Experiment 40

“Trifid with a free cube: the 24 known letters as equations on the cube's coordinates, solved exactly”

Rerun of experiment 39, whose summary line crashed when no cube survived, so its record carries an error instead of a result; the solver and bounds are unchanged. Experiment 33 searched free cubes by trial and error and plateaued at 17 of 24. Here the cube is not searched but solved: every known letter ties three cube coordinates to three carved letters' coordinates, the ties are merged, and the remaining freedom is enumerated under the rule that all 27 symbols have different addresses.

revertedstatus
experiment 39parents
32 stime, all phases
0candidates
0/24known letters
1

The experiment method

What the method does with the 24 known letters.

In trifidtrifid A cipher that turns each letter into three small digits, like floor, row and seat in a 27-seat cube, writes several letters' digits in rows, and regroups them, so each new letter is made from pieces of three old ones. every letter has a three-digit address in a 27-seat cube, and the cipher shuffles the digits of a few letters at a time. Experiment 33 tried to find the cube by guessing and nudging, and could not. This time we do algebra instead: each known letter says 'this carved letter's first digit equals that message letter's second digit', and so on. Chaining all those equalities leaves only a little freedom, and the rule that no two letters share an address closes most of it. Every cube that survives is tried, and the one whose message reads most like Englishthe English score A number saying how much a stretch of letters looks like English, computed from how common each run of four letters is in English text. wins, if any does.

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 ACCENTevery period from 2 to 15 and every alignment: does any cube agree with all 24 known letters?period 2period 3period 4period 5period 6period 7period 8period 9period 10period 11period 12period 13period 14period 15one alignment: no cube fitsthe unknown is the whole cube: each known letter ties its three digits to a celllayer 1KRYPTOSABlayer 2CDEFGHIJLlayer 3MNQUVWXZ?each letter becomes three digits: layer, row, column (EAST shown)E2 1 3A1 3 2S1 3 1T1 2 2119 settings, 0 cubes: trifid is dead for any cubeOUT: NO READING; NO CUBE IS CONSISTENT WITH THE KNOWN LETTERSEASTNORTHEASTBERLINCLOCK
the solve

For each period and alignment the known letters become equations between cube coordinates; a search fills the cube cell by cell and backs up at every contradiction.

the result

119 settings: not one admits a cube, and none hit the search's limits, so the answer is exact. Shuffled coppers admit none either.

the verdict

Trifid is dead at periods 2 to 15 for any cube, not just for cubes built from words.

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. Period 5, blocks from position 1: the block 21-25 holds the known letters E A S T at 22-25.
  2. Carved letters of that block: J Q S S E, each with a three-digit address.
  3. The rule says: the first digit of J is the first digit of the block's first letter, the second digit of J is the first digit of its second letter E, the third digit of J the first digit of A, and so on round the block.
  4. Write each such tie as 'this digit equals that digit'.
  5. Chain the ties: if J's second digit equals E's first, and E's first equals K's third, then J's second equals K's third.
  6. Give each chain a value 0, 1 or 2 so that no two letters end with the same address.
  7. Decrypt and read.
the worksheet: the first steps by hand, on squared paper
SOLVING FOR THE CUBE: EACH KNOWN LETTER FIXES THE CELL ITS DIGITS POINT TOKRYPTOSABlayer 1CDEFGHIJLlayer 2MNQUVWXZ?layer 3E213A132S131T122N312O123R112T122layerrowcolumnWrite the three digits of each letter in acolumn under it, as here. For a block ofletters, read the digits across row by row,regroup them in threes and look each three upin the cube.Working backwards from the carved letters,every known letter fixes which cube cell itsdigits point to; a cube must agree with all ofthem at once.
4

Worked example, generated

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

period 5, blocks from position 1: the block 21-25 carries E A S T at 22-25 against the carved WFLRV

