case cracked. Pilkins let it slip that the target was Mount Rushmore — more specifically: Teddy Roosevelt’s mustache. Turns out he was just pulling my leg.

But I did learn a few things. The remaining two letters are operators — that is, they somehow modify a nearby character.

One of the operators is the number 0. It adds a 0 to the digit directly preceding it. So a 5 followed by a 0 becomes 50.

Not sure about the other operator.

Day 6 (2018-02-01): Today, the big bosses Squiggles and Marlene gathered all of us up into the meeting room. They had grave looks on their faces. I knew I was toast.

Squiggles said, “Fellow Tinklers, I come here to tell you that we have a rat in this organization.”

I eyed the door. It was blocked by two hulking bodyguards. I held my breath, waiting for the inevitable.

Squiggles slowly walked toward a cupboard and fidgeted with something inside it.

He turned around, holding a cage. With a pet rat inside it.

“His name is Milton. He is our new mascot.”

The crew cheered and then crowded around Milton. They made ewwing and awing noises.

I just can’t handle the stress. This will be my last clue.

The other operator is “-“, a negative symbol. It turns the character directly after it into a negative number. So if you have the string 15-32, then 1+5-3+2=5.

Adios


*(Team of Incurable Ne’er-do-wells, Knaves, Lollygaggers, Egregious Rascals, and Scallywags)

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 20: Overclocking

Update 2018/01/22: Bill Briere made quick work of this one and solved it in less than a day. He gets one point on the Leaderboard. I will post the solution soon.


Here’s Challenge 20 for January 21, 2018. It’s a picture puzzle:

Be sure to click the image to enlarge it. To win, you must provide the plaintext AND describe how the cipher works.

Here are a few hints:

  1. It reads left to right. When you get to the end of a row, continue to the beginning of the next row.
  2. An X in the plaintext could be an X or a Z.
  3. The following symbol is a D:


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 19: Google Earth Treasure Hunt 3

Update 2018-01-20 : Congratulations to Puzzling.Stackexchange user Gareth McCaughan for solving Challenge 19 in less than a day. He gets one point on the Leaderboard. I will post the solution soon.


Here’s Challenge #19 for January 20, 2018. It’s another Google Earth Treasure Hunt!

While exploring a cave near my house, I found a small safe. Aside from some rust, it doesn’t appear to be that old. I have reason to believe there’s something valuable inside, but I don’t know the combination. There’s a poem engraved on the side that may be the key. Can you help me solve it?

This Google Earth Treasure Hunt is a little different from the past two. Instead of hopping from location to location, you’ll need to find one specific place on Google Earth. And you do need to use Google Earth. I can’t guarantee the results if you use Google Maps Satellite View or other map software. Also, keep in mind the real place may have changed since the satellite image was taken, so go by Google Earth.

Once you pinpoint it, use the additional hints in the poem to work out the seven digits of my combination lock. I’m hoping for riches!


Here’s the poem:

This lock will open for those who
can brave the turns of English yew
The shortest route* to tower find
inside the longest of its kind
When paths diverge, record a score
where left is 1 and right is 4
And when four paths present to thee
the inside ones, mark 2 and 3
To open safe and riches win
the 7 numbers, enter in


* There may be multiple solutions, but only one is the shortest route in distance.

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.

SOLUTION to Challenge 17: Aurous Coleoptera

Congratulations to Bill Briere for solving Challenge 17 in less than an hour after I posted it!  He gets one point on the Leaderboard!

If you haven’t tried your hand at solving it, read this first.

Here’s original description and ciphertext:

It’s a relatively simple one. The title Aurous Coleoptera should be the only hint you need. To win, be the first to provide the solution AND the source of the quote.

6;95:]80028†‡?2;8†]48;48(4?95*6*38*?6;:-5*-‡*);(?-;5*8*6395‡1;4876*†]46-44?95*6*38*?6;:95:*‡;2:.(‡.8(5..06-5;6‡*(8)‡0¶8


“Aurous” is a synonym for gold. “Coleoptera” is the insect order of beetles. Combine them to get “The Gold Beetle” or “The Gold Bug.”

The plaintext is: “It may well be doubted whether human ingenuity can construct an enigma of the kind which human ingenuity may not, by proper application, resolve.”

