“Trifid with a free cube: the 24 known letters as equations on the cube's coordinates, solved exactly”
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.
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.
the idea
Instead of trying cubes, solve for one: each known letter, with its carved letter, ties three digits to a cell of the cube, and the cube must satisfy all 24 ties at once.
what happened
The search ran but the code failed on its last line, so the record holds no result. Experiment 40 is the same design, fixed and rerun.
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.
Period 5, blocks from position 1: the block 21-25 holds the known letters E A S T at 22-25.
Carved letters of that block: J Q S S E, each with a three-digit address.
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.
Write each such tie as 'this digit equals that digit'.
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.
Give each chain a value 0, 1 or 2 so that no two letters end with the same address.
Decrypt and read.
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.
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
For each period and alignment, list the blocks and, for every known plaintext letter in a block, the three coordinate equalities it implies.
Merge equalities with union-find over the 81 coordinate variables.
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.
Decrypt the text under each surviving cube, score the free letters as English, keep the best.
Repeat on two shuffles.
8
Result and bounds
The harness's record: candidates, gates, loss, and the negative result this experiment leaves behind.