Experiment 13

“Gromark and its cousins: must the 13 known shifts in a row obey a lagged-Fibonacci rule?”

New root in family 5. The key is a running number stream from a short primer, as in the ACA Gromark; the cribs give 13 consecutive key values, which a recurrence must reproduce.

revertedstatus
a new rootparents
89 stime, all phases
0candidates
0/24known letters
1

The experiment method

What the method does with the 24 known letters.

GromarkGromark A cipher whose secret is a short number that grows into a long stream of digits by adding pairs of earlier digits; each message letter is pushed along the alphabet by the next digit. is a cipher where the secret is a short number, say five digits, that grows into a long stream of digits by addingVigenere The three rules of arithmetic a table can use: add the key letter to the message letter, subtract the message letter from the key letter, or subtract the key letter from the message letter. pairs of earlier digits, like a Fibonacci sequence. Each letter of the messagethe message The 97 letters of English that Jim Sanborn wrote and enciphered. is pushed along the alphabet by the next digit. The 13 known letters of EASTNORTHEAST tell us 13 pushes in a row, and in a Fibonacci-style stream each push must equal the sum of two earlier ones. We test every pair of positions the rule might add, in base 10a number base How many different digits a counting system uses before it carries: ten for ordinary numbers, sixteen for hexadecimal, twenty for the Maya. and in base 26, and with every alphabet the puzzle's wordsthe puzzle's words A list of 62 words tied to the sculpture: the earlier secret words, place names, the names of the people involved, the words in the Morse code, and the themes Sanborn has named. can build. A rule that fits the known pushes would be found out at once.

Words in the margin are defined on the glossary page.

1.1

The method at work

The carved letters go in at the top; the reading comes out at the bottom.

OBKRUOXOGHULBSOLIFBBWEASTNORTHEASTOTWTQSJQSSEKZZWATJKLUDIAWINFBBERLINCLOCKWGDKZXTJCDIGKUHUAUEKCARIN: THE COPPER, ALL 97 LETTERS, THE 24 KNOWN PLACES IN ACCENTknownEASTNORTHEASTcarvedFLRVQQPRNGKSSshift817937612812901413shift 5 back + 4 back250121013182020outlined: not a single digit; shift 2, 17, already fails base 10base 26: 8 of 8 checks failOUT: NO READING; NO RULE SURVIVESEASTNORTHEASTBERLINCLOCK
the shifts

Under EASTNORTHEAST each known letter and its carved letter give a shift: plain alphabet for the message, the sculpture's for the carving. Thirteen shifts in a row.

the rule

A Gromark key builds itself: each shift is the sum of two earlier ones. Drawn for the rule 5 back plus 4 back. In base 10 every shift must be a single digit; in base 26 each sum must match.

the bound

All 66 rules with lags up to 12, bases 10 and 26, over 184,512 alphabet choices and three ways of combining: 0 survivors with six or more checks.

2

Best candidate against the known letters

This experiment produced no candidate: the search ruled everything out before a full text was built. Its product is the bounded negative in section 7.

3

Walkthrough, by hand

Redo the method on a few letters with pencil and paper.

  1. Take the standard alphabet for the plaintext and the KRYPTOS alphabet for the ciphertext, Vigenere.
  2. Position 22: plaintext E is index 4 in the standard alphabet; ciphertext F is index 11 in KRYPTOS; the shift is 11 - 4 = 7.
  3. Position 23: A is 0, L is 18; shift 18.
  4. Continue to position 34 for 13 shifts in a row.
  5. Try the Gromark rule, a = 5, b = 4: the sixth shift must equal the first plus the second, mod 10, and all shifts must be digits. The shift 18 already fails base 10.
  6. In base 26, check the sixth against the first plus the second mod 26, the seventh against the second plus the third, and so on: eight checks.
  7. One failure kills the rule under this alphabet pair; try the next pair.
the worksheet: the first steps by hand, on squared paper
THE SHIFTS UNDER EASTNORTHEAST, THEN THE GROMARK CHECK: EACH SHIFT THE SUM OF THE SHIFTS 5 AND 4 BACKplaceknownplain no.carvedcarved no.shift22E4F12823A0L171724S18R1925T19V22326N13Q20727O14Q20628R17P31229T19R1830H7N191231E4G13932A0K0033S18S61434T19S613shift no.5 back + 4 backactual68 + 17 = 256717 + 9 = 01289 + 3 = 12893 + 7 = 1012107 + 6 = 139116 + 12 = 1801212 + 8 = 2014138 + 12 = 2013Shift 2 is 17, not a digit, so base 10fails before any check. In base 26 all 8checks fail.
4

