Experiment 19

“Substitution first, a route last: do the 24 crib images stay one letter to one letter under any route?”

New family, opened by 0018's finding that only substitution followed by transposition reproduces K4's width-21 repeats. The cribs now constrain the ciphertext letters the route sends them to, not the letters at 22-34 and 64-74.

revertedstatus
experiment 18parents
3 mintime, all phases
0candidates
0/24known letters
1

The experiment method

What the method does with the 24 known letters.

Suppose each letter of the messagethe message The 97 letters of English that Jim Sanborn wrote and enciphered. was first swapped for another letter, and only then were the letters shuffled into a new order by a grida grid Writing the message into rows of some width and reading it out in another order, column by column for instance, which shuffles the letters without changing them.. The 24 letters we know are then scattered somewhere in the carved textthe carved letters The 97 letters cut into the copper at the CIA, shown at the top of every experiment page., and each grid tells us where. One rule must hold: one carved letter cannot stand for two different message letters. We check that rule for every grid we have, using the 24 known lettersthe known letters Sanborn has confirmed that positions 22 to 34 of the message read EASTNORTHEAST and positions 64 to 74 read BERLINCLOCK., and count the grids that pass. If none pass, this whole way of building the cipher is out, however the swapping was done.

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 ACCENTwith no route, each known letter sits over its own carved letter; a one-for-one substitution needs every copy over the same oneEoverF22G31Y65E sits over F, G and Y: three letters for one, so no single substitution, unless a route moves themAoverL23K32CoverM70P73LoverV67Z71NoverQ26T69OoverQ27F72RoverP28P66SoverR24S33ToverV25R29S34copies over different carved letterscopies over the same oneOUT: NO READING; NO ROUTE MAKES THE KNOWN LETTERS ONE-TO-ONEEASTNORTHEASTBERLINCLOCK
the test

If the message were first put through a simple substitution, each letter always becoming the same other letter, and then moved by a route, every copy of a known letter would land on the same carved letter, and different letters on different ones.

drawn for no route

Without moving anything, the three Es of the known text sit over three different carved letters. Each route moves them somewhere else; the test asks whether any route lines them up.

the bound

23,931 routes: none passes, for a one-for-one substitution or for one with several cipher letters per plain letter. Shuffled coppers pass none either.

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 route: write in rows of 7 and read down the columns.
  2. The message letter at position 22, the E of EAST, is written at row 4, column 1, and read out 4th: so the carved letter at 4, R, is the swapped E.
  3. Position 31, the other E, lands at 32: the carved K.
  4. Under a plain swap E must always become the same letter, so R and K already conflict: this route is dead for a plain swap.
  5. Under a swap with spare letters, E may have two symbols, so R and K is allowed, but R must then never be the image of another letter.
  6. Position 66, the R of BERLIN, lands at 38: the carved T; no conflict.
  7. Go through all 24; one clash kills the route.
the worksheet: the first steps by hand, on squared paper
THE KNOWN LETTERS GROUPED: WHERE EACH COPY SITS, AND THE CARVED LETTER UNDER ITletterplacescarvedA23 32L KB64NC70 73M PE22 31 65F G YH30NI68TK74KletterplacescarvedL67 71V ZN26 69Q TO27 72Q FR28 66P PS24 33R ST25 29 34V R SA tick means every copy of the letter sitsover the same carved letter. A route thatworked would turn every cross into a tick andkeep the carved letters of different rowsapart.
4

Worked example, generated

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

route: columnar width 7 read down the columns; images of the known letters

plain A: carved images F Z
plain B: carved images H
plain C: carved images K Q
plain E: carved images R S
plain H: carved images B
plain I: carved images P
plain K: carved images K
plain L: carved images J U
plain N: carved images I X
plain O: carved images K V
plain R: carved images T U
plain S: carved images K N
plain T: carved images K U W

homophonic condition: fails; clashes: K from S and T, K from T and O, U from R and T, K from O and C, U from T and L, K from C and K
monoalphabetic condition: fails
5

Hypothesis, for a cryptanalyst

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

Zodiac 340 was a homophonic substitution followed by a transposition, and its width-19 repeats were the trace (knowledge/04-structure.md). K4's eleven width-21 repeats have p about 0.005 under every substitution-last generator and p near 1 under transposition of English (0018). Against: Sanborn's secondhand 'one-to-one mapping' of NYPVTT to BERLIN (knowledge/02-sanborn-said.md, 2011 and 2013 dinners), which says the letters at 64-69 are BERLIN's own images; this family contradicts that reading and is tested because the statistic is the stronger evidence. Model: each plaintext letter goes to one of its symbols (monoalphabetic, or homophonic over the same 26 symbols), then a route T permutes positions. The image of crib position i is ct[T(i)]. Constraint: a ciphertext letter has one plaintext source, so two crib positions with different plaintext letters must have different images; under the stricter monoalphabetic reading, equal plaintext letters must also have equal images. Routes: the 23,930 of 0016 plus the identity. For each route both conditions are checked; survivors are counted against the same test on shuffled K4. A monoalphabetic survivor gets a hill-climb over the unknown letters and a candidate. Bounds: complete for this route set.

6

Prediction

Written before the run; the result decides it.

Under the homophonic condition, fewer than one route in a hundred survives, about the same rate as on shuffled K4. Under the monoalphabetic condition no route survives. No candidate passes.

7

Method

  1. Build the route set.
  2. For each route, find the ciphertext image of each crib position.
  3. Homophonic condition: no ciphertext letter is the image of two different plaintext letters.
  4. Monoalphabetic condition: in addition, equal plaintext letters have equal images.
  5. Count survivors of each; repeat on five shuffles of K4.
  6. For monoalphabetic survivors, hill-climb the unknown mappings by English score and submit the best text.
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

23931 routes: 0 pass the homophonic condition, 0 the monoalphabetic one; shuffled K4 gives [0, 0, 0, 0, 0] and [0, 0, 0, 0, 0].

What was searched, as the harness recorded it.

family
substitution (mono or homophonic over 26 symbols) then transposition
routes
23,931
homophonic survivors
0
homophonic survivor routes
none
monoalphabetic survivors
0
monoalphabetic survivor routes
none
shuffled k4 survivors
  1. homophonic
    0
    monoalphabetic
    0
  2. homophonic
    0
    monoalphabetic
    0
  3. homophonic
    0
    monoalphabetic
    0
  4. homophonic
    0
    monoalphabetic
    0
  5. homophonic
    0
    monoalphabetic
    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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