Experiment 28

“Gromark in bases 2 to 9: the even bases that reproduce the width-21 repeats”

Bases 2 to 9 instead of 10 and 26. 0027 showed Gromark base 2 is the only simulated construction that reproduces K4's width-21 repeats, and Bean found base-8 primers that fit the cribs. Every forced shift must now be below the base, a filter of (base/26) to the 24th power per alphabet triple.

revertedstatus
experiment 13parents
7 stime, all phases
0candidates
0/24known letters
unclocked
1

The experiment method

What the method does with the 24 known letters.

Yesterday's workshop found one kind of fake that shares a real oddity of K4the carved letters The 97 letters cut into the copper at the CIA, shown at the top of every experiment page.: a 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. cipher whose growing number stream uses only the digits 0 and 1. Such streams repeat their rhythm every 21 letters, which is just what K4 shows. A Gromark in base 2 pushes each letter along by 0 or 1 only; in base 8, by 0 to 7. We already know the 24 pushes that the known lettersthe known letters Sanborn has confirmed that positions 22 to 34 of the message read EASTNORTHEAST and positions 64 to 74 read BERLINCLOCK. demand under each alphabet, so we simply check whether any alphabet keeps all 24 pushes small enough, and then whether they follow the 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. rule.

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 ACCENTthe shift each known letter needs on the sculpture's table, one bar per known place1E10A21S18T1N15O2R23T5H2E19A0S2T11B17E2R5L15I11N9C8L7O20C0Kbase 10: digits 0 to 9; shiftsabove this line are impossiblebest over 184,512 alphabet triples: some shift of at least 15, so bases up to 9 are deadOUT: NO READING; NO SMALL-BASE RUNNING NUMBER FITSEASTNORTHEASTBERLINCLOCK
Gromark

A Gromark key is a running number: each digit is the sum of two earlier digits, in some base. Each digit shifts one letter along the alphabet, so a base-b key can only shift by 0 to b-1.

the bars

Drawn for the sculpture's table: many known letters need a shift of 10 or more. On some table built from the puzzle's words the shifts might all be small.

the bound

Over 184,512 combinations of alphabets the smallest possible largest shift is 15. Bases 2 to 9 are dead before any rule is tried; base 10 and up were tested in experiments 46 and later.

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 BY HAND ON THE SCULPTURE'S TABLE: POSITION OF CARVED LETTER MINUS POSITION OF KNOWN LETTERplaceknownpositioncarvedpositionshift22E11F12123A7L171024S6R12125T4V221826N19Q20127O5Q201528R1P3229T4R123Positions are in the sculpture's alphabetKRYPTOSABCDEFGHIJLMNQUVWXZ, counted from0. The shift is the carved position minusthe known one, wrapping at 26.A tick means a single digit could makethat shift. One cross is enough to ruleout a base-10 key on this table.
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
largest shift 17: base 8 needs all below 8, so this alphabet pair is out at once
5

Hypothesis, for a cryptanalyst

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

Rogot, Deavours and Blackman: a Gromark key in an even base with a five-digit primer has an underlying period-21 structure, and Bean found base-8 primers that fit the cribs with period 84 (knowledge/04-structure.md, 05-attacks.md). 0027 confirmed that base 2 reproduces the width-21 repeats (p 0.34) where nothing else does. 0013 tested bases 10 and 26 only. Here the bases are 2 to 9, every lag pair 1 <= b < a <= 12, under the same 248-alphabet pool for plain and cipher and three modes. In base B every one of the 24 forced shifts must be below B, which alone has chance (B/26)^24 per triple, from 10^-27 at base 2 to 10^-11 at base 9; a triple passing that and six recurrence checks would be decisive. For each survivor the recurrence runs from the 22-34 window to a full key stream, the text is decrypted and submitted. Bounds: complete for these bases, rules and alphabets; Bean's free annealed alphabets remain his.

6

Prediction

Written before the run; the result decides it.

No alphabet triple keeps all 24 forced shifts below 9, let alone below 2, so no base from 2 to 9 reaches the recurrence test. Zero candidates pass.

7

Method

  1. Build the alphabet pool.
  2. For each plain alphabet, cipher alphabet and mode, compute the forced shift at each of the 24 crib positions and record the largest.
  3. For each base B from 2 to 9, keep triples whose largest shift is below B.
  4. For those, check every lag pair's recurrence on both crib runs.
  5. Extend survivors to a full stream, 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; smallest largest-shift over all triples 15; bases 2-9 x 66 rules. 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
2, 3, 4, 5, 6, 7, 8, 9
smallest largest shift over all triples
15
alphabets
248
alphabet triples
184,512
modes
3
triples with all shifts below 9
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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