substitution cipher

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.

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.

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.

SOLUTION to Challenge 11: The Right Type

Congratulations to Puzzling.Stackexchange user Kayzeroshort for being the first to solve Challenge 11! It took him 12 days, which means he is awarded 12 points on the Leaderboard.

I’ll reveal the secrets below, but if you haven’t tried your hand, read this first.

The ciphertext:

YGWCSD QDCHVF WCLEZJ CHCTJQ YUEOYG YSE,GY KYJXDZ KYCYEP IFKWGW
TIHOEP CHRTJR DOCRTU YGY,HB PEDWUC GWU.QD ;WRTDR BESDOY ZDMNUC
SFQDCH RKWCH. POXHQO QFYOY. YPSAYR QFYHYG Y,HGIT YGRT

The original post:

Today’s cipher is a method I created based on a known method, although it is likely that someone else thought of it before me.

I’ll give you three clues:

    1. The method is a spinoff of the famous method used to encode Challenge 10.

    2. The punctuation in the ciphertext encodes letters in the plaintext. (There is no punctuation in the solution.)

    3. The following text is not a deciphering tool but a clue that will point you in the right direction. What came after this?

      3 5 7 9 N O P Q R S T U V W X Y Z
      2 4 6 8 . A B C D E F G H I J K L M

The solution, as explained* by kayzeroshort on Puzzling.Stackexchange:

The lines …

3 5 7 9 N O P Q R S T U V W X Y Z

2 4 6 8 . A B C D E F G H I J K L M

… are the characters used on an early typewriter designed by Christopher Latham Sholes. He went on to invent the first typewriter with a QWERTY keyboard layout, the same layout used on modern day computer keyboards.

That clue leads to  your own computer keyboard being the codebook. Using your keyboard, you can follow the rules of the Playfair cipher to decipher the code. The grid would look like this:

QWERTYUIOP
ASDFGHJKL;
ZXCVBNM,./

The deciphered text is

T H E X A S E A N D F R E X D O M A N D B E A U T Y W I T H W H I C H T H I S M A C H I N E W O R K S I S T R U L Y W O N D E R F U L E V E R Y T H I N G N O W S E E M S T O M E A S P E R F E C T A S I T C A N B E M A D E A N D I F E X L N O I N S P I R A T I O N T O A ; T E R A N Y T H I N G F U R T H E R

or

THE EASE AND FREEDOM AND BEAUTY WITH WHICH THIS MACHINE WORKS IS TRULY WONDERFUL. EVERYTHING NOW SEEMS TO ME AS PERFECT AS IT CAN BE MADE AND I FEEL NO INSPIRATION TO ALTER ANYTHING FURTHER.

As with the Playfair cipher, to get the final deciphered text from the raw decipher, replace X’s with the previous letter.

The plaintext is an excerpt from an 1870 letter by Christopher Latham Sholes about his new typewriter. It was some of the first text ever written on the QWERTY typewriter.

The puzzle title THE RIGHT TYPE is a hint alluding to “typewriter.”

* lightly edited by me.

 


With CodeAWeek.com, I hope to release one cipher, puzzle, or mystery every 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 10: No Cheating

Congratulations to Stackexchange user M Oehm, who was the first to solve NO CHEATING. It took him one day, which means he is awarded 1 point on the Leaderboard.

I’ll reveal the secrets below, but if you haven’t tried your hand, read this first.

The ciphertext:

WXYCNW LAHWNZ WONSNI LEXAPE OSMWOW RCSRQC AZ

And the original puzzle description:

Today’s cipher is a famous cryptographic method known for its economy, ingenuity, and difficulty in cracking. Its inventor created all sorts of cool things, including musical instruments, a timepiece especially useful at the North Pole, and the earliest ancestor of the Oculus Rift. Aside from this cipher, which does not bear his name, he is most famous for co-inventing a means of communicating at long distances.

The key to unlocking the mystery is his name. Good luck!

The solution:

The title “NO CHEATING” was meant to lead you to the Playfair cipher, created by Charles Wheatstone in 1854. He invented all sorts of cool things, including the English Concertina, the Polar Clock, the Stereoscope. He co-invented telegraphy, a means of communicating at long distances.

The key is CHARLESWHEATSTONE. You make a 5 by 5 square and fill it with the letters of the alphabet, starting with the key.  This gives the following square:

C H A R L
E S W T O
N B D F G
I K M P Q
U V X Y Z

Since our square only has 25 spaces, we must do away with one letter. In this case, that letter was J.

We follow the rules of the Playfair Cipher to get the message:

AM UR DE RH AS GU ST BE EN CO MX IT TE DA TS AL TH IL LX

If we take out the padded Xs and convert the G* to a J, we get the fully decoded message:

“A murder has just been committed at Salt Hill.”

… which is the first line of an early telegraph that helped catch a murderer. It was the first arrest ever made using the technology.

* Usually J is encoded is I, but I made a mistake and coded it as G here.


With CodeAWeek.com, I hope to release one cipher, puzzle, or mystery every 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.