Wolfram Computation Meets Knowledge

The Solution of the Zodiac Killer’s 340-Character Cipher

Editor’s note: The Zodiac Killer (an unidentified American serial killer active during the 1960s and 70s) sent numerous taunting letters to the press in the San Francisco area with regard to a local murder spree. In these letters, the killer took responsibility for the crimes and threatened to commit further murders. He also included three ciphers, each containing one-third of a 408-character cryptogram. The killer claimed that this cryptogram would reveal his identity when deciphered. The killer sent the fourth and final cipher (discussed in this blog post) to the San Francisco Chronicle after the 408-character cryptogram, deciphered in 1969, did not reveal the killer’s identity.

The Solution of the Zodiac Killer’s 340-Character Cipher

In 2020, Melbourne, Australia, had a 112-day lockdown of the entire city to help stop the spread of COVID-19. The wearing of masks was mandatory and we were limited to one hour a day of outside activity. Otherwise, we were stuck in our homes. This gave me lots of time to look into interesting problems I’d been putting off for years.

I was inspired by a YouTube video by David Oranchak, which looked at the Zodiac Killer’s 340-character cipher (Z340), which is pictured below. This cipher is considered one of the holy grails of cryptography, as at the time the cipher had resisted attacks for 50 years, so any attempts to find a solution were truly a moonshot.

The Zodiac Killer

The Zodiac Killer’s 340-character cipher

In his presentation, David explored the idea that the cipher is both a homophonic substitution cipher and a transposition cipher. Highly efficient programs for solving homophonic substitution ciphers exist, the best of which is AZdecrypt. Experiments suggest that AZdecrypt can solve all homophonic substitution ciphers of the same length and symbol distribution as the Z340. However, AZdecrypt cannot be used to solve the Z340 because when you run it on the Z340, it does not produce a solution. Perhaps solving the Z340 is a case of finding by trial and error the correct transposition, then using AZdecrypt to solve the homophonic substitution cipher.

David outlined one particular transposition, which was discovered independently and posted to zodiackillersite.com by user “daikon” and Jarl van Eycke (the author of AZdecrypt): the “period-19,” which had some interesting statistical properties that would suggest that they were closer to the correct transposition. Just for fun, I decided to plot this transposition using Mathematica:


   Table[1 + Mod[19 i, 340] -> i, {i, 0, 339}] // SparseArray // 
    Normal, 17] // drawTransposition;
Magnify[%, 0.5]

However, this looked nothing like daikon and Jarl’s period-19 transposition. To my surprise, it turned out their transposition used a period-18 when wrapping around (periodically) vertically.

Period-19 transposition

While this transposition was visually interesting, it didn’t strike me as a very natural pencil-and-paper construction. It should be noted that Z340 was created in 1969, and therefore was almost certainly constructed using pencil and paper.

I saw a connection between the period-19 transposition and a 1,2-decimation of the cipher. That is, starting from the top-left corner and moving one vertical step, then two horizontal steps, wrapping periodically both horizontally and vertically, like the cipher is wrapped around a torus. This transposition takes similar diagonals to the period-19 transposition:


decimate2D[array_, {n_, m_}] := Module[{d0, d1},
  {d0, d1} = Dimensions[array]; 
  (Table[array[[Mod[n i, d0] + 1, Mod[m i, d1] + 1]], {i, 0, 
      d0 d1 - 1}]) /; 
   CoprimeQ[d0, n] && CoprimeQ[d1, m] && CoprimeQ[d0, d1]

z12decimation = 
  Partition[decimate2D[Partition[Range[0, 340 - 1], 17], {1, 2}], 17];

Running the 1,2-decimation transposition of the Z340 through AZdecrypt did not produce a solution.

One way to investigate the likelihood of finding the correct transposition of a homophonic substitution cipher is by counting the number of repeating bigrams (pairs of symbols). In Mathematica, it’s easy to write up this code for arbitrary -grams:

countngrams[l_List, n_Integer] := 
 Total[Map[Last, Tally[Partition[l, n, 1]], 1] - 1]

countngrams[l_List, n_Integer] := 
 Total[Map[Last, Tally[Partition[l, n, 1]], 1] - 1]

Then we can construct a large number of Z340-like ciphers and compare their bigram count distribution to a large number of random shuffles of the Z340:

