To add to our list of challenge ciphers (Bellaso’s, d’Agapeyeff’s, Feynman’s, etc), here’s one I hadn’t seen before from Helen Fouché Gaines’ (1956) “Cryptanalysis: A Study of Ciphers and Their Solution”, which I found courtesy of Greg Ross’s Futility Closet website:-
VQBUP PVSPG GFPNU EDOKD XHEWT IYCLK XRZAP
VUFSA WEMUX GPNIV QJMNJ JNIZY KBPNF RRHTB
WWNUQ JAJGJ FHADQ LQMFL XRGGW UGWVZ GKFBC
MPXKE KQCQQ LBODO QJVEL.
The cipher is the last in a series of exercises at the end of a chapter titled “Investigating the Unknown Cipher,” and she gives no hint as to its source. Of the exercises, she writes, “There is none in which the system may not be learned through analysis, unless perhaps the final unnumbered cryptogram.” The solution says simply “Unsolved.”
If you look at the book itself (p.217), all Gaines says is “Here is one which nobody has been able to decrypt:“. Hence it is not at all clear whether this is a composed challenge cipher (i.e. designed to confound) or an accidental challenge cipher (i.e. one found in the wild but never yet solved). I suspect the latter… but perhaps someone will know for sure either way.
Incidentally, the 1968 comment on this mentioned in the Futility Closet post is online here (it’s on p.5): just so you know, the authors there offer an [entirely fictional, I expect] “Nicodemus J. Grumbow award” for anyone solving it.
As far as the ciphertext itself goes, it has a flattish distribution (Q appears 9 times, while T & Y appear only twice each, all 26 letters are used), with a standard deviation of 1.52144, i.e. much flatter than a normal alphabet would present.
It has no repeated trigrams, while QJ & PN appear three times (DO, GW, QL, GG, VQ, PV, NU, NI and XR each appear twice). There are seven doubled letter-pairs, all appearing once only each (PP, GG, JJ, RR, WW, GG, QQ). There are a few visible patterns in the text that vaguely suggest some kind of structuring (JAJGJ, QCQQ, QLQ and QQL), but all of which might just be random.
As a result, it doesn’t appear to be a monoalphabetic substitution, nor a (conventional) polyalphabetic substitution (as there seems to be no obvious cycles, loops, or repeats). The cipher text is 125 characters long, which (as a mathematician) makes me idly wonder whether this was partly enciphered using some kind of a 5x5x5 three-dimensional transposition cipher, the sort of thing a Bond villain would gloat about in his/her evil monologue. I don’t believe for a minute that this is the case, of course, but I thought I’d mention it all the same. 🙂
Any thoughts? Is there anything that suggests to you what kind of a cipher this might be?
popup sass sos SSI (Social Security Insurance)
🙂
The spacing between QJ is 22, and 44. Maybe the key is 11 letters in length.
That is assuming some sort of polyalphabetic cipher, Vignere maybe.
Going to take another look at this tonight to see what I can come up with. Thanks for bringing this one to my attention Nick 🙂
Stu: ooh, I didn’t notice that at all – good catch, Stu! 🙂 A length-11 polyalpha would be well worth checking for…
If you divide it into pairs 18 of the 24 recurring digrams remain intact, this suggests pair for pair substitution to me, possibly some kind of Playfair with 1 or 3 nulls tacked on the end to complete the blocks of 5.
Tony
If you break it into pairs into columns of 5, you get recurring digrams in the same columns. (XR in 1, PN in 2 and QL in 3)
VQ BU PP VS PG
GF PN UE DO KD
XH EW TI YC LK
XR ZA PV UF SA
WE MU XG PN IV
QJ MN JJ NI ZY
KB PN FR RH TB
WW NU QJ AJ GJ
FH AD QL QM FL
XR GG WU GW VZ
GK FB CM PX KE
KQ CQ QL BO DO
QJ VE L
This cipher reminds me so much of Kryptos K4 that I can’t bring myself to look at it.
Jim: 🙂
NICK & ALL. I have solved this exercise in ciphering. I t was a fest of skill that I had to use a POLYALPHABETIC KEY, & POLYMORHIC PATTERNS. A PATTERN DICTIONARY is formed. Here is my deciphered work: BRAVA, TRUST, STARS, ERASE or ALIVE, SHEEP, DRILL, POLIL or BONIN, WORDS, TREES. STARS, OVERO, TODAY, ABAFT, EERIE, EERIE, TRAIN or TOAST, TOAST or THESE, THREE, TRAIN, SHELL, STEPS, ANANN, TESTS or STAGE,TESTS then NOTES(IF STAGE then NORTH). Thanks.
