“Affine Hill ciphers of size 2 to 5, solved from every known letter, not only whole blocks”
0015 and 0044 solved Hill matrices from whole crib blocks and used the partial blocks only as checks, which left the 4x4 affine case open (it needs five whole blocks). Written as decryption, p = B c + w, every known plaintext letter gives one linear equation in one row of B, because the carved block is always known. So each row of an n x n affine Hill is solved from all the known letters in that row position, partial blocks included, for n = 2 to 5.
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a new rootparents
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1
The experiment method
What the method does with the 24 known letters.
A Hill ciphera Hill cipher A cipher that turns letters into numbers, takes them two, three or four at a time, and mixes each group with a fixed table of numbers. cuts the messagethe message The 97 letters of English that Jim Sanborn wrote and enciphered. into blocks of a few letters and mixes each block with a small table of numbers. Experiments 15 and 44 solved 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. from blocks where all letters are known. Turning the method around, every single known letter gives one equation, even in a block that is only partly known. With that, even the 4-letter version with an extra shift, which had too few whole blocks before, can be solved exactly.
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 cipher
A Hill cipher mixes each block of a few letters with a small table of numbers, sometimes plus a shift. Earlier runs solved it only from blocks whose letters were all known, too few for the 4-letter version with a shift.
the method
Every single known letter gives one equation for one row of the table, even in a partly known block. Solved mod 2 and mod 13, each row either has a solution or does not.
the bound
Plain and affine tables of size 2 to 4 are dead at every alignment and numbering. At size 5 the plain table is dead; the affine one is mostly open, too few equations, as on the scrambled coppers.
2
Experiment lineage
What this experiment was built on, from the lineage page: 1 experiment 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.
Cut the copper into blocks of 4 from place 1: OBKR, UOXO, and so on; place 22 is the second letter of block 6 (places 21-24, WFLR).
Number the letters, A = 0 ... Z = 25.
Place 22 holds E = 4, the second letter of its block, so row 2 of the table times the block's four numbers, plus shift 2, equals 4 mod 26.
Every known letter in a block's second position gives such an equation for row 2: five unknowns, about six equations.
Solve each row; if all rows have solutions and the table can be undone, the setting fits.
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.
4x4 affine Hill, blocks from place 1, letters A = 0: equations for row 2 (the second letter of each block)
block places carved block numbers known second letter
21-24 WFLR [22, 5, 11, 17] E = 4
25-28 VQQP [21, 16, 16, 15] N = 13
29-32 RNGK [17, 13, 6, 10] H = 7
33-36 SSOT [18, 18, 14, 19] T = 19
65-68 YPVT [24, 15, 21, 19] R = 17
69-72 TMZF [19, 12, 25, 5] C = 2
73-76 PKWG [15, 10, 22, 6] K = 10
row 2 has five unknowns (four table entries and a shift); each line above is one equation
6
Hypothesis, for a cryptanalyst
Family, key space, the constraint under test, and the clues it reads.
Family 8 open item 'the 4x4 affine' (families.md); the HILL column of the tableau (C21, Bauer 2016-17). Model: blocks of n carved letters c (n = 2-5), message block p = B c + w mod 26, B invertible mod 26, w any vector (w = 0 is plain Hill). Row r of (B, w) has n + 1 unknowns; each block whose r-th message letter is known gives one equation. Solved exactly mod 2 and mod 13 (all solutions enumerated when a row is underdetermined, up to a cap) and joined by the Chinese remainder theorem; a setting survives if every row has a solution and some choice makes B invertible. Block alignments 0 to n-1 (the last partial block left plain); numberings: standard or KRYPTOS order, 0- or 1-based. Ten shuffled coppers.
7
Prediction
Written before the run; the result decides it.
No setting with at least one check survives for n = 2-4, plain or affine; n = 5 affine has no checks (6 unknowns per row, about 5 equations) and is reported as untestable. Zero candidates.
8
Method
For each n, alignment and numbering, group the known letters by block and row position.
For each row, solve the equations mod 2 and mod 13.
Join the rows, require B invertible mod 26, and count the spare equations (checks).
Decrypt survivors.
Repeat on ten shuffled coppers.
9
Result and bounds
The harness's record: candidates, gates, loss, and the negative result this experiment leaves behind.