Histogram[{randomShuffle340, z340CiphersBigrams}, {1}, 
 ChartBaseStyle -> EdgeForm[Thin], Frame -> True, Axes -> False, 
 FrameLabel -> {Style["repeating bigram count", 12], 
   Style["frequency", 12]}, 
 ChartLegends -> {"random shuffles of Z340", 
   "random Z340-like ciphers"}, 
 ChartStyle -> {RGBColor[0.514366, 0.731746, 0.415503], 
   RGBColor[0.996414, 0.825742, 0.330007]}]

The mean number of bigrams for the random shuffles was 19.8, and for random Z340-like ciphers was 34.5. The Z340 has 25 repeating bigrams, while the daikon and Jarl period-19 transposition and the 1,2-decimation has 37 repeating bigrams. Thus, statistically, we thought we were on the right track.

After the 1,2-decimation transposition did not produce a solution we decided to work on a large search through candidate transpositions. It was difficult to find out what transpositions had been tested in the past, so I decided to enumerate all reasonable 1- and 2-step transpositions, sort them by their bigram count and run them through AZdecrypt. Some of these transpositions, for example, included:

Examples of 1- and 2-step transpositions

I also included all proper one-dimensional and two-dimensional decimation transpositions. For one-dimensional enumerations, we have the following 128 proper decimations:

Select[Range[340], CoprimeQ[#, 340]&]

Select[Range[340], CoprimeQ[#, 340] &]
% // Length

For two-dimensional enumerations, we have the following 128 proper decimations:

Join @@ Outer

Join @@ Outer[List, Select[Range[19], CoprimeQ[#, 20] &], Range[16]]
% // Length

For example, the 3,4-decimation generates the following transposition:


Partition[decimate2D[Partition[Range[0, 339], 17], {3, 4}], 17] // 
   invert // drawTransposition;
Magnify[%, 0.5]

Using AZdecrypt, we tested all row–major, column–major, alternating row–column, alternating column–row, inward spirals, outward spirals, diagonals and proper one-dimensional and two-dimensional decimation transpositions. This experiment didn’t yield anything looking like a solution, so we tested all pairs of transpositions. Then we considered testing all 3-tuples of transpositions; however, this would require testing 155,929,364,660,224 candidate ciphers. Naively checking one a second would take over five million years. So we limited our experiments to decimations which would be reasonable to write out by hand and then only tested candidates with a high bigram count. Once again, this search turned up nothing.

Perhaps there’s another step we are missing? Given the way the Zodiac Killer’s 408-character cipher (pictured below) was sent in three equally sized sections, we conjectured the Z340 was constructed from a number of distinct segments, then encrypted with a transposition and a homophonic substitution:

The Zodiac Killer's 408-character cipher

The Zodiac Killer's 408-character cipher

We considered splitting the cipher horizontally into two and three segments, vertically into two and three segments, and both horizontally and vertically into and segments. For example:

Examples of splitting the cipher

Then we used Reduce to compute all possible segments, which resulted in proper two-dimensional decimations:


Reduce[a > 0 && b > 0 && c > 0 && d > 0 && a + b == 17 && 
  c + d == 20 && GCD[a, c] == 1 && GCD[a, d] == 1 && GCD[b, c] == 1 &&
   GCD[b, d] == 1, {a, b, c, d}, Integers]

Given the high bigram count of the 1,2-decimation transposition, we started our search with two-dimensional decimations, with each segment having the same (single) transposition. As we had seen so many times before, this experiment didn’t turn up anything.

The next search for compositions of multiple transpositions and all combinations of transpositions for all sections would be a significantly larger undertaking. So we decided to reanalyse the results from the initial search.

Out of the 650,000 transpositions we tested, one contained a few particularly interesting segments of plaintext:


This was even more interesting as the transposition that produced this candidate decryption was the 1,2-decimation, with the cipher split into three vertical segments (pictured below):

Cipher split into three vertical segments

Investigating this result further, David used our 9,9,2-vertical segment, 1,2-decimation transposition and AZdecrypt to crib the phrases “HOPE YOU ARE,” “TRYING TO CATCH ME” and “THE GAS CHAMBER.” With these cribs locked in place, AZdecrypt found the following solution of the first segment:

The plaintext solution for the first segment of the Zodiac Killer

Eureka! After 51 years, we had decrypted some of the Z340. This was a very special moment. The discovery of the 9,9,2-vertical segment, 1,2-decimation transposition and the power of AZdecrypt for solving homophonic substitution ciphers had produced a partial decrypt of the Z340.

