“F14.1: two different keyword alphabets from the whole dictionary, one on each side of the table, with a repeating key (free, then a key word): every pair of 208,064 alphabets matched exactly through an index, key lengths 1-26, both forms; K4 against 50 shuffles, plants at six lengths”
F14.1 (issue #93), the first F14 item. 0109 put one dictionary keyword alphabet on both sides (or one side with the standard or KRYPTOS alphabet); 0020 paired the 248 puzzle words. Here the plain and cipher alphabets are any two of the 208,062 dictionary keyword alphabets plus standard and KRYPTOS (4.3 x 10^10 ordered pairs). At key length L the 24 known letters give checks only between places in the same key slot, and each check splits into a cipher-side difference (the cipher alphabet's places of the two carved letters) and a plain-side one (the plain alphabet's places of the two known letters), so every cipher alphabet is indexed by its difference vector and every plain alphabet looked up: all pairs in about 0.4 million lookups per length and form, exactly. Where pairs fit, the key letters the known letters force are checked against every dictionary and puzzle word of length L, every start, key letters numbered on the standard, KRYPTOS, plain or cipher alphabet with any indicator shift.
Scheidt said he first disguised the Englishthe message The 97 letters of English that Jim Sanborn wrote and enciphered., then used the tablethe table The right-hand half of the sculpture, 26 rows of the alphabet each shifted one place further, starting from the word KRYPTOS.. A simple way to do both by hand: scramble the alphabet with one secret word, and set the table's other alphabet with a second word, then use a repeating key wordthe secret word The word or string a cipher needs besides the method.. That is two scrambled alphabets, which earlier tests left open because there are too many to try one by one. But if each scrambled alphabet comes from an English word, there are only about 200,000 of each, and the clue wordsthe known letters Sanborn has confirmed that positions 22 to 34 of the message read EASTNORTHEAST and positions 64 to 74 read BERLINCLOCK. tell us, for each key length, a few facts that link the two. We file every alphabet by those facts so that matching pairs fall out of a lookup, then check whether the key the clues force is an English word.
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
F14.1: Scheidt's mask as a keyword alphabet before a keyed table: two scrambled alphabets, each from an English word, and a key word.
the check
Every pair of 208,064 alphabets, matched exactly through an index of the gaps the known letters ask for; the forced key checked against every word.
the bound
No key word on K4 or any shuffle; plants 12/12. K4's 5 pairs at length 16 replicated against 1,000 fresh shuffles: p 0.009, not a flag (0245).
2
Experiment lineage
What this experiment was built on, from the lineage page: 7 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
Laid out as the copper carries K4: OBKR ends the line that finishes K3, then three lines of 31. Left, as carved, the 24 known places in accent. Right, the candidate: a known place it gets right is an accent cell, one it misses is outlined. This is the candidate with most known letters right; none passed every gate.
in: the copper, always the same
as carved, 97 letters; the 24 known places in accent
ECDMRIPFEIMEHNLSSTTRTVDOHW?OBKR
UOXOGHULBSOLIFBBWFLRVQQPRNGKSSO
TWTQSJQSSEKZZWATJKLUDIAWINFBNYP
VTTMZFPKWGDKZXTJCDIGKUHUAUEKCAR
OBKR ends the line that finishes K3, then three lines of 31
out: the candidate
every place read; known places in accent, misses outlined
ECDMRIPFEIMEHNLSSTTRTVDOHW?KVXC
CXHPKGLNYPWRRYQFBEASTNORTHEASTE
IBHVVNNPTPYTMBJGPLYWHAVISBUCBER
LINCLOCKIHOYTJNLRJRFTDMPDAPYANP
checked: the known letters in place, and the reading turns back into the copper
4
How the score is made
The loss of the best candidate, part by part. Zero is solved.
Known letters missed
0 of 24, ten points each
0.00
English
the 73 free letters score -7.91; English text scores -3.6
4.31
Key
267 bits to write it down (key words are cheap, arbitrary letters 4.7 bits each); each 60 beyond 60 costs a point
3.45
Loss
of 240, the worst possible
7.76 3.2%
5
Walkthrough, by hand
Redo the method on a few letters with pencil and paper.
Choose a key length, say 19, and list the pairs of known places 19 apart: places 26 and 64 hold N and B, carved Q and N.
In the added form the same key letter took N to Q and B to N, so on the cipher alphabet Q sits as far after N as N sits after B on the plain alphabet.
For every cipher alphabet write down how far Q is from N (and the same for every other pair): that is its file number.
For every plain alphabet work out the file number it needs and look it up: the alphabets filed there are the only partners.
For each partner pair read off the key letters the known letters force and check whether they spell an English word from some starting letter.
the worksheet: the first steps by hand, on squared paper
6
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.
7
Hypothesis, for a cryptanalyst
Family, key space, the constraint under test, and the clues it reads.