Worked example, generated

The same steps run by the code on the real letters.

plain alphabet standard, cipher alphabet KRYPTOS, Vigenere: forced shifts at 22-34

pos  plain  cipher  shift
 22    E      F       8
 23    A      L      17
 24    S      R       9
 25    T      V       3
 26    N      Q       7
 27    O      Q       6
 28    R      P      12
 29    T      R       8
 30    H      N      12
 31    E      G       9
 32    A      K       0
 33    S      S      14
 34    T      S      13

shifts: 8 17 9 3 7 6 12 8 12 9 0 14 13
Gromark rule a=5 b=4 base 10: all digits? False; base 26 check at i=5: 6 vs 25
5

Hypothesis, for a cryptanalyst

Family, key space, the constraint under test, and the clues it reads.

Gromark (Gillogly 1999, 2004; Bean, HistoCrypt 2021, knowledge/05-attacks.md): a primer of a digits expands by k[i] = k[i-a] + k[i-b] mod B, and each plaintext letter shifts by k[i] between a plain and a cipher alphabet. Bean exhausted base 10 with a = 5, b = 4 under free annealed alphabets. Here the rule is the variable: every lag pair 1 <= b < a <= 12 and bases 10 and 26, under plain and cipher alphabets drawn from the standard, KRYPTOS and the four Friedman mixings of every vocabulary word, with the three tableau modes. Positions 22-34 force 13 consecutive shifts and 64-74 force 11; the recurrence is checked on both runs, giving 24 - 2a checks per rule. In base 10 every forced shift must also be a digit. A survivor with at least six passing checks has chance probability below 26 to the minus six per trial; with about 700,000 trials per rule, chance survivors should be rare. For each survivor the recurrence runs backward and forward from the 22-34 window to a full key stream, the text is decrypted and submitted. Bounds: complete for the stated rules, bases and alphabet pools; free alphabets outside the vocabulary remain Bean's.

6

Prediction

Written before the run; the result decides it.

In base 10 almost no alphabet pair makes all 24 forced shifts digits, and none of those also satisfies a recurrence. In base 26, a handful of (rule, alphabet) survivors appear at the long lags where only a few checks apply, none with six or more checks. No decrypted candidate is English; zero pass.

7

Method

  1. Build the alphabet pool: standard, KRYPTOS and the four mixings of each vocabulary word.
  2. For each plain alphabet, cipher alphabet and mode, compute the forced shift at each of the 24 crib positions.
  3. For base 10, drop any triple with a shift above 9.
  4. For each lag pair (a, b) check k[i] = k[i-a] + k[i-b] mod B on both crib runs; keep triples with every check passing and at least six checks.
  5. For each survivor, extend the stream over all 97 positions from the 22-34 window, confirm it reproduces 64-74, decrypt and submit.
8

Result and bounds

The harness's record: candidates, gates, loss, and the negative result this experiment leaves behind.

0candidates returned
0passed every gate
0/24known letters, best
–lowest loss
nonechampion before

184512 alphabet triples x 66 rules x 2 bases. 0 survivors with >= 6 checks; 0 full-stream candidates submitted.

What was searched, as the harness recorded it.

family
running numerical key, k[i] = k[i-a] + k[i-b] mod base
rules
66 lag pairs, 1 <= b < a <= 12
bases
10, 26
alphabets
248
alphabet triples
184,512
modes
3
base10 all digit triples
0
min checks
6
survivors
none
n survivors
0
n survivors consistent with second run
0

How this experiment sits among the others is on the lineage page; its family's open hypotheses are on the families page.

Rendered from the record on 2026-10-18 12:26 UTC; updated after every experiment.
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