This is a quote from the character Legrand in Edgar Allan Poe’s classic short story The Gold Bug. The cipher is, of course, “The Gold Bug cipher,” a simple substitution cipher that Poe created for his story. When The Gold Bug was originally published in 1843, it helped popularize cryptography and inspired many a young codebreaker.

The solid gold bug in the story was a scarab-like beetle.

You can make your own Gold Bug cipher by using Dcode’s excellent online tool.


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 17: Aurous Coleoptera

Update 2018-01-15 : Bill Briere made quick work of Challenge 17, and he gets one point on the Leaderboard! Read the solution here.

Here’s Challenge #17 for January 15th, 2018.

It’s a relatively simple one. The title should be the only hint you need. To win, be the first to provide the solution AND the source of the quote.

6;95:]80028†‡?2;8†]48;48(4?95*6*38*?6;:-5*-‡*);(?-;5*8*6395‡1;4876*†]46-44?95*6*38*?6;:95:*‡;2:.(‡.8(5..06-5;6‡*(8)‡0¶8


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.

SOLUTION to Challenge 3: A Simple Calculation

Congratulations to Euchre Mutt for being the first person to solve Challenge 3!  The puzzle remained unsolved for a whopping 134 days … a site record! … which means he is awarded 134 points on the Leaderboard.

If you haven’t tried your hand at solving it, read this first.

Here’s original description and ciphertext:

With this cipher, I hope to add a new encryption method to the playbook. If you formulate the correct hypothesis, you just might take away the prize.

Some clues:

  1. It’s a substitution cipher.
  2. The ciphertext is 102 characters. The plaintext solution is 34 characters.
  3. What is 102 divided by 34, and what does that tell you?
  4. The plaintext solution is a quote from a famous mathematician.

The ciphertext is:

LCFBPDLED HCJJKAHFF TSNIGANRK GMBHJMQCO XETSSRVLZ CFDZOWVLP JJKFSRPCK BXFIVPWJO RSLYNXFSD HRHAOKQQK IJAEPFBUI QPZ


The key to solving this cipher lies in a very simple mathematical formula. It’s so simple that you’re going to kick yourself for not figuring out.

First, number the alphabet, where A=1, B=2, C=3, etc.

Now, space the ciphertext into blocks of 3 characters and apply the following formula to each block: A+B-C

The result of that formula is one plaintext character. Apply the same formula to every block of three characters, and you’ll discover the solution.

For example, if we apply it to the first three characters “LCF” …

L+C-F=

12+3-6=9

9=I

Continue down the line to get the solution:

“In math, you’re either right or you’re wrong”

… which is a quote by NASA mathematician Katherine Johnson. Johnson was profiled in the book and film Hidden Figures.

There are a couple of hints in the challenge description, which I’ll put in bold here:

With this cipher, I hope to add a new encryption method to the playbook. If you formulate the correct hypothesis, you just might take away the prize.


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.

SOLUTION to Challenge 13: Fractionating

Congratulations to Euchre Mutt for being the first person to solve Challenge 13!  It remained unsolved for 45 days, which means he is awarded 45 points on the leaderboard.

If you haven’t tried your hand at solving it, read this first.

Here’s original description and ciphertext:

Today, I wanted to learn about ciphers that use fractionation. I decided on a method created by a French amateur cryptographer who spent his days in working in a customs warehouse.

Three hints:

  1. J=I
  2. You need to figure out the keyword and the period.
  3. “Now Morn her rosie steps in th’ Eastern Clime, Advancing, sow’d the earth with Orient Pearle.”

And the ciphertext is:

xmwuv hnmhv hderq
tbkht newsv hnmhv
fpeks ngifq rlqv


The solution:

This puzzle uses the bifid cipher, invented by Félix Delastelle around 1901. Delastelle was a bonded warehouseman by day and an amateur cryptographer by night. The bifid cipher uses a numbered grid in a very clever way to fractionate the plaintext, making it harder to crack.

The third clue is a line from Milton’s Paradise Lost.  That should help you figure out that the keyword is MILTON. Incidentally, this is the opening line from Book 5, which should lead you to a block length of 5.

The decoded text reads:  “Who overcomes by force, hath overcome but half his foe, ” another quote from Paradise Lost.

You can use this online Biphid Decoder to test all this out.


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.