I’ve been working on Helen Fouché Gaines’s 1939 unsolved cipher (with the help of Astra 6) and I think I found a pretty strong solution.
The key I ended up getting is MARK, using a progressing Variant Beaufort/Vigenere-type system that Gaines actually teaches in the book. The key moves forward one letter after every 4 characters:
MARK → NBSL → OCTM → PDUN → …
There also seems to be an extra J in the ciphertext. There are two Js next to each other across the 5-letter groups (QJMNJ JNIZY), and if one of those is treated as an extra character, the whole message seems to line up.
The direct output, without me correcting the weird letters, comes out roughly:
“HQ SECQND DIZ.
REQORTS FROM ALL OVSERVITION POSTS ON OUR RIGHT FLANK
INDICETE THOT ENEOY IS COZTINUOUSLY
KTTEMPTING TO OVERLYP OUR LINE
SEND GID ZZZZ”
The obvious reading seems to be:
“HQ SECOND DIV.
REPORTS FROM ALL OBSERVATION POSTS ON OUR RIGHT FLANK
INDICATE THAT ENEMY IS CONTINUOUSLY ATTEMPTING TO
OVERLAP OUR LINE.
SEND AID.
ZZZZ”
Interesting part is that the weird letters don’t really look random. Basically, there are 12 letters that are off. For 10 of them, the ciphertext is exactly what you get if whoever encrypted it used math the opposite way at that spot, so adding the key instead of subtracting it. One more of the letters can also be explained if only the progression part was done backwards. So that leaves basically one letter that I still don’t have a good explanation for.
I still wouldn’t say this is 100% done since I don’t know why those specific letters would have been encrypted the opposite way, but so far it seems like the only likely solution to Helens final cipher.
Independent Analysis of Elias Badalov’s Candidate Solution to Helen Fouché Gaines’s Final Unnumbered Cryptogram
16 September 2026
Attribution and purpose of this report
The central discovery discussed in this report belongs to Elias Badalov.
Badalov identified the key MARK, proposed the four-character progressing-key system
MARK → NBSL → OCTM → PDUN → …
and obtained the underlying military plaintext from Helen Fouché Gaines’s final unnumbered cryptogram.
Badalov also identified the suspicious doubled J in the ciphertext and observed that most of the remaining incorrect letters behave as though the opposite Vigenère/Variant-Beaufort arithmetic had been used.
I have taken Badalov’s proposed solution as the starting hypothesis and gone on from his work, concentrating on testing it rather than searching for a different solution.
I should also make clear that most of the computational and cryptanalytic work described below was carried out with ChatGPT, which I call Maxwell. I supplied the questions, hypotheses and directions for the investigation, while Maxwell performed most of the calculations, exhaustive searches, comparisons and historical cross-checking.
The intention of this report is therefore not to claim independent discovery of the solution.
Its purpose is to ask a different question:
How well does Elias Badalov’s MARK solution survive independent reproduction, historical comparison and attempts to falsify it?
The answer, so far, is: remarkably well.
1. The cryptogram
Gaines concludes Elementary Cryptanalysis with an unnumbered cryptogram preceded by the statement:
“Here is one which nobody has ever been able to decrypt.”
The printed ciphertext contains 125 letters in 25 groups of five:
VQBUP PVSPG GFPNU EDOKD XHEWT
IYCLK XRZAP VUFSA WEMUX GPNIV
QJMNJ JNIZY KBPNF RRHTB WWNUQ
JAJGJ FHADQ LQMFL XRGGW UGWVZ
GKFBC MPXKE KQCQQ LBODO QJVEL
The conspicuous central sequence is:
QJMNJ JNIZY
which contains two adjacent Js across a five-letter group boundary.
Source: Project Gutenberg edition of Gaines’s book.
2. Badalov’s proposed cipher system
Number the alphabet:
A = 0, B = 1, … Z = 25
The initial four-letter key is:
MARK
After each complete four-character block, every key letter advances by one alphabet position:
MARK
NBSL
OCTM
PDUN
QEVO
…
So the progression is:
MARK → NBSL → OCTM → PDUN → …
The proposed cipher is Variant Beaufort.
