“F14.5: 0244's two dictionary keyword alphabets and key word, with one carved letter lost (Scheidt's '98 characters'): the 23 distinct places a lost letter changes the key's alignment, every alphabet pair, key lengths 1-26; K4 against 20 shuffles per variant, a planted lost-letter message per variant”
F14.5 (issue #93). Scheidt, 2005: 'I think there are 98 characters left' (C18(b)); A4, F1.6 and 0086 did this for earlier key sources. The enciphering ran over 98 letters and one carved letter was lost: the key index of a carved place at or after the loss is one more than its place. The known words stay where Sanborn put them on the copper. With a free key only the alignment between known places matters, so a loss before 22 or after 74 (and a '?' counted before OBKR) changes nothing that 0244 did not test; a loss at 23-34 or 65-74 splits a known run, and any loss at 35-64 moves BERLINCLOCK's key slots by one against EASTNORTHEAST's: 23 distinct variants. 0244's exact index search and key-word check run unchanged on each, with the key index shifted.
One of the creators once said there were 98 characters, not 97. If Sanborn dropped one letter when cutting the copper, as he did in an earlier section, then from that letter on the secret key would be off by one step, and every exact test so far would have missed it. We rerun the two-scrambled-alphabets test for every place where a lost letter would change anything.
1.1
The method at work
The carved letters go in at the top; the reading comes out at the bottom.
the idea
F14.5: Scheidt's '98 characters': if one carved letter was lost, the key slips by one from that place on.
the check
0244's exact search on the 23 distinct places a lost letter changes anything, with K4's shuffles and a planted message per place.
the bound
No key word on K4 or any shuffle in any variant; plants 23 of 23. A '?' counted before OBKR changes nothing 0244 did not test.
2
Experiment lineage
What this experiment was built on, from the lineage page: 8 experiments before it, back to the root of its line. Its number is in bold.
keptreverted, in its family's colour×error or timeoutreverted under today's rulesbuilt on a parent in the same familybuilt on a parent in another family
3
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 under Result and bounds below.
4
Walkthrough, by hand
Redo the method on a few letters with pencil and paper.
Suppose Sanborn enciphered 98 letters and left one out when cutting, say between EASTNORTHEAST and BERLINCLOCK.
Then the key letter under each letter of BERLINCLOCK is the one for the next place.
Shift BERLINCLOCK's key places by one and rerun the alphabet-pair lookup of 0244.
Do the same for a letter lost inside either known word, at each of its places.
Compare K4 with shuffles treated the same way, and check that planted messages with a lost letter are found.
the worksheet: the first steps by hand, on squared paper
5
Worked example, generated
The same steps run by the code on the real letters.
Key length 19: known places in the same key slot, and what they ask of the two alphabets (added form)
places known carved the gap on the cipher alphabet must equal the gap on the plain alphabet
26,64 NB QN gap(Q to N) on the cipher alphabet = gap(N to B) on the plain alphabet
27,65 OE QY gap(Q to Y) on the cipher alphabet = gap(O to E) on the plain alphabet
28,66 RR PP gap(P to P) on the cipher alphabet = gap(R to R) on the plain alphabet
29,67 TL RV gap(R to V) on the cipher alphabet = gap(T to L) on the plain alphabet
30,68 HI NT gap(N to T) on the cipher alphabet = gap(H to I) on the plain alphabet
31,69 EN GT gap(G to T) on the cipher alphabet = gap(E to N) on the plain alphabet
32,70 AC KM gap(K to M) on the cipher alphabet = gap(A to C) on the plain alphabet
33,71 SL SZ gap(S to Z) on the cipher alphabet = gap(S to L) on the plain alphabet
34,72 TO SF gap(S to F) on the cipher alphabet = gap(T to O) on the plain alphabet
Every cipher alphabet is filed by its gaps; every plain alphabet looks up the gaps it needs.
6
Hypothesis, for a cryptanalyst
Family, key space, the constraint under test, and the clues it reads.
C18(b) and C17 (Scheidt 2005 '98 characters left'; Sanborn 'whether it's 97, whether it's 100'; K2 lost a letter for visual balance, 01-facts.md), F14.1's keyed mask (C4(a), C5). A lost letter misaligns the key from that place on, which can hide a fitting key from every exact test run on the 97 as carved (A1, A4). Pre-registered: in no variant does K4 match a key word; fitting-pair counts as the copies'. A plant (98-letter Carter, two random dictionary alphabets, a random dictionary key word of 16-24 letters, one carved letter dropped at the variant's place) is found when its true alphabets and key word are recovered under its own variant.
7
Prediction
Written before the run; the result decides it.
No key word on K4 in any of the 23 variants; fitting pairs at the copies' rate; plants found 23 of 23.
8
Method
Variants: a letter lost at carved place j = 23..34, 35 (standing for 35-64), 65..74; key index of carved place q is q for q < j and q + 1 from j on.
For each variant: 0244's search (208,064 keyword alphabets, every ordered pair matched through the gap index, key lengths 1-26, both forms; key words of length L from the dictionary and the puzzle, 4 numberings, any shift).
K4 and 20 shuffles per variant; one plant per variant (98 letters enciphered, the letter at j dropped).
Any K4 key-word match decrypted with its variant and submitted.
9
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
23 lost-letter variants x 4.3e10 alphabet pairs x lengths 1-26: K4 key-word matches 0; cells with K4 above all 20 copies 13 of 1196; plants found 23/23