Scheidt's mask in its keyed form (C4(a), C5; 03-scheidt-said.md: 'I masked the English language', 'executed with the correct key word/s', 'building layers... mix and match'): a keyword monoalphabetic substitution before a keyed tableau, which is one Quagmire IV (plain alphabet = the mask's keyword alphabet, cipher alphabet = the table's). With free alphabets this is F1.2, consistent at lengths 8, 13, 16, 19, 20, 23, 24, 26 at the copies' rate (0123, 0124; prior-attempts database: 'satisfiable at random-text rates'), and 0136's bound leaves it to English. Two keyword alphabets are about 35 bits, not 176, so the known letters decide them: expected chance fits per form, 4.3 x 10^10 / 26^c with c same-slot checks, are under 0.01 at every length to 22, 5 at 23, 3,600 at 24 (where a key word of 24 letters, 5 in the dictionary, decides), and 2.5 x 10^6 and 1.7 x 10^9 at 25 and 26 (no dictionary word that long: counted only). Pre-registered: K4 fits no pair at 1-22 or matches no key word at 23-24 beyond the copies' counts; a plant (two random dictionary alphabets and a random dictionary key word) is found when its true pair fits and its key word is recovered.
8
Prediction
Written before the run; the result decides it.
K4 fits no alphabet pair at lengths 1-22 (copies: none or almost none); at 23 a handful of pairs on K4 and copies alike, none with a key word; at 24 thousands of pairs on both, none with a dictionary key word; 25-26 counted, as the copies. Plants found at every planted length (19, 20, 21, 22, 23, 24). Zero candidates.
9
Method
Alphabets: the keyword-mixed alphabet of every dictionary word of 4+ letters (0109's 208,062 distinct), plus standard and KRYPTOS.
For each text, key length L = 1-26 and form (added: c = CA[p + k]; Beaufort: c = CA[k - p]): the same-slot pairs of known places; each cipher alphabet's vector of place differences of the carved letters, indexed; each plain alphabet's vector of the known letters' differences (negated for Beaufort) looked up. 25-26: counts only.
For each fitting pair at 1-24: the forced key at the covered slots, checked against every rotation of every dictionary word (3-24) and puzzle word of length L, key letters numbered on standard, KRYPTOS, the plain or the cipher alphabet, any shift. 3b. Where every key slot holds a known place (lengths to 13, and 15, 16), a fitting pair fixes the whole key and reading: those readings are kept for K4 and every copy and scored alike (added after a dry run showed 5 fitting pairs on K4 at 16 and none on 3 copies).
K4 and 50 shuffles (same search); 12 plants (2 per length 19-24) with their own search.
Any K4 key-word fit decrypted in full and submitted.
10
Result and bounds
The harness's record: candidates, gates, loss, and the negative result this experiment leaves behind.
5candidates returned
0passed every gate
24/24known letters, best
3.2%lowest loss
nonechampion before
All 5 candidates, lowest loss first.
1failed at baseline24/24 known lettersloss 3.2%
ECDMRIPFEIMEHNLSSTTRTVDOHW?KVXC
CXHPKGLNYPWRRYQFBEASTNORTHEASTE
IBHVVNNPTPYTMBJGPLYWHAVISBUCBER
LINCLOCKIHOYTJNLRJRFTDMPDAPYANP
2failed at baseline24/24 known lettersloss 3.3%
ECDMRIPFEIMEHNLSSTTRTVDOHW?KVXC
CXZPKHLNYOWRRYQFBEASTNORTHEASTE
IBZVVNNPTPFTMBJHPLYWZAVISBACBER
LINCLOCKIZOFTJNLRJRFTDMPDAPFANP
3failed at baseline24/24 known lettersloss 3.3%
ECDMRIPFEIMEHNLSSTTRTVDOHW?KVXC
CXZPKHLNYOWRRYQFBEASTNORTHEASTE
IBZVVNNPTPFTMBJHPLYWZAVISBACBER
LINCLOCKIZOFTJNLRJRFTDMPDAPFANP
4failed at baseline24/24 known lettersloss 3.4%
ECDMRIPFEIMEHNLSSTTRTVDOHW?JVXC
AXPNJHLMZLWRRZQFDEASTNORTHEASTE
IDPVVMNNTNGTKBHGNYZWPAWISBBCBER
LINCLOCKIPOGTHMLRHSFTDKQDANGAMN
5failed at baseline24/24 known lettersloss 3.5%
ECDMRIPFEIMEHNLSSTTRTVDOHW?CGBD
EWKNHHYPZKVEOJYGFEASTNORTHEASTT
IFZBBKNCTVFSMVFVMYJGLZAICEDIBER
LINCLOCKIKMFSLRMROKBDFJQNUVFDZI
208,064 alphabets, 43,290,628,096 ordered pairs per form, key lengths 1-26. K4: first fit at L 16; key-word matches K4 0, copies max 0; plants found 12/12
What was searched, as the harness recorded it.
item
F14.1
alphabets
208,064
alphabet pairs per form
43,290,628,096
alphabet source
keyword-mixed alphabet of every dictionary word of 4+ letters (0109) plus standard and KRYPTOS
key lengths
1, 26
key word check lengths
1, 24
key word rotations
3
3,384
4
18,210
5
48,505
6
103,593
7
165,249
8
238,028
9
290,340
10
308,065
11
285,659
12
245,436
13
194,025
14
136,710
15
88,845
16
54,016
17
30,838
18
15,174
19
8,132
20
3,960
21
1,722
22
902
23
391
24
120
key numbering
standard, KRYPTOS, the plain alphabet or the cipher alphabet, any indicator shift 0-25