In simple form:
ciphertext = plaintext – key
or:
C = P – K (mod 26)
Decipherment is therefore:
P = C + K (mod 26)
This is not an artificial construction invented specifically to fit the cryptogram.
Gaines explicitly teaches Variant Beaufort and describes progressing keywords of precisely this general type. In Chapter XX, for example, her keyword CULPEPER advances to DVMQFQFS, then EWNRGRGT, and so on.
Therefore the progression
MARK → NBSL → OCTM → PDUN …
is fully compatible with a system Gaines actually teaches.
3. Reproduction of Badalov’s raw decryption
The printed ciphertext contains:
… QJMNJ JNIZY …
Removing either one of the two adjacent Js produces the same 124-character sequence.
No other adjustment to the alignment is required.
Applying progressing Variant Beaufort with initial key MARK gives exactly:
HQ SECQND DIZ.
REQORTS FROM ALL OVSERVITION POSTS ON OUR RIGHT FLANK
INDICETE THOT ENEOY IS COZTINUOUSLY
KTTEMPTING TO OVERLYP OUR LINE
SEND GID ZZZZ
This independently reproduces Badalov’s result.
Out of the 124 deciphered letters, 112 are already exactly correct.
The obvious plaintext is:
HQ SECOND DIV.
REPORTS FROM ALL OBSERVATION POSTS ON OUR RIGHT FLANK
INDICATE THAT ENEMY IS CONTINUOUSLY ATTEMPTING TO
OVERLAP OUR LINE.
SEND AID.
ZZZZ
So before making any corrections to the anomalous letters, the cipher produces:
112 correct letters out of 124 = 90.3% exact.
This is important.
The proposed plaintext is not being reconstructed from generally meaningless output. Almost the entire grammatical structure and message appears directly from one short key and one fixed cipher rule.
4. The twelve anomalous positions
There are twelve discrepancies between the direct decryption and the natural plaintext.
For example:
SECQND instead of SECOND
DIZ instead of DIV
REQORTS instead of REPORTS
OVSERVITION instead of OBSERVATION
INDICETE instead of INDICATE
THOT instead of THAT
ENEOY instead of ENEMY
COZTINUOUSLY instead of CONTINUOUSLY
KTTEMPTING instead of ATTEMPTING
OVERLYP instead of OVERLAP
GID instead of AID
The striking point is that these bad letters are not arbitrary.
For 10 of the 12 discrepancies, the actual printed ciphertext letter is exactly what would result if the operator used the opposite arithmetic:
Normal Variant Beaufort:
C = P – K
Opposite Vigenère operation:
C = P + K
In other words, ten of the strange characters are exactly what would be produced by using Vigenère instead of Variant Beaufort at that character.
The remaining two also have very small explanations.
At one position, reversing only the direction of the progression component produces the printed letter exactly.
At another, the progression has just advanced. Using the immediately previous progression value for that one character produces the printed ciphertext exactly.
Thus all twelve discrepancies have simple cryptographic-operation descriptions.
What remains unknown is why those particular positions were affected.
5. Gaines’s recommended working procedure
Gaines gives an interesting practical recommendation for progressing-key ciphers.
She suggests carrying out the encipherment in two stages:
produce an ordinary periodic cryptogram;
apply the progression afterward.
Her worked example separates:
the initial keyword,
the plaintext,
the primary cryptogram,
the progression,
and the final cryptogram.
That makes the proposed MARK construction historically appropriate not only mathematically but also in terms of Gaines’s actual working method.
There is, however, an important caution.
Some of the anomalous characters would require coordinated changes in more than one stage if the operator followed the two-stage procedure literally.
For that reason it is premature to claim that all twelve discrepancies are simply careless mistakes.
They might be:
encipherment errors,
an unrecognized secondary convention,
production errors,
or some combination of these.
6. The doubled J
The printed ciphertext contains 125 characters.
The proposed plaintext contains 124 characters.
The question of which character must be removed can be tested without assuming in advance that the suspicious J is the answer.
Every possible one-character deletion was tested.
The best results are:
Deleted position Character Exact letters
55 J 112/124
56 J 112/124
54 N 111/124
57 N 111/124
53 M 110/124
58 I 110/124
Positions 55 and 56 are the two adjacent Js.
Since they are identical and adjacent, removing either produces the same sequence.
Thus the cipher model independently identifies precisely the doubled-J location originally noticed by Badalov.
No other deletion gives an equal result.
However, this does not tell us why the extra J is present.
