“Gromark with any mixed alphabet, in number bases other than ten”
Experiment 69 closed base-10 Gromark for any alphabet. Scheidt mentioned base 16 and Mayan base 20 and 60, and the Berlin clock counts in fives. Here the running digits are in base 2 to 16 (except 10) and base 26, with every two-term rule up to five digits back and every primer, and the alphabet is again left completely free.
Experiment 69 showed that 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 key is a growing stream of digits plus a scrambled alphabet, cannot have made the copper if the digits run from 0 to 9. Scheidt has talked about other number systemsa number base How many different digits a counting system uses before it carries: ten for ordinary numbers, sixteen for hexadecimal, twenty for the Maya., base 16 and the Maya's base 20, and the Berlin clock counts in fives. So we run the same exact test with digits in every base from 2 to 16 and in base 26, again with any scrambled alphabet at all.
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.
the idea
Experiment 69 closed Gromark in base 10 with a free alphabet. This runs the same exact test in every other base up to 16, and in base 26.
the check
In base 2 with primer 10110, places 26 and 27 both carve Q but land in slots 13 and 15: dead whatever the alphabet.
the bound
15,135,782 primers across 146 base-and-rule settings, four forms: none consistent on K4 or on any shuffled copper. Two-term Gromark is exhausted.
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.
Base 2, each digit the sum of the digits five and four back, mod 2, primer 1 0 1 1 0.
Places 26 and 27 both carve Q, from known N (13) and O (14).
With the form M[P + d] = c: 13 + 0 = 13 and 14 + 1 = 15, two different slots, which cannot both hold Q.
This primer is dead.
Repeat for every primer, rule, base and form.
the worksheet: the first steps by hand, on squared paper
4
Worked example, generated
The same steps run by the code on the real letters.
base 2, digits 5 and 4 back, primer 10110
stream 1011011011011011011011011011011011
place known carved digit slot (M[P+d]=c)
26 N Q 0 13
27 O Q 1 15
22 E F 1 5
23 A L 0 0
5
Hypothesis, for a cryptanalyst
Family, key space, the constraint under test, and the clues it reads.
Family 5, free-alphabet Gromark beyond base 10 (C4(b): Scheidt's remarks on 'base 16' and Mayan bases, 03-scheidt-said.md; C12(b): the Mengenlehreuhr counts in fives and ones, sources/nyt-2010-berlin-clue.txt). Key digits d[i] = d[i-a] + d[i-b] mod B, 1 <= a < b <= 5 (10 rules), every primer of b digits, bases B = 2-9 and 11-16 (and 26 for rules with b <= 4); the digit is added to the letter index mod 26 as in Gromark. Four forms as in 0069 (M[P+d]=c, M[P-d]=c, M[C-d]=p, M[C+d]=p; M an unknown mixed alphabet). Same-letter crib pairs fix the digit difference to one or two values (d_k - d_l = need or need - 26 within -(B-1)..B-1), which filters almost every primer; the rest get the full bijection check. Exhaustive for these bases and rules. K4 and ten shuffled coppers.
6
Prediction
Written before the run; the result decides it.
No primer under any base, rule and form gives a consistent alphabet on K4; the ten shuffled coppers give none either, except possibly in base 26 where the digit differences are least constrained, and there no more often on K4 than on the shuffles. Zero candidates pass.
7
Method
For each base and rule, generate the digit stream from every primer and keep the digits at the 24 known places.
For each form, list same-letter pairs and the digit differences they allow.
Keep primers that meet every pair, then check the full one-to-one condition.
Repeat on ten shuffled coppers.
8
Result and bounds
The harness's record: candidates, gates, loss, and the negative result this experiment leaves behind.
146 of 146 base-rule settings, 15,135,782 primers x 4 forms. Consistent primers by base, K4 (controls): base 2: 0 (0-0); base 3: 0 (0-0); base 4: 0 (0-0); base 5: 0 (0-0); base 6: 0 (0-0); base 7: 0 (0-0); base 8: 0 (0-0); base 9: 0 (0-0); base 11: 0 (0-0); base 12: 0 (0-0); base 13: 0 (0-0); base 14: 0 (0-0); base 15: 0 (0-0); base 16: 0 (0-0); base 26: 0 (0-0).
What was searched, as the harness recorded it.
family
Gromark, free mixed alphabet, bases 2-9, 11-16 and 26
rules
two-term, lags up to 5 (base 26: up to 4)
settings done
146
settings planned
146
primers
15,135,782
forms
M[P+d]=c, M[P-d]=c, M[C-d]=p, M[C+d]=p
by base
2
k4
0
controls
0, 0, 0, 0, 0, 0, 0, 0, 0, 0
primers
196
3
k4
0
controls
0, 0, 0, 0, 0, 0, 0, 0, 0, 0
primers
1,278
4
k4
0
controls
0, 0, 0, 0, 0, 0, 0, 0, 0, 0
primers
5,008
5
k4
0
controls
0, 0, 0, 0, 0, 0, 0, 0, 0, 0
primers
14,650
6
k4
0
controls
0, 0, 0, 0, 0, 0, 0, 0, 0, 0
primers
35,460
7
k4
0
controls
0, 0, 0, 0, 0, 0, 0, 0, 0, 0
primers
75,166
8
k4
0
controls
0, 0, 0, 0, 0, 0, 0, 0, 0, 0
primers
144,448
9
k4
0
controls
0, 0, 0, 0, 0, 0, 0, 0, 0, 0
primers
257,418
11
k4
0
controls
0, 0, 0, 0, 0, 0, 0, 0, 0, 0
primers
690,910
12
k4
0
controls
0, 0, 0, 0, 0, 0, 0, 0, 0, 0
primers
1,061,136
13
k4
0
controls
0, 0, 0, 0, 0, 0, 0, 0, 0, 0
primers
1,575,418
14
k4
0
controls
0, 0, 0, 0, 0, 0, 0, 0, 0, 0
primers
2,272,228
15
k4
0
controls
0, 0, 0, 0, 0, 0, 0, 0, 0, 0
primers
3,196,350
16
k4
0
controls
0, 0, 0, 0, 0, 0, 0, 0, 0, 0
primers
4,399,360
26
k4
0
controls
0, 0, 0, 0, 0, 0, 0, 0, 0, 0
primers
1,406,756
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.