81 coordinate variables merge into 21 classes under all the ties from the 24 known letters
backtracking over the classes with the all-different rule visited 4 nodes and found 0 cube(s)
5

Hypothesis, for a cryptanalyst

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

Trifid over 27 symbols (A-Z and '?') with period p: the three coordinate rows of a block are concatenated and regrouped, so the k-th coordinate of ciphertext letter j equals coordinate m div p of plaintext letter m mod p, with m = 3j + k. Each known plaintext letter therefore gives three equalities between coordinate variables of symbols, 81 variables in all; a union-find merges them into classes, and the bijection, 27 distinct triples with each axis value used exactly nine times, is enforced by backtracking over class values with counting prunes. Every surviving cube decrypts the text with all 24 known letters right by construction, so the English score of the other 73 letters decides, against the same procedure on shuffled K4. Periods 2-15, every alignment. A planted trifid text must be recovered exactly. Bounds: complete for trifid with any cube at periods 2-15, within the node cap of the search.

6

Prediction

Written before the run; the result decides it.

At every period and alignment the equalities either contradict the bijection or leave a handful of cubes, none of whose decryptions read better than -5.5 per quadgram; shuffled K4 behaves the same. Zero candidates pass.

7

Method

  1. For each period and alignment, list the blocks and, for every known plaintext letter in a block, the three coordinate equalities it implies.
  2. Merge equalities with union-find over the 81 coordinate variables.
  3. Backtrack over the classes' values in {0,1,2}, pruning when two symbols would share an address or an axis value would exceed nine symbols.
  4. Decrypt the text under each surviving cube, score the free letters as English, keep the best.
  5. Repeat on two shuffles.
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
34champion before

119 settings; 0 admit at least one cube; 0 hit a cap. No cube is consistent with the known letters at any setting; shuffled K4 [None, None]. 0 candidates.

What was searched, as the harness recorded it.

family
trifid, free 27-symbol cube, periods 2-15, every alignment, solved by constraint propagation and backtracking
settings
119
settings with cubes
0
settings capped
0
node cap
3,000,000
solution cap
5,000
per setting
  1. period
    2
    offset
    0
    classes
    20
    nodes
    1
    cubes
    0
    capped
    no
    best english
    None
    best hits
    None
  2. period
    2
    offset
    1
    classes
    21
    nodes
    4
    cubes
    0
    capped
    no
    best english
    None
    best hits
    None
  3. period
    3
    offset
    0
    classes
    18
    nodes
    1
    cubes
    0
    capped
    no
    best english
    None
    best hits
    None
  4. period
    3
    offset
    1
    classes
    21
    nodes
    4
    cubes
    0
    capped
    no
    best english
    None
    best hits
    None
  5. period
    3
    offset
    2
    classes
    21
    nodes
    1
    cubes
    0
    capped
    no
    best english
    None
    best hits
    None
  6. period
    4
    offset
    0
    classes
    18
    nodes
    1
    cubes
    0
    capped
    no
    best english
    None
    best hits
    None
  7. period
    4
    offset
    1
    classes
    22
    nodes
    1
    cubes
    0
    capped
    no
    best english
    None
    best hits
    None
  8. period
    4
    offset
    2
    classes
    20
    nodes
    1
    cubes
    0
    capped
    no
    best english
    None
    best hits
    None
  9. period
    4
    offset
    3
    classes
    20
    nodes
    1
    cubes
    0
    capped
    no
    best english
    None
    best hits
    None
  10. period
    5
    offset
    0
    classes
    21
    nodes
    4
    cubes
    0
    capped
    no
    best english
    None
    best hits
    None
  11. period
    5
    offset
    1
    classes
    23
    nodes
    1
    cubes
    0
    capped
    no
    best english
    None
    best hits
    None
  12. period
    5
    offset
    2
    classes
    19
    nodes
    1
    cubes
    0
    capped
    no
    best english
    None
    best hits
    None
and 107 more
best
none
shuffled control best
None, None

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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