It could be:
a deliberately inserted null,
an encipherment error,
a manuscript/transcription error,
or a typesetting error.
Gaines’s book contains a substantial number of documented cryptographic errors. Project Gutenberg’s proofreaders even found one exercise with seven letters missing.
For that reason the cautious conclusion is:
The MARK solution requires one extra character at the doubled-J location, but the origin of that extra character remains unknown.
7. Exhaustive four-letter-key test
The proposed plaintext was then frozen.
Rather than simply trying other English words, the entire four-letter keyspace was tested.
The search included:
every possible one-character deletion from the 125-letter ciphertext;
every possible four-letter key;
the same +1 progression after every four characters;
the same Variant Beaufort operation.
There are:
26^4 = 456,976
possible four-letter keys.
With 125 possible deletions this gives:
57,122,000 deletion/key combinations.
The global optimum is:
delete one of the doubled Js + key MARK = 112/124 exact letters
No different four-letter key achieves more than:
87/124
Several different keys tie at that second-best score, so there is no meaningful unique runner-up.
The important point is the gap:
MARK scores 112.
The best non-MARK keys score only 87.
That is a difference of 25 exact characters.
MARK is therefore not simply one of many roughly equivalent four-letter keys.
8. Matched shuffled controls
A further question is whether an exhaustive search of millions of possibilities could routinely find apparently impressive results by chance.
To test this, the letters of the proposed 124-character plaintext were randomly shuffled.
This preserves exactly the same letter frequencies but destroys the message.
For each shuffled plaintext, the complete four-letter-key and deletion search was performed again.
A total of 10,000 shuffled controls were tested.
The results were approximately:
median best score: 17/124
90th percentile: 19/124
99th percentile: 21/124
maximum of all 10,000 controls: 23/124
The proposed plaintext with MARK scores:
112/124
No shuffled control comes remotely close.
This is not a formal probability that Badalov’s solution is correct.
The plaintext was originally discovered using the ciphertext, so it is not an independently specified hypothesis.
What the test does establish is narrower:
Once Badalov’s proposed plaintext is frozen, its compatibility with MARK and the progressing Variant system is extraordinarily specific.
9. Held-out recovery of MARK
Another test asks whether MARK can be recovered from one portion of the message and then used successfully on a different portion.
Using only the first 20 characters, the four initial key letters were inferred independently.
Despite three anomalous characters occurring in this short section, the recovered key is:
MARK
That frozen key then predicts:
95 of the remaining 104 characters exactly.
The test can also be run in reverse.
Using positions 21 through 124 to derive the key again produces:
MARK
That independently recovered key then predicts 17 of the first 20 characters, with the three failures corresponding to the already identified anomalies.
Therefore MARK is not dependent upon one particularly favorable section of the cryptogram.
Independent portions of the text recover the same key.
10. A particularly important control: Gaines exercise 147
One of the strongest historical comparisons came from another cryptogram in Gaines’s own book.
Exercise 147 appears in the same chapter on progressing-key systems.
Reconstructing its corrected ciphertext shows that it is a progressing Vigenère cipher using the initial four-letter key:
LOVE
The key advances by one alphabet position after every four characters.
Its plaintext begins:
TO LIVE CONTENT WITH SMALL MEANS
TO SEEK ELEGANCE RATHER THAN LUXURY
AND REFINEMENT RATHER THAN FASHION …
Project Gutenberg records four corrections to the originally printed exercise 147 ciphertext:
UVSYC → UBSYC
IXYDY → IXYPY
VXOOK → NXOOK
NNHMM → NNHWM
The interesting part comes when the original wrong letters are restored.
Three of those four wrong ciphertext characters are exactly what results if the opposite cipher operation is used.
In exercise 147 the correct system is Vigenère:
C = P + K
But at three erroneous positions the printed character is exactly the result of Variant Beaufort:
C = P – K
For example:
Plaintext L, effective key Q:
Correct Vigenère gives B.
Variant gives V.
The original erroneous ciphertext contains V.
Another example:
Plaintext R, effective key W:
Correct Vigenère gives N.
Variant gives V.
The erroneous printed ciphertext again contains V.
And:
Plaintext R, effective key F:
Correct Vigenère gives W.
Variant gives M.
The original erroneous ciphertext contains M.
So in exercise 147:
three of four documented errors are exact Vigenère/Variant sign reversals.
This is unusually relevant to Badalov’s solution.
