“Porta: the reciprocal half-alphabet tableau, which the 24 conventions never covered”
A different tableau. Porta splits the alphabet into two halves and always sends a letter to the other half; the key letter chooses one of 13 reciprocal alphabets. Tested with the alphabet in standard, KRYPTOS and every vocabulary-mixed order.
PortaPorta An old cipher that cuts the alphabet into two halves and always swaps a letter into the other half, with the secret word choosing one of thirteen swap patterns. is an old cipher that cuts the alphabet in two halves and always swaps a letter into the other half: a letter from the first half always lands in the second, and back. That gives a quick check: every known letter must have been swapped across the halves. With the ordinary alphabetthe plain alphabet A to Z in the usual order., E and its carved partner F are both in the first half, so Porta fails at the very first known letter. But if the alphabet was first reordered by a secret word, the halves change, so we try every reordering the puzzle's wordsthe puzzle's words A list of 62 words tied to the sculpture: the earlier secret words, place names, the names of the people involved, the words in the Morse code, and the themes Sanborn has named. allow.
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.
Porta
Porta is a table cipher in which every letter of the first half of the alphabet becomes one of the second half, and the other way round, whatever the key.
the test
So each known letter and the letter carved over it must sit in opposite halves. Tried for 248 letter orders; the best still has 7 places where they share a half.
the verdict
One shared half is enough to rule an order out, so Porta is dead before any key is tried.
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.
Write the alphabet in two rows: A to M above, N to Z below.
Position 22 is plaintext E and carved F; both are in the top row.
In Porta a top-row letter always becomes a bottom-row letter, so this cannot be Porta on the plain alphabet.
Now reorder the alphabet by a word, say PALIMPSEST: PALIMSETBCDFG above, HJKNOQRUVWXYZ below.
E and F are again both above: dead.
Try every reordering the puzzle's words give; a reordering that separates all 24 pairs would then get its key letters worked out one position at a time.
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.
halves test on the standard order: A-M above, N-Z below
pos plain half carved half swapped across?
22 E top F top NO
23 A top L top NO
24 S bottom R bottom NO
25 T bottom V bottom NO
26 N bottom Q bottom NO
64 B top N bottom yes
65 E top Y bottom yes
66 R bottom P bottom NO
5
Hypothesis, for a cryptanalyst
Family, key space, the constraint under test, and the clues it reads.
Porta (1563; in Friedman and Hitt, so in what Scheidt taught, C5) is reciprocal: with the alphabet A split into halves H1 and H2, key letter k selects j = index(k) div 2, and p in H1 at index i goes to H2[(i + j) mod 13], p in H2 at index i goes to H1[(i - j) mod 13]. Our 24 conventions are Vigenere, Beaufort and variant Beaufort over full alphabets and do not contain it. Two tests. First the halves: every crib plaintext letter must sit in the opposite half from its ciphertext letter; under the standard order E and F are both in A-M, so standard Porta dies at position 22 at once, and each of the 248 vocabulary-mixed orders passes with chance 2 to the minus 24. Second, for any order that passes, the forced j at each crib position gives a period sweep 1-52 by the divisor rule, with the forced key fragment matched against the vocabulary. Both the ACA direction and its mirror are tried. Bounds: complete for Porta over this alphabet pool.
6
Prediction
Written before the run; the result decides it.
No alphabet order passes the halves test for all 24 cribs; the best passes about 18 of 24. So no period sweep runs and zero candidates pass.
7
Method
Build the alphabet pool.
For each order, split into halves and count crib positions whose plaintext and ciphertext letters are in the same half.
For any order with zero such positions, derive the forced alphabet number j at each crib and sweep periods 1-52 by the divisor rule in both directions.
Match forced fragments against the vocabulary; decrypt and submit matches.
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
248 alphabet orders; 0 pass the halves test (best has 7 same-half positions); 0 (order, direction, period) survivors; 0 matches.
What was searched, as the harness recorded it.
family
Porta over standard, KRYPTOS and vocabulary-mixed alphabet orders
alphabets
248
halves test
crib positions whose plaintext and ciphertext share a half; zero required
best orders
same half positions
7
alphabet
CARTER/alt_vertical
same half positions
7
alphabet
CARTER/columnar
same half positions
7
alphabet
CLOCK/alt_vertical
same half positions
7
alphabet
HILL/alt_vertical
same half positions
8
alphabet
SECRET/keyed_columnar
same half positions
8
alphabet
WORLDCLOCK/alt_vertical
same half positions
8
alphabet
WORLDCLOCK/columnar
same half positions
8
alphabet
WORLDCLOCK/keyed_columnar
same half positions
9
alphabet
ABSCISSAE/keyed_columnar
same half positions
9
alphabet
CARNARVON/keyword
orders passing halves
none
period survivors
none
n period survivors
0
vocab matches
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.