What about the remaining two segments? It was possible that we had found just one correct vertical split of 9 rows and the remaining 11 rows required a different segmentation, or it was possible that a different transposition was needed, or even a different key for the substitution cipher, or any two combinations of these possibilities, or even all three possibilities. Our work was far from over.

David discovered that we could use the key from the first segment on the last segment to produce the following plaintext—without any transposition:

Plaintext without transposition

Including some spaces, the cipher’s candidate plaintext gives:

The cipher’s candidate plaintext gives

Then, reversing a few words:

The candidate plaintext with some reversed words

This seemed to be a pretty reasonable decryption of the last segment.

What about the second segment? If we crib all the legible text from the first section we get the following decryption:

Cribbed legible text

Some parts of this kind of made sense, but we certainly weren’t there yet. We asked Jarl van Eycke, the author of AZdecrypt, to help us with this segment. He made the following brilliant observations:

  • The LIFEIS plaintext (the second segment) is read left to right.
  • The LIFEIS plaintext is excluded from the 1,2-decimation transposition and read left to right.
  • Numerous spelling mistakes are corrected if “H” on row six is moved to the fourth column.
  • Apply the 1,2-decimation transposition, skipping the positions containing “LIFEIS”:

LIFEIS plaintext

Then the second segment becomes:

The second segment becomes

The cipher key and transposition that we discovered for the Z340 cipher are given by:

The cipher key for Zodiac Killer

The transposition for Zodiac Killer

Fifty-one years after the Zodiac Killer mailed this cipher to the San Francisco Chronicle, we had a solution. David submitted this solution to the FBI Cryptanalysis and Racketeering Records Unit (CRRU) on Saturday, December 5, 2020. Shortly after, the FBI was able to officially confirm the validity of our solution.

The FBI confirmation of the validity of our solution

Essentially all my work on the Z340 was done in Mathematica. I used the Spartan high-performance computing cluster at the University of Melbourne to eliminate candidate transpositions using zkdecrypto and David used AZdecrypt. Otherwise, all the statistical analysis of the Z340 and the creation and analysis of the millions of candidate transpositions was done using Mathematica. The reason for my use of Mathematica is simple; it is by far the most time-efficient language I could use for such a task.

Sam Blake has a PhD in mathematics from Monash University in Australia and is a research fellow at the University of Melbourne. He’s been an avid Mathematica user since 2004 and worked for Wolfram Research for four years. He is broadly interested in research related to numerical modeling, data science, symbolic computation, steganography and cryptography.

Get full access to the latest Wolfram Language functionality with a Mathematica 12.2 or Wolfram|One trial.


Join the discussion

!Please enter your comment (at least 5 characters).

!Please enter your name.

!Please enter a valid email address.


  1. Thanks Sam, that was a brilliant piece of sleuthing on yours and David’s part. It is especially intriguing how the use of Mathematica’s visualizations allowed you to determine the between the appropriate transpositions and substitutions. It is also significant that after being able to translate 3 phrases from the first crib you were quickly able to reduce the number of possibilities by fixing those specific translations. This was also part of the success with Turing’s original Bombe translator when he recognized that “Heil Hitler” and other stock phrases were being used in the Enigma transmissions from the German high command.

    Again congrats!


  2. Very nice analysis! Thanks for sharing. Have you looked at this function: https://resources.wolframcloud.com/FunctionRepository/resources/LatticePointsArrangement

    Perhaps useful to generate the various grids.

  3. Thank you for sharing this.

  4. “Be. Sure. To. Drink. Your. Ovaltine.”

  5. If we didn’t have the computing power that we have today, how long would it have taken to de-crypt these messages?

    • Hi David,

      I honestly do not know how this could have been solved without modern computing resources. My only conclusion is that Zodiac had a far better grasp of creating these ciphers than he had the ability to reason with a pencil and paper estimation just how difficult this cipher would be to solve.



  6. Does anyone not find the contrast between the mathematical prowess and linguistic control somewhat jarring? An accomplice perhaps?

    • I honestly would not doubt for a minute that there was an accomplice. Could he have done this on his own, yes, would it have been difficult to do on his own, also yes. It is just boggling that he was willing to put this much time into a cipher, he thought was easy enough, for us only to figure out 50+ years later.

  7. That is truly amazing. I can’t believe how in depth his patterns were when it came to writing these messages. Backwards, then forwards, then the spelling of paradice.