11. Why exercise 147 matters
Before finding exercise 147, the ten opposite-arithmetic characters in the final cryptogram were one of the biggest reasons for caution.
Why would someone repeatedly switch between Variant Beaufort and Vigenère arithmetic?
Exercise 147 shows that something remarkably similar already occurs in another Gaines progressing-key exercise.
The structural comparison is close:
Exercise 147
initial key: LOVE
key length: 4
progression: +1 every four characters
normal operation: Vigenère
several published errors equal exact Variant results
Final unnumbered cryptogram
proposed initial key: MARK
key length: 4
progression: +1 every four characters
normal operation: Variant Beaufort
ten anomalous letters equal exact Vigenère results
This does not prove that the same person made the same mistake.
The published errata cannot tell us whether exercise 147’s bad characters originated with Gaines, with an encipherer, in a manuscript, or during printing.
Nevertheless, it establishes an extremely relevant internal precedent.
Repeated opposite-arithmetic letters are demonstrably present elsewhere in Gaines’s own progressing-key exercises.
12. Exploratory patterns that are not part of the core claim
During the investigation several additional patterns were discovered.
The printed cryptogram has:
125 = 5 × 5 × 5
characters.
This permits it to be arranged naturally as a 5×5×5 cube.
A substantial concentration of anomaly positions appears on one simple diagonal plane.
There is also a strong period-8 concentration produced by the interaction between the four-letter key and the key progression.
These patterns were statistically unusual in exploratory controls.
However, no documented Gaines mechanism has yet been found that naturally produces them.
For that reason they should not currently be used as evidence for the main solution.
They should simply remain recorded as possible future leads.
Trying to modify the cipher now to force those patterns into a complete explanation would risk post-hoc overfitting.
13. What the independent analysis supports
The following results can be reproduced directly.
1. The proposed key is MARK
The progression is:
MARK → NBSL → OCTM → PDUN → …
This is the kind of progressing keyword Gaines explicitly teaches.
2. The basic operation is Variant Beaufort
In simple form:
C = P – K
This is a cipher operation Gaines discusses directly.
3. Removing one of the doubled Js gives 124 characters
The resulting ciphertext decrypts with MARK to 112 correct plaintext characters out of 124.
4. The deletion point is independently recovered
When every possible single-character deletion is tested, the best result occurs at the doubled-J position.
5. MARK is the global four-letter optimum
Across all 57,122,000 one-deletion/four-letter-key combinations, MARK at the doubled-J location gives the highest score.
6. Randomized controls do not reproduce the result
Among 10,000 shuffled-plaintext controls, the highest score was only 23/124.
MARK scores 112/124.
7. MARK can be recovered independently from different sections
The beginning of the message recovers MARK.
The remainder of the message independently recovers MARK.
8. Ten of the twelve anomalous letters are exact opposite-arithmetic results
They are precisely what Vigenère would produce where Variant Beaufort was expected.
9. The same general error occurs elsewhere in Gaines
Three documented errors in exercise 147 are exact Variant-for-Vigenère substitutions in another four-letter progressing-key cryptogram.
14. What the analysis does not establish
Several questions remain open.
The extra J
The model clearly identifies an extra character at the doubled-J location.
But we do not know whether that J is:
intentional,
a null,
an encipherment mistake,
or a production error.
The twelve anomalous characters
Their mathematical behavior is clear.
Their cause is not.
There is presently no demonstrated rule predicting exactly where the twelve deviations should occur.
Statistical interpretation
The exhaustive search and shuffled controls show that the frozen Badalov hypothesis is highly specific.
They do not provide a formal probability that the solution is correct because the original plaintext and key were discovered from the cryptogram itself.
A secondary cipher mechanism
No convincing historically documented second mechanism has been found that explains all twelve anomalies.
It may be that no such mechanism exists and that the discrepancies are simply errors.
15. Overall assessment
Badalov’s candidate solution requires remarkably few ingredients:
Initial key: MARK
Progression: advance each key letter by one after every four characters
Cipher: Variant Beaufort
Alignment: remove one character at the doubled-J location
With those choices fixed, the direct result contains:
112 correct plaintext letters out of 124
and forms an unmistakably coherent military message.
The remaining twelve characters are not random-looking corruptions.
Ten are exact outputs of the closely related Vigenère operation.
The other two have simple progression-level explanations.
MARK is overwhelmingly better than other four-letter keys, can be recovered independently from different parts of the message, and performs vastly beyond matched randomized controls.
Most importantly, the proposed construction is historically appropriate.
Gaines explicitly teaches both Variant Beaufort and this type of progressing keyword.
And another four-letter progressing-key exercise in Gaines’s own book contains documented erroneous letters that fall exactly on the opposite Vigenère/Variant operation.
Conclusion
The discovery of the MARK solution belongs to Elias Badalov.
The purpose of this investigation was not to rediscover the cipher but to take Badalov’s proposal seriously enough to try to break it.
So far, those attempts have strengthened rather than weakened its central claim.
The most cautious summary is:
Elias Badalov’s candidate solution to Helen Fouché Gaines’s final unnumbered cryptogram is a progressing Variant Beaufort cipher with initial key MARK, advancing the key by one alphabet position after every four characters, with one additional character at the doubled-J location. The direct decipherment yields 112 of 124 plaintext characters exactly. Ten of the twelve remaining discrepancies are exact Vigenère-for-Variant substitutions, while the other two have simple progression-stage explanations. Exhaustive key/deletion searches, randomized controls, held-out key recovery and comparison with Gaines’s own progressing-key exercises provide substantial independent support for the proposal.
I would presently describe this as a:
strong candidate solution with substantial independent validation
rather than as a formally proved solution.
The most useful next step is not further adjustment of the cipher.
It is independent reproduction and attempted falsification by other cryptanalysts.
Acknowledgements
Elias Badalov deserves credit for discovering the MARK key, the progressing-key solution, the proposed plaintext, the doubled-J issue and the principal Vigenère/Variant anomaly pattern.
The subsequent validation work described here was directed by Russell , with most of the calculations, exhaustive searches, cryptanalytic comparisons and historical cross-checking carried out using ChatGPT (“Maxwell”).
The investigation deliberately adopted a freeze-and-test approach: once Badalov’s hypothesis was specified, the key, plaintext and principal cipher mechanism were held fixed wherever possible rather than repeatedly adjusted to improve the result.
Principal source
Helen Fouché Gaines, Elementary Cryptanalysis, American Photographic Publishing Co., 1939.
Digitized by Project Gutenberg as eBook 75074:
https://www.gutenberg.org/ebooks/75074
Russell,
Is that military message an invented, or an historical one that was encrypted when first written, or which was subsequently invented and encrypted to provide an example of that method?
I ask because if the encryption were original it might accurately reflect the original sender’s spoken English, in which case one might suggest that English was not his or her first language.
Elias Badalov’s solution is easy to confirm, just put the modified Ciphertext
with one ‘J’ removed -124 letters- into a Vigenere app, like Rumkin or Cryptool with this Vigenere key :
“MARKN BSLOC TMPDU NQEVO RFWPS GXQTH
YRUIZ SVJAT WKBUX LCVYM DWZNE XAOFY
BPGZC QHADR IBESJ CFTKD GULEH VMFIW
NGJXO HKYPI LZQJM ARKNB SLOCT MPDUN QEVO”
Weirdly, the Vigenere app must be in “Encrypt” mode for this to work.
With only the details to work out, Elias’s solution looks solid.
For a Vigenere setting of Decrypt to work on the modified Gaines ciphertext,
the key must start with ‘OAJQ’ and decrement by 1, like so:
OAJQ
NZIP
MYHO
LXGN
KWFM …
Nice work Elias.
Thanks — these are very useful comments.
D.N. O’Donovan: I don’t know the provenance of the plaintext. As far as I can find, Gaines gives no information about where the final cryptogram came from, so at present I don’t think we can say whether it was a genuine historical message or one constructed as a cryptographic exercise. That is certainly something worth investigating.
xiaozhuan: Yes — and I think your Vigenère-app test is a particularly nice independent confirmation.
The reason Encrypt mode works with the progressing MARK key is that Variant Beaufort decipherment here is effectively:
P = C + K
which is exactly the arithmetic performed by Vigenère encryption.
And your OAJQ observation is the same result from the other direction. OAJQ is the modulo-26 negative of MARK, so Vigenère Decrypt with OAJQ decreasing by one every four characters is mathematically equivalent to Variant Beaufort with MARK increasing by one.
So you’ve reproduced the central MARK mechanism independently using an ordinary Vigenère implementation. That’s exactly the sort of external check I was hoping for.
Thanks for checking it.
So happy that a solution was found to this one. I love seeing how easily something can remain “unsolvable” given a very simplistic encoding scheme. Congrats to Elias (